Well-structured Class 10 ICSE Maths Solutions S Chand Chapter 15 Three Dimensional Solids Ex 15(b) facilitate a deeper understanding of mathematical principles.
S Chand Class 10 ICSE Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(b)
Question 1.
Find the volume of the cylinders whose radii and heights are given below. Take π to be \(\frac { 22 }{ 7 }\).
(i) r = 7 cm, h = 8 cm
(ii) r = 7 cm, h = 12 cm
(iii) r = 14 cm, h = 16 cm
(iv) r = 21 cm, h = 40 cm
Solution:
(i) Radius of a cylinder (r) = 7 cm and height (h) = 8 cm
∴ Volume = πr2h = \(\frac { 22 }{ 7 }\) × 7 × 7 × 8 cm3 = 1232 cm3
(ii) Radius of cylinder (r) 7 cm and height (h) = 12 cm
∴ Volume = πr2h = \(\frac { 22 }{ 7 }\) × 7 × 7 × 12 cm3 = 1848 cm3
(iii) Radius of the cylinder (r)= 14 cm and height (h)= 16 cm
∴ Volume = πr2h = \(\frac { 22 }{ 7 }\) × 14 × 14 × 16 = 9856 cm3
(iv) Radius of the cylinder (r) = 21 cm and height (h) = 40 cm
∴ Volume = πr2h = \(\frac { 22 }{ 7 }\) × 21 × 21 × 40 cm3 = 55440 cm3
Question 2.
Find the diameter of the circular cylinders if:
(a) Volume is 44 cm3, height 3.5 cm;
(b) Volume 385 cm3, height 1 dm.
Solution:
(a) Volume of a cylinder = 44 cm3
Height (h) = 3.5 cm = \(\frac { 35 }{ 10 }\)cm
∴ Radius = \(\sqrt{\frac{\text { Volume }}{\pi h}}\)
= \(\sqrt{\frac{44 \times 7 \times 10}{22 \times 35}}\) = √4 = 2 cm
∴ Diameter = 2 × r = 2 × 2 = 4 cm
(b) Volume of the cylinder = 385 cm2
and height (h) = 1 dm = 10 cm
∴ Radius = \(\sqrt{\frac{\text { Volume }}{\pi h}}\)
= \(\sqrt{\frac{385 \times 7}{22 \times 10}}\) = \(\sqrt{\frac{49}{4}}\) = \(\frac{7}{2}\) cm
∴ Diameter = 2r = \(\frac{7}{2}\) × 2 = 7 cm
Question 3.
Find the height of the circular cylinders if:
(a) Volume is 66 cm3, radius 2 cm;
(b) Volume 4 litres, radius 5 cm.
Solution:
(a) Volume of a cylinder = 66 cm3 and radius (r) = 2 cm
∴ Height = \(\frac{\text { Volume }}{\pi r^2}\)
= \(\frac{66 \times 7}{22 \times 2 \times 2}\) = \(\frac{21}{4}\)cm = 5.25 cm
(b) Volume of a cylinder = 4 litres = 4000 cm3
Radius (r) = 5 cm
∴ Height = \(\frac{\text { Volume }}{\pi r^2}\)
= \(\frac{4000 \times 7}{22 \times 5 \times 5}\)cm = \(\frac{560}{11}\)cm = 50.9 cm
Question 4.
A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m3.
Solution:
Diameter of a wooden pole = 20 cm
∴ Radius (r) = \(\frac{20}{2}\) = 10 cm
and height (h) = 7 m = 700 cm
∴ Volume = πr2h = \(\frac{22}{7}\) × 10 × 10 × 7 × 100 cm3
= 220000 cm3 = \(\frac{220000}{1000000}\) m3
Weight of 1 m3 = 225 kg
Total weight = \(\frac{220000}{1000000}\) × 225 kg
= \(\frac{198}{4}\) = 49\(\frac{1}{2}\)kg = 49.5 kg
Question 5.
Find the volume of metal in the hollow pipe of internal radius 3 cm, metal 1 cm thick and length 6 cm.
Solution:
Length of metal hollow pipe (h) = 6 cm
Internal radius (r) = 3 cm
Thickness of pipe = 1 cm
∴ Outer radius (R) = 3 + 1 = 4 cm
∴ Volume of the metal used = πh(R2 – r2)
= \(\frac{22}{7}\) × 6 × (42 – 32)
= \(\frac{22}{7}\) × 6 × 7 × 1 = 132 cm3
Question 6.
