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Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises.

Sales-Tax (VAT)

Question 1.
A man purchased a pair of shoes for ₹809.60 which includes 8% rebate on the marked price and then 10% sales tax on the remaining price. Find the marked price of the pair of shoes.
Solution:

Question 2.
The catalogue price of an article is ₹36,000. The shopkeeper gives two successive discounts of 10% each. He further gives an off-season discount of 5 % on the balance. If sales-tax at the rate 10% is charged on the remaining amount, find :
(i) the sales tax charged
(ii) the selling price of the article including sales-tax.
Solution:

Question 3.
A sells an article to B for ₹80,000 and charges sales-tax at 8%. B sells the same article to C for ₹1,12,000 and charges sales- tax at the rate of 12%. Find the VAT paid by B in this transaction.
Solution:

Question 4.
The marked price of an article is ₹1,600. Mohan buys this article at 20% discount and sells it at its marked price. If the sales-tax at each stage is 6%; find :
(i) the price at which the article can be bought,
(ii) the VAT (value added tax) paid by Mohan.
Solution:

Question 5.
A sells an old laptop to B for ₹12,600; B sells it to C for ₹14,000 and C sells the same laptop to D for ₹16,000.
If the rate of VAT at each stage is 10%, find the VAT paid by :
(i) B
(ii) C
Solution:

Banking

Question 6.
Ashok deposits ₹3200 per month in a cumulative account for 3 years at the rate of 9% per annum. Find the maturity value of this account
Solution:

Question 7.
Mrs. Kama has a recurring deposit account in Punjab National Bank for 3 years at 8% p.a. If she gets ₹9,990 as interest at the time of maturity, find:
(i) the monthly instalment.
(ii) the maturity value of the account.
Solution:

Question 8.
A man has a 5 year recurring deposit account in a bank and deposits ₹240 per month. If he receives ₹17,694 at the time of maturity, find the rate of interest.
Solution:

Question 9.
Sheela has a recurring deposit account in a bank of ₹2,000 per month at the rate of 10% per anum. If she gets ₹83,100 at the time of maturity, find the total time (in years) for which the account was held.
Solution:

Question 10.
A man deposits ₹900 per month in a recurring account for 2 years. If he gets 1,800 as interest at the time of maturity, find the rate of interest .
Solution:

Shares And Dividend

Question 11.
What is the market value of 4 \(\frac { 1 }{ 2 }\) % (₹100) share, when an investment of ₹1,800 produces an income of ₹72 ?
Solution:

Question 12.
By investing ₹10,000 in the shares of a company, a man gets an income of ₹800; the dividend being 10%. If the face-value of each share is ₹100, find :
(i) the market value of each share.
(ii) the rate per cent which the person earns on his investment.
Solution:

Question 13.
A man holds 800 shares of ₹100 each of a company paying 7.5% dividend semiannually.
(i) Calculate his annual dividend.
(ii) If he had bought these shares at 40% premium, what percentage return does he get on his investment ?
Solution:

Question 14.
A man invests ₹10,560 in a company, paying 9% dividend, at the time when its ₹100 shares can be bought at a premium of ₹32. Find:
(i) the number of shares bought by him;
(ii) his annual income from these shares and
(iii) the rate of return on his investment .
Solution:

Question 15.
Find the market value of 12% ₹25 shares of a company which pays a dividend of ₹1,875 on an investment of ₹20,000.
Solution:

Linear Inequations

Question 16.
The given diagram represents two sets A and B on real number lines.

(i) Write down A and B in set builder notation.
(ii) Represent A ∪ B, A ∩ B, A’ ∩ B, A – B and B – A on separate number lines.
Solution:

Question 17.
Find the value of x, which satisfy the inequation:

Graph the solution set on the real number line .
Solution:

Question 18.
State for each of the following statements whether it is true or false :
(a) If (x – a) (x – b) < 0, then x < a, and x < b.
(b) If a < 0 and b < 0, then (a + b)² > 0.
(c) If a and b are any two integers such that a > b, then a² > b².
(d) If p = q + 2, then p > q.
(e) If a and b are two negative integers such that a < b , then \(\frac { 1 }{ a }\) < \(\frac { 1 }{ b }\)
Solution:

Question 19.
Given 20 – 5x < 5(x + 8), find the smallest value of x when :
(i) x \(\epsilon\) I
(ii) x \(\epsilon\) W
(iii) x \(\epsilon\) N
Solution:

Question 20.
If x \(\epsilon\) Z, solve : 2 + 4x < 2x – 5 < 3x. Also, represent its solution on the real number line.
Solution:

Quadratic Equation

Question 21.

Solution:

Question 22.

Solution:

Question 23.
Find the value of k for which the roots of the following equation are real and equal k²x² – 2 (2k -1) x + 4 = 0
Solution:

Question 24.

Solution:

Question 25.
If -5 is a root of the quadratic equation 2x² +px -15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.
Solution:

Problems On Quadratic Equations

Question 26.
x articles are bought at ₹(x – 8) each and (x – 2) some other articles are bought at ₹(x – 3) each. If the total cost of all these articles is ₹76, how many articles of first kind were bought ?
Solution:

Question 27.
In a two digit number, the unit’s digit exceeds its ten’s digit by 2. The product of the given number and the sum of its digits is equal to 144. Find the number.
Solution:

Question 28.
The time taken by a person to cover 150 km was 2.5 hours more than the time taken in return journey. If he returned at a speed of 10 km/hour more than the speed of going, what was the speed per hour in each direction ?
Solution:

Question 29.
A takes 9 days more than B to do a certain piece of work. Together they can do the work in 6 days. How many days will A alone take to do the work ?
Solution:

Question 30.
A man bought a certain number of chairs for ₹10,000. He kept one for his own use and sold the rest at the rate ₹50 more than he gave for one chair. Besides getting his own chair for nothing, he made a profit of ₹450. How many chairs did he buy ?
Solution:

Question 31.
In the given figure; the area of unshaded portion is 75% of the area of the shaded portion. Find the value of x.

Solution:

Ratio And Proportion

Question 32.

Solution:

Question 33.
If a:b = 2:3,b:c = 4:5 and c: d = 6:7, find :a:b :c :d.
Solution:

Question 34.

Solution:

Question 35.
Find the compound ratio of:
(i) (a-b) : (a+b) and (b²+ab): (a²-ab)
(ii) (x+y): (x-y); (x²+y²): (x+y)² and (x²-y²)²: (x⁴-y⁴)
(iii) (x²– 25): (x²+ 3x – 10); (x²-4): (x²+ 3x+2) and (x + 1): (x² + 2x)
Solution:

Question 36.
The ratio of the prices of two fans was 16: 23. Two years later, when the price of the first fan had risen by 10% and that of the second by Rs. 477, the ratio of their prices became 11: 20. Find the original prices of two fans.
Solution:
The ratio of prices of two fans = 16 : 23
Let the price of first fan = 6x
then price of second fan = 23x

Remainder And Factor Theorems

Question 37.
Given that x + 2 and x – 3 are the factors of x³ + ax + b, calculate the values of a and b. Also find the remaining factor.
Solution:

Question 38.
Use the remainder theorem to factorise the expression 2x³ + 9x² + 7x – 6 = 0 Hense, solve the equation 2x³ + 9x² + 7 x – 6 = 0
Solution:

Question 39.
When 2x³ + 5x² – 2x + 8 is divided by (x – a) the remainder is 2a³ + 5a². Find the value of a.
Solution:

Question 40.
What number should be added to x³ – 9x² – 2x + 3 so that the remainder may be 5 when divided by (x – 2) ?
Solution:

Question 41.
Let R₁ and R₂ are remainders when the polynomials x³ + 2x² – 5ax – 7 and x³ + ax² – 12x + 6 are divided by (x +1) and (x – 2) respectively. If 2R₁ + R₂ = 6; find the value of a.
Solution:

Matrices

Question 42.

Solution:

Question 43.

Solution:

Question 44.

Solution:

Question 45.

Solution:

Arithmetic Progression (A.P.)

Question 46.
Find the 15th term of the A.P. with second term 11 and common difference 9.
Solution:

Question 47.
How many three digit numbers are divisible by 7 ?
Solution:

Question 48.
Find the sum of terms of the A.P.: 4,9,14,…….,89.
Solution:

Question 49.
Daya gets pocket money from his father every day. Out of the pocket money, he saves ₹2.75 on first day, ₹3.00 on second day, ₹3.25 on third day and so on. Find:
(i) the amount saved by Daya on 14th day.
(ii) the amount saved by Daya on 30th day.
(iii) the total amount saved by him in 30 days.
Solution:

Question 50.
If the sum of first m terms of an A.P. is n and sum of first n terms of the same A.P. is m. Show that sum of first (m + n) terms of it is (m + n).
Solution:

Geometric Progression (GP.)

Question 51.
3^rd term of a GP. is 27 and its 6th term is 729; find the product of its first and 7^th terms.
Solution:

Question 52.
Find 5 geometric means between 1 and 27.
Solution:

Question 53.
Find the sum of the sequence 96 – 48 + 24…. upto 10 terms.
Solution:

Question 54.
Find the sum of first n terms of:
(i) 4 + 44 + 444 + …….
(ii) 0.7 + 0.77 + 0.777 + …..
Solution:

Question 55.
Find the value of 0.4\(\mathring { 2 } \)\(\mathring { 3 } \).
Solution:

Reflection

Question 56.
Find the values of m and n in each case if:
(i) (4, -3) on reflection in x-axis gives (-m, n)
(ii) (m, 5) on reflection in y-axis gives (-5, n-2)
(iii) (-6, n+2) on reflection in origin gives (m+3, -4)
Solution:

Question 57.
Points A and B have the co-ordinates (-2,4) and (-4,1) respectively. Find :
(i) The co-ordinates of A’, the image of A in the line x = 0.
(ii) The co-ordinates of B’, the image of Bin y-axis.
(iii) The co-ordinates of A”, the image of A in the line BB’,
Hence, write the angle between, the lines A’A” and B B’. Assign a special name to the figure B’ A’ B A”
Solution:

Question 58.
Triangle OA₁B₁ is the reflection of triangle OAB in origin, where A, (4, -5) is the image of A and B, (-7, 0) is the image of B.
(i) Write down the co-ordinates of A and B and draw a diagram to represent this information.
(ii) Give the special name to the quadrilateral ABA₁ B₁. Give reason.
(iii) Find the co-ordinates of A₂, the image of A under reflection in x-axis followed by reflection in y-axis.
(iv) Find the co-ordinates of B₂, the image of B under reflection in y-axis followed by reflection in origin.
(v) Does the quadrilateral obtained has any line symmetry ? Give reason.
(vi) Does it have any point symmetry ?
Solution:

Section And Mid-Point Formulae

Question 59.
In what ratio does the point M (P, -1) divide the line segment joining the points A (1,-3) and B (6,2) ? Hence, find the value of p.
Solution:

Question 60.
A (-4,4), B (x, -1) and C (6,y) are the vertices of ∆ABC. If the centroid of this triangle ABC is at the origin, find the values of x and y.
Solution:

Question 61.
A (2,5), B (-1,2) and C (5,8) are the vertices of a triangle ABC. Pand Q are points on AB and AC respectively such that AP: PB=AQ: QC = 1:2.
(a) Find the co-ordinates of points P and Q
(b) Show that BC = 3 x PQ.
Solution:

Question 62.
Show that the points (a, b), (a+3, b+4), (a -1, b + 7) and (a – 4, b + 3) are the vertices of a parallelogram.
Solution:

Equation Of straight Line

Question 63.
Given points A(l, 5), B (-3,7) and C (15,9).
(i) Find the equation of a line passing through the mid-point of AC and the point B.
(ii) Find the equation of the line through C and parallel to AB.
(iii) The lines obtained in parts (i) and (ii) above, intersect each other at a point P. Find the co-ordinates of the point P.
(iv) Assign, giving reason, a special names of the figure PABC.
Solution:

Question 64.
The line x- 4y=6 is the perpendicular bisector of the line segment AB. If B = (1,3); find the co-ordinates of point A.
Solution:

Question 65.
Find the equation of a line passing through the points (7, -3) and (2, -2). If this line meets x- axis at point P and y-axis at point Q; find the co-ordinates of points P and Q.
Solution:

Question 66.
A (-3,1), B (4,4) and C (1, -2) are the vertices of a triangle ABC. Find:
(i) the equation of median BD,
(ii) the equation of altitude AE.
Solution:

Question 67.
Find the equation of perpendicular bisector of the line segment joining the points (4, -3) and (3,1).
Solution:

Question 68.
(a) If (p +1) x + y = 3 and 3y – (p -1) x = 4 are perpendicular to each other find the value of p.
(b) If y + (2p +1) x + 3 = 0 and 8y – (2p -1) x = 5 are mutually prependicular, find the value of p.
Solution:

Question 69.
The co-ordinates of the vertex A of a square ABCD are (1, 2) and the equation of the diagonal BD is x + 2y = 10. Find the equation of the other diagonal and the coordinates of the centre of the square.
Solution:

Similarity

Question 70.
M is mid-point of a line segment AB; AXB and MYB are equilateral triangles on opposite sides of AB; XY cuts AB at Z. Prove that AZ = 2ZB.
Solution:

Question 71.
In the given figure, if AC = 3cm and CB = 6 cm, find the length of CR.

Solution:

Question 72.
The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point O. If BO : OD = 4:7; find:
(i) ∆AOD : ∆AOB
(ii) ∆AOB : ∆ACB
(iii) ∆DOC : ∆AOB
(iv) ∆ABD : ∆BOC
Solution:

Question 73.
A model of a ship is made to a scale of 1 : 160. Find :
(i) the length of the ship, if the length of its model is 1.2 m.
(ii) the area of the deck of the ship, if the area of the deck of its model is 1.2 m².
(iii) the volume of the ship, if the volume of its model is 1.2m³.
Solution:

Question 74.
In trapezium ABCD, AB || DC and DC = 2 AB. EF, drawn parallel to AB cuts AD in F and BC in E such that 4 BE = 3 EC. Diagonal DB intersects FE at point G Prove that: 7 EF = 10 AB.

Solution:

Loci

Question 75.
In triangle ABC, D is mid-point of AB and CD is perpendicular to AB. Bisector of ∠ABC meets CD at E and AC at F. Prove that:
(i) E is equidistant from A and B.
(ii) F is equidistant from AB and BC.
Solution:

Question 76.
Use graph paper for this questions. Take 2 cm = 1 unit on both axes.
(i) Plot the points A (1,1), B (5,3) and C (2,7)
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from AB and AC.
(iv) Locate the point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure and record the length PA in cm.
Solution:

Circles

Question 77.
In the given figure, ∠ADC = 130° and BC = BE. Find ∠CBE if AB ⊥ CE.

Solution:

Question 78.
In the given figure, ∠OAB=30° and ∠OCB= 57°, find ∠BOC and ∠AOC.

Solution:

Question 79.
In the given figure, O is the centre of the circle. If chord AB = chord AC, OP⊥ AB and OQ⊥ AC; show that: PB=QC.
Solution:

Question 80.
In the given figure, AB and XY are diameters of a circle with centre O. If ∠APX=30°, find:
(i) ∠AOX
(ii) ∠APY
(iii) ∠BPY
(iv) ∠OAX

Solution:

Question 81.
(a) In the adjoining figure; AB = AD, BD = CD and ∠DBC = 2 ∠ABD.
Prove that: ABCD is a cyclic quadrilateral.

(b) AB is a diameter of a circle with centre O, Chord CD is equal to radius OC. AC and BD produced intersect at P. Prove that ∠APB = 60°.

Solution:

Tangents And Intersecting Chords

Question 82.
In the given figure, AC=AB and ∠ABC=72°.
OA and OB are two tangents. Determine:
(i) ∠AOB
(ii) angle subtended by the chord AB at the centre.

Solution:

Question 83.
In the given figure, PQ, PR and ST are tangents to the same circle. If ∠P = 40° and ∠QRT = 75°, find a, b and c.

Solution:

Question 84.
In the given figure, ∠ABC = 90° and BC is diameter of the given circle. Show that:
(i) AC x AD =AB²
(ii) AC x CD = BC²

Solution:

Question 85.
In the given figure; AB, BC and CA are tangents to the given circle. If AB = 12 cm, BC = 8 cm and AC=10 cm, find the lengths of AD,BE = CF.

Solution:

Question 86.
(a) AB and CD are two chords of a circle intersecting at a point P inside the circle. If:
(i) AB = 24 cm, AP = 4 cm and PD = 8 cm, determine CP.
(ii) AP = 3 cm, PB = 2.5 cm and CD = 6.5 cm determine CP.
(b) AB and CD are two chords of a circle intersecting at a point P outside the circle. If:
(i) PA = 8 cm, PC – 5 cm and PD = 4 cm, determine AB.
(ii) PC = 30 cm, CD = 14 cm and PA = 24 cm, determine AB.
Solution:

Construction

Question 87.
Construct a triangle ABC in which AC = 5 cm, BC = 7 cm and AB = 6 cm.
(i) Mark D, the mid point of AB.
(ii) Construct a circle which touches BC at C and passes through D.
Solution:

Question 88.
Using ruler and compasses only, draw a circle of radius 4 cm. Produce AB, a diameter of this circle up to point X so that BX = 4cm. Construct a circle to touch AB at X and to touch the circle, drawn earlier externally.
Solution:

Mensuration

Question 89.
A cylindrical bucket 28 cm in diameter and 72 cm high is full of water. The water is emptied into a rectangular tank 66 cm long and 28 cm wide. Find the height of the water level in the tank.
Solution:

Question 90.
A tent is of the shape of right circular cylinder upto height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner surface of the tent at Rs. 4 per sq. metre, if the radius of the base is 14 metres.
Solution:

Question P.Q.
In the given figure, diameter of the biggest semi-circle is 108cm, and diameter of the smallest circle is 36 cm. Calculate the area of the shaded portion.

Solution:

Question 91.
A copper wire of diameter 6 mm is evenly wrapped on the cylinder of length 18 cm and diameter 49 cm to cover the whole surface. Find:
(i) the length
(ii) the volume of the wire
Solution:

Question 92.
A pool has a uniform circular cross-section of radius 5 m and uniform depth 1.4m. It is filled by a pipe which delivers water at the rate of 20 litres per sec. Calculate, in minutes, the time taken to All the pool. If the pool is emptied in 42 min. by another cylindrical pipe through which water flows at 2 m per sec, calculate the radius of the pipe in cm.
Solution:

Question 93.
A test tube consists of a hemisphere and a cylinder of the same radius. The volume of water required to fill the whole tube is 2849/3cm³ and 2618/3cm³ of water are required to fill the tube to a level which is 2 cm below the top of the tube. Find the radius of the tube and the length of its cylinderical part.
Solution:

Question 94.
A sphere is placed in an inverted hollow conical vessel of base radius 5 cm and vertical height 12 cm. If the highest point of the sphere is at the level of the base of the cone, find the radius of the sphere. Show that the volume of the sphere and the conical vessel are as 40 : 81.
Solution:

Question 95.
The difference between the outer and the inner curved surface areas of a hollow cylinder, 14cm. long is 88sq. cm. Find the outer and the inner radii of the cylinder given that the volume of metal used is 176 cu. cm.
Solution:

Trigonometry

Question 96.

Solution:

Question 97.
If tan A = 1 and tan B = √3 ; evaluate :
(i) cos A cos B – sin A sin B
(ii) sin A cos B + cos A sin B
Solution:

Question 98.
As observed from the top of a 100 m high light house, the angles, of depression of two ships approaching it are 30° and 45°. If one ship is directly behind the other, find the distance between the two ships.
Solution:

Question 99.

Solution:

Question 100.
From the top of a light house, it is observed that a ship is sailing directly towards it and the angle of depression of the ship changes from 30° to 45° in 10 minutes. Assuming that the ship is sailing with uniform speed; calculate in how much more time (in minutes) will the ship reach to the light house.
Solution:
Let LM be the height of light house = h
Angle of depression changes from 30° to 45° in 10 minutes.

Statistics 

Question 101.
Calculate the mean mark in the distribution given below :

Also state (i) median class (ii) the modal class.
Solution:

Question 102.
Draw an ogive for the following distribution :

Use the ogive drawn to determine :
(i) the median income,
(ii) the number of employees whose income exceeds Rs. 190.
Solution:

Question 103.
The result of an examination are tabulated below :

Draw the ogive for above data and from it determine :
(i) the number of candidates who got marks less than 45.
(ii) the number of candidates who got marks more than 75.
Solution:

Probability

Question 104.
A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. If a ball is drawn from the bag, without looking into it, find the probability that the ball drawn is
(i) yellow
(ii) red
(iii) blue
Solution:

Question 105.
A bag contains 6 red balls, 8 blue balls and 10 yellow balls, all the balls being of the same size. If a ball is drawn from the bag, without looking into it, find the probability that the ball drawn is
(i) yellow
(ii) red
(iii) blue
(iv) not yellow
(v) not blue
Solution:

Question 106.
Two dice are thrown at the same time. Write down all the possible outcomes. Find the probability of getting the sum of two numbers appearing on the top of the dice as :
(i) 13
(ii) less than 13
(iii) 10
(iv) less then 10
Solution:

Question 107.
Five cards : the ten, jack, queen, king and ace. of diamonds are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen ?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace ? (b) a queen ?
Solution:

Question 108.
(i) A lot of 20 bulbs contains 4 defective bulbs, one bulb is drawn at random, from the lot. What is the probability that this bulb is defective ?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises  are helpful to complete your math homework.

