Table of contents  
Exercise 11.1  
Exercise 11.2  
Exercise 11.3  
Exercise 11.4 
Q1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area (Assume π = 22 / 7)
Ans:
Diameter of the base of the cone is 10.5 cm. To find the radius, we need to divide the diameter by 2.
Radius of the base of the cone, r = diameter / 2 = 10.5 / 2 = 5.25 cm
The slant height of the cone is given as 10 cm. Let's denote it as l.
Slant height of the cone, l = 10 cm
Now, we can find the curved surface area (CSA) of the cone using the formula:
CSA = π r l
where π (pi) is approximately equal to 22 / 7.
CSA = (22 / 7) × 5.25 × 10
CSA = 165 cm²
Hence, the curved surface area of the cone is 165 cm².
Q2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. (Assume π = 22 / 7)
Solution:
The radius of the cone, r = 24 / 2 m = 12m
Slant height, l = 21 m
To find the total surface area of a cone, we use the formula: Total Surface area of the cone = πr(l + r)
Total Surface area of the cone = (22 / 7) × 12 × (21 + 12) m^{2}
= (22 / 7) × 12 × 33 m^{2}
= 1244.57 m^{2}
Thus, the total surface area of the cone is 1244.57 m^{2}.
Q3. Curved surface area of a cone is 308 cm^{2} and its slant height is 14 cm. Find
(i) radius of the base and
(ii) total surface area of the cone. (Assume π = 22 / 7)
Ans: Slant height of cone, l = 14 cm
Let the radius of the cone be r.
(i) We know, CSA of cone = πrl
Given: Curved surface area of a cone is 308 cm^{2}
(308) = (22 / 7) × r × 14
308 = 44r
r = 308 / 44 = 7cm
The radius of the cone base is 7 cm.
(ii) To find the total surface area of the cone, we need to add the curved surface area (CSA) and the area of the base (πr^{2}).
Total surface area of cone = CSA of cone + Area of base (πr^{2})
Total surface area of cone = 308 + (22 / 7) × 7^{2} = 308 + 154
Therefore, the total surface area of the cone is 462 cm^{2}.
Q4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m^{2} canvas is Rs 70. (Assume π = 22 / 7)
Ans: Let ABC be a conical tent with vertex A, base center O, and base circumference point B.
Height of conical tent (AO), h = 10 m
Radius of conical tent (BO), r = 24m
Let the slant height of the tent (AB) be l.
(i) In right triangle ABO, we have
AB^{2} = AO^{2} + BO^{2} (using Pythagoras theorem)
l^{2} = h^{2} + r^{2}
= (10)^{2} + (24)^{2}
= 100 + 576
= 676
l = √676 = 26m
Therefore, the slant height of the tent is 26 m.
(ii) The curved surface area (CSA) of the conical tent can be calculated using the formula:
CSA = πrl
= (22/7) × 24 × 26 m^{2}
=13728/7 m^{2}
Now, let's calculate the cost of the canvas required to make the tent.
Cost of 1 m^{2} canvas = Rs 70
Cost of (13728 / 7) m^{2} canvas is equal to Rs (13728 / 7) × 70 = Rs 137280
Therefore, the cost of the canvas required to make such a tent is Rs 137280.
Q5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. [Use π = 3.14]
Ans: Height of conical tent, h = 8m
Radius of base of tent, r = 6m
Slant height of tent, l^{2} = (r^{2} + h^{2})
l^{2} = (62 + 82) = (36 + 64) = (100)
or l = 10
Again, CSA of conical tent = πrl
= (3.14 × 6 × 10) m^{2}
= 188.4m^{2}
Let the length of tarpaulin sheet required be L
As 20 cm will be wasted, therefore,
Effective length will be (L  0.2m).
Breadth of tarpaulin = 3m (given)
Area of sheet = CSA of tent
[(L – 0.2) × 3] = 188.4
L  0.2 = 62.8
L = 63
Therefore, the length of the required tarpaulin sheet will be 63 m.
Q6. The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of Rs. 210 per 100 m^{2}. (Assume π = 22 / 7).
Ans: Slant height of conical tomb, l = 25m
Base radius, r = diameter / 2 = 14 / 2 m = 7m CSA of conical tomb = πrl= (22 / 7) × 7 × 25 = 550
CSA of conical tomb = 550m^{2}
Cost of whitewashing 550 m^{2} area, which is Rs (210 × 550) / 100
= Rs.1155
Therefore, cost will be Rs. 1155 while whitewashing tomb.
Q.7. A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. (Assume π = 22 / 7)
Ans: Radius of conical cap, r = 7 cm
Height of conical cap, h = 24cm Slant height, l^{2} = (r^{2} + h^{2})
= (7^{2} + 24^{2})
= (49 + 576)
= (625)
Or l = 25 cm
CSA of 1 conical cap = πrl
= (22 / 7) × 7 × 25
= 550
CSA of 10 caps = (10 × 550) cm^{2} = 5500 cm^{2}
Therefore, the area of the sheet required to make 10 such caps is 5500 cm^{2}.