The sum of the radius of the base and the height of a solid cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 cm2, find the volume of the cylinder. (Take π = \(\frac{22}{7}\))
Solution:
Sum of the radius of the base and height of a solid cylinder = 37 cm
Total surface area = 1628 cm2
∴ 2πr(h + r) = 1628
⇒ 2 × \(\frac{22}{7}\) × r × 37 = 1628
r = \(\frac{1628 \times 7}{2 \times 22 \times 37}\) = 7 cm
∴ Height = 37 – 7 = 30 cm
Now volume = πr2h
= \(\frac{22}{7}\) × 7 × 7 × 30 cm3 = 4620 cm3
Question 7.
A cylindrical tank has capacity 6160 cu m. Find its depth if the diameter of its base is 28 m. Also, find the area of the inside curved surface of the tank. (Take π = \(\frac{22}{7}\) )
Solution:
Capacity of a cylindrical tank = 6160 cu m Diameter of its base = 28 m
∴ Radius (r) = \(\frac{28}{2}\) = 14 m
∴ Depth (h) = \(\frac{\text { Volume }}{\pi r^2}\)
= \(\frac{6160 \times 7}{22 \times 14 \times 14}\) = 10 m
Area of curved surface of its inner side = 2πrh
= 2 × \(\frac{22}{7}\) × 14 × 10 = 880 m2
Question 8.
The area of the curved surface of a cylinder is 4400 cm2, and the circum-ference of its base is 110 cm. Find
(i) the height of the cylinder.
(ii) the volume of the cylinder.
\(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Curved surface area of a cylinder = 4400cm2
Circumference its base = 110 cm
∴ 2πrh = 4400
and 2πr = 110 cm ⇒ r = \(\frac{110 \times 7}{2 \times 22}\) = \(\frac{35}{2}\)cm
(i) ∴ Height (h) = \(\frac{4400}{110}\) = 40 cm
(ii) Volume = πr2h = \(\frac{22}{7}\) × \(\frac{35}{2}\) × \(\frac{35}{2}\) × 40 = 38500 cm2
Question 9.
(i) How many cubic metres of earth must be dug out to make a well 20 metres deep and 2 metres in diameter? \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
(ii) If the inner curved surface of the well in part (i) above is to be plastered at the rate of ₹5 per m2, find the cost of Plastering \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Depth of a well (h) = 20 m
Diameter = 2 m
∴ Radius (r) = \(\frac{2}{2}\) = 1 m
(i) ∴ Volume of earth dugout = πr2h
= \(\frac{22}{7}\) × 1 × 1 × 20 = \(\frac{440}{7}\) m3
= 66\(\frac{6}{7}\) m3
(ii) Area of inner curved surface = 2πrh
= 2 × \(\frac{22}{7}\) × 1 × 20 = \(\frac{880}{7}\) m2
Rate of Plastering the inner surface = ₹5 per m2
∴ Total cost = \(\frac{880}{7}\) × 5
= ₹ \(\frac{4400}{7}\) = ₹628.57
Question 10.
A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm2 (sq cm). Find
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place. (Take n to be 3.14)
Solution:
Diameter of a cylinder = 20 cm
∴ Radius (r) = \(\frac{20}{2}\) = 10 cm
Curved surface area = 1000 cm2
(i) ∴ Height = \(\frac{\text { Curved surface area }}{\text { 2πr }}\)
= \(\frac{1000}{2 \times 3.14 \times 10}\) = \(\frac{100}{6.28}\)cm = 15.9 cm
(ii) Volume = πr2h
= 3.14 × 10 × 10 × 15.9 cm3
= 314 × 15.9 = 4992.6 cm3
Question 11.
An electric geyser is cylindrical in shape, having a diameter of 35 cm and height of 1.2 m. Neglecting the thickness of its walls, calculate
(i) its outer lateral surface area;
(ii) its capacity in liters \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Diameter of a geyser = 35 cm
∴ Radius (r) = \(\frac{35}{2}\)cm
and height (h) = 1.2 m = 120 cm
(i) ∴ Outer lateral surface area
= 2nrh = 2 × \(\frac{22}{7}\) × \(\frac{35}{2}\) × 120 cm2
= 13200 cm2
(ii) Capacity = πr2h
= \(\frac{22}{7}\) × \(\frac{35}{2}\) × \(\frac{35}{2}\) × 120 cm3 = 115500 cm3
= \(\frac{115500}{1000}\) litres = 115.5 litres
Question 12.