Selina Concise Mathematics Class 10 ICSE Solutions Similarity (With Applications to Maps and Models)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models)

Similarity Exercise 15A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In the figure, given below, straight lines AB and CD intersect at P; and AC // BD. Prove that:
(i) ∆APC and ∆BPD are similar.
(ii) If BD = 2.4 cm AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.

Solution:
(i)

(ii)

Question 2.
In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that:
(i) ∆APB is similar to ∆CPD
(ii) PA × PD = PB × PC
Solution:
(i)

(ii)

Question 3.
P is a point on side BC of a parallelogram ABCD. IfDPproduced meets AB produced at point L, prove that:
(i) DP: PL = DC: BL.
(ii) DL: DP=AL: DC.
Solution:
(i)

(ii)

Question 4.
In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO=2DO; show that:
(i) ∆AOB is similar to ∆COD.
(ii) OA × OD – OB × OC.
Solution:
(i)

(ii)

Question 5.
In ∆ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that:
(i) CB: BA=CP: PA
(ii) AB × BC = BP × CA
Solution:
(i)

(ii)

Question 6.
In ∆ABC; BM ⊥ AC and CN ⊥ AB; show that:
\(\frac{\mathbf{A B}}{A C}=\frac{B M}{C N}=\frac{A M}{A N}\)
Solution:

Question 7.
In the given figure, DE//BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.
(i) Write all possible pairs of similar triangles.
(ii) Find lengths of ME and DM.

Solution:
(i)

(ii)

Question 8.
In the given figure, AD =AE and AD² = BD × EC
Prove that: triangles ABD and CAE are similar.

Solution:

Question 9.
In the given figure, AB // DC, BO = 6 cm and DQ = 8 cm; find: BP × DO.

Solution:

Question 10.
Angle BAC of triangle ABC is obtuse and AB =AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively. If PQ = 15 cm and PR=9 cm; find the length of PB
Solution:

Question 11.
State, true or false:
(i) Two similar polygons are necessarily congruent.
(ii) Two congruent polygons are necessarily similar.
(iii) All equiangular triangles are similar.
(iv) All isosceles triangles are similar.
(v) Two isosceles-right triangles are similar.
(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.
(vii) The diagonals of a trapezium, divide each other into proportional segments.
Solution:
(i) False
(ii) True
(iii) True
(iv) False
(v) True
(vi) True
(vii) True

Question 12.
Given = ∠GHE = ∠ DFE = 90°, DH = 8, DF = 12, DG = 3x + 1 and DE = 4x + 2.

Find; the lengths of segments DG and DE.
Solution:

Question 13.
D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that CA² = CB × CD.
Solution:

Question 14.
In the given figure, ∆ABC and ∆AMP are right angled at B and M respectively. Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Proe that ∆ABC ~ ∆AMP
(ii) Find AB and BC.
Solution:

Question 15.
Given : RS and PT are altitudes of A PQR prove that:
(i)∆PQT ~ ∆QRS,
(ii) PQ × QS = RQ × QT.
Solution:

Question 16.
Given : ABCD is a rhombus, DPR and CBR are straight lines

Prove that: DP × CR = DC × PR.
Solution:

Question 17.
Given: FB = FD, AE ⊥ FD and FC ⊥ AD. Prove : \(\frac{\mathbf{F B}}{\mathbf{A D}}=\frac{\mathbf{B C}}{\mathbf{E} \mathbf{D}}\)
Solution:

Question 18.
In ∆ PQR, ∠ Q = 90° and QM is perpendicu¬lar to PR, Prove that:
(i) PQ² = PM × PR
(ii) QR² = PR × MR
(iff) PQ² + QR² = PR²
Solution:

Question 19.
In ∆ ABC, ∠ B = 90° and BD × AC.
(i) If CD = 10 cm and BD = 8 cm; find AD.
(ii) If AC = 18 cm and AD = 6 cm; find BD.
(iii) If AC = 9 cm, AB = 7 cm; find AD.
Solution:

Question 20.
In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N.

Find the lengths of PN and RM.
Solution:

Question 21.
In quadrilateral ABCD, diagonals AC and BD intersect at point E. Such that
AE : EC = BE :’ED.
Show that ABCD is a parallelogram
Solution:

Question 22.
In ∆ ABC, AD is perpendicular to side BC and AD² = BD × DC.
Show that angle BAC = 90°

Solution:

Question 23.
In the given figure AB // EF // DC; AB ~ 67.5 cm. DC = 40.5 cm and AE = 52.5 cm.

(i) Name the three pairs of similar triangles.
(ii) Find the lengths of EC and EF.
Solution:

Question 24.
In the given figure, QR is parallel to AB and DR is parallel to QB.

Prove that— PQ² = PD × PA.
Solution:

Question 25.
Through the mid-point M of the side CD o£. a parallelogram ABCD, the line BM is drawn ‘ intersecting diagonal AC in L and AD produced in E.
Prove that : EL = 2 BL.
Solution:

Question 26.
In the figure given below P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC.

(i) Calculate the ratio PQ : AC, giving reason for your answer.
(ii) In triangle ARC, ∠ ARC = 90° and in triangle PQS, ∠ PSQ = 90°. Given QS = 6 cm, calculate the length of AR. [1999]
Solution:

Question 27.
In the right angled triangle QPR, PM is an altitude.

Given that QR = 8 cm and MQ = 3.5 cm. Calculate, the value of PR., [2000]
Given— In right angled ∆ QPR, ∠ P = 90° PM ⊥ QR, QR = 8 cm, MQ = 3.5 cm
Calculate— PR
Solution:

Question 28.
In the figure given below, the medians BD and CE of a triangle ABC meet at G.
Prove that—
(i) ∆ EGD ~ ∆ CGB
(ii) BG = 2 GD from (i) above.
Solution:

Similarity Exercise 15B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In the following figure, point D divides AB in the ratio 3:5. Find:

Also, if:
(iv) DE = 2.4 cm, find the length of BC.
(v) BC = 4.8 cm, find the length of DE.

Solution:
(i).

(ii)

(iii)

(iv)

(v)

Question 2.
In the given figure, PQ//AB;
CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find:
(i) \(\frac{\mathbf{C P}}{\mathbf{P A}}\)
(ii) PQ
(iii) If AP=x, then the value of AC in terms of x.

Solution:
(i)

(ii)

(iii)

Question 3.
A line PQ is drawn parallel tp the side BC of AABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA=6.0 cm and AQ = 4.2 cm, find the length of AP.
Solution:

Question 4.
In ∆ABC, D and E are the points on sides AB and AC respectively.
Find whether DE // BC, if:
(i) AB=9 cm, AD=4 cm, AE=6 cm and EC = 7.5 cm.
(ii) AB=63 cm, EC=11.0 cm, AD=0.8 cm and AE = 1.6 cm.
Solution
(i).

(ii).

Question 5.
In the given figure, ∆ABC ~ ∆ADE. If AE: EC = 4 :7 and DE = 6.6 cm, find BC. If ‘x’ be the length of the perpendicular fromA to DE, find the length of perpendicular from

A to DE find the length of perpendicular from A to BC in terms of ‘x’.
Solution:

Question 6.
A line segment DE is drawn parallel to base BC of AABC which cuts AB at point D and AC at point E. If AB = 5 BD and EC=3.2 cm, find the length of AE.
Solution:

Question 7.
In the figure, given below, AB, Cd and EFare parallel lines. Given AB = 7.5 cm, DC =y cm, EF=4.5 cm, BC=x cm and CE=3 cm, calculate the values of x and y.

Solution:

Question 8.
In the figure, given below, PQR is a right- angle triangle right angled at Q. XY is parallel to QR, PQ = 6 cm, P Y=4 cm and PX : XQ = 1:2. Calculate the lengths of PR and QR.

Solution:

Question 9.
In the following figure, M is mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E. Prove that PE = 2PD.

Solution:

Question 10.
The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. IfAE=4 cm, AF = 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.

Solution:

Similarity Exercise 15C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
(i) The ratio between the corresponding sides of two similar triangles is 2 is to 5. Find the ratio between the areas of these triangles.
(ii) Areas of two similar triangles are 98 sq. cm and 128 sq. cm. Find the ratio between the lengths of their corresponding sides.
Solution:

Question 2.
A line PQ is drawn parallel to the base BC, of ∆ ABC which meets sides AB and AC at points P and Q respectively. If AP = \(\frac{1}{3}\) PB; find the value of:

Solution:

Question 3.
The perimeters of two similar triangles are 30 cm and 24cm. If one side of first triangle is 12cm, determine the corresponding side of the second triangle.
Solution:

Question 4.
In the given figure AX : XB = 3 : 5

Find :
(i) the length of BC, if length of XY is 18 cm.
(ii) ratio between the areas of trapezium XBCY and triangle ABC.
Solution:

Question 5.
ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP : AB.
Given— In ∆ ABC, PQ || BC in such away that area APQ = area PQCB
To Find— The ratio ol’ BP : AB

Solution:

Question 6.
In the given triangle PQR, LM is parallel to QR and PM : MR = 3 : 4

Solution:

Question 7.
The given diagram shows two isosceles triangles which are similar also. In (he given dia¬gram, PQ and BC are not parallel:
PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ.

Calculate—
(i) the length of AP
(ii) the ratio of the areas of triangle APQ and triangle ABC.
Solution:

Question 8.
In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC =1:2. DP produced meets AB produced at Q. Given the area of triangle CPQ = 20 cm2.
Calculate—
(i) area of triangle CDP
(ii) area of parallelogram ABCD [1996]

Solution:

Question 9.
In the given figure. BC is parallel to DE. Area of triangle ABC = 25 cm².
Area of trapezium BCED = 24 cm² and DE = 14 cm. Calculate the length of BC.
Also. Find the area of triangle BCD.

Solution:

Question 10.
The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 3 : 5.

Find:
(i) ∆ APB : ∆ CPB
(ii) ∆ DPC : ∆ APB
(iii) ∆ ADP : ∆ APB
(iv) ∆ APB : ∆ ADB
Solution:

Question 11.
In the given figure, ARC is a triangle. DE is parallel to BC and \(\frac{A D}{D B}=\frac{3}{2}\).
(i) Determine the ratios \(\frac{A D}{A B}, \frac{D E}{B C}\).
(ii) Prove that ∆DEF is similar to ∆CBF. Hence, find \(\frac{E F}{F B}\).
(iii) What is the ratio of the areas of ∆DEF and ∆BFC?

Solution:

Question 12.
In the given figure, ∠B = ∠E, ∠ACD = ∠BCE, AB=10.4 cm and DE=7.8 cm. Find the ratio between areas of the ∆ABC and ∆DEC.

Solution:

Question 13.
Triangle ABC is an isosceles triangle in which AB = AC = 13 cm and BC = 10 cm. AD is perpendicular to BC. If CE = 8 cm and EF ⊥ AB, find:

Solution:

Similarity Exercise 15D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A’ B’ C’. Calculate:
(i) the length of AB, if A’ B’ = 6 cm.
(ii) the length of C’ A’ if CA = 4 cm.
Solution:
(i)

(ii)

Question 2.
A triangle LMN has been reduced by scale factor 0.8 to the triangle L’ M’ N’. Calculate:
(i) the length of M’ N’, if MN = 8 cm.
(ii) the length of LM, if L’ M’ = 5.4 cm.
Solution:
(i)

(ii)

Question 3.
A triangle ABC is enlarged, about the point O as centre of enlargement, and the scale factor is 3. Find:
(i) A’ B’, if AB = 4 cm.
(ii) BC, if B’ C’ = 15 cm.
(iii) OA, if OA’= 6 cm.
(iv) OC’, if OC = 21 cm.
Also, state the value of:

Solution:
(i)

(ii)

(iii)

(iv)

Question 4.
A model of an aeroplane is made to a scale of 1:400. Calculate:
(i) the length, in cm, of the model; if the length of the aeroplane is 40 m.
(ii) the length, in m, of the aeroplane, if length of its model is 16 cm.
Solution:

(ii)

Question 5.
The dimensions of the model of a multistorey building are 1.2 m × 75 cm × 2 m. If the scale factor is 1:30; find the actual dimensions of the building.
Solution:

Question 6.
On a map drawn to a scale of 1: 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and angle ABC = 90°.
Calculate:
(i) the actual lengths of AB and BC in km.
(ii) the area of the plot in sq. km.
Solution:

(ii)

Question 7.
A model of a ship of made to a scale 1 : 300
(i) The length of the model of ship is 2 m. Calculate the lengths of the ship.
(ii) The area of the deck ship is 180,000 m². Calculate the area of the deck of the model.
(iii) The volume of the model in 6.5 m³. Calculate the volume of the ship. (2016)
Solution:

Question 7(old).

Solution:

Question 8.
An aeroplane is 30 in long and its model is 15 cm long. If the total outer surface area of the model is 150 cm², find the cost of painting the outer surface of the aeroplane at the rate of ₹ 120 per sq. m. Given that 50 sq. m of the surface of the aeroplane is left for windows.
Solution:

Similarity Exercise 15E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5cm and BC = 18 cm.

Find:

Solution:
(i)

(ii)

Question 2.
In the following figure, ABCD to a trapezium with AB//DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP=6 cm and BE = 15 cm.
Calculate:
(i) EC
(ii) AF
(iii) PE

Solution:
(i)

(ii)

(iii)

Question 3.
In the following figure, AB, CD and EF are perpendicular to the straight line BDF.

If AB = x and CD = z unit and EF = y unit, prove that : \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\).
Solution:

Question 4.
Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that:
\(\frac{A B}{P Q}=\frac{A D}{P M}\).
Solution:

Question 5.
Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles,
prove that: \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
Solution:

Question 6.
Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that: \(\frac{A B}{P Q}=\frac{A D}{P M}\).
Solution:

Question 7.
In the following figure, ∠AXY = ∠AYX. If \(\frac{\mathbf{B X}}{\mathbf{A X}}=\frac{\mathbf{C} \mathbf{Y}}{\mathbf{A} \mathbf{Y}}\), show that triangle ABC is isosceles.

Solution:



Question 8.
In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown.

Prove that: \(\frac{A B}{B C}=\frac{P Q}{Q R}\)
Solution:

Question 9.
In the following figure, DE //AC and DC //AP. Prove that: \(\frac{B E}{E C}=\frac{B C}{C P}\)

Solution:

Question 10.
In the figure given below, AB//EF// CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.
Calculate:
(i) EF
(ii) AC

Solution:
(i)

(ii)

Question 11.
In ∆ABC, ∠ABC = ∠DAC. AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ∆ACD is similar to ∆BCA.
(ii) Find BC and CD.
(iii) Find area of ∆ACD: area of ∆ABC. (2014)
Solution:
(i)

(ii)

(iii)

Question 12.
In the given triangle P, Q and R are the midpoints of sides AB, BC and AC respectively. Prove that triangle PQR is similar to triangle ABC.

Solution:

Question 13.
In the following figure, AD and CE are medians of ∆ ABC. DF is drawn parallel to CE. Prove that:
(i) EF = FB;
(ii) AG : GD = 2 : 1

Solution:

Question 14.
The two similar triangles are equal in area. Prove that the triangles are congruent.
Solution:

Question 15.
The ratio between the altitudes of two similar triangles is 3 : 5; write the ratio between their:
(i) medians
(ii) perimeters
(iii) areas
Solution:

Question 16.
The ratio between the areas of two similar triangles is 16 : 25. Find the ratio between their:
(i) perimeters
(ii) altitudes
(iii) medians.
Solution:

Question 17.
The following figure shows a triangle PQR in which XY is parallel to QR. If PX: XQ = 1:3 and QR = 9 cm, find the length of XY.
Further, if the area of ∆ PXY = x cm²; find in terms of x, the area of :
(i) triangle PQR.
(ii) trapezium XQRY.

Solution:

Question 18.
On a map, drawn to a scale of 1 : 20000, a rectangular plot of land ABCD has AB = 24 cm, and BC = 32 cm. Calculate :
(i) The diagonal distance of the plot in kilometre
(ii) The area of the plot in sq. km.
Solution:

Question 19.
The dimensions of the model of a multistoreyed building are lm by 60 cm by 1.20 m. If the scale factor is 1 : 50,. find the actual
dimensions of the building. Also, find :
(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is 50 sq cm.
(ii) the space (volume) inside a room of the model, if the space inside the corresponding room of the building is 90m³.
Solution:

Question 20.
In ∆ABC, ∠ACB = 90° and CD ⊥ AB. Prove that : \(\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}=\frac{\mathrm{BD}}{\mathrm{AD}}\).
Solution:

Question 21.
A triangle ABC with AB = 3 cm, BC = 6 cm and AC = 4 cm is enlarged to ∆DEF such that the longest side of ∆DEF = 9 cm. Find the scale factor and hence, the lengths of the other sides of ∆DEF.
Solution:

Question 22.
Two isosceles triangles have equal vertical angles. Show that the triangles are similar.
If the ratio between the areas of these two triangles is 16 : 25, find the ratio between their corresponding altitudes.
Solution:

Question 23.
In ∆ABC, AP: PB = 2 :3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA.

Find:
(i) area ∆APO: area ∆ABC.
(ii) area ∆APO: area ∆CQO.
Solution:

Question 24.
The following figure shows a triangle ABC in which AD and BE are perpendiculars to BC and AC respectively.

Show that:
(i) ∆ADC – ∆BEC
(ii) CA × CE = CB × CD
(iii) ∆ ABC – ∆DEC
(iv) CD × AB = CA × DE
Solution:

Question 25.
In the given figure, ABC is a triangle-with ∠EDB = ∠ACB.
Prove that ∆ABC ~ ∆EBD.
If BE=6 cm, EC = 4 cm,
BD = 5 cm and area of
∆BED = 9 cm². Calculate the
(i) length of AB
(ii) area of ∆ABC

Solution:

Question 26.
In the given figure, ABC is a right-angled triangle with ZBAC = 90°.
(i) Prove ∆ADB ~ ∆CDA.
(ii) If BD = 18 cm, CD = 8 cm, find AD.
(iii) Find the ratio of the area of ∆ADB is to area of ∆CDA.

Solution:

Question 27.
In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm. Calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.

Solution:
(i)

(ii)

(iii)

Question 28.
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:

(i) ∆ADE ~ ∆ACB.
(ii) If AC = 13 cm, BC = 5 cm and AE=4 cm. Find DEandAD.
(iii) Find, area of ∆ADE : area of quadrilateral BCED. (2015)
Solution:
(i)

(ii)

(iii)

Question 29.
Given: AB // DE and BC // EF. Prove that:
(i) \(\frac{\mathrm{AD}}{\mathrm{DG}}=\frac{\mathrm{CF}}{\mathrm{FG}}\)
(ii) ∆DFG ~ ∆ACG.

Solution:

Question 30.

More Resources for Selina Concise Class 10 ICSE Solutions

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Selina Concise Biology Class 10 ICSE Solutions The Nervous System

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 9 The Nervous System. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 9 The Nervous System

Exercise 1

Solution A.1.
(b) neurolemma

Solution A.2.
(d) Pons – consciousness

Solution A.3.
(b) Contains both sensory and motor fibres

Solution B.1.
(a) Cerebrospinal fluid
(b) Synapse
(c) Cerebrum
(d) Hypothalamus

Solution B.2.
(a) Stimulus: Receptor:: Impulse: Effectors
(b) Cerebrum: Diencephalon:: Cerebellum: Medulla oblongata
(c) Receptor: Sensory nerve:: Motor nerve: Effector

Solution B.3.
(a) Sensory
(b) Maintaining posture and equilibrium
(c) Spinal cord

Solution C.1.
(a) Corpus Callosum – It is located located in the forebrain. It connects two cerebral hemispheres and transfers information from one hemisphere to other.
(b) Central canal – It is located in centre of the spinal cord. It is in continuation with the cavities of the brain. It is filled with cerebrospinal fluid and acts as shock proof cushion. In addition, it also helps in exchange of materials with neurons.

Solution C.2.
(a) False
(b) False
(c) True
(d) True

Solution C.3.
(a)

Cerebrum	Cerebellum
The cerebrum controls all voluntary actions. It enables us to think, reason, plan and memorize.	The cerebellum on the other hand maintains balance of the body and coordinates muscular activity.

(b)

Sympathetic Nervous System	Parasympathetic Nervous System
Sympathetic nervous system prepares the body for violent action against the abnormal condition.	Parasympathetic nervous system is concerned with re-establishing normal conditions after the violent act is over.

(c)

Sensory Nerve	Motor Nerve
Sensory nerve brings impulses from the receptors i.e. sense organs to the brain or spinal cord.	Motor nerve carries impulse from the brain or spinal cord to effector organs such as muscles or glands.

(d)

Medulla Oblongata	Cerebellum
Medulla oblongata controls the  activities of internal organs and many other involuntary actions	The cerebellum on the other hand maintains balance of the body and coordinates muscular activity.