Q8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m^{2}, what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) = 1.02).
Ans: Given:
Radius of cone, r = diameter / 2 = 40 / 2 cm = 20cm = 0.2 m
Height of cone, h = 1m
Slant height of cone is l, and l^{2} = (r^{2} + h^{2})
Using given values, l^{2} = (0.22 + 12)
= (1.04)
Or l = 1.02
Slant height of the cone is 1.02 m
Now,
CSA of each cone = πrl
= (3.14 × 0.2 × 1.02)
= 0.64056m^{2}
CSA of 50 such cones = (50 × 0.64056)
CSA of 50 such cones = 32.028 m^{2}
Again,
Cost of painting 1 m^{2} area = Rs 12 (given)
Cost of painting 32.028 m^{2} area
= Rs (32.028 × 12)
= Rs.384.336
= Rs.384.34 (approximately)
Therefore, the cost of painting all these cones is Rs. 384.34.
Q1. Find the surface area of a sphere of radius:
(i) 10.5cm
(ii) 5.6cm
(iii) 14cm
(Assume π = 22 / 7)
Ans: Formula: Surface area of sphere (SA) = 4πr^{2}
(i) Radius of sphere, r = 10.5 cm
SA = 4 × (22 / 7) × 10.52 = 1386
Surface area of sphere is 1386 cm^{2}
(ii) Radius of sphere, r = 5.6cm
Using formula, SA = 4 × (22 / 7) × 5.62 = 394.24
Surface area of sphere is 394.24 cm^{2}
(iii) Radius of sphere, r = 14cm
SA = 4πr^{2}
= 4 × (22 / 7) × (14)^{2}
= 2464
Surface area of sphere is 2464 cm^{2}.
Q2. Find the surface area of a sphere of diameter:
(i) 14cm
(ii) 21cm
(iii) 3.5cm
(Assume π = 22 / 7)
Ans: (i) Radius of sphere, r = diameter / 2 = 14 / 2 cm = 7 cm
Formula for Surface area of sphere = 4πr2
= 4 × (22 / 7) × 7^{2 }= 616
Surface area of a sphere is 616 cm^{2}
(ii) Radius (r) of sphere = 21/2 = 10.5 cm
Surface area of sphere = 4πr^{2}
= 4 × (22 / 7) × 10.52 = 1386
Surface area of a sphere is 1386 cm^{2}
Therefore, the surface area of a sphere having diameter 21cm is 1386 cm^{2}
(iii) Radius(r) of sphere = 3.5 / 2 = 1.75 cm Surface area of sphere = 4πr2
= 4 × (22 / 7) × 1.752 = 38.5
Surface area of a sphere is 38.5 cm^{2}.
Q3. Find the total surface area of a hemisphere of radius 10 cm. [Use π = 3.14]
Ans: Radius of hemisphere, r = 10cm
Formula: Total surface area of hemisphere = 3πr^{2}
= 3 × 3.14 × 10^{2} = 942
The total surface area of given hemisphere is 942 cm^{2}.
Q4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Ans: Let r_{1} and r_{2} be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So
r_{1} = 7cm
r_{2} = 14 cm
Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)
= 4r_{1}^{2} / 4r_{2}^{2}
= (r_{1} / r_{2})^{2}
= (7 / 14)^{2} = (1 / 2)^{2} = 1 / 4
Therefore, the ratio between the surface areas is 1 : 4.
Q5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tinplating it on the inside at the rate of Rs 16 per 100 cm^{2}. (Assume π = 22 / 7)
Ans: Inner radius of hemispherical bowl, say r = diameter / 2 = (10.5) / 2 cm = 5.25 cm
Formula for Surface area of hemispherical bowl = 2πr^{2}
= 2 × (22 / 7) × (5.25)^{2} = 173.25
Surface area of hemispherical bowl is 173.25 cm^{2}
Cost of tinplating 100 cm^{2} area = Rs 16
Cost of tinplating 1 cm^{2} area = Rs 16 /100
Cost of tinplating 173.25 cm^{2} area = Rs. (16 × 173.25) / 100 = Rs 27.72
Therefore, the cost of tinplating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm^{2} is Rs 27.72.
Q6. Find the radius of a sphere whose surface area is 154 cm^{2}. (Assume π = 22/7)
Ans: Let the radius of the sphere be r.
Surface area of sphere = 154 (given)
Now,
4πr^{2} = 154
r^{2 }= (154 × 7) / (4 × 22) = (49 / 4)
r = (7 / 2) = 3.5
The radius of the sphere is 3.5 cm.
Q7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Ans: If diameter of earth is said d, then the diameter of moon will be d / 4 (as per given statement)
Radius of earth = d / 2
Radius of moon = 1 / 2 × d / 4 = d / 8
Surface area of moon = 4π(d / 8)^{2}
Surface area of earth = 4π(d / 2)^{2}
Ratio of their Surace areas = 4π(d / 8)^{2} / 4π (d / 2)^{2 }= 4 / 64 = 1 / 16
The ratio between their surface areas is 1 : 16.