How many cylindrical glasses, diameter 8 cm, height 15 cm, can be filled from cylindrical vessel, diameter 30 cm, height 80 cm, full of milk. (There is no need to substitute for π).
Solution:
Diameter of a cylindrical vessel = 30 cm
∴ Radius (R) = \(\frac{30}{2}\) = 15 cm
and height (H) = 80 cm
∴ Volume = πR2H
= π(15)2 × 80 = π × 225 × 80 cm3 Diameter of cylindrical glass = 8 cm
∴ Radius (r) = \(\frac{8}{2}\) = 4 cm
and height (h) = 15 cm
∴ Volume of glass = πr2h
= π(4)2 × 15 = π × 4 × 4 × 15 cm3
= 240π cm3
∴ Number of glasses = \(\frac{\text { Volume of vessel }}{\text { volume of glass }}\) = \(\frac{225 \times 80 \pi}{240 \pi}\) = 75 glasses
Question 13.
A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic metre of gravel are required to gravel the path to a depth of 7 cm?
Solution:
Width of circular path = 2 m
Diameter of circular pond = 40 m
Inner radius (r) = \(\frac{40}{2}\) = 20 m
Outer radius (R) = 20 + 2 = 22 m
∴ Area of path = π[R2 – r2]
= \(\frac{22}{7}\) [222 – 202]
= \(\frac{22}{7}\) × 42 × 2 = 264 m2
Depth of path = 7 cm = \(\frac{7}{100}\) m
∴ Volume of gravel required = 264 × \(\frac{7}{100}\) m3
= \(\frac{1848}{100}\) = 18.48 m3
Question 14.
An iron block is in the form of a cylinder of diameter 0.5 m and length 3.5 m. This block is to be rolled into the form of a bar having square section of side 25 cm. Find the length of the bar.
Solution:
Diameter of iron cylindrical block = 0.5 m
∴ Radius (r) = \(\frac{0.5}{2}\) = 0.25 m
= 25 cm
Length = 3.5 m = 350 cm
Volume of block = πr2h
= \(\frac{22}{7}\) × 25 × 25 × 350 cm3
= 687500 cm3
∴ Volume of bar = \(\frac{687500}{25 \times 25}\) = 1100 cm = 11 m
Question 15.
A swimming pool 70 m long, 44 m wide, 3 m deep is fdled by water issuing from a pipe of diameter 14 cm, at 2 m per second. How many hours does it take to fill the bath?
Solution:
Length of swimming pool (l) = 70 m
Breadth (b) = 44 m
and depth (d) = 3 m
∴ Volume of water in it = 70 × 44 × 3 m3 = 9240 m3
Diameter of the pipe = 14 cm
∴ Radius (r) = \(\frac{14}{2}\) = 7 cm = 0.07 m
Length of water flow
= \(\frac{\text { Volume of water }}{\text { Area of mouth of pipe }}\)
= \(\frac{9240}{\pi r^2}\) = \(\frac{9240 \times 7}{22 \times 0.07 \times 0.07}\)m
= \(\frac{60}{0.0001}\) = 60 × 10000 m
Speed of water flow = 2 m per second
∴ Time taken = \(\frac{60 \times 10000}{2}\)seconds
= \(\frac{60 \times 10000}{2 \times 60 \times 60}\)hrs. = \(\frac{250}{3}\)hrs = 83\(\frac{1}{3}\)hours
Question 16.
What length of solid cylinder 2 cm in diameter must be taken to be cast into a hollow cylinder of external diameter 12 cm, 0.25 cm thick, 15 cm long?