(e)

Cerebrum	Spinal Cord
The grey matter containing cytons lies in the cortex (outer region) while the white matter containing axons lies in the medullary region (inner region).	The grey matter containing cytons lies in the medullary region i.e. inner side while the white matter containing axons lies in the cortex i.e. the outer region.

Solution C.4.
(a) Cerebellum maintains balance of the body and coordinates muscular activity.
(b) Myelin sheath acts like an insulation and prevents mixing of impulses in the adjacent axons.

Solution C.5.
(a) Synapse: It is a gap between the axon terminal of one neuron and the dendrites of the adjacent neuron. It transmits nerve impulse from one neuron to another neuron.
(b) Association Neuron: It interconnects sensory and motor neurons.
(c) Medullary sheath: It provides insulation and prevents mixing of impulses in the adjacent axons.
(d) Medulla Oblongata: It controls activities of internal organs such as peristalsis, breathing and many other involuntary actions.
(e) Cerebellum: It maintains balance of the body and coordinates muscular activity.
(f) Cerebrospinal Fluid: It acts like a cushion and protects the brain from shocks.

Solution C.6.
(a) Sensory, motor and mixed nerves
(b) Somatic and autonomic nervous system
(c) Natural and conditioned reflexes
(d) Sensory, motor and association neurons
(e) Gray and white matter

Solution C.7.
(a) Stimulus — receptor — sensory neuron — central nervous system — motor neuron — effector — response
(b) Resting — depolarization — repolarization
(c) Dendrites — Dendron — perikaryon — nucleus — axon — axon endings
(d) Cerebrum — diencephalon — mid-brain — cerebellum — pons — medulla oblongata

Solution D.1.
(a) Reflex action  is an autonomic, quick and involuntary action in the body brought about by a stimulus.
(b)

Solution D.2.
The advantages of having a nervous system are as follows:

1. Keeps us informed about the outside world through sense organs.
2. Enables us to remember, think and reason out.
3. Controls and harmonizes all voluntary muscular activities such as running, holding, writing
4. Regulates involuntary activities such as breathing, beating of the heart without our thinking about them.

Solution D.3.
The brain and the spinal cord lie in the skull and the vertebral column respectively. They have an important role to play because all bodily activities are controlled by them. A stimulus from any part of the body is always carried to the brain or spinal cord for the correct response. A response to a stimulus is also generated in the central nervous system. Therefore, the brain and the spinal cord are called the central nervous system.

Solution D.4.
Reflex actions are involuntary actions which occur unknowingly. Voluntary actions on the other hand are performed consciously.
Picking up an apple and eating it is an example of voluntary action whereas withdrawal of hand on touching a hot object is an example of reflex action.

Solution D.5.

Solution D.6.

Organ	Sympathetic System	Parasympathetic System
e.g. Lungs	Dilates bronchi and bronchioles	Constricts bronchi and bronchioles
1. Heart	Accelerates heartbeat	Retards heartbeat
2. Pupil	Dilates	Constricts
3. Salivary gland	Inhibits the secretion of saliva causing the drying of the mouth	Stimulates the release of saliva

Solution E.1.
Salivation is an example of conditioned reflex that develops due to experience or learning. Saliva starts pouring when you chew or eat food. Therefore, this reflex will occur not just on the sight or smell of food. The brain actually needs to remember the taste of food. Boy B started salivating because he must have tasted that food prior unlike boy A.

Solution E.2.

Situation	Organ/body part	Change/action	Part of autonomic nervous system involved
1. You have entered a dark room	Eye	Pupil dilates	Sympathetic
2. Your body is consuming lot of glucose while running a race	Liver	Glycogen is converted into glucose in liver	Sympathetic
3. You are chewing a tasty food	Salivary gland	Salivation increases	Parasympathetic
4. You are running a race	Adrenal gland	Release of adrenaline and noradrenaline increases	Sympathetic
5. You are retiring to bed for sleep	Heart	Heart rate slows down	Parasympathetic
6. You are shivering in intense cold	Body hairs	Hair raised	Sympathetic

Solution E.3.
Fill in the following information in the diagram.

1. Central Nervous System
2. Autonomic
3. 12
4. spinal
5. 31
6. dilates
7. constricts
8. liver

Exercise 2

Solution A.1.
(b) Cornea

Solution A.2.
(b) Cochlea

Solution A.3.
(c) Eustachian tube, tympanum and utriculus

Solution A.4.
(a) Retina

Solution B.1.
(a) Rhodopsin
(b) Eustachian tube
(c) Hammer
(d) Dura mater
(e) Eustachian tube
(f) Cornea
(g) Auditory nerves
(h) Rods and cones
(i) Hypermetropia

Solution B.2.
(a) Cones: Iodopsin:: rods: rhodopsin
(b) Sound: ear drum:: dynamic balance: semi-circular canals

Solution B.3.

Solution C.1.
(a) Myopia results when the eye ball is lengthened from front to back or the lens is too curved.
Hyperopia results from either too shortening of the eyeball from front to back or when the lens is too flat.
(b) Rods are sensitive to dim light but do not respond to colour.
cones are sensitive to bright light and are responsible for colour vision.
(c) cochlea is responsible for hearing; it can perceive the senses of hearing.
Semicircular canals are responsible for perceiving the senses to maintain the body balance.
(d) Rod cells contain rhodopsin whereas the cone cells contain iodopsin.
(e) Dynamic balance is when the body is in motion whereas static balance is positional balance with respect to gravity.

Solution C.2.
(a) False
Correct statement: Deafness is caused due to rupturing of the eardrum.
(b) False
Correct statement: Semicircular canals are concerned with dynamic balance.

Solution C.3.
(a) Fovea centralis is located at the back of the eye almost at the centre of the eyeball. It is the region of the brightest vision and also of the colour vision.
(b) Organ of corti is located in the inner ear. It contains sensory cells which process hearing.

Solution C.4.
(a) True
(b) False/ Ciliary muscles regulate the size of the lens.
(c) True
(d) False/The auditory nerve responsible for sound as well as for the body balance.
(e) True
(f) False/ flavour is a combination of taste and smell.
(g) False/ short-sightedness is myopia and hyperopia is long-sightedness.
(h) True

Solution C.5.
(a) Auditory canal, tympanum, ear ossicles, oval window, cochlea
(b) Conjunctiva, cornea, lens, retina, optic nerve

Solution C.6.
(a) Organ of Corti and hearing
(b) Olfactory nerve and smell
(c) Retina and vision
(d) Taste bud and taste

Solution C.7.
(a) Lacrimal gland is a tear gland located at the upper sideward portion of the eye orbit. Its secreation lubricates the surface of the eye, washes aways the dust particles and kills germs
(b) Yellow spot is the region of brightest vision and contains maximum sensory cells whereas a blind spot contains no sensory cells and this is the point of no vision.
(c) Presbyopia is an age-old eye defect. In this condition, the lens loses flexibility resulting in far-sightedness.
Cataract is also very common in old people, the cornea becomes opaque and the vision is cut down even to blindness.
(d) The process of focusing the eye at different distances is called the power of accommodation.
(e) The image formed on the retina is inverted and real.

Solution C.8.
An optical illusion is the life-like continuous movement on the screen. Television is an example of optical illusion, where the scanning beam of a picture frame of the TV camera moves so rapidly on the viewing screen of the TV set that our eyes cannot keep pace with it.

Solution C.9.
(a) Oval window is located in the middle ear. It helps in setting the fluid in the cochlear canals into vibration.
(b) Cochlea is located in the inner ear. It helps in transmitting impulses to the brain via the auditory nerve.
(c) Semicircular canals are located in the inner ear. These help in maintaining the dynamic equilibrium of the body.
(d) Utriculus is located in the inner ear. It joins the semi-circular canals to cochlea. It also helps in maintaining static balance of the body.

Solution C.10.

Solution C.11.

Structure	Function
Yellow Spot	Region of the brightest vision
Auditory nerve	Transfers impulse from inner ear to brain
Ciliary muscle	Helps to change the focal length of the eye lens
Spinal cord	Conducts impulses
Oval window	Sets fluid in cochlear canal into vibration
Semicircular canals	Dynamic equilibrium

Solution D.1.
While reading a book, the lens is more convex or rounded due to contraction of ciliary muscles because the book is usually read from a short distance. When we raise our head and look at a distant object, the ciliary muscles relax to build the tension on the suspensory ligament so that they can stretch the lens. This change in the curvature of the lens makes us focus on distant object.

Solution D.2.
The brain sees the vivid picture of the dream through the eyes. Our eyes have actually never seen the vivid picture. This is an example of optical illusion. The area of dream is controlled by the cerebrum of the central nervous system. So sometime we can remember the vivid picture seen in the dream.

Solution D.3.
If we look at a bright object and then close our eyes, the sensation of light persists for a short period. This is known as persistence image or the after image. It lasts for one-tenth of a second. Therefore by closing the eyes and gently pressing them with your palms, you see some specs of brilliant light.

Solution D.4.
Adaptation is the ability to adjust vision in bright and dark areas. When we enter a dark room from bright light, the rhodopsin pigment broken down in bright light is regenerated. It dilates the pupil and allows more light to enter the eyes. This is called dark adaptation. On the other hand, if we enter bright area from a dark room, the rhodopsin pigment is bleached. This constricts the pupil and reduces the light entering the eyes. This is called ‘light adaptation‘

Accommodation is the process of focusing the eye at different distances. This is mainly brought about by a change in the curvature of the lens. When the ciliary muscles contract, the lens becomes thicker and we are able to focus a nearby object. On the other hand when the ciliary muscles relax, the lens remains stretched i.e. the normal condition and we are able to focus on distant object.

Solution D.5.
Our eyes are designed to focus at a great variety of distances. To focus constantly at a short distance can make the lens focusing muscles fatigued. Therefore, we do not enjoy watching a movie from a very short distance from the screen in cinema hall.

Solution D.6.

Defect of vision	Cause	Corrective measure
Myopia	Lengthening of eye ball from front to back or the lens is too curved.	Using suitable concave lens
Hyperopia	Shortening of eye ball from front to back or the lens is too flat.	Using suitable convex lens
Astigmatism	Uneven curvature of the cornea	Using suitable cylindrical lenses
Presbyopia	Loss of flexibility of lens	Using suitable convex lens
Cataract	Lens turning opaque	Surgery or use of convex lens or implantation of plastic lens.
Colour blindness	Genetic defect	No control measure
Squint	Formation of cross-eye	Surgery and suitable exercise

Solution D.7.
The three ear ossicles are: Malleus (hammer), Incus (anvil) and Stapes (stirr up).
The last ear ossicle, stapes, vibrates and transmits the vibration to the oval window.
The role of other two ear ossicles is to magnify the vibration of stapes as a result of their lever like action.

Solution D.8.
The process of focusing the eye at different distances is called the power of accommodation. The ciliary muscles are responsible for the power of accommodation.

Solution E.1.
a. The ability of the eye to focus sharply on things which are near to the eye as well as far off is known as the power of accommodation.
b. Shape of the eye:
Near vision – flattened
Distant – rounded or more convex
c. Ciliary muscles and suspensory ligament
d. In the dark: Cells – rod cells, Pigment – rhodopsin
In the light: Cells – cone cells, Pigment – iodopsin

Solution E.2.
a. The middle ear or membranous labyrinth has two structures inside it, the cochlea and the semi-circular canals.
b. Malleus, incus and stapes
c. Static balance – Utriculus and sacculus (inner ear)
Hearing – Internal ear
Dynamic balance – Semi-circular canals (inner ear)
d. Collectively they are termed as ossicles.

Solution E.3.
(a) Cornea is comparable to the lens cover of the camera.
The iris and pupil act like the aperture of a camera.
(b) The cornea is the eye’s main focusing element. It takes widely diverging rays of light and bends them through the pupil; the rays are further converged by the lens.

Solution E.4.
(a) Myopia
(b) The two possible reasons for myopia are either the eye ball is lengthened from front to back or the lens is too curved.
(c) 1 – vitreous humour, 2 – blind spot, 3-lens, 4-pupil
(d) Concave lens

Solution E.5.
(i) Ear
(ii) m – malleus, i – incus and s – stapes respectively. These are collectively called as ear ossicles.
(iii) Cochlea. The vibrating movements in the hair of the sense cells of cochlea transmit the impulse for hearing to the brain via auditory nerve.
(iv) Tympanic membrane. It vibrates and then sets the ear ossicles into vibration in the process of hearing.

Solution E.6.
(i) Ear ossicles
(ii) A – Cochlea, B – Semicircular canals, C – Ear ossicles
(iii) Cochlea helps in transmitting impulses to the brain via the auditory nerve. Semicircular canals help in maintaining dynamic equilibrium of the body.
(iv) Organ of Corti

Solution E.7.

Solution E.8.
(a) Myopia
(b) A-Normal eye, B-Myopia
(c) Looking glasses with the concave lens are required here.

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Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation (Histograms, Frequency Polygon and Ogives)

Graphical Representation Exercise 23 – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.
Construct a frequency distribution table for the number given below, using the class intervals 21-30, 31-40 … etc.
75, 67, 57, 50, 26, 33, 44, 58, 67, 75, 78, 43, 41, 31, 21, 32, 40, 62, 54, 69, 48, 47, 51, 38, 39, 43, 61, 63, 68, 53, 56, 49, 59, 37, 40, 68, 23, 28, 36, 47
Use the table obtained to draw:
(i) a histogram (ii) an ogive
Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

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Selina Concise Biology Class 10 ICSE Solutions Population- The Increasing Numbers and Rising Problems

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 12 Population The Increasing Numbers and Rising Problems. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 12 Population – The Increasing Numbers and Rising Problems

Exercise 1

Solution A.1.
(d) Use of antibiotics and prophylactic vaccinations

Solution A.2.
(a) Per 1000 people per year

Solution B.1.
Demography

Solution B.2.
(i) Tubectomy (For Female)
(ii) Vasectomy (For Male)

Solution C.1.
(a) True
(b) False

Solution C.2.
It is approximately 7 billion.

Solution C.3.

Solution C.4.
(a) False.
Dog was the first domesticated animal.

(b) True.
Rapidly growing industries made human life more and more comfortable, with greater opportunities of jobs and with more production of food. All this favoured population rise.

(c) False.
Present human population growth is following geometrical progression.

(d) False.
Birth Rate (natality) is the number of live births per 1000 people of population per year.

(e) False.
Vasectomy is the surgical method of contraception in human males while tubectomy is the surgical method used in females.

Solution C.5.
The rate of growth of population of the world is 1.092% (this rate results in about 145 net additions to the worldwide population every minute or 2.4 every second {2011 estimates}) and for India is 1.344%.

Solution C.6.
By law the minimum age is 21 years for boys and 18 years for girls.

Solution C.7.
Two advantages of small family are:

1. Parents can give more attention to their children.
2. Small family helps every country in controlling the growth of population.

Solution C.8.

1. Illiteracy:
   Most of the rural population which forms the bulk of our society is still illiterate, ignorant and superstitious.
   They also do not know the functioning of the human reproductive system.
2. Traditional Beliefs:
   Among the people from lower strata of the society, children are regarded as a gift of God and a sign of prosperity.
   Therefore, they make no effort to avoid pregnancy.

Solution C.9.
Population growth is not the only threat humanity is facing, but it will be a major contributor to the crises that await us and the planet in the coming century. Overpopulating the planet puts us all at risk of extreme environmental and social consequences that we are beginning to witness today. The extreme growth in human population is mortally taxing the Earth and its resources. Each individual person has a unique impact on the planet’s environment. Some people may be relatively less damaging than others, but no living individual is without an ecological footprint. In other words, each person needs basic resources and almost all people aspire to utilize significantly more resources than are required by their basic needs. As a result, the Earth is attempting to impose its own checks on human population. We can witness these “checks” in the form of widespread disease and the emergence of new disease strains, food and water shortages, poor harvests and violent and destructive weather caused by climate change. While it should be obvious that the Earth is a finite sphere and cannot endure infinite growth by any single species, we should also remember that Earth’s current web-of-life is the result of billions of years of complex evolution. It is irreplaceable. When we look forward to the next 40 years, the most significant population increases will take place in the areas of our world where natural resources and the infrastructure of modernity are already the scarcest. 95% of the human population growth is occurring in countries already struggling with poverty, illiteracy and civil unrest. It will further stress, the already strained ecological systems and worsen poverty in much of the developing world, thus aggravating threats to international security.

Thus, the statement made by an author ‘some great author has said that a population explosion is far more dangerous than an atomic explosion’ is true.

Solution C.10.
Poverty and population have been closely linked ever since the world faced changes due to the major revolutions. Poverty has its own effects on the population and vice versa. Poverty prevails because of illiteracy and traditional beliefs in the economically weaker strata. Since illiteracy and traditional beliefs prevail the people from this stratum, they regard children as gift of God and a sign of prosperity. They consider children to be helping hands in increasing the family income, hence they keep producing more children forgetting that their current situation would do no good for the children and they would add more to this already overburdened poverty strata. Hence, the population keeps on rising and so does poverty. As the population increases the quality of life goes down.

Solution C.11.

1. Tool making revolution.
2. Agricultural revolution.
3. Scientific industrial revolution.

Solution C.12.
According to census, the Indian population in 1981 was 685 million and it was 846 million in 1991.

Solution C.13.
Yes, there could be a corresponding operation made in women. The name of the surgical procedure in females is ‘tubectomy’. In tubectomy, the abdomen is opened and the fallopian tubes (oviducts) are cut or ligated i.e. tied with nylon thread to close the passage of the egg.

Solution C.14.
Family welfare centres are those places where any help or advice about family planning is available free of cost. These places could be any hospitals, dispensaries, etc. The inverted red triangle is the symbol of family welfare in India.

Solution C.15.
Below are some of the advantages of having a small family:

1. Financial condition of family is deeply related to the size of the family. A living cost of a large family is surely much higher than a small family. A large family has more expenses on cloth, toys, education and food whereas expenses in small family are very low.
2. Parents can easily fulfill the needs of one or two children. They can provide them best education and look after them very well whereas when there are many children to look after parents just cannot fulfill even the basic needs of the children properly. Therefore, as a result, children suffer, the parents suffer and the nation suffers.
3. A child in a small family receives more support from their parents than in a large family. In large family, parents have many children to look after, so they cannot give their best support to everyone whereas in small family parents can give more support to children as they have only one or two children to look after.
4. Family size also affects the health, especially that of the mother and the child. Frequent pregnancies can cause illness to both the mother and the children. It can disrupt the health of the women. It puts mother and baby’s health at risk. So having a small family definitely leads to healthy and happy family.

Solution D.1.

1. Food: The first and most important need of the humans (or any living organism) is food. But with the production of food rising by arithmetic progression and population growing by geometric progression i.e. the number at each step is being multiplied. At the same time growing population is increasing the use of more and more agricultural land to build houses. Thus it is evident that food would be running short for the unchecked rising population.
2. Water: Availability of clean and germ-free water for drinking purposes would be more and more scarce with increase in population; the reason would be mainly, the pollution of rivers, ponds, lakes etc.
3. Land: Man is bringing more and more land under cultivation and also using up land for building more residential colonies, factories and industries. Usable land would thus become less and less available.

Solution D.2.

1. The orthodox view, to have at least one son especially in Indian society, should be modified with education. People should be educated that their greed for a son can lead to numerous children in the household which would worsen both their family’s health and wealth. They should focus on proper upbringing of the child, be it a son or a daughter.
2. Married couples should be educated to delay the birth of their first child, to space the second with a sufficient interval for proper upbringing and to stop the third. They should also be educated to adopt family planning methods by which they can prevent pregnancy after two children. These include devices for both men and women, for example: Condoms, intrauterine devices (IUD) and oral pills.

Solution D.3.
For developing countries like India, population explosion is a curse and is damaging the development of the country and its society. The developing countries are already facing a lack in their resources, and with the rapidly increasing population, the resources available per person are reduced further, leading to increased poverty, malnutrition, and other large population-related problems. The literal meaning of population is “the whole number of people or inhabitants in a country or region”, and the literal meaning of population explosion is “a pyramiding of numbers of a biological population”. As the number of people in a pyramid increases, so do the problems related to the increased population. Some of the reasons for this population explosion are poverty, better medical facilities, and immigration from the neighboring countries. The population in India continues to increase at an alarming rate. The effects of this population increase are evident in the increasing poverty, unemployment, air and water pollution, and shortage of food, health resources and educational resources.

Solution E.1.
(a) B; 1981; 1991.
(b) B; 1981; 1991.
(c) B; 1971; 1981.

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Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 10 Electro-magnetism. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Physics Chapter 10 Electro-magnetism

Exercise 10(A)

Solution 1.

Experiment:
In Fig , AB is a wire lying in the north- south direction and connected to a battery through a rheostat and a tapping key. A compass needle is placed just below the wire. It is observed that

1. When the key is open i.e., no current passes through the wire, the needle shows no deflection and it points in the N-S direction (i.e. along the earth’s magnetic field). In this position, the needle is parallel to the wire as shown in Fig. (a).
   
2. When the key is pressed, a current passes in the wire in the direction from A to B (i.e. From south to north) and the north pole(N) of the needle deflects towards the west [Fig. (b)].
   
3. When the direction of current in the wire is reversed by reversing the connections at the terminals of the battery, North Pole (N) of the needle deflects towards the east [Fig. (c)].
   