Q8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume π = 22 / 7)
Ans: Given:
Inner radius of hemispherical bowl = 5cm
Thickness of the bowl = 0.25 cm
Outer radius of hemispherical bowl = (5 + 0.25) cm = 5.25 cm
Formula for outer CSA of hemispherical bowl = 2πr^{2}, where r is radius of hemisphere
= 2 × (22 / 7) × (5.25)^{2} = 173.25
Therefore, the outer curved surface area of the bowl is 173.25 cm^{2}.
Q9. A right circular cylinder just encloses a sphere of radius r (see Fig). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in(i) and (ii).
Ans: (i) Surface area of sphere = 4πr^{2}, where r is the radius of sphere
(ii) Height of cylinder, h = r + r = 2r
Radius of cylinder = r
CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h)
= 4πr^{2}
(iii) Ratio between areas = (Surface area of sphere) / CSA of Cylinder)
= 4r^{2} / 4r^{2}
= 1 / 1
Ratio of the areas obtained in (i) and (ii) is 1 : 1.
Q1. Find the volume of the right circular cone with
(i) radius 6cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm (Assume π = 22 / 7)
Ans: Volume of cone = (1 / 3) πr^{2}h cube units
Where r be radius and h be the height of the cone
(i) Radius of cone, r = 6 cm
Height of cone, h = 7cm
Say, V be the volume of the cone, we have
V = (1 / 3) × (22 / 7) × 36 × 7
= (12 × 22)
= 264
The volume of the cone is 264 cm^{3}.
(ii) Radius of cone, r = 3.5cm
Height of cone, h = 12cm
Volume of cone = (1 / 3) × (22 / 7) × 3.52 × 7 = 154
Hence, The volume of the cone is 154 cm^{3}.
Q2. Find the capacity in litres of a conical vessel with
(i) radius 7cm, slant height 25 cm
(ii) height 12 cm, slant height 12 cm (Assume π = 22 / 7)
Ans: (i) Radius of cone, r = 7 cm
Slant height of cone, l = 25 cm
or h = 24
Height of the cone is 24 cm
Now,
Volume of cone, V = (1/3) πr^{2}h (formula)
V = (1 / 3) × (22 / 7) × 72 × 24
= (154 × 8)
= 1232
So, the volume of the vessel is 1232 cm3
Therefore, capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm^{3})
= 1.232 Liters.
(ii) Height of cone, h = 12 cm
Slant height of cone, l = 13 cm
r = 5
Hence, the radius of cone is 5 cm.
Now, Volume of cone,
V = (1 / 3)πr^{2}h V = (1 / 3) × (22 / 7) × 52 × 12 cm^{3}
= 2200 / 7
Volume of cone is 2200 / 7 cm^{3}
Now, Capacity of the conical vessel= 2200 / 7000 litres (1L = 1000 cm^{3})
= 11 / 35 litres.
Q3. The height of a cone is 15cm. If its volume is 1570cm^{3}, find the diameter of its base. (Use π = 3.14)
Ans: Height of the cone, h = 15 cm
Volume of cone = 1570 cm3
Let r be the radius of the cone
As we know: Volume of cone, V = (1 / 3) πr^{2}h
So, (1 / 3) πr^{2}h = 1570
(1 / 3) × 3.14 × r^{2} × 15 = 1570
r^{2} = 100
r = 10
Radius of the base of cone 10 cm.
Q4. If the volume of a right circular cone of height 9cm is 48πcm3, find the diameter of its base.
Ans: Height of cone, h = 9cm
Volume of cone = 48π cm^{3}
Let r be the radius of the cone.
As we know: Volume of cone, V = (1 / 3) πr^{2}h
So, 1/3 π r^{2}(9) = 48 π
r^{2 }= 16
r = 4
Radius of cone is 4 cm.
So, diameter = 2 × Radius = 8
Thus, diameter of base is 8cm.
Q5. A conical pit of top diameter 3.5m is 12m deep. What is its capacity in kiloliters? (Assume π = 22 / 7)
Ans: Diameter of conical pit = 3.5 m
Radius of conical pit, r = diameter / 2 = (3.5 / 2)m = 1.75m
Height of pit, h = Depth of pit = 12m
Volume of cone, V = (1 / 3) πr2h
V = (1 / 3) × (22 / 7) × (1.75)^{2} × 12 = 38.5
Volume of cone is 38.5 m3
Hence, capacity of the pit = (38.5 × 1) kiloliters = 38.5 kiloliters.
Q6. The volume of a right circular cone is 9856cm^{3}. If the diameter of the base is 28cm, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone (Assume π = 22 / 7)
Ans: Volume of a right circular cone = 9856 cm^{3}
Diameter of the base = 28 cm
(i) Radius of cone, r = (28 /2) cm = 14 cm
Let the height of the cone be h
Volume of cone, V = (1 / 3) πr^{2}h
(1 / 3) πr^{2}h = 9856
(1 / 3) × (22 / 7) × 14 × 14 × h = 9856
h = 48
The height of the cone is 48 cm.
Slant height of the cone is 50 cm.
(iii) curved surface area of cone = πrl
= (22 / 7) × 