Solution:
External diameter of a hollow cylinder = 12 cm
∴ Radius (R) = \(\frac{12}{2}\) = 6 cm
Thickness of sheet = 0.25 cm
∴ Inner radius (r) = 6 – 0.25 = 5.75 cm and length (h)= 15 cm
∴ Volume of metal used = πh(R2 -r2)
= \(\frac{22}{7}\) × 15 × (62 – 5.752) cm3
= \(\frac{330}{7}\) (11.75 × 0.25) cm3
= \(\frac{330}{7}\) × 11.75 × 0.25 cm3
Now volume of solid cylinder
= \(\frac{330}{7}\) × 11.75 × 0.25 cm3
Diameter of cylinder = 2 cm
∴ Radius = \(\frac{2}{2}\) = 1 cm
∴ Length of cylinder = \(\frac{\text { Volume }}{\pi r^2}\) = \(\frac{330 \times 11.75 \times 0.25}{7 \times \pi \times 1 \times 1}\)
= \(\frac{330 \times 1175 \times 25 \times 7}{7 \times 100 \times 100 \times 22}\) = \(\frac{705}{16}\) = 44.0625 cm
Question 17.
A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm, and its length is 21 cm. The metal every where is 0.4 cm thick. Calculate the volume of the metal to 1 place od decimal. \(\left(\pi=\frac{22}{7}\right)\)
Solution:
Internal diameter of the open tube = 11.2cm
∴ Radius (r) = \(\frac{11.2}{2}\) = 5.6 cm
Length (h) = 21 cm
Thickness of tube = 0.4 cm
∴ External radius (R) = 5.6 + 0.4 = 6.0 cm
∴ Volume = πh(R2 – r2)
= \(\frac{22}{7}\) × 21 × (62 – 5.62) cm3
= 306.24 cm3 = 306.2 cm3
Question 18.
Water flows along a pipe of radius 0.6 cm at 8 cm per second. This pipe is draining the water from a tank which holds 1000 litres of water when full. How long would it take to completely empty the tank?
Solution:
Water in the tank = 1000 l = 1 m3
∴ Volume of water = 1000 × 1000 m3 = 1000000 cm3
Radius of pipe = 0.6 cm
Area of mouth of pipe = πr2
= \(\frac { 22 }{ 7 }\) × 0.6 × 0.6 = \(\frac { 792 }{ 700 }\) cm2
∴ Length of flow of water in pipe = \(\frac{1000000 \times 7 \times 10 \times 10}{22 \times 6 \times 6}\)
= \(\frac{700000000}{792}\)cm
speed of water from 8 cm per sec.
∴ Time taken = \(\frac{700000000}{792 \times 8}\)sec.
= \(\frac{700000000}{792 \times 8 \times 60 \times 60}\)hr
= \(\frac{1104780}{60 \times 60}\)hr = 30.69 hours
Question 19.
A cylindrical bucket 28 cm in diameter; 72 cm high and full of water, is emptied into a rectangular tank 66 cm long, 28 cm wide. Find the height of the water-level in the tank. \(\left(\pi=\frac{22}{7}\right)\)
Solution:
Diameter of cylindrical bucket = 28 cm
∴ Radius (r) = \(\frac { 28 }{ 2 }\) = 14 cm
Height (h) = 72 cm
∴ Volume of water in it = πr2h
= \(\frac { 22 }{ 7 }\) × 14 × 14 × 72 cm3 = 44352 cm3
Water in the tank = 44352 cm3
Length of rectangular tank = 66 cm
Breadth = 28 cm
∴ Height = \(\frac{\text { Volume }}{l \times b}\) = \(\frac{44352}{66 \times 28}\) = 24 cm
Question 20.
There is some water in a cylindrical vessel of base diameter 14 cm. When an iron- cube is entirely immersed in it, the height of the water rises by 8\(\frac { 9 }{ 14 }\) cm. Find the length of the edge of the cube \(\left(\pi=\frac{22}{7}\right)\)
Solution:
Diameter of cylindrical vessel = 14 cm
∴ Radius (r) = \(\frac { 14 }{ 2 }\) = 7 cm
Height of water = 8\(\frac { 9 }{ 14 }\) = \(\frac { 121 }{ 14 }\) cm
∴ Volume of water = πr2h
= \(\frac { 22 }{ 7 }\) × 7 × 7 × \(\frac { 121 }{ 14 }\)cm3 = 1331 cm3
Now volume of cube = 1331 cm3
Question 21.
If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one?
Solution:
Let radius of the base of a cylinder = r and height = h
∴ Volume πr2h
If radius is halved, then
Radius (r1) = \(\frac { r }{ 2 }\)
Height = h
Volume = π(r1)2h
∴ Volume = π\(\left(\frac{r}{2}\right)^2\) h = \(\frac{\pi r^2 h}{4}\)
∴ Ratio in the volume is reduced to the original cylinder = \(\frac{\pi r^2 h}{4}: \pi r^2 h\)
= 1 : 4
Question 22.