4. If the compass needle is placed just above the wire, the North Pole (N) deflects towards east when the direction of current in wire is from A to B [Fig. (d)], but the needle deflects towards west as in fig (e), if the direction of current in wire is from B to A.
   

The above observations of the experiment suggest that a current carrying wire produces a magnetic field around it.

Solution 2.

Solution 3.

(a) On decreasing the current the magnetic field lines become rarer.
(b) The direction of magnetic field lines will get reversed.

Solution 4.

Right hand thumb rule determines the direction of magnetic field around a current carrying wire.
It states that if we hold the current carrying conductor in right hand such that the thumb points in the direction of flow of current, then the fingers encircle the wire in the direction of the magnetic fields lines.

Solution 5.

(a) The direction of magnetic field at a point just underneath is towards east.
(b) Right hand thumb rule.

Solution 6.

A current carrying conductor produces a magnetic field around it and the magnetic needle in this magnetic field experience a torque due to which it deflects to align itself in the direction of magnetic field.

Solution 7.

Solution 8.

Face of the coil exhibit North polarity.

Solution 9.

(i) Along the axis of coil inwards.
(ii) Along the axis of coil outwards.

Solution 10.

Solution 11.

Right hand thumb rule: If we hold the current carrying conductor in right hand such that the thumb points in the direction of flow of current, then the fingers encircle the wire in the direction of the magnetic fields lines.

Solution 12.

(a) A – North pole, B – South pole.
(b) The magnet will be repelled because the end of the solenoid near the north pole of magnet becomes the north pole as current at this face is anticlockwise and the two like poles repel.

Solution 13.

(a) A – North pole, B – South pole.
(b) The north pole of compass needle will deflect towards west.
Reason: The end A of the coil behaves like north pole which repels north pole of compass needle towards west.

Solution 14.

The magnetic field due to a solenoid can be made stronger by using:

1. By increasing the number of turns of winding in the solenoid.
2. By increasing the current through the solenoid.

Solution 15.

A current carrying freely suspended solenoid at rest behaves like a bar magnet. It is because a current carrying solenoid behaves like a bar magnet.

Solution 16.

The needle of the compass will rest in in the direction of magnetic field due to solenoid at that point.

Solution 17.

Magnetic fielddue to a solenoid carrying current increases if a soft iron bar is introduced inside the solenoid.

Solution 18.

(A) When current flows in a wire, it creates magnetic field around it.
(B) On reserving the direction of current in a wire, the magnetic field produced by it gets reversed.
(C) A current carrying solenoid behaves like a bar magnet
(D) A current carrying solenoid when freely suspended, it always rest in north-south direction.

Solution 19.

Solution 20.

(a) X-north pole, Y –south pole.
(b) By reducing resistance of circuit by mean of rheostat to increase current.

Solution 21.

(a) Solenoid is a cylindrical coil of diameter less than its length.
(b) The device so obtained is electromagnet.
(c) It is used in electric relay.

Solution 22.

Solution 23.

An electromagnet is a temporary strong magnet made from a piece of soft iron when current flows in the coil wound around it. It is an artificial magnet.

The strength of magnetic field of an electromagnet depends on:

1. Number of turns: The strength of magnetic increases on increasing the number of turns of winding in the solenoid.
2. Current: The strength of magnetic field increases on increasing the current through the solenoid.

Solution 24.

At A-south pole and at B-north pole.

Solution 25.

The strength of an electromagnet can be increased by following ways:

1. By increasing the number of turns of winding in the solenoid.
2. By increasing the current through the solenoid.

Solution 27.

1. An electromagnet can produce a strong magnetic field.
2. The strength of the magnetic field of an electromagnet can easily be changed by changing the current in its solenoid.

Solution 28.

Electromagnet	Permanent magnet
It is made up of soft iron	It is made up of steel.
The magnetic field strength can be changed.	The magnetic field strength cannot be changed.
The electromagnets of very strong field can be made.	The permanent magnets are not so strong.

Solution 29.

The soft iron bar acquires the magnetic properties only when an electric current flows through the solenoid and loses the magnetic properties as the current is switched off. That’s why soft iron is used as the core of the electromagnet in an electric bell.

Solution 30.

If an a.c. source is used in place of battery, the core of electromagnet will get magnetized, but the polarity at its ends will change. Since attraction of armature does not depend on the polarity of electromagnet, so the bell will still ring on pressing the switch.

Solution 31.

Solution 32.

The material used for making the armature of an electric bell is soft iron which can induce magnetism rapidly.

Solution 1 (MCQ).

The presence of magnetic field at a point can be detected by a compass needle.
Note: In the presence of a magnetic field, the needle of compass rests only in the direction of magnetic field and in the absence of any magnetic field, the needle of compass can rest in any direction. In the earth’s magnetic field alone, the needle rests along north-south direction.

Solution 2 (MCQ).

By reversing the direction of current in a wire, the magnetic field produced by it gets reversed in direction.
Hint: On reversing the direction of current in a wire, the polarity of the faces of the wire also reverses. Thus, the direction of magnetic field produced by it also gets reversed.

Exercise 10(B)

Solution 1.

The magnitude of force on a current carrying conductor placed in a magnetic field depends on:

1. On strength of magnetic field B.
2. On current I in the conductor.
3. On length of conductor.

Magnitude of force on a current carrying conductor placed in a magnetic field depends directly on these three factors.

Solution 2.

(a) When current in the conductor is in the direction of magnetic field force will be zero.
(b) When current in the conductor is normal to the magnetic field.

Solution 3.

Direction of force is also reversed.

Solution 4.

Fleming’s left hand rule: Stretch the forefinger, middle finger and the thumb of your left hand mutually perpendicular to each other. If the forefinger indicates the direction of magnetic field and the middle finger indicates the direction of current, then the thumb will indicate the direction of motion of conductor.

Solution 6.

Unit is: Newton/ampere x meter (or NA^-1m^-1).

Solution 7.

(a) The coil will experience a torque due to which it will rotate.
(b) The coil will come to rest when their plane become normal to the magnetic field.
(c) (i) When plane of a oil is parallel to the magnetic field,
(ii) When plane of coil is normal to the magnetic field.
(d) The instrument which makes use of the principle stated above is d.c. motor.

Solution 8.

(a) The coil begins to rotate in anticlockwise direction.
(b) This is because, after half rotation, the arms AB and CD get interchanged, so the direction of torque on coil reverses. To keep the coil rotating in same direction, commutator is needed to change the direction of current in the coil after each half rotation of coil.

Solution 9.

Electric motor: An electric motor is a device which converts the electrical energy into the mechanical energy.
Principle: An electric motor (dc motor) works on the principle that when an electric current is passed through a conductor placed normally in a magnetic field, a force acts on the conductor as a result of which the conductor begins to move and mechanical energy is obtained.

Solution 10.

Solution 11.

Electrical energy converts into mechanical energy.

Solution 12.

The speed of rotation of an electric motor can be increased by:

1. Increasing the strength of current.
2. Increasing the number of turns in the coil.

Solution 13.

Electric motor is used in electrical gadgets like fan, washing machine, juicer, mixer, grinder etc.

Solution 1 (MCQ).

In an electric motor, the energy transformation is from electrical to mechanical.
Note: An electric motor is a device which converts electrical energy into mechanical energy.

Exercise 10(C)

Solution 1.

(a) Electromagnetic induction: whenever there is change in number of magnetic field lines associated with conductor, an electromotive force is developed between the ends of the conductor which lasts as long as the change is taking place.

1. When the magnet is stationary there is no deflection in galvanometer. The pointer read zero. [Fig. (a)]
2. When the magnetwith north pole facing the solenoid is moved towards the solenoid, the galvanometer shows a deflection towards the right showing that a current flows in the solenoid in the direction as shown in [Fig (b)]
3. As the motion of magnet stops, the pointer of the galvanometer comes to the zero position [Fig (c)]. This shows that the current in the solenoid flows as long as the magnet is moving.
4. If the magnet is moved away from the solenoid, the current again flows in the solenoid, but now in a direction opposite to that shown in [Fig. (b)] and therefore the pointer of the galvanometer deflects towards left[ Fig. (d)].
5. If the magnet is moved away rapidly i.e. with more velocity, the extent of deflection in the galvanometer increases although the direction of deflection remains the same.It shows that more current flows now.
6. If the polarity of the magnet is reversed and then the magnet is brought towards the solenoid, the current in solenoid flows in the direction opposite to that shown in Fig (b) and so the pointer of galvanometer deflect towards left [Fig. (e)].

Solution 2.

Faraday’s formulated two laws of electromagnetic induction:

1. Whenever there is a change in the magnetic flux linked with a coil, an e.m.f. is induced. The induced e.m.f. lasts so long as there is a change in the magnetic flux linked with the coil.
2. The magnitude of the e.m.f. induced is directly proportional to the rate of change of the magnetic flux linked with the coil. If the rate of change of magnetic flux remains uniform, a steady e.m.f. is induced.

Solution 3.

Magnitude of induced e.m.f depend upon:

1. The change in the magnetic flux.
2. The time in which the magnetic flux changes.

Solution 4.

(a) Mechanical energy changes to the electrical energy.
(b) Phenomenon is called electromagnetic induction.

Solution 5.

(a) When there is a relative motion between the coil and the magnet, the magnetic flux linked with the coil changes. If the north pole of the magnet is moved towards the coil, the magnetic flux through the coil increases as shown in above figure. Due to change in the magnetic flux linked with the coil, an e.m.f. is induced in the coil. This e.m.f. causes a current to flow in the coil if the circuit of the coil is closed.
(b) The source of energy associated with the current obtained in part (a) is mechanical energy.

Solution 6.

1. When the magnet is stationary there is no deflection in galvanometer. The pointer read zero. [Fig (a)]
2. When the magnetwith north pole facing the solenoid is moved towards the solenoid, the galvanometer shows a deflection towards the right showing that a current flows in the solenoid in the direction as shown in Fig (b)
3. As the motion of magnet stops, the pointer of the galvanometer to the zero position [Fig (c)]. This shows that the current in the solenoid flows as long as the magnet is moving.
4. If the magnet is moved away from the solenoid , the current again flows in the solenoid , but now in the direction opposite to that shown in [Fig. (b)] and therefore the pointer of the galvanometer deflects towards left [Fig. (d)].
5. If the magnet is moved away rapidly i.e. with more velocity, the extent of deflection in the galvanometer increases although the direction of deflection remains the same.It shows that more current flows now.
6. If the polarity of the magnet is reversed and then the magnet is brought towards the solenoid, the current in solenoid flows in the direction opposite to that shown in Fig (b) and so the pointer of galvanometer deflect towards left [Fig. (e)].

(b) Magnitude of induced e.m.f depend upon:

1. The change in the magnetic flux.
2. The time in which the magnetic flux changes.

(c) The direction of induced e.m.f depends on whether there is an increase or decrease in the magnetic flux.

Solution 7.

The current induced in a closed circuit only if there is change in number of magnetic field lines linked with the circuit.

Solution 8.

1. Yes.
2. Yes.
3. Yes.
4. No.

Solution 9.

Fleming’s right hand rule determines the direction of current induced in the conductor.

Solution 10.

Fleming’s right hand rule: Stretch the forefinger, middle finger and the thumb of your right hand mutually perpendicular to each other. If the forefinger indicates the direction of magnetic field and the thumb will indicates the direction of motion of conductor, then the middle finger indicates the direction of induced current.

Solution 11.

Lenz’s law: It states that the direction of induced e.m.f. (or induced current) is such that it always tends to oppose the cause which produces it.

Solution 12.

When a coil has a large number of turns, then magnitude of induced e.m.f. in the coil become more and then by Lenz’s law it will oppose more.

Solution 13.

So that the mechanical energy spent in producing the change, is transformed into the electrical energy in form of induced current.

Solution 14.

Lenz’s law implies the law of conservation of energy. It shows that the mechanical energy spent in doing work, against the opposing force experienced by the moving magnet, is transformed into the electrical energy due to which current flows in the solenoid.

Solution 15.

The pointer of galvanometer deflects. The deflection last so long as the coil moves.

(a) Deflection becomes twice.
(b) Deflection becomes thrice

Solution 16.

1. The pointer of galvanometer deflects towards left. The deflection lasts so long as the coil moves.
2. (a) Deflection becomes twice (b) Deflection becomes thrice.

Solution 17.

(a)

1. When switch is closed suddenly, the galvanometer needle deflects for a moment.
2. If switch is kept closed then galvanometer needle returns to zero.
3. If switch is opened again then galvanometer needle deflects again but in opposite to the direction of deflection in case (a).

(b) This can be explained by Faraday’s law which states that whenever there is change in the magnetic flux linked with a coil, an e.m.f. is induced. The induced e.m.f. lasts as long as there is a change in the magnetic flux linked with the coil.

Solution 18.

An A.C. generator works on the principle of ‘electromagnetic induction’.
Statement: Whenever a coil is rotated in a magnetic field, the magnetic flux linked with the coil changes, and therefore, an EMF is induced between the ends of the coil. Thus, a generator acts like a source of current if an external circuit containing load is connected between the ends of its coil.

Solution 19.

The number of rotations of the coil in one second or the speed of rotation of the coil.

Solution 20.

Solution 21.

(a)In an a.c generator, if the speed at which the coil rotates is doubled, the frequency is also doubled.
(b) Maximum output voltage is also doubled.

Solution 22.

Two ways in an a.c generator to produce a higher e.m.f. are:

1. By increasing the speed of rotation of the coil.
2. By increasing the number of turns of coil.

Solution 23.

Mechanical energy changes into the electrical energy.

Solution 24.

Two dissimilarities between D.C. motor and A.C. generator:

A.C. Generator	D.C. Motor
1. A generator is a device which converts mechanical energy into electrical energy.	1. A D.C. motor is a device which converts electrical energy into mechanical energy.
2. A generator works on the principle of electromagnetic induction.	2. A D.C. works on the principle of force acting on a current carrying conductor placed in a magnetic field.

Similarity: Both in A.C generator and D.C motor, a coil rotates in a magnetic field between the pole pieces of a powerful electromagnet.

Solution 25.

The voltage of a.c. can be stepped up by the use of step-up transformer at the power generating station before transmitting it over long distances. It reduces the loss of electrical energy as heat in the transmission line wires. On the other hand, if d.c. is generated at the power generating station, its voltage cannot be increased for transmission, and so due to passage of high current in the transmission line wires, there will be a huge loss of electrical energy as heat in the line wires.

Solution 26.

The purpose of the transformer is to step up or step down the a.c. voltage.
No, a transformer cannot be used with a direct current source.

Solution 28.

Solution 29.

Solution 30.

The device is step up transformer.
It works on the principle of electromagnetic induction.

Solution 31.

Step-up transformer: The step-up transformer is used to change a low voltage alternating e.m.f. to a high voltage alternating e.m.f. of same frequency.

Working: When the terminals of primary coil are connected to the source of alternating e.m.f., a varying current flows through it which also produces a varying magnetic field in the core of the transformer. Thus, the magnetic field lines linked with the secondary coil vary and induce an e.m.f. in the secondary coil. The induced e.m.f. varies in the same manner as the applied e.m.f. in the primary coil varies, and thus, has the same frequency as that of the applied e.m.f.
The magnitude of e.m.f. induced in the secondary coil depends on the ‘turns ratio’ and the magnitude of the applied e.m.f.
For a transformer,

Two characteristics of the primary coil as compared to its secondary coil:

1. The number of turns in the primary coil is less than the number of turns in the secondary coil.
2. A thicker wire is used in the primary coil as compared to that in the secondary coil.

Solution 32.

Working: In a step down transformer, the number of turns in secondary coil are less than the number of turns in the primary coil i.e., turns ratio N_S/N_P<1.
As E_s/E_p = N_S/N_P.
So E_s/E_ps is less than E_p.

Two uses of step down transformer are:

1. With electric bells
2. At the power sub-stations to step-down the voltage before its distribution to the customers.

Solution 34.

A is the primary coil. B is the secondary coil.
We have drawn laminated core in the diagram.
The material of this part is soft iron.
This transformer a step-down transformer because the number of turns in primary coil is much greater than that in the secondary coil.

Solution 35.

(a) Soft iron core is used. The core is made up from the thin laminated sheets of soft iron of T and U shape, placed alternately one above the other and insulated from each other by paint or varnish coating over them.

Solution 36.

The secondary windings of a transformer in which the voltage is stepped down are usually made up of thicker than the primary because more current flows in the secondary coil. The use of thicker wire reduces its resistance and therefore the loss of energy as heat in the coil.

Solution 37.

To reduce the energy losses due to eddy currents.

Solution 38.

1. In a step-up transformer, the number of turns in the primary is less than the number of turns in the secondary.
2. The transformer is used in alternating current circuits.
3. In a transformer, the frequency of A.C. voltage remain same.

Solution 39.

Solution 40.

The energy loss in a transformer is called ‘copper loss’.

Copper losses: Primary and secondary coils of a transformer are generally made of copper wire. These copper wires have resistance. When current flows through these wires, a part of the energy is lost in the form of heat. This energy lost through the windings of the transformer is known as copper loss.

This loss can be minimized by using thick wires for the windings. Use of thick wire reduces its resistance and therefore reduces the loss of energy as heat in the coil.

Solution 41.

Step up transformer	Step down transformer
It increases the a.c. voltage and decrease the current.	It decreases the a.c. voltage and increase the current.
The wire of primary coil is thicker than that in the secondary coil.	The wire in the secondary coil is thicker than that in the primary coil.

Solution 42.

Soft iron is used in all.

Solution 1 (MCQ).

Fleming’s right hand rule
Statement: According to Fleming’s right hand rule, if we stretch the thumb, middle finger and forefinger of our right hand mutually perpendicular to each other such that the forefinger indicates the direction of magnetic field and thumb indicates the direction of motion of conductor, then the middle finger will indicate the direction of induced current.

Solution 2 (MCQ).

N_S > N_P
Hint: Since a step-up transformer is used to change a low voltage alternating e.m.f. to a high voltage alternating e.m.f. of same frequency, the number of turns in the secondary coil is more than the number of turns in the primary coil, i.e. N_S > N_P.

Numericals

Solution 2.

Solution 3.

Solution 4.

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Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding

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Selina ICSE Solutions for Class 10 Chemistry Chapter 2 Chemical Bonding

Exercise Intext 1

Solution 1.

Atoms lose, gain or share electrons to attain noble gas configuration.

Solution 2.

(a) A chemical bond may be defined as the force of attraction between any two atoms, in a molecule, to maintain stability.
(b) The chemical bond formed between two atoms by transfer of one or more electrons from the atom of a metallic electropositive element to an atom of a non-metallic electronegative element.
(c) The chemical bond formed due to mutual sharing of electrons between the given pairs of atoms of non-metallic elements.

Solution 3.

Conditions for formation of Ionic bond are:

1. The atom which changes into cation should possess 1, 2 or 3 valency electrons. The other atom which changes into anion should possess 5, 6 or 7 electrons in the valence shell.
2. A high difference of electronegativity of the two atoms is necessary for the formation of an Ionic bond.
3. There must be an overall decrease in energy i.e., energy must be released.
   For this an atom should have low value of Ionisation potential and the other atom should have high value of electron affinity.
4. Higher the lattice energy, greater will be the case of forming an ionic compound.

Solution 4.

It will form a cation: M³⁺
M₂(SO₄)₃
M(NO₃)₃
M₃(PO₄)₃
M₂(CO₃)₃
M(OH)₃

Solution 5.

Atoms combine with other atoms to attain stable octet or noble gas configuration.

Solution 6.

Ionic compounds are generally formed between metals and non-metals as metals always lose electrons to form cations while non-metals gain electrons forming anions to complete their octet. These oppositely charged ions are held together by electrostatic force of attraction and hence results in an ionic compound.

Solution 7.

Solution 8.

(a) X has 7 electrons in its outermost shell and Y has only one electron in its outermost shell so Y loses its one electron and X gains that electron to form an ionic bond.
(b) The formula of the compound would be XY.

Solution 9.

Solution 10.

(a) Sodium atom and sodium ion

1. Sodium atom has one electron in M shell while sodium ion has 8 electrons in L shell.
2. Sodium atom is neutral while sodium ion is positively charged.
3. Sodium atom is highly reactive while its ion is inert.
4. Sodium atom is poisonous while sodium ion is non-poisonous.

(b) Chlorine atom and chlorine ion

1. Chlorine atom has 7 electrons in its M shell while Chloride ion has 8 electrons in the same shell.
2. Chlorine atom is neutral while chloride ion is negatively charged.
3. Chlorine atom is highly reactive while its ion is inert.
4. Chlorine gas is poisonous while chloride ion is non-poisonous.

Solution 11.

Fluoride ion is negatively charged while neon atom is neutral.

Solution 12.

(a) Transfer of electron(s) is involved in the formation of an electrovalent bond. The electropositive atom undergoes oxidation, while the electronegative atom undergoes reduction. This is known as a redox process.

Oxidation: In the electronic concept, oxidation is a process in which an atom or ion loses electron(s).
Zn → Zn²⁺ + 2e^–

Reduction: In the electronic concept, the reduction is a process in which an atom or ion accepts electron(s).
Cu²⁺ + 2e^–→ Cu

(b)

1. Zn → Zn²⁺ + 2e^– (Oxidation)
   Pb²⁺ + 2e^–  → Pb (Reduction)
2. Zn → Zn²⁺ + 2e^– (Oxidation)
   Cu²⁺ + 2e^–→ Cu (Reduction)
3. Cl₂ + 2e^–→ 2Cl^– (Reduction)
   2Br^–→ Br₂ + 2e^– (Oxidation)
4. Sn²⁺→ Sn⁴⁺ + 2e^– (Oxidation)
   2Hg²⁺ + 2e^–→ Hg₂ (Reduction)
5. Cu⁺→ Cu²⁺ + e^– (Oxidation)
   Cu⁺ + e^– → Cu (Reduction)

(c)

2K + Cl₂→2KCl

1. Oxidation: In the electronic concept, oxidation is a process in which an atom or ion loses electron(s).
   K → K⁺ + e^–
2. Reduction: In the electronic concept, the reduction is a process in which an atom or ion accepts electron(s).
   Cl₂ + 2e^–→ 2Cl^–
3. Oxidising agent
   An oxidising agent oxidises other substances either by accepting electrons or by providing oxygen or an electronegative ion, or by removing hydrogen or an electropositive ion.
   Cl₂ + 2e^–→ 2Cl^–
4. Reducing agent
   A reducing agent reduces other substances either by providing electrons or by providing hydrogen or an electropositive ion, or by removing oxygen or an electronegative ion.
   K → K⁺ + e^–

Exercise Intext 2

Solution 1.

(i) Both atoms should have four or more electrons in their outermost shells, i.e., non-metals.
(ii) Both the atoms should have high electronegativity.
(iii) Both the atoms should have high electron affinity and high ionisation potential.
(iv) Electronegativity difference between the two atoms should be zero or negligible.
(v) The approach of the atoms towards one another should be accompanied by decrease of energy.

Solution 2.

(a) A is a non-metal; B is a metal while C is a chemically inert element.
(b) BA

Solution 3.

(a) (i) E (ii) B
(b) C₂D
(c) A and C are metals while B, D and E are non -metals.

Solution 3(2017).

Solution 4.

(a) Ionic compounds are formed as a result of transfer of one or more electrons from the atom of a metallic electropositive element to an atom of a non-metallic electronegative element.
A polar covalent compound is the one in which there is an unequal distribution of electrons between the two atoms.

(b) Ionic compounds, made up of ions, are generally crystalline solids with high melting and boiling points.
They are soluble in water and good conductors of electricity in aqueous solution and molten state.
Covalent compounds, made up of molecules, can exist as soft solids or liquids or gases with low melting and boiling points. They are generally insoluble in water and poor conductors of electricity.

(c) Polar covalent compounds are formed between 2 non-metal atoms that have different electronegativities and therefore have unequal sharing of the bonded electron pair. Non-polar compounds are formed when two identical non-metals equally share electrons between them.

Solution 5.

(a) X⁺
(b) X will be a strong reducing agent as it will have the tendency to donate its valence electron.

Solution 6.

Covalent compounds are said to be polar when shared pair of electrons are unequally distributed between the two atoms. For example in HCl, the high electronegativity of the chlorine atom attracts the shared electron pair towards itself. As a result, it develops a slight negative charge and hydrogen atom develops a slight positive charge. Hence, a polar covalent bond is formed.

Solution 7.

During the formation of a non-polar covalent bond between two similar atoms or dissimilar atoms, the atoms involved in sharing share the electrons equally. The molecule of methane has four carbon-hydrogen single covalent bonds. It is a non-polar covalent compound as the electrons are shared by the carbon and hydrogen atoms equally and hence the shared pair lies between the atoms at an equal distance from both carbon and hydrogen atom.

Solution 7.

b. Methane is a covalent compound and is non-polar in nature. This is because the shared pair of electrons is equally distributed between the two atoms. So, no charge separation takes place and the molecule is symmetrical and electrically neutral.

Solution 8.

(a) Properties of Ionic Compounds:

1. Ionic compounds usually exist in the form of crystalline solids.
2. Ionic compounds have high melting and boiling points.
3. Ionic compounds are generally soluble in water but insoluble in organic solvents.
4. They are good conductors of electricity in the fused or in aqueous solution state.

(b) Properties of Covalent Compounds:

1. The covalent compounds exist as gases or liquids or soft solids.
2. The melting and boiling points of covalent compounds are generally low.
3. Covalent compounds are insoluble in water but dissolve in organic solvents.
4. They are non-conductors of electricity in solid, molten or aqueous state.

Solution 9.

(a)

1. A reaction in which oxidation and reduction occur simultaneously is called an oxidation-reduction, or simply, a redox reaction.
2. Redox reactions involve the transfer of electrons between two chemical species.
3. The reaction in which electron is gained is called a reduction reaction and the reaction in which electron is lost is called oxidation reaction.
4. The compound that loses an electron is said to be oxidized, the one that gains an electron is said to be reduced.

(c)
(i) Potassium undergoes oxidation as it loses an electron and forms a cation.
(ii) Chlorine undergoes reduction as it gains an electron and forms chloride anion.
(iii) Potassium acts a reducing agent and gets oxidised.
(iv) Chlorine acts an oxidizing agent and gets reduced.

Solution 9.

Solution 10.

(a) Electrovalent compounds in the solid state do not conduct electricity because movement of ions in the solid state is not possible due to their rigid structure. But these compounds conduct electricity in the molten state. This is possible in the molten state since the electrostatic forces of attraction between the oppositely charged ions become weak. Thus, the ions move freely and conduct electricity.

(b) The atoms of covalent compounds are bound tightly to each other in stable molecules, but the molecules are generally not very strongly attracted to other molecules in the compound. On the other hand, the atoms (ions) in electrovalent compounds show strong attractions to other ions in their vicinity. This generally leads to low melting points for covalent solids, and high melting points for electrovalent solids.

(c) Electrovalent compounds dissolve in polar solvents like water because the forces of attraction between positive and negative charges become weak in water. But since covalent compound are made up of molecules, they do not ionize in water and hence do not dissolve in water.

(d) Since it takes a lot of energy to break the positive and negative charges apart from each other, the ionic compounds are so hard. But on applying stress, Ions of the same charge are brought side-by-side and so the opposite ions repel each other and crystal breaks into pieces.

(e) Since polar covalent compounds are made up of charged particles, they conduct electricity in aqueous solution.

Solution 10.

Dipole molecule is a molecule that has both, slight positive and slight negative charge.
For example, in HCl hydrogen has a slight positive charge and chlorine has a slight negative charge. The dipole moment of HCl molecule is 1.03 D and may be represented as:

Solution 11.

a.

i. Y = 9
ii. Z = 12

b. Ionic bond with molecular formula ZY₂.

Solution 12.

MgCl2 – Electrovalent compound	CCl4 – Covalent compound
They are hard crystalline solids consisting of ions.	These are gases or liquids or soft solids.
They have high melting and boiling points.	They have low melting and boiling points.
They conduct electricity in the fused or aqueous state.	They do not conduct electricity in the solid, molten or aqueous state.
These are soluble in inorganic solvents but insoluble in organic solvents.	These are insoluble in water but dissolve in organic solvents.

Solution 13.

Potassium chloride is an electrovalent compound and conducts electricity in the molten or aqueous state because the electrostatic forces of attraction weaken in the fused state or in aqueous solution.

Polar covalent compounds like hydrogen chloride ionise in their solutions and can act as an electrolyte. So, both can conduct electricity in their aqueous solutions.

Solution 14.

a. HCland NH₃

b. HCl + H₂O → H₃O⁺ + Cl^–
NH₃ + H₂O →NH₄⁺ + OH^–

Solution 15.

Formula of compound when combined with sulphur – MSFormula of compound when combined with chlorine –MCl₂

Solution 16.

(c) If the compound formed between A and B is melted and an electric current is passed through the molten compound, then element A will be obtained at the cathode and B at the anode of the electrolytic cell.

Exercise 1

Solution 1.

The bond formed between two atoms by sharing a pair of electrons, provided entirely by one of the combining atoms but shared by both is called a coordinate bond. It is represented by an arrow starting from the donor atoms and ending in the acceptor atom.

Conditions:

1. One of the two atoms must have at least one lone pair of electrons.
2. Another atom should be short of at least a lone pair of electrons.

The two lone pair of electrons in the oxygen atom of water is used to form coordinate bond with the hydrogen ion which is short of an electron resulting in the formation of the hydronium ion.

H₂O + H⁺ H₃O⁺ Over here the hydrogen ion accepts one lone pair of electrons of the oxygen atom of water molecule leading to the formation of a coordinate covalent bond.

Solution 2.

A pair of electrons which is not shared with any other atom is known as a lone pair of electrons. It is provided to the other atom for the formation of a coordinate bond.

A pair of electrons which is shared between two atoms resulting in the formation of a covalent bond is called a shared pair.

Solution 3.

a. Polar covalent bond
b. Ionic bond
c. O and H are bonded with a single covalentbond and oxygen possesses a single negative charge in the hydroxyl ion.
d. Covalent bond
e. Coordinate bond
f. Electrovalentbond, dative bond (or coordinate bond) and covalent bond

Solution 4.

Solution 5.

Mg

Solution 6.

	Sodium	Phosphorus	Carbon
Formula of chloride	NaCl	PCl5	CCl4
Nature of bonding	Ionic	Covalent	Covalent
Physical state of chloride	Solid	Solid	Liquid

Solution 7.

a.
CaO- 1 calcium atom + 1 oxygen atom
Cl₂ – 2 chlorine atoms
H₂O – 2 hydrogen atoms + 1 oxygen atom
CCl₄ – 1 carbon atom + 4 chlorine atoms

b.  

Ca – will donate two electrons
O – will accept two electrons
Cl – will accept one electron, so two Cl atoms will share an electron pair.
C – will accept four electrons by sharing electrons pairs with hydrogen forming covalent bonds.
H – will donate one electron by sharing an electron pair with carbon.

Solution 8.

(a) Unequal, polar
(b) Middle, equally
(c) Electrovalent, electrostatic

Solution 9.

a. 

1. C
2. C
3. D

b. 

1. Y is getting reduced.
2. Y is positive and it will migrate towards negative electrode that is cathode.

Solution 10.

Solution 1 (2004).

Solution 1 (2005).

(a) (i) C (ii) C (iii) D
(b) (i)reduced (ii) negative
(c) (i) H₃O⁺ ions

(ii) Like dissolves like. Since carbon tetrachloride is non-polar and water is polar compound, carbon tetrachloride does not dissolve in water.
(iii) Solid
(iv) No as ionic bonds can only be made by transfer of electrons from a metal to non metal.

Solution 1 (2006).

(a) (i) B (ii) A
(b) (i) Reduction
(ii) Oxidation
(iii) Reduction

Solution 1 (2007).

(i) Ions
(ii) Electrons are shared between the atoms of two or more elements
(iii) Two
(iv) Magnesium is oxidized and chlorine is reduced

Solution 1 (2008).
(a)
(i) D
(b)
(i) Covalent bond
(ii) Coordinate bond.

Solution 1 (2009).

Solution 1 (2010).

a. Oxidation

b.
i. ionic bond
ii. covalent and oordinate bond
iii. covalent bond

Solution 1 (2011).

c. HCl is a covalent compound formed by sharing one electron between chlorine and hydrogen. Because chlorine is more electronegative than hydrogen, the shared pair of electrons shifts towards the chlorine atom. So, a partial negative charge (δ^–) develops on chlorine and a partial positive charge (δ⁺) develops on hydrogen. Hence, the covalent bond is polar in nature.

Solution 1 (2012).

Solution 1 (2013).

a. Dative or coordinate bond
b. B Ammonium chloride
c. C Are insoluble in water
d.

Carbon tetrachloride	Sodium chloride
It is insoluble in water but dissolves in organic solvents.	It is soluble in water but insoluble in organic solvents.
It is a non-conductor of electricity due to the absence of ions.	It does not conduct electricity in the solid state but conducts electricity in the fused or aqueous state.

Solution 1 (2014).

a. B
b. D
c. Ionisation
d. Their constituent particles are molecules. These exist as gases or liquids or soft solids because they have weak forces of attraction between their molecules.

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Selina Concise Physics Class 10 ICSE Solutions Spectrum

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Selina ICSE Solutions for Class 10 Physics Chapter 6 Spectrum

Exercise 6(A)

Solution 1.

The deviation produced by the prism depends on the following four factors:

1. The angle of incidence – As the angle of incidence increases, first the angle of deviation decreases and reaches to a minimum value for a certain angle of incidence. By further increasing the angle of incidence, the angle of deviation is found to increase.
2. The material of prism (i.e., on refractive index) – For a given angle of incidence, the prism with a higher refractive index produces a greater deviation than the prism which has a lower refractive index.
3. Angle of prism- Angle of deviation increases with the increase in the angle of prism.
4. The colour or wavelength of light used- Angle of deviation increases with the decrease in wavelength of light.

Solution 2.

The deviation caused by a prism increases with the decrease in the wavelength of light incident on it.

Solution 3.

Speed of light increases with increase in the wavelength.

Solution 4.

Red colour travels fastest and Blue colour travels slowest in glass.

Solution 5.

Colour of light is related to its wavelength.

Solution 6.

(i) 4000 Å to 8000 Å
(ii) 400 nm to 800 nm

Solution 7.

(i) For blue light, approximate wavelength = 4800 Å
(ii) For red light, approximate wavelength = 8000 Å

Solution 8.

Seven prominent colours of the white light spectrum in order of their increasing frequencies:
Red, Orange, Yellow, Green, Blue, Indigo, Violet

Solution 10.

Green, Yellow orange and red have wavelength longer than blue light.

Solution 11.

A glass prism deviates the violet light most and the red light least.

Solution 12.

(a) In vacuum, both have the same speeds.
(b) In glass, red light has a greater speed.

Solution 13.

The phenomenon of splitting of white light by a prism into its constituent colours is known as dispersion of light.

Solution 14.

When white light is incident on the first surface of a prism and enters in glass, light of different colours due to different speeds in glass, is refracted or deviated through different angles. Thus the dispersion of white light into its constituent colours takes place at the first surface of prism. Thus the cause of dispersion is the change in speed of light with wavelength or frequency.

Solution 15.

When white light is incident on the first surface of a prism and enters in glass, light of different colours due to different speeds in glass, is refracted or deviated through different angles. Thus the dispersion of white light into its constituent colours takes place at the first surface of prism.

On the second surface, only refraction takes place and different colours are deviated through different angles. As a result, the colours get further separated on refraction at the second surface (violet being deviated the most and red the least).

Solution 16.

The colour band obtained on a screen on passing white light through a prism is called the spectrum.

Solution 17.

(a) Violet, Indigo, Blue, Green, Yellow, Orange, Red.
(b) No, different colours have different widths in the spectrum.
(c) (i) Violet colour is deviated the most. (ii) Red colour is deviated the least.

Solution 18.

Solution 19.

• Constituent colours of white light are seen on the screen after dispersion through the prism.
  
• When a slit is introduced in between the prism and screen to pass only the light of green colour, only green light is observed on the screen.
• From the observation, we conclude that prism itself produces no colour.

Solution 20.

• If a monochromatic beam of light undergoes minimum deviation through an equi-angular prism, then the beam passes parallel to the base of prism.
• White light splits into its constituent colours i.e., spectrum is formed.
• We conclude that white light is polychromatic.

Solution 1 (MCQ).

Both deviation and dispersion.
Hint: When a white light ray falls on the first surface of a prism, light rays of different colours due to their different speeds in glass get refracted (or deviated) through different angles. Thus, the dispersion of white light into its constituent colours takes place at the first surface of prism.

Solution 2 (MCQ).

The colour of the extreme end opposite to the base of the prism is red.
Hint: The angle of deviation decreases with the increase in wavelength of light for a given angle of incidence. Since the red light has greatest wavelength, it gets deviated the least and is seen on the extreme end opposite to the base of prism.

Numericals

Solution 1.

Solution 2.

Exercise 6(B)

Solution 1.

(a) Five radiations in the order of their increasing frequencies are:
Infrared waves, Visible light, Ultraviolet, X-rays and Gamma rays.
(b) Gamma rays have the highest penetrating power.

Solution 2.

(a) Gamma rays, X-rays, infrared rays, micro waves, radio waves.
(b) Microwave is used for satellite communication.

Solution 3.

(a) Gamma ray.
(b) Gamma rays have strong penetrating power.

Solution 5.

(a) X-rays are used in the study of crystals.
(b) It is also used to detect fracture in bones.

Solution 8.

4000 Å to 8000 Å

Solution 9.

(i) Infrared
(ii) Ultraviolet

Solution 10.

The part of spectrum beyond the red and the violet ends is called the invisible spectrum as our eyes do not respond to the spectrum beyond the red and the violet extremes.

Solution 11.

(a) infrared radiation
(b) ultra violet radiation

Solution 12.

(i) Ultraviolet rays-wavelength range 100Å to 4000Å

(ii) Visible light-wavelength range 4000Å to 8000Å

(iii) Infrared radiations-wavelength range 8000Å to 10⁷Å

Solution 13.

(i) Infrared radiations are longer than 8 x 10^-7m.
(ii) ultraviolet radiations are shorter than 4 x 10^-7 m.

Solution 14.

Solution 15.

(i) Microwaves are used for satellite communication.
(ii) Ultraviolet radiations are used for detecting the purity of gems, eggs, ghee etc.
(iii) Infrared radiations are used in remote control of television and other gadgets.
(iv) Gamma rays are used in medical science to kill cancer cells.

Solution 16.

Solution 17.

(a) A- Gamma rays, B-infrared radiations
(b) Ratio of speeds of these waves in vacuum is 1:1 as all electromagnetic waves travel with the speed of light in vacuum.

Solution 18.

All heated bodies such as a heated iron ball, flame, fire etc., are the sources of infrared radiations.
The electric arc and sparks give ultraviolet radiations.

Solution 19.

Infrared radiations are the electromagnetic waves of wavelength in the range of 8000Å to 10⁷Å.

Detection: If a thermometer with a blackened bulb is moved from the violet end towards the red end, it is observed that there is a slow rise in temperature, but when it is moved beyond the red region, a rapid rise in temperature is noticed. It means that the portion of spectrum beyond the red end has certain radiations which produce a strong heating effect, but they are not visible. These radiations are called the infrared radiations.

Use: The infrared radiations are used for therapeutic purposes by doctors.

Solution 20.

The electromagnetic radiations of wavelength from 100Å to 4000Å are called the ultraviolet radiations.

Detection: If the different radiations from the red part of the spectrum to the violet end and beyond it, are made incident on the silver-chloride solution, it is observed that from the red to the violet end, the solution remains unaffected. However just beyond the violet end, it first turns violet and finally it becomes dark brown. Thus there exist certain radiations beyond the violet end of the spectrum, which are chemically more active than visible light, called ultraviolet radiations.

Use: Ultraviolet radiations are used for sterilizing purposes.

Solution 21.

• Ultraviolet radiations travel in a straight line with a speed of 3 x 10⁸ m in air (or vacuum).
• They obey the laws of reflection and refraction.
• They affect the photographic plate.

Solution 22.

• Ultraviolet radiations produce fluorescence on striking a zinc sulphide screen.
• They cause health hazards like cancer on the body.

Solution 23.

• Infrared radiations travel in straight line as light does, with a speed equal to 3 x 10⁸m/s in vacuum.
• They obey the laws of reflection and refraction.
• They do not cause fluorescence on zinc sulphide screen.

Solution 24.

Solution 25.

Solution 26.

Solution 27.

1. Infrared radiations are used in photography in fog because they are not much scattered by the atmosphere, so they can penetrate appreciably through it.
2. Infrared radiations are used as signals during the war as they are not visible and they are not absorbed much in the medium.
3. Infrared lamps are used in dark rooms for developing photographs since they do not affect the photographic film chemically, but they provide some visibility.
4. Infrared spectrum can be obtained only with the help of a rock-salt prism since the rock-salt prism does not absorb infrared radiations whereas a glass prism absorbs them.
5. A quartz prism is used to obtain the spectrum of the ultraviolet radiations as they are not absorbed by quartz, whereas ordinary glass absorbs the ultraviolet light.
6. Ultraviolet bulbs have a quartz envelope instead of glass as they are not absorbed by quartz, whereas ordinary glad absorbs the ultraviolet light.

Solution 1 (MCQ).

Gamma rays

Solution 2 (MCQ).

Carbon arc-lamp

Solution 3 (MCQ).

Infrared radiation
Hint: Infrared radiations produce strong heating effect.

Numericals

Solution 1.

(a) Frequency =500MHz =500 x 10⁶Hz
Wavelength= 60 cm=0.6 m
Velocity of wave= frequency x wavelength
= 500x 10⁶ x 0.6=3 x 10⁸m/s
(b) Electromagnetic wave is travelling through air.

Solution 2.

Exercise 6(C)

Solution 1.

When white light from sun enters the earth’s atmosphere, the light gets scattered i.e., the light spreads in all directions by the dust particles, free water molecules and the molecules of the gases present in the atmosphere. This phenomenon is called scattering of light.

Solution 2.

The intensity of scattered light is found to be inversely proportional to the fourth power of wavelength of light. This relation holds when the size of air molecules is much smaller than the wavelength of the light incident.