A rectangular sheet of paper 22 cm long and 12 cm broad can be formed into the curved surface of a cylinder in two ways. Find the difference between the volumes of the two cylinders which can be formed.
Solution:
Length of sheet = 22 cm
and breadth = 12 cm
If it is folded length wire, then Circumference of so formed cylinder = 22cm
∴ Radius (r) = \(\frac{C}{2 \pi}\) = \(\frac{22 \times 7}{2 \times 22}\) = \(\frac{7}{2}\) cm
and height (h) = 12 cm
Volume = πr2h = \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 12 cm3 = 462 cm3
If it is folded breadth wire, then
Circumference = 12 cm
∴ Radius = \(\frac{12}{2 \pi}\) = \(\frac{12 \times 7}{2 \times 22}\) = \(\frac{2}{11}\)cm
and height (h) = 22 cm
∴ volume = πr2h = \(\frac{22}{7}\) × \(\frac{21}{11}\) × \(\frac{21}{11}\) × 22 = 252 cm3
Difference volumes = 462 – 252 = 210 cm3
Question 23.
A 20 m deep well with diameter 7 m is dug up and the earth from digging is spread evenly to form a platform 22 m × 14 m. Determine the height of the platform.
Solution:
Depth of a well = 20 m
Diameter = 7 m
∴Radius = \(\frac{7}{2}\) m
∴ Volume of earth so dugout = πr2h
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 20 m3 = 770 m3
Now volume of the platform = 770 m3
Size of the platform = 27 m × 14 m
∴ Height = \(\frac{\text { Volume }}{l \times b}\) = \(\frac{770}{22 \times 14}\) = \(\frac { 5 }{ 2 }\)m = 2.5 m
Question 24.
The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 min in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one fifth of a litre?
(Answer correct to the nearest 100 words) \(\left(\pi=\frac{22}{7}\right)\)
Solution:
Length of cylindrical barrel of a pen = 7 cm and diameter – 5 mm
∴ Radius = \(\frac{5}{2}\)mm = \(\frac{5}{20}\) = \(\frac{1}{4}\)cm
∴ volume of link in it = πr2h
= \(\frac{22}{7}\) × \(\frac{1}{4}\) × \(\frac{1}{4}\) × 7 = \(\frac{11}{8}\) cm3
Volume of link in the bottle = \(\frac{1}{5}\)l = \(\frac{1000}{5}\) = 200 cm3
∴ Total number of barrels = 200 ÷ \(\frac{11}{8}\)
= \(\frac{200 \times 8}{11}\) = \(\frac{1600}{11}\)
words written in one barrel = 310
Total number of words = \(\frac{1600}{11}\) × 310
= 45090 words
= 45100 words (approx)
Question 25.
A closed rectangular box 40 cm long, 30 cm wide and 25 cm deep, has the same volume as that of a cylindrical tin of radius 17.5 cm. Calculate the height of the cylindrical tin correct to 1 decimal place. (Take π to be 3.14)
Solution:
Length of a rectangular box (l) = 40 cm
Breadth (b) = 30 cm
and height (h) = 25 cm
Volume = lbh = 40 × 30 × 25 cm3 = 30000 cm3
Radius of cylindrical tin = 17.5 cm
Volume = 30000 cm3
Height = \(\frac{\text { Volume }}{\pi r^2}\) = \(\frac{30000}{3.14 \times 17.5 \times 17.5}\)cm = \(\frac{30000 \times 10 \times 10}{3.14 \times 175 \times 175}\) = 31.197 cm = 31.2 cm
Question 26.
Earth taken out on digging a circular tank of diameter 17.5 m is spread all around the tank uniformly to a width of 4 m, to form an embankment of height 2 m. Calculate the depth of the circular tank, correct to two decimal places.
Solution:
Diameter of a circular tank = 17.5 m
∴ Radius = \(\frac{17.5}{2}\) = 8.75 m
width of embankment = 4 m
∴ Outer radius = 8.75 + 4 = 12.75 m
and height (h) = 2 m
∴ Volume of the embankment
= πh(R2 – r2) = \(\frac{22}{7}\) × 2 × (12.752 – 8.752)
= \(\frac{44}{7}\)(21.5 × 4) m3
= \(\frac{44}{7}\) × 86 m3 = \(\frac{3784}{7}\) m3
∴ Volume of earth dugout of the tank = \(\frac{3784}{7}\) m3
∴ Height = \(\frac{\text { Volume }}{\pi r^2}\) = \(\frac{3784 \times 7}{7 \times 8.75 \times 8.75 \times 22}\)m
= 2.246 = 2.25 m
Question 27.