Solution 3.

Violet colour is scattered the most and red the least as the intensity of scattered light is found to be inversely proportional to the fourth power of wavelength of light.

Solution 6.

Since the wavelength of red light is the longest in the visible light, the light of red colour is scattered the least by the air molecules of the atmosphere and therefore the light of red colour can penetrate to a longer distance. Thus red light can be seen from the farthest distance as compared to other colours of same intensity. Hence it is used for danger signal so that the signal may be visible from the far distance.

Solution 7.

On the moon, since there is no atmosphere, therefore there is no scattering of sun light incident on the moon surface. Hence to an observer on the surface of moon (space), no light reaches the eye of the observer except the light directly from the sun. Thus the sky will have no colour and will appear black to an observer on the moon surface.

Solution 8.

Scattering property of light is responsible for the blue colour of the sky as the blue colour is scattered the most due to its short wavelength.

Solution 9.

As the light travels through the atmosphere, it gets scattered in different directions by the air molecules present in its path. The blue light due to its short wavelength is scattered more as compared to the red light of long wavelength. Thus the light reaching our eye directly from sun is rich in red colour, while the light reaching our eye from all other directions is the scattered blue light. Therefore, the sky in direction other than in the direction of sun is seen blue.

Solution 10.

At the time of sunrise and sunset, the light from sun has to travel the longest distance of atmosphere to reach the observer. The light travelling from the sun loses blue light of short wavelength due to scattering, while the red light of long wavelength is scattered a little, so is not lost much. Thus blue light is almost absent in sunlight reaching the observer, while it is rich in red colour.

Solution 11.

At noon, the sun is above our head, so we get light rays directly from the sun without much scattering of any particular colour. Further, light has to travel less depth of atmosphere; hence the sky is seen white.

Solution 12.

The clouds are nearer the earth surface and they contain dust particles and aggregates of water molecules of sizes bigger than the wavelength of visible light. Therefore, the dust particles and water molecules present in clouds scatter all colours of incident white light from sun to the same extent and hence when the scattered light reaches our eye, the clouds are seen white.

Solution 1 (MCQ).

Blue colour
Hint: When light of certain frequency falls on that atom or molecule, this atom or molecule responds to the light, whenever the size of the atom or molecule comparable to the wavelength of light. The sizes of nitrogen and oxygen molecules in atmosphere are comparable to the wavelength of blue light. These molecules act as scattering centers for scattering of blue light. This is also the reason that we see the sky as blue.

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Selina Concise Physics Class 10 ICSE Solutions Current Electricity

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 8 Current Electricity. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Physics Chapter 8 Current Electricity

Exercise 8(A)

Solution 1.

Current is defined as the rate of flow of charge.
I=Q/t
Its S.I. unit is Ampere.

Solution 2.

Electric potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point. Its unit is the volt.

Solution 3.

The potential difference between two points is equal to the work done in moving a unit positive charge from one point to the other.
It’s S.I. unit is Volt.

Solution 4.

One volt is the potential difference between two points in an electric circuit when 1 joule of work is done to move charge of 1 coulomb from one point to other.

Solution 7.

In a metal, the charges responsible for the flow of current are the free electrons. The direction of flow of current is conventionally taken opposite to the direction of motion of electrons.

Solution 8.

It states that electric current flowing through a metallic wire is directly proportional to the potential difference V across its ends provided its temperature remains the same. This is called Ohm’s law.
V = IR

Solution 11.

Solution 12.

Ohmic Resistor: An ohmic resistor is a resistor that obeys Ohm’s law. For example: all metallic conductors (such as silver, aluminium, copper, iron etc.)

Solution 13.

Solution 14.

1. Ohmic resistor obeys ohm’s law i.e., V/I is constant for all values of V or I; whereas Non-ohmic resistor does not obey ohm’s law i.e., V/I is not same for all values of V or I.
2. In Ohmic resistor, V-I graph is linear in nature whereas in non-ohmic resistor, V-I graph is non-linear in nature.

Solution 16.

In the above graph, T₁ > T₂. The straight line A is steeper than the line B, which leads us to conclude that the resistance of conductor is more at high temperature T₁ than at low temperature T₂. Thus, we can say that resistance of a conductor increases with the increase in temperature.

Solution 17.

Solution 18.

Resistance of a wire is directly proportional to the length of the wire.
R ∝ l
The resistance of a conductor depends on the number of collisions which the electrons suffer with the fixed positive ions while moving from one end to the other end of the conductor. Obviously the number of collisions will be more in a longer conductor as compared to a shorter conductor. Therefore, a longer conductor offers more resistance.

Solution 19.

With the increase in temperature of conductor, both the random motion of electrons and the amplitude of vibration of fixed positive ions increase. As a result, the number of collisions increases. Hence, the resistance of a conductor increases with the increase in its temperature.
The resistance of filament of a bulb is more when it is glowing (i.e., when it is at a high temperature) as compared to when it is not glowing (i.e., when it is cold).

Solution 20.

Iron wire will have more resistance than copper wire of the same length and same radius because resistivity of iron is more than that of copper.

Solution 21.

1. Resistance of a wire is directly proportional to the length of the wire means with the increase in length resistance also increases.
   R ∝ l
2. Resistance of a wire is inversely proportional to the area of cross-section of the wire. If area of cross-section of the wire is more, then resistance will be less and vice versa.
   R ∝ 1/A
3. Resistance increases with the increase in temperature since with increase in temperature the number of collisions increases.
4. Resistance depends on the nature of conductor because different substances have different concentration of free electrons.Substances such as silver, copper etc. offer less resistance and are called good conductors; but substances such as rubber, glass etc. offer very high resistance and are called insulators.

Solution 22.

The resistivity of a material is the resistance of a wire of that material of unit length and unit area of cross-section.
Its S.I. unit is ohm metre.

Solution 23.

Solution 24.

Metal < Semiconductor < Insulator

Solution 26.

Manganin

Solution 28.

‘Copper or Aluminium’ is used as a material for making connection wires because the resistivity of these materials is very small, and thus, wires made of these materials possess negligible resistance.

Solution 30.

Manganin is used for making the standard resistor because its resistivity is quite large and the effect of change in temperature on their resistance is negligible.

Solution 31.

Generally fuse wire is made of an alloy of lead and tin because its resistivity is high and melting point is low.

Solution 32.

A wire made of tungsten is used for filament of electric bulb because it has a high melting point and high resistivity.

A nichrome wire is used as a heating element for a room heater because the resistivity of nichrome is high and increase in its value with increase in temperature is high.

Solution 33.

A superconductor is a substance of zero resistance at a very low temperature. Example: Mercury at 4.2 K.

Solution 34.

Superconductor

Solution 1 (MCQ).

Nichrome is an ohmic resistance.
Hint: Substances that obey Ohm’s law are called Ohmic resistors.

Solution 2 (MCQ).

For carbon, resistance decreases with increase in temperature.
Hint: For semiconductors such as carbon and silicon, the resistance and resistivity decreases with the increase in temperature.

Numericals

Solution 1.

Solution 9.

Solution 11.

Solution 12.

Exercise 8(B)

Solution 1.

e.m.f.: When no current is drawn from a cell, the potential difference between the terminals of the cell is called its electro-motive force (or e.m.f.).
Terminal voltage: When current is drawn from a cell, the potential difference between the electrodes of the cell is called its terminal voltage.
Internal Resistance: The resistance offered by the electrolyte inside the cell to the flow of electric current through it is called the internal resistance of the cell.

Solution 2.

e.m.f. of cell	Terminal voltage of cell
1. It is measured by the amount of  work done in moving a unit positive charge in the complete circuit inside and outside the cell.	1. It is measured by the amount of work done in moving a unit positive charge in the circuit outside the cell.
2. It is the characteristic of the cell i.e., it does not depend on the amount of current drawn from the cell	2. It depends on the amount of current drawn from the cell. More the current is drawn from the cell, less is the terminal voltage.
3. It is equal to the terminal voltage when cell is not in use, while greater than the terminal voltage when cell is in use.	3. It is equal to the emf of cell when cell is not in use, while less than the emf when cell is in use.

 

Solution 3.

Internal resistance of a cell depends upon the following factors:

1. The surface area of the electrodes: Larger the surface area of the electrodes, less is the internal resistance.
2. The distance between the electrodes: More the distance between the electrodes, greater is the internal resistance.

Solution 4.

Solution 5.

(a) Terminal voltage is less than the emf : Terminal Voltage < e.m.f.
(b) e.m.f. is equal to the terminal voltage when no current is drawn.

Solution 6.

When the electric cell is in a closed circuit the current flows through the circuit. There is a fall of potential across the internal resistance of the cell. So, the p.d. across the terminals in a closed circuit is less than the p.d. across the terminals in an open circuit by an amount equal to the potential drop across the internal resistance of the cell.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

(a) series
(b) parallel
(c) parallel
(d) series

Solution 11.

For the same change in I, change in V is less for the straight line A than for the straight line B (i.e., the straight line A is less steeper than B), so the straight line A represents small resistance, while the straight line B represents more resistance. In parallel combination, the resistance decreases while in series combination, the resistance increases. So A represents the parallel combination.

Solution 1 (MCQ).

In series combination of resistances, current is same in each resistance.
Hint: In a series combination, the current has a single path for its flow. Hence, the same current passes through each resistor.

Solution 2 (MCQ).

In parallel combination of resistances, P.D. is same across each resistance.
Hint: In parallel combination, the ends of each resistor are connected to the ends of the same source of potential. Thus, the potential difference across each resistance is same and is equal to the potential difference across the terminals of the source (or battery).

Solution 3 (MCQ).

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.

Solution 12.

Solution 13.

Solution 14.

Solution 15.

Solution 16.

Solution 17.

Solution 18.

Solution 19.

Solution 20.

Solution 21.

Solution 23.

Solution 24.

Solution 25.

Solution 26.

Solution 27.

Solution 28.

Solution 29.

Solution 30.

Exercise 8(C)

Solution 1.

Solution 2.

Solution 3.

Solution 4.

The S.I. unit of electrical energy is joule.
1Wh = 3600 J

Solution 5.

The power of an appliance is 100 W. It means that 100 J of electrical energy is consumed by the appliance in 1 second.

Solution 6.

The S.I. unit of electrical power is Watt.

Solution 7.

(i) The household unit of electricity is kilowatt-hour (kWh).
One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour.
(ii) The voltage of the electricity that is generally supplied to a house is 220 Volt.

Solution 8.

(i) Electrical power is measured in kW and
(ii) Electrical energy is measured in kWh.

Solution 9.

One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour.
Its value in SI unit is 1kWh = 3.6 x 10⁶J

Solution 10.

Kilowatt is the unit of electrical power whereas kilowatt-hour is the unit of electrical energy.

Solution 11.

Solution 12.

An electrical appliance such as electric bulb, geyser etc. is rated with power (P) and voltage (V) which is known as its power rating. For example: If an electric bulb is rated as 50W-220V, it means that when the bulb is lighted on a 220 V supply, it consumes 50 W electrical power.

Solution 13.

It means that if the bulb is lighted on a 250 V supply, it consumes 100 W electrical power (which means 100J of electrical energy is converted in the filament of bulb into the light and heat energy in 1 second).

Solution 14.

Solution 15.

Solution 16.

When current is passed in a wire, the heat produced in it depends on the three factors:

1. on the amount of current passing through the wire,
2. on the resistance of wire and
3. on the time for which current is passed in the wire.

• Dependence of heat produced on the current in wire: The amount of heat H produced in the wire is directly proportional to the square of current I passing through the wire,  i.e., H ∝ I²
• Dependence of heat produced on the resistance of wire: The amount of heat H produced in the wire is directly proportional to the resistance R of the wire, i.e., H ∝ R
• Dependence of heat produced on the time: The amount of heat H produced in the wire is directly proportional to the time t for which current is passed in the wire, i.e., H ∝ t

 

Solution 1 (MCQ).

Solution 2 (MCQ).

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

When one lamp is connected across the mains, it draws 0.25 A current, while if two lamps are connected in series across the mains, current through each bulb becomes

(i.e., current is halved), hence heating (= I²Rt) in each bulb becomes one-fourth, so each bulb appears less bright.

Solution 11.

Solution 12.

Solution 13.

Solution 14.

Solution 15.

Solution 16.

Solution 17.

Solution 18.

Solution 19.

Solution 20.

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Selina Concise Chemistry Class 10 ICSE Solutions Periodic Table, Periodic Properties and Variations of Properties

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 1 Periodic Table, Periodic Properties and Variations of Properties. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 1 Periodic Table, Periodic Properties and Variations of Properties

Exercise Intext 1

Solution 1.

(a) The modern periodic law states that “The properties of elements are the periodic functions of their atomic number.” Henry Moseley put forward the modern periodic law.
(b) A tabular arrangement of the elements in groups (vertical columns) and periods (horizontal rows) highlighting the regular trends in properties of elements is called a Periodic Table. Modern Periodic table has 7 periods and 18 groups.

Solution 2.

Valency is the combining capacity of the atom of an element. It is equal to the number of electrons an atom can donate or accept or share. It is just a number and does not have a positive or negative sign.
Group 1elements have 1 electron in their outermost orbital, while Group 7 elements have 7 electrons in their outermost orbital.
Valency depends on the number of electrons in the outermost shell (i.e. valence shell).
If the number of electrons present in the outermost shell is 1, then it can donate one electron while combining with other elements to obtain a stable electronic configuration.
If the number of electrons present in the outermost shell is 7, then its valency is again 1 (8 – 7 = 1) as it can accept 1 electron from the combining atom.
In a given period, the number of electrons in the valence (outermost) shell increases from left to right. But the valency increases only up to Group 14, where it becomes 4, and then it decreases, that is, it becomes 1 in Group 17.

Solution 3.

The horizontal rows are known as periods and vertical columns in the periodic table are known as groups.

Solution 4.

Periodicity is observed due to the similar electronic configuration..

Solution 5.

(i) Though the number of shells remain the same, number of valence electrons increases by one, as we move across any given period from left to right.
(ii) While going from top to bottom in a group, the number of shells increases successively i.e. one by one but the number of valence electrons remains the same.

Solution 6.

(a) Elements in the same group have equal valency.
(b) Valency depends upon the number of valence electrons in an atom.
(c) Copper and zinc are transition elements.
(d) Noble gases are placed at the extreme right of the periodic table.

Solution 7.

(a) Sodium and Potassium
(b) Calcium and Magnesium
(c) Chlorine and Bromine
(d) Neon and Argon
(e) Iron and Cobalt
(f) Cerium and Europium
(g) Uranium and Neptunium

Solution 8.

(a) The properties that reappear at regular intervals, or in which there is a gradual variation at regular intervals, are called periodic properties and the phenomenon is known as the periodicity of elements.
(b) The third period elements, Na, Mg, Al, Si, P and Cl summarize the properties of their respective groups and are called typical elements.
(c) The elements of the second period show resemblance in properties with the elements of the next group of the third period leading to a diagonal relationship. Such elements are called bridge elements.

Solution 9.

Beryllium and magnesium will show similar chemical reactions as calcium. Since these elements belong to same group 2 and also have two electrons in their outermost shell like calcium.

Solution 10.

1. Metals: Lithium, Beryllium, Sodium, Magnesium, Aluminium, Potassium, Calcium
2. Metalloids: Boron, Silicon
3. Non-metals: Hydrogen, Helium, Carbon, Nitrogen, Oxygen, Fluorine, Neon, Phosphorous, Sulphur, Chlorine, Argon

Solution 11.

(i) Properties: Non-metallic, highest electronegativity in the respective periods, highest ionisation potentials in the respective periods, highest electron affinity in the respective periods
 (ii) Salt-forming; hence, the common name is halogens.

Solution 12.

The main characteristic of the last element in each period of the periodic table is they are inert or chemically unreactive.
The general name of such elements is ‘Noble gases’.

Solution 13.

According to atomic structure, the number of valence electrons determines the first and the last element in a period.

Solution 14.

Elements	Valency	Formula of oxides
Na	1	Na2O
Mg	2	MgO
Al	3	Al2O3
Si	4	SiO2
P	5	P2O5
S	2	SO2
Cl	1	Cl2O

Solution 15.

(i) Noble gases
(ii) Representative elements
(iii) Transition elements
(iv) Halogens
(v) Alkaline Earth metals

Solution 16.

(i) 30
(ii) It belongs to group 12 and fourth period.
(iii) It is a metal.
(iv) The name assigned to this group is IIB.

Solution 17.

(i) Electronic configuration of P: 2,8,5
(ii) 15^th Group and 3^rd Period.
(iii) Valency of P = 8 – 5 = 3
(iv) Phosphorus is a non-metal.
(v) It is an oxidizing agent.
(vi) Formula with chlorine = PCl₃

Exercise Intext 2

Solution 1.

Atomic size is the distance between the centre of the nucleus of an atom and its outermost shell.
It’s measured in Angstrom and Picometre.

Solution 2.

(i) The atomic size of an atom increases when we go down a group from top to bottom.
(ii) It increases as we move from right to left in a period.

Solution 3.

Second Period: Fluorine <Neon< Oxygen< Nitrogen < Carbon < Boron< Beryllium < Lithium.
Third Period: Chlorine < Argon < Sulphur < Phosphorus < Silicon < Aluminum < Magnesium < Sodium.

Solution 4.

(i) The size of Neon is bigger compared to fluorine because the outer shell of neon is complete(octet).As a result, the effect of nuclear pull over the valence shell electrons cannot be seen. Hence the size of Neon is greater than fluorine.
(ii) Since atomic number of magnesium is more than sodium but the numbers of shells are same, the nuclear pull is more in case of Mg atom. Hence its size is smaller than sodium.

Solution 5.

(i) An atom is always bigger than cation since cation is formed by the loss of electrons; hence protons are more than electrons in a cation. So the electrons are strongly attracted by the nucleus and are pulled inward.
(ii) An anion is bigger than an atom since it is formed by gain of electrons and so the number of electrons are more than protons. The effective positive charge in the nucleus is less, so less inward pull is experienced. Hence the size expands.
(iii) Fe ²⁺ is bigger than Fe³⁺ since Fe ²⁺ has more number of electrons than Fe³⁺ and hence the inner pull by nucleus is less strong on it as compared to the pull on Fe³⁺.

Solution 6.

1. In increasing metallic character: F < O < N < C < B < Be < Li
2. In decreasing non-metallic character: Cl > S > P > Si > Al > Mg > Na

Solution 7.

(i) Across a period, the chemical reactivity of elements first decreases and then increases.
(ii) Down the group, chemical reactivity increases as the tendency to lose electrons increases down the group.

Solution 7.

The periodic variation in electronic configuration as one move sequentially in increasing order of atomic number produces a periodic variation in properties.
As the elements are arranged in increasing order of atomic number, the metals with tendency to lose electrons are placed on the left and the metallic character decreases from left to right and increases down a group and non-metals with tendency to gain electrons are placed automatically on the right and the non-metallic character increase across a period and decreases down a group.

Solution 8.

(i) The metallic character decreases as we go from left to right in a period.
(ii) It increases as we go down a group.

Solution 9.

(i) The element from the 17^th group has 7 electrons in its outermost shell.
(ii) The name of the element is chlorine.
(iii) Chlorine belongs to the halogen family.
(iv) The element has ₁₃Y²⁷ three electrons in its outermost shell which it can donate; hence, its valency is three. While the valency of chlorine is 1. Thus, ₁₃Y²⁷ which is Aluminium can donate three electrons, and chlorine can accept 1 electron to get the stable electronic configuration.
Therefore, the formula of the compound is AlCl₃.

Solution 10.

(i) Yes, these elements belong to the same group but are not from the same period.
(ii) We know that m.p. decreases on going down the group. Hence, from the above table, the elements can be ordered according to their period as follows:

Elements	B	C	A
m.p.	180.0	97.0	63.0

The metallic character increases as one moves down the group.
Hence, the order of the given elements with increasing metallic character is as follows:
B

Solution 10.

The melting and boiling points of metals decrease on going down the group.
Example: Observe the trend in group 1 elements given in the following table:

Solution 11.

Correct option: (ii) Potassium

Solution 12.

Correct option: (iii) I^–

Solution 13.

(i)  Barium will form ions most readily as the outermost valence electron which experiences the least force of attraction by positively charged nucleus can be given away readily to form cations.
(ii) All Group II elements have two valence electrons.

Solution 14.

Solution 15.

1. Group = 1
2. Period = 4
3. Valence electrons = 1
4. Valency = 1
5. Metal

Solution 16.

(i) It belongs to group II and has 2 valence electrons, so it is a metal.
(ii) Barium is placed below calcium in the group. Since the reactivity increases below the group, barium is more reactive than calcium.
(iii) It needs to lose 2 valence electrons to complete its octet configuration, so its valency is 2.
(iv) The formula of its phosphate will be Ba₃ (PO₄)2.
(v) As we move from left to right in a period, the size decreases, so it will be smaller than caesium.

Solution 17.

Sincethe size of the atom increases down the group, the ionic radii will also increase. Hence, the order of increasing atomic numbers in the group is Z < Y < X.

Solution 18.