Water is flowing at the rate of 7 metres per second through a circular pipe whose internal diameter is 2 cm into a cylindrical tank, the radius of whose base is 40 cm. Determine the increase in the water level in \(\frac{1}{2}\) hour.
Solution:
Speed of waterflow = 7 m per sec.
Internal diameter of pipe = 2 cm
∴ Radius (r) = \(\frac{2}{2}\) = 1 cm
Radius of the base of the circular tank = 40 cm
Time = \(\frac{1}{2}\) hour
Waterflow in \(\frac{1}{2}\) hour (h) =\(\frac{7 \times 60 \times 60}{2}\)m
= 12600 m = 12600 × 100 = 1260000 cm
∴ Volume of water = πr2h
= \(\frac{22}{7}\) × 1 × 1 × 126000 cm3
= 3960000 cm3
Now volume of water in the tank = 3960000 cm3
Radius = 40 cm
∴ Height = \(\frac{\text { Volume }}{\pi r^2}\)
= \(\frac{3960000 \times 7}{22 \times 40 \times 40}\) = \(\frac{6300}{8}\) cm = 787.5 cm
Question 28.
Water flow through a cylindrical pipe of internal diameter 7 cm at 36 km/hr. Calculate time in minutes it would take to fill a cylindrical tank, the radius of whose base is 35 cm and height is 1 m.
Solution:
Diameter of circular end of pipe = 7 cm
Radius (r1) of circular end of pipe = \(\frac{7}{2}\) cm
= 3.5 cm or 0.035 m
Area of cross section = π × r2
= π × (0.35)2 m2
= 1225 × 10-6 π m2
Speed of water = 36 km/h
= 36 × \(\frac{1000}{60}\) = 600 m
Volume of water that flows in 1 minute from pipe = 600 × 1225 × 10-6 π m2
Volume of water that flows in t minutes from pipe = t × 735 × 10-3 π m2
Radius (r2) of cicular end of cylindrical tank = 35 cm = 0.35 m
Depth (h2) of cylindrical tank = 1 m
Let the tank be filled completely in t minutes.
Volume of water filled in tank in t minutes is equal to the volume of water flowed in t minutes from the pipe.
Volume of water that flows in t times = Volume of water in tank
⇒ t × 735 × 10-3 π m2 = π × (r2)2 × h2
⇒ t = 0.35 × 0.35 × 1 × \(\frac{1000}{735}\)
⇒ t = 0.166 minutes
Question 29.
Find the number of coins, 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
Diameter of a coin = 1.5 cm
∴ Radius (r) = \(\frac{1.5}{2}\) = 0.75 cm
and thickness = 0.2 cm
∴ Volume of one coin = πr2h
= π × 0.75 × 0.75 × 0.2 cm3 = 0.11257π cm3
Diameter of cylinder = 4.5 cm
∴ Radius = \(\frac{4.5}{2}\) = 2.25 cm
and height = 10 cm
∴ Volume = π × 2.25 × 2.25 × 10 cm3 = 50.625π cm3
∴ Number of coins = \(\frac{50.625 \pi}{0.1125 \pi}\) = 450
Question 30.
If 1 cubic cm of cast iron weighs 21 g, then find the weight of a cast iron pipe of length 1 m with a bore of 3 cm in which the thickness of the metal is 1 cm.
Solution:
Weight of 1 cm3 = 21 g
Length of pipe (h) = 1 m = 100 cm
Internal diameter = 3 cm
∴ Radius (r) = \(\frac{3}{2}\) = 1.5 cm
Thickness of metal = 1 cm
∴ External radius (R) = 1.5 + 1 = 2.5 cm
Now volume of iron = πh(R2 – r2)
= \(\frac{22}{7}\) × 100(2.52 = 1.52) cm3
= \(\frac{2200}{7}\)(4 × 1) cm3
= \(\frac{8800}{7}\) cm3
∴ Total weight of the metal = \(\frac{8800}{7}\) × 21 g
\(\frac{8800 \times 3}{1000}\) kg = 26.4 kg