(i) All groups do not contain both metals and non-metals. Group I and II contain only metals.
(ii) Atoms of elements in the same group have same number of valence electrons. They have same number of electrons present in their outermost shell.
(iii) The non-metallic character increases across a period with increase in atomic number. This is because across the period, the size of atom decreases and the valence shell electrons are held more tightly.
(iv) On moving from left to right in a period, the reactivity of elements first decreases and then increases, while in groups, chemical reactivity of metals increases going down the group whereas reactivity of non-metals is decreases down the group.

Solution 19.

(i) A metal of valency one = 19
(ii) A solid non-metal of period 3 = 15
(iii) A rare gas = 2
(iv) A gaseous element with valency 2 = 8
(v) An element of group 2 = 4

Solution 20.

(i) The properties of the elements are a periodic function of their atomic number.
(ii) Moving across a period of the periodic table, the elements show increasing non-metallic character.
(iii) The elements at the bottom of a group would be expected to show more metallic character than the elements at the top
(iv) The similarities in the properties of a group of elements are because they have the same number of outer electrons.

Solution 21(i).

An anion is formed by the gain of electrons. In the chloride ion, the number of electrons is more than the number of protons. The effective positive charge in the nucleus is less, so the less inward pull is experienced. Hence, the size expands.

Solution 21(ii).

The inert gas argon is the next element after chlorine in the third period.

In a period, the size of an atom decreases from left to right due to an increase in nuclear charge with an increase in the atomic number. However, the size of the atoms of inert gases is bigger than the previous atom of halogen in the respective period. This is because the outer shell of inert gases is complete. They have the maximum number of electrons in their outermost orbit; thus, electronic repulsions are maximum. Hence, the size of the atom of an inert gas is bigger.

Solution 21(iii).

Ionisation potential of the element increases across a period because the atomic size decreases due to an increase in the nuclear charge, and thus, more energy is required to remove the electron(s).

Solution 21(iv).

Alkali metals are strong reducing agents because they lose electrons easily to complete their octet.

Solution 22(i).

Neon (Atomic number = 10)
Electronic configuration = 1s²2s²2p⁶

Solution 22(ii).

Electronic configuration = 2, 8, 3
Hence, atomic number = 13
The element having atomic number 13 is Aluminium.

Solution 22(iii).

The element has a total of three shells; hence, the element belongs to the third period.
Five valence electrons indicate that the element belongs to the fifth group (VA).
Hence, the element is phosphorus.

Solution 22(iv).

The element has a total of four shells; hence, the element belongs to the fourth period.
Two valence electrons indicate that the element belongs to the second group (IIA).
Hence, the element is calcium.

Solution 22(v).

Twice as many electrons in its second shell as in its first shell indicates electronic configuration 1s²2s².
From the electronic configuration, the total number of electrons is 4.
We know that
Number of electrons = Number of protons = Atomic number
The element with atomic number 4 is beryllium.

Solution 23(i).

Period 1:
Number of elements = 2
Hydrogen, helium

Period 2:
Number of elements = 8
Lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, neon

Period 3:
Number of elements = 8
Sodium, magnesium, aluminium, silicon, phosphorus, sulphur, chlorine, argon

Solution 23(ii).

A common feature of the electronic configuration of the elements at the end of Period 2 and Period 3 is that the atoms have 8 electrons in their outermost shell.

Solution 23(iii).

If an element is in Group 17, it is likely to be non-metallic in character, while with one electron in its outermost energy level (shell), then it is likely to be metallic.

Solution 1.

(i) Electron affinity
(ii) Atomic size
(iii) Metallic character
(iv) Non-metallic character
(v) Ionization energy

Exercise Intext 3

Solution 1.

(a) The energy required to remove an electron from a neutral isolated gaseous atom and convert it into a positively charged gaseous ion is called Ionization energy or ionization potential.

(b) M(g)+ I.E   →   M⁺(g) + e^–
M can be any element
It is measured in electron volts per atom. Its S.I unit kJmol^-1.

Solution 2.

Ionisation potential values depend on

1. Atomic size: The greater the atomic size, the lesser the force of attraction. Electrons of the outermost shell lie further away from the nucleus, so their removal is easier and the ionisation energy required is less.
2. Nuclear charge: The greater the nuclear charge, greater is the attraction for the electrons of the outermost shell. Therefore, the electrons in the outermost shell are more firmly held because of which greater energy is required to remove them.

Solution 3.

(a) Ionization energy increases as we move from left to right across a period as the atomic size decreases.
(b) Ionization energy decreases down a group as the atomic size increases.

Solution 4.

Helium has the highest ionization energy of all the elements while cesium has the lowest ionization energy.

Solution 5.

Second period: Lithium

Second period: Neon > Fluorine > Oxygen > Nitrogen > Carbon > Boron > Beryllium > Lithium
Third Period: Argon> Chlorine > Sulphur > Phosphorus > Silicon > Aluminum >Magnesium > Sodium

Solution 6(2010).

(a) Electron affinity is the energy released when a neutral gaseous atom acquires an electron to form an anion.
(b) Second period: Lithium
Second period: Lithium<Boron < Carbon < Oxygen < Fluorine
Neon, Nitrogen and Beryllium do not follow the trend.

Solution 7.

Electron affinity values generally increase across the period left to right and decrease down the group top to bottom.

Solution 8(a).

Electronegativity is the tendency of an atom in a molecule to attract the shared pair of electrons towards itself.
Electronegativity is a dimensionless property; hence, it has no unit.

Solution 8(b).

Correct option – (i).
The element with least electronegativity is lithium.

Solution 9.

(a) On moving across a period, nuclear pull increases because of the increase in atomic number, and thus, the atomic size decreases. Hence, elements cannot lose electrons easily. Hence, Group 17 elements are strong non-metals, while Group 1 elements are strong metals.

(b)  On moving across a period, nuclear pull increases because of the increase in atomic number, and thus, the atomic size decreases. Hence, elements cannot lose electrons easily. Hence, Group 17 elements are strong non-metals, while Group 1 elements are strong metals. Down a group, the atomic size increases and the nuclear charge also increases. The effect of an increased atomic size is greater as compared to the increased nuclear charge. Therefore, metallic nature increases as one moves down a group, i.e. they can lose electrons easily.

(c) The atomic size of halogens is very small. The smaller the atomic size, the greater the electron affinity, because the effective attractive force between the nucleus and the valence electrons is greater in smaller atoms, and so the electrons are held firmly.

(d) The reducing property depends on the ionisation potential and electron affinity of the elements. In a period, from left to right in a horizontal row of the periodic table, the atomic size decreases and the nuclear charge increases, so the electron affinity and ionisation energy both increase. Hence, the tendency to lose electrons decreases across the period from left to right and thus the reducing property also decreases across the period from left to right.

The electron affinity and ionisation potential decreases along the group from top to bottom. Hence, the tendency to lose electrons increases, and thus, the reducing property also increases along the group from top to bottom.

(e) In a period, the size of an atom decreases from left to right. This is because the nuclear charge, i.e. the atomic number increases from left to right in the same period, thereby bringing the outermost shell closer to the nucleus. Therefore, considering the third period given above, it has been found that sodium is the largest in size, while chlorine is the smallest.

Solution 10.

(i) Ionization energy
(ii) Metallic character
(iii) Electronegativity

Solution 11.

(a) G (due to the smallest atomic size).
(b) G and O as both have outermost electronic configuration np⁵.
(c) A and I as both have outermost electronic configuration ns¹.
(d) D (1s²2s²2p²)
(e) I as alkali metals have least ionisation energy. Also, ionisation energy decreases with an increase in the atomic size that    decreases on moving down the group.
(f) O, as halogens have the least atomic size.

Solution 12.

(a) Thallium. Because the metallic character increases down the group, thallium will have the most metallic character.
(b) Boron. Electronegativity decreases down the group as the size increases; hence, boron will be the most electronegative atom.
(c) Three. The number of electrons present in the valence shell is the same for each group. Hence, all these elements and thallium will have 3 valence electrons.
(d) BCl₃
(e) Since metallic character decreases from left to right and non-metallic character increases from left to right, elements in the group to the right of this boron group will be less metallic in character.

Solution 9.

(a) As we move from left to right the increase in atomic number and decrease in size results in a greater nuclear pull. As a result, the ability to attract the electrons increases, and so does the electron affinity.
But noble gases have complete stable octet configuration, hence their electron affinity is lower than halogens.
Hence halogens on extreme right have highest electron affinity in a period.

(b) Chlorine is smaller than sulphur with a bigger atomic number. Since its nuclear pull is more, hence its electron affinity is higher than sulphur.

Solution 10.

Since size of chlorine is bigger than fluorine hence the electrons being farther away from the nucleus experience a lesser force of attraction, hence electron negativity of chlorine is less than fluorine.

Solution 11.

Electronegativity measures an atom’s tendency to attract shared pair of electrons towards itself.
Its S.I unit is Pauling unit.

Solution 12.

(i) The element fluorine has the highest electronegativity and Caesium has the lowest electronegativity.
(ii) The nature of oxides changes from basic to acidic as we move from left to right in third period. Hence sodium forms most basic oxide while oxide of Aluminum is amphoteric and oxides of phosphorus, sulphur and chlorine are progressively acidic.

Exercise 1

Solution a(2015)

(i) Lithium
Reason: Electronegativity increases from left to right. Lithium is present on the left side of the periodic table; hence, it will be the least electronegative element.

Solution b(2015)

(i) Ba metal will form ions readily because the ionisationenergy decreases down the group as the size increases.
(ii) On moving down the group, the number of electrons in the outermost shell, i.e.valence electrons remain the same. So, the valency in a group remains the same, i.e. 2.

Solution a(2009)

Correct option is A. Lithium
In a period from left to right, electron affinity decreases as the non-metallic character increases.

Solution a(2009)

Correct option is A. Lithium

In a period from left to right, electron affinity decreases as the non-metallic character increases.

Solution b(2009)

1. The most electronegative is J.
2. Valence electrons present in G are 5.
3. B contains 1 valence electron and H contains 6 valence electrons. So, the valency of B is +1 and the valency of H is – 2.
4. In the compound between F and J, the type of bond formed will be covalent.
5. The electron dot structure for the compound formed between C and K is
   

Solution a(2010)

The number of electrons in the valence shell of a halogen is 7.
Correct option: D

Solution b(2010)

Electronegativity across the period increases.

Solution c(2010)

Non-metallic character down the group decreases.

Solution d(2010)

Atomic number of an element is 16.

1. It belongs to Period 3.
2. The number of valence electrons in the element is 6.
3. The element is a non-metal.

Solution a(2011)

The oxidising power of elements depends on the tendency to gain electrons which increases from left to right along a period due to increase in nuclear pull.

Solution (b)2011

1. Across a period, the ionisation potential increases.
2. Down the group, electron affinity decreases.

Solution c(i)(2011)

In the periodic table, alkali metals are placed in Group I. So, the correct option is A.

Solution c(ii)(2011)

The correct option is C.

The elements of halogen family are non-metallic in nature.

Solution (d)2011

Three shells indicate that the element belongs to the third period.

Three valence electrons indicate that the element belongs to the third group.

Solution a(2012)

Correct option: (D) Argon

Solution b(2012)

1. Because the atomic radius decreases across a period. Due to this, attraction between the nucleus and the electron increases. This results in an increase in the ionisation potential.
2. Alkali metals are good reducing agents because they have a greater tendency to lose electrons.

Solution c(2012)

Electronic configuration of E with atomic number 19 = 1s²2s²2p⁶3s²3p⁶4s¹
E is a metal.

Electronic configuration of F with atomic number 8 =
1s²2s²2p⁴
F is a non-metal.

Electronic configuration of G with atomic number 17 =
1s²2s²2p⁶3s²3p⁵
G is a non-metal.

Solution d(2012)

A metal present in Period 3, Group I of the periodic table is sodium.

Solution a(2013)

Correct option: (D) Fluorine

Solution b(2013)

1. I
2. R
3. M
4. 5
5. T
6. Y
7. Ionic bond will be formed and the molecular formula is A₂H.

Solution c(2013)

The element which has the highest ionisation potential is helium (He).

Solution a (2014)

1. Correct option: D (atomic radius decreases and nuclear charge increases)
2. Correct option: A (3 shells and 2 valence electrons)

Solution b (2014)

(a) An element Z having atomic number 16 is Sulphur.

(i) Sulphur belongs to Period 3 and Group 16.
(ii) Sulphur is a non-metal.

Solution e (2014)

Electron affinity

Solution f (2014)

A: (ii)
B: (i)

Solution d (2014)

1. Ionic bond exists between M and O.
2. 1 electron is present in the outermost shell of M.
3. M belongs to Group 1 in the modern periodic table.

Solution 1(2008)

B

Solution 2

1. A covalent oxide of a metalloid. – SiO₂ (Si is a metalloid)
2. An oxide which when dissolved in water forms acid. – SO₂ (SO₂ + H₂O → H₂SO₃)
3. A basic oxide. – Na₂O (Na₂O + H₂O → 2NaOH)
4. An amphoteric oxide. – Al₂O₃ (shows both acidic and basic properties)

Solution 3 (2015)

a. Mg,Cl, Na, S, Si (increasing order of atomic size) –
Cl < S < Si < Mg < Na
99 pm < 104 pm < 117 pm < 160 pm < 186 pm

b. Cs, Na, Li, K, Rb (increasing metallic character)
Li < Na < K < Rb < Cs

c. Na, K,Cl, S, Si (increasing ionisation potential) –
Cl < S < Si < Na < K
1256 < 999 < 786 < 496 < 419

d. Cl, F, Br, I (increasing electron affinity) –
I < Br < F < Cl
-295 KJ mol^-1 < -324 KJ mol^-1 < -327.9 KJ mol^-1 < -349 KJ mol^-1

e. Cs, Na, Li, K,Rb (decreasing electronegativity) –
Li > Na > K = Rb > Cs
1.0 > 0.9 > 0.8 = 0.8 > 0.7

Solution 4

(a) An element with atomic number 9 and 35
(b) An element with atomic number 9.

Solution 5

The ionisation energy is the minimum energy required to remove the outermost electron from a gaseous neutral atom to form a cation.
Position in a group: X will be above Y ( because of ionisation energy decreases down the group )
Position in a period: X will be the right side of Y ( because ionisation energy increases from left to right)

Solution 6

(a) Cl < Cl¯
(b) Mg²⁺ < Mg⁺ < Mg
(c) O < N < P

Solution 7

(a) Cl
Metals have low ionisation energy and non-metals have high ionisation energy. Also, across the period, ionisationenergy tends to increase. The elements P, Na and Cl belong to the third period. Na – Group 1, P – Group 15 and Cl – Group 17.

(b) Ne
Inert gases have zero electron affinity because of their stable electronic configuration.

(c) He
Ionisation energy decreases with an increase in the atomic size, i.e. it decreases as one moves down a group. Ne, He and Ar are inert gases. He – Period 1, Ne – Period 2 and Ar – Period 3.

Solution 8

Na, Mg, Al, Si, P, S, Cl

Solution 9 (2016)

(a)  The element below sodium in the same group would be expected to have a lower electro-negativity than sodium, and the element above chlorine would be expected to have a higher ionisation potential than chlorine.
(b) On moving from left to right in a given period, the number of shells remains the same.
(c) On moving down a group, the number of valence electrons remains the same.
(d) Metals are good reducing agents because they are electron donors.

Solution 10

(a) Increases
(b) Increases
(c) Increases
(d) Decreases
(e) Increases

Solution 11

(a) Period 2
(b) Nitrogen (N), between carbon and oxygen
(c) Be< N< F
(d) Fluorine

Solution 1

(a) Na and F
(b) Argon
(c) C, N, O and F are non-metals present in period 2 while Na, Mg and Al are metals in period 3.
(d) Silicon
(e) Argon
(f) Mg
(g) Fluorine
(h) K

Solution 1.

(a) The total number of electron shells in an atom determines the period to which the element belongs, and the valence electrons determine the group to which it will belong. So with the help of electronic configuration we can figure out the period and group number of an element.

Elements with one and two valence electrons belong to group 1 and 2 respectively, while to determine the group number of elements with 3 to 8 valence electrons, we add 10 to their valence electrons.

For example an element X has atomic number 15
Its configuration will be:
K shell has 2 electrons, L will have 8, and the remaining 5 will be placed in M shell
Since it has three shells it belongs to period 3 and with 5 valence electrons the element will be placed in five plus ten, that is the 15^th group
So with the help of electronic configuration we can figure out the period and group number of an element.

(b) Atomic number = Number of protons
Hence, number of protons in K atom = 19
Number of neutrons = Mass number – Atomic number
Hence, number of neutrons in K atom = 39-19 = 20
Number of electrons = Number of protons
Hence, number of electrons= 19
And electronic configuration of K atom = 2, 8, 8, 1
Since K atom has 4 shells, hence it belongs to fourth period.
With one valence electron, it belongs to group 1
Number of protons in P atom = 15
Number of neutrons in P atom = 31-15 = 16
Number of electrons in P atom = 15
And electronic configuration of P atom = 2, 8, 5
Since it has three shells, it belongs to period 3 and with 5 valence electrons Phosphorus is found in five plus ten that is 15^th group.

Solution 2.

(a) Fluorine, chlorine and bromine are non-metals with seven valence electrons. They are highly electronegative elements with valency of one. They exist as diatomic molecules. They form ionic compounds with alkali metals.
(b) They are known as halogens. The term means salt forming and therefore compounds containing these elements are called salts.

Solution 3.

The last element in each period of the periodic table is a gaseous element with its valence shell completely filled. Except for helium with complete duplet configuration, rest all the 5 gases have complete octet configuration.
These group 18 elements are commonly referred to as noble gases.

Solution 4.

The electronic configuration of an element determines its position in Modern Periodic table.
The element with one valence electron is the first while the element with 8 valence electrons is placed in the 18th group of a period.

Solution 5.

(i) The number of valence electrons increases by one as we move from left to right in a period.
The group number 1 and 2 have 1 and 2 valence electrons respectively while group 13 to 18 have group number minus 10 = valence electrons. So,group 13 to 18 have 3, 4, 5, 6, 7 and 8 valence electrons respectively.

(ii) Valency is determined by the number of valence electrons. For elements belonging to group 1, 2 and 13, the valency is equal to the number of valence electrons, so their valency is 1, 2 and 3 respectively.
Since the elements in group 14 to 17 needs to gain electrons to complete their octet configuration. Their valency is 8 minus the number of valence electrons. So their valencies are 4, 3, 2 and 1 respectively.

Solution 6.

(a) Periods
(b) Increases
(c) Decreases

Solution 7.

(a) Since it belongs to group II, it has 2 valence electrons and hence it is a metal.
(b) Barium is placed below calcium in the group. Since, it has more number of shells; it is easier for it to lose its valence electrons to complete its octet configuration. Hence it is more reactive than calcium.
(c) It needs to lose its 2 valence electrons to complete its octet configuration; therefore its valency is also 2.
(d) The formula of its phosphate will be (Ba)₃ (PO₄)₂
(e) As we move from left to right in a period, the size decreases, therefore, it will be smaller than Cesium.

Solution 8.

(a) The number of valence electrons increases by one as we move across any given period.
Therefore as we move from Lithium to Neon in period 2, the valence electrons will increase from 1 to 7.

(b) The metallic character decreases as we move from left to right while the non-metallic character increases.
On going from left to right in a period, the chemical reactivity of elements first decreases and then increases.
For example in period 3, Sodium is the most reactive metal and Chlorine is the most reactive non-metal and Silicon is least reactive.

(c) The oxides of metals are basic and that of non-metals are acidic in general. Therefore since metallic strength decreases and non-metallic strength increases on moving from left to right across a period, the strength of basic oxides decreases, while the strength of acidic oxides increases.
For example, sodium forms a basic oxide, while sulphur and phosphorus form acidic oxides.

Solution 9.

(a) Noble gases- H and P
(b) Halogens- G and O
(c) Alkali metals – A and I
(d) D and L have valency of 4
(e) I with atomic number 11.
(f) Cl has the least atomic size in period 3 with atomic number 17.

Solution 10.

As we move down a group, the numbers of shells increases and hence the atomic size increases.
Therefore, Z will have the smallest atomic number followed by Y, while X will have the largest atomic number.
So the elements in order of increasing atomic number will be Z<Y<X.

Solution 11.

(a) Since, the distance of the valence electrons from the nucleus keeps on increasing down the group, therefore, the ionization energy keeps on decreasing. Hence the reactivity of alkali metals increases from lithium to francium.

(b) As we move down a group, the size keeps on increasing, so it becomes more difficult for atoms to attract electrons. Thus reactivity of halogens decreases from Fluorine to Astatine.

Solution 12.

(a) Since it belongs to period 3 it has 3 shells, K, L and M. The outermost M shell will have 2 valence electrons as it is placed in group II.
(b) With 2 valence electrons, its valency will be 2.
(c) Since it has electronic configuration of 2, 8, 2, its atomic number is 12 and hence X is Magnesium.
(d) It is a metal.

Solution 13.

(a) Group 1since the valence electrons is 1.
(b) With 4 shells T belong to period 4.
(c) Number of electrons = 2+8+8+1=19
(d) T needs to lose one electron to complete its octet hence its valency is 1.
(e) Since it has one valence electron, it is a metal.

Solution 14.

(a) Group 1: Lithium< Sodium< Potassium< Rubidium < Caesium< Francium
Group 17: Fluorine < Chlorine < Bromine< Iodine < Astatine

(b) Group 1: Francium
Group 17: Astatine< Iodine< Bromine< Chlorine< Fluorine

(c) Group 1: Francium< Cesium< Rubidium< Potassium< Sodium< Lithium
Group 17: Astatine< Iodine< Bromine< Chlorine< Fluorine

(d) Group 1: Francium
Group 17: Astatine

(e) Group 1: Lithium>Sodium> Potassium> Rubidium> Cesium> Francium
Group 17: Fluorine > Chlorine> Bromine > Iodine > Astatine

Solution 15.

(a) atomic number
(b) period, non-metallic character
(c) more
(d) number of outer electrons

Solution 16.

(a) Anion is formed by the gain of electrons. Thus the numbers of electrons are more than protons. The effective positive charge in the nucleus is less, so less inward pull is experienced. Hence the size expands. So the size of an atom is greater than the size of parent atom.

(b) Since Argon has stable octet configuration, so due to the inter- electronic repulsions the effect of nuclear pull over the valence shell electrons cannot be seen which results in the bigger size.

(c) Since size of Bromine is bigger than chlorine, so it becomes more difficult for Br atoms to attract electrons. Thus, Cl is more reactive than Br.

Solution 17.

(a) Neon
(b) Aluminum
(c) Phosphorus
(d) Calcium
(e) Carbon

Solution 18.

(a) Na and F
(b) Argon
(c) C, N, O and F are non-metals present in period 2 while Na, Mg and Al are metals in period 3.
(d) Silicon
(e) Argon
(f) Mg
(g) Fluorine
(h) K

Solution 19.

(a) Element with atomic number 9 and 35
(b) Element with atomic number 9.

Solution 20.

(a) Period 1 has 2 elements while period 2 and period 3 have 8 elements each.
(b) Hydrogen and helium
(c) The elements at the end of period 2 and Period 3 have 8 electrons in its outermost shell.
(d) Non metallic, metallic.

Solution 21.

Position in a group: X and Y
Position in a period: Y and X

Solution 22.

Period no. = no. of shells, so n=3
From the formula M₂ O₃ its valency is 3.
Since it is a metal, its valence shell has 3 electrons.
So its electronic configuration is 2,8,3
Atomic number=13
Hence the metal is Aluminum with valency 3.

Solution 23.

(a) Since the elements in a group have same number of valence electrons, they can either contain metals or non-metals like alkali and alkaline metals have only metals whereas halogens are non-metals.

(b) No two elements have the same number of electrons instead atoms of the same elements in the same group have the same number of valence electrons.

(c) Non-metals have the tendency to gain electrons to attain stable configuration and therefore are said to be electronegative. As we move from left to right the increase in atomic number and decrease in size results in a greater nuclear pull. As a result the non-metallic character increases across a period.

(d) On moving from left to right in a period, the reactivity first decreases and then increases since the tendency to lose electrons first decreases on going from left to right and then from P to Cl, tendency to gain electrons increases, so reactivity increases then. In case of a group, reactivity increases on going down since the tendency to lose electrons increases but for non-metals, reactivity decreases on going down the group as the tendency to gain electrons decreases down the group.

Solution 24.

(a) Cl < Cl¯
(b) Mg²⁺ < Mg⁺ < Mg
(c) O < N < P

Solution 25.

(a) Cl
Metals have low ionisation energy and non-metals have high ionisation energy. Also, across the period, ionisation energy tends to increase. The elements P, Na and Cl belong to the third period. Na – Group 1, P – Group 15 and Cl – Group 17.

(b) Ne
Inert gases have zero electron affinity because of their stable electronic configuration.

(c) He
Ionisation energy decreases with an increase in the atomic size, i.e. it decreases as one moves down a group. Ne, He and Ar are inert gases. He – Period 1, Ne – Period 2 and Ar – Period 3.

Solution 26.

(a) (iv) Argon
(b) (iii) Calcium
(c) (iii) Helium

Solution 1 (2003).

(a) (Al)₂(SO₄)₃
(b) Covalent bonding
(c) Same number of valence electrons
(d) Helium
(e) 8
(f) Electron affinity
(g) Decreases, atomic number, number of shells

Solution 1 (2004).

(a) Na, Mg, Al, Si, P, S, Cl
(b)
(i) Lower, higher
(ii) remains the same
(iii) remains the same

Solution 1 (2005).

(a) Increases
(b) Increases
(c) Increases
(d) Decreases
(e) Increases

Solution 1 (2006).

(a) Period 2
(b) Nitrogen (N), between carbon and oxygen
(c) Carbon
(d) Be < N < F
(e) Fluorine

Solution 1 (2007).

Solution 1 (2008).

B.

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Selina Concise Biology Class 10 ICSE Solutions Health Organisations

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 14 Health Organisations. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 14 Health Organisations

Exercise 1

Solution A.1.
(a) May 8

Solution A.2.
(c) Geneva

Solution B.1.

1. A
2. B
3. A
4. B
5. A
6. A

Column I (Activity)	Column II (Organisation)
(i)To extend relief to victims of earthquake.	Red Cross
(ii)To lay pharmaceutical standards for Important drugs.	WHO
(iii)Arranging ambulance in emergencies.	Red Cross
(iv)To suggest quarantine measures.	WHO
(v)Training of midwives.	Red Cross
(vi)Procuring and supplying blood for transfusion.	Red Cross

Solution B.2.
Geneva

Solution B.3.
(a) WHO: World Health Organisation
(b) UNO: United Nations Organisation

Solution C.1.
(i) Sanitation – Removal and proper disposal of garbage, sewage and other wastes, elimination of breeding places of flies, mosquitoes, etc.
(ii) Supply of safe drinking water.
(iii) Keeping statistical records – Apart from the registration of births and deaths, to maintain the information about the health and diseases of the people in their area need regularly.

Solution C.2.

1. Food and water borne diseases:
   The contaminated food and water cause several diseases. Water borne diseases occur due to contaminated water from hand pumps or mixing of untreated sewage with river water.
2. Insect and air-borne diseases:
   Lack of cleanliness leads to breeding of houseflies, mosquitoes which are the carries of certain diseases.
3. Lack of medical facilities:
   Lack of medical facilities especially in rural areas, leads to unavoidable deaths and damage to health. Lack of knowledge and superstitions beliefs also delay timely treatment which may result in serious consequences.

Solution C.3.
Functions of WHO:

• To promote and support projects for research on diseases.
• To collect and supply information about the occurrence of diseases of epidemic nature such as cholera, plague, yellow fever, etc.

Solution C.4.
Functions of Red Cross:

• To extend relief and help to the victims of any calamity – flood, fire, famine, earthquakes, etc.
• To procure and supply blood for the needy victims of war and other calamities.
• To extend all possible first-aid in any accident.
• To arrange for ambulance services in all emergencies.

Solution C.5.
Functions of World Health Organisation (WHO):

• To promote and support projects for research on diseases.
• To collect and supply information about the occurrence of diseases of epidemic nature such as cholera, plague, yellow fever, etc.
• To lay pharmaceuticals standards for important drugs, to ensure purity and size of the dose.
• To organize campaigns for the control of epidemic (widespread) and endemic (local) diseases.

Solution C.6.

1. To extend relief and help to victims of any calamity
2. To procure and supply blood for the needy victims of war or calamity
3. To extend all possible first aid in any accident
4. To educate people in accident prevention
5. To arrange for ambulance services in emergencies
6. To look after maternal and child welfare centres
7. To train midwives
   (any four)

Solution D.1.

• The World Health Organization (WHO) is a specialized agency of the United Nations (UN) that is concerned with international public health.
• It was established on 7 April 1948, with headquarters in Geneva, Switzerland and is a member of the United Nations Development Group.
• There were several reasons for the formation of WHO:
• Member countries of the UNO focused on the need for creating an international body to look after the health problem of people of the world.
• This was particularly felt in the field of research on the causes and cures of the diseases.
• The combined efforts in this direction were to give better and faster results.
• The poor and developing countries were to benefit quickly.

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Selina Concise Physics Class 10 ICSE Solutions Force

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 1 Force. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Physics Chapter 1 Force

Exercise 1(A)

Solution 1.

(a) When the body is free to move it produces translational motion.
(b) When the body is pivoted at a point, it produces rotational motion.

Solution 2.

The moment of force is equal to the product of the magnitude of the force and the perpendicular distance of the line of action of force from the axis of rotation.
S.I. unit of moment of force is Newton metre (Nm).

Solution 3.

Moment of a force is a vector.

Solution 4.

Moment of force about a point depends on the following two factors:

1. The magnitude of the force applied and,
2. The distance of line of action of the force from the axis of rotation.

Solution 5.

When the body is pivoted at a point, the force applied on the body at a suitable point rotates the body about the axis passing through the pivoted point.
The direction of rotation can be changed by changing the point of application of force. The given figure shows the anticlockwise and clockwise moments produced in a disc pivoted at its centre by changing the point of application of force F from A to B.

Solution 6.

Moment of force about a given axis = force x perpendicular distance of force from the axis of rotation.

Solution 7.

Moment of force depends on the distance of line of action of the force from the axis of rotation. Decreasing the perpendicular distance from the axis reduces the moment of a given force.

Solution 9.

If the turning effect on the body is anticlockwise, moment of force is called anticlockwise moment and it is taken as positive while if the turning effect on the body is clockwise, moment of force is called clockwise moment and is taken negative.

Solution 10.

It is easier to open a door by applying the force at the free end of it because larger the perpendicular distance , less is the force needed to turn the body.

Solution 11.

The stone of hand flour grinder is provided with a handle near its rim so that it can be rotated easily about the iron pivot at its centre by a small force applied at the handle.

Solution 12.

It is easier to turn the steering wheel of a large diameter than that of a small diameter because less force is applied on steering of large diameter which is at a large distance from the centre of rim.

Solution 13.

A spanner (or wrench) has a long handle to produce larger turning moment so that nut can easily be turned with a less force.

Solution 14.

Solution 15.

Solution 16.

(a) Resultant force acting on the body = F-F=0 moment of forces = 0 i.e., no motion of the body
(b) The forces tend to rotate the body about the mid-point between two forces, Moment of forces= Fr

Solution 17.

At A and B, two equal and opposite forces each of magnitude F are applied. The two forces rotate the bar in anticlockwise direction.

Solution 18.

Two equal and opposite parallel forces not acting along the same line, form a couple. A couple is always needed to produce the rotation. For example, turning a key in a lock and turning a steering wheel.

Solution 19.

The moment of a couple is equal to the product of the either force and the perpendicular distance between the line of action of both the forces. S.I unit of moment of couple is Nm.

Solution 20.

At A and B, two equal and opposite forces each of magnitude F are applied. The two forces rotate the bar in anticlockwise direction. The perpendicular distance between two forces is AB which is called the couple arm.
Moment of force F at the end A
= F x OA(anticlockwise)
Moment of force F at the end B
= F x OB(anticlockwise)
Total moment of couple =F x OA + F x OB
= F x (OA +OB)= F x AB
= F x d(anticlockwise)
=Either force x perpendicular distance between the two forces (or couple arm)
Thus, Moment of couple = Force x Couple arm

Solution 21.

When a number of forces acting on a body produce no change in its state of rest or of motion, the body is said to be in equilibrium.

Solution 22.

(i) When a body remains in the state of rest under the influence of the applied forces, the body is in static equilibrium. For example a book lying on a table is in static equilibrium.
(ii) When a body remains in the same state of motion (translational or rotational), under the influence of the applied forces, the body is said to be in dynamic equilibrium. For example, a rain drop reaches the earth with a constant velocity is in dynamic equilibrium.

Solution 23.

For a body to be in equilibrium:

1. The resultant of all the forces acting on the body should be equal to zero.
2. The resultant moment of all the forces acting on the body about the point of rotation should be zero.

Solution 24.

According to the principle of moments, if the algebraic sum of moments of all the forces acting on the body about the axis of rotation is zero, the body is in equilibrium. A physical balance (or beam balance) works on the principle of moments.

Solution 25.

Solution 1 (MCQ).

The moment of a force about a given axis depends on both on the force and its perpendicular distance from the axis.
Hint: Moment of force = Force x Perpendicular distance

Solution 2 (MCQ).

The body will have rotational as well as translational motion.

Numericals

Solution 1.

Moment of force= force x perpendicular distance of force from point O
Moment of force= F x r
5Nm= 10 x r
R= 5/10 =0.5 m

Solution 2.

Length, r=10cm =0.1m
F= 5N
Moment of force= F x r= 5 x 0.1 = 0.5 Nm

Solution 3.

Given , F= 2N
Diameter=2m
Perpendicular distance between B and O =1m
(i)Moment of force at point O
= F x r
= 2 x 1=2Nm (clockwise)
(ii)Moment of force at point A= F x r
= 2 x 2=4Nm (clockwise)

Solution 4.

Given AO=2m and OB=4m

(i) Moment of force F₁(=5N) at A about the point O
=F₁ x OA
=5 x 2= 10Nm (anticlockwise)

(ii) Moment of force F₂(=3N) at B about the point O
= F₂ x OB
=3 x 4=12 Nm(clockwise)

(iii) Total moment of forces about the mid-point O
= 12- 10=2Nm(clockwise)

Solution 5.

Given, AB=4m hence, OA=2m and OB =2m
Moment of force F(=10N) at A about the point O
= F x OA= 10 x 2= 20Nm (clockwise)
Moment of force F (=10N) at point B about the point O
= F x OB= 10 x 2 =20Nm (clockwise)
Total moment of forces about the mid-point O=
=20 +20= 40Nm(clockwise)

Solution 6.

(i) Perpendicular distance of point A from the force F=10 N at B is 0.5m , while it is zero from the force F=10N at A
Hence, moment of force about A is
= 10 N x 0.5m=5Nm(clockwise)

(ii) Perpendicular distance of point B from the force F=10 N at A is 0.5m, while it is zero from the force F=10N at B
Hence, moment of force about B is
= 10 N x 0.5m=5Nm(clockwise)

(iii) Perpendicular distance of point O from either of the forces F=10N is 0.25 m
Moment of force F(=10N) at A about O= 10N x 0.25m
=2.5Nm(clockwise)
And moment of force F(=10N) at B about O
=10N x 0.25m=2.5Nm(clockwise)
Hence, total moment of the two forces about O
=0.25 + 0.25=5Nm (clockwise)

Solution 7.

Solution 8.
Let the 50gf weight produce anticlockwise moment about the middle point of metre rule .i.e, at 50cm.
Let a weight of 100gf produce a clockwise moment about the middle point. Let its distance from the middle be d cm. Then,
according to principle of moments,
Anticlockwise moment = Clockwise moment
50gf x 50 cm=100gf x d

Solution 9.

(i) Weight mg (W) of rule produces an anti-clockwise moment about the knife edge O. In order to balance it, 20gf must be suspended at the end B to produce clockwise moment about the knife edge O.

Solution 10.

Anticlockwise moment= 40gf x 40 cm
Clockwise moment= 80gf x d cm
From the principle of moments,
Anticlockwise moment= Clockwise moment
40gf x 40 cm =80gf x d

Solution 11.

(i) Anticlockwise moment= 40gf x (50-10)cm
=40gf x 40cm=1600 gf x cm
Clockwise moment= 20gf x (90- 50) =20gf x 40cm
=800 gf x cm
Anticlockwise moment is not equal to clockwise moment. Hence the metre rule is not in equilibrium and it will turn anticlockwise.

(ii) To balance it, 40gf weight should be kept on right hand side so as to produce a clockwise moment about the middle point. Let its distance from the middle be d cm. Then,
clockwise moment= 20gf x 40cm + 40gf x d cm
From the principle of moments,
Anticlockwise moment= Clockwise moment
40 gf x 40 cm= 20gf x 40 + 40 x d cm
1600-800=40gf x dcm

Solution 12.

From the principle of moments,
Anticlockwise moment= Clockwise moment
20kgf x 2m =40kgf x d

Solution 13.

Solution 14.

(i) Total anticlockwise moment about O
= 150gf x 40 cm=6000gf cm

(ii) Total clockwise moment about O,
=250gf x 20 cm= 5000gf cm

(iii) The difference of anticlockwise and clockwise moment= 6000- 5000= 1000gf cm

(iv) From the principle of moments,
Anticlockwise moment= Clockwise moment
To balance it, 100gf weight should be kept on right hand side so as to produce a clockwise moment about the O. Let its distance from the point O be d cm. Then,
150gf x 40 cm=250gf x 20 cm +100gf x d
6000gf cm= 5000gf cm + 100gf x d
1000gf cm =100 gf x d

Solution 15.

Solution 16.

Solution 17.

(i) From the principle of moments,
Clockwise moment= Anticlockwise moment
100g x (50-40) cm= mx(40-20) cm
100g x 10 cm = m x 20 cm = m =50 g

(ii) The rule will tilt on the side of mass m (anticlockwise), if the mass m is moved to the mark 10cm.

(iii) Anticlockwise moment if mass m is moved to the mark 10 cm= 50g x (40-10)cm =50 x 30=1500g cm
Clockwise moment=100g x (50-40) cm= 1000g cm
Resultant moment= 1500g cm -1000g cm= 500g cm (anticlockwise)

(iv) From the principle of moments,
Clockwise moment= Anticlockwise moment
To balance it, 50g weight should be kept on right hand side so as to produce a clockwise moment .Let its distance from fulcrum be d cm. Then,
100g x (50-40) cm + 50g x d =50g x (40-10)cm
1000g cm + 50g x d =1500 g cm
50 g x d= 500g cm
So, d =10 cm
By suspending the mass 50g at the mark 50 cm, it can be balanced.

Exercise 1(B)

Solution 1.

Centre of gravity is the point about which the algebraic sum of moments of weights of particles constituting the body is zero and the entire weight of the body is considered to act at this point.

Solution 2.

Yes, the centre of gravity can be situated outside the material of the body. For example, centre of gravity of ring.

Solution 3.

The position of centre of gravity of a body of given mass depends on its shape i.e., on the distribution of mass in it. For example: the centre of gravity of a uniform wire is at its mid-point. But if this wire is bent into the form of a circle, its centre of gravity will then be at the centre of circle.

Solution 4.

The position of centre of gravity of a
(a) rectangular lamina is at the point of intersection of its diagonals.
(b) cylinder is at the mid point on the axis of cylinder.

Solution 5.

(a) Centre of gravity of a triangular lamina is situated at the point of intersection of its medians.
(b) Centre of gravity of a circular lamina is situated at the centre of circular lamina.

Solution 6.

Centre of gravity of a uniform ring is situated at the centre of ring.

Solution 7.

Solution 8.

Take a triangular lamina. Make three fine holes at a, b, c near the edge of triangular lamina. Now suspend the given lamina along with a plumb line from hole ‘a’. Check that the lamina is free to oscillate about the point of suspension. When lamina has come to rest, draw straight line ad along the plumb line. Repeat the experiment by suspending the lamina through hole ‘b’ and then through hole ‘c’ for which we get straight lines be and cf respectively. It is noticed that the lines ad, be and cf intersect each other at a common point G which is the position of centre of gravity of triangular lamina i.e. the point of intersection of medians.

Solution 9.

(i) False. The position of centre of gravity of a body of given mass depends on its shape i.e., on the distribution of mass in it.
(ii) True.

Solution 10.

Solution 11.

Solution 1 (MCQ).

At its geometrical centre

Exercise 1(C)

Solution 1.

When a particle moves with a constant speed in a circular path, its motion is said to be the uniform circular motion. For example : Revolution of earth around sun is an example of uniform circular motion.

Solution 2.

Solution 3.

Yes, uniform circular motion has an accelerated motion with a constant speed.

Solution 4.

Motion of a cyclist on a circular track is an example of motion in which speed remains uniform, but the velocity changes.

Solution 5.

When the object moves in a circular path with uniform speed, it means that its magnitude of velocity does not change, only its direction changes continuously. Hence, it is considered as uniformly accelerated motion.

Solution 6.

Uniform linear motion	Uniform circular motion
The body moves along a straight line.	The body moves along a circular path.
Speed and direction both remain constant.	Speed is constant, but direction changes continuously.
It is not an accelerated motion.	It is an accelerated motion.

Solution 7.

Centripetal force is required for circular motion. It is always directed towards the centre of circle.

Solution 8.

Force acting on a body which is in circular motion is called centripetal force. It acts towards the centre of circular path.

Solution 9.

A planet moves around the sun in a nearly circular path for which the gravitational force of attraction on the planet by the sun provides the necessary centripetal force required for circular motion.

Solution 10.

(a) They act in opposite directions.
(b) No, centrifugal force is not the force of reaction of centripetal force.

Solution 11.

No, centrifugal force is a fictitious force.

Solution 12.

a. On standing outside the disc, we find that the pebble is moving on a circular path. On standing at the centre of the disc, we find that the pebble is stationary placed just in front of us.

Solution 13.

Force of tension in the thread provides the centripetal force.

Solution 15.

(a) False
(b) True
(c) True
(d) False

Solution 1 (MCQ).

Speed
Hint: Speed is scalar but velocity and acceleration are vector quantities. So, speed remains constant but velocity and acceleration change with the change in direction, and in circular motion the direction of motion changes at every point.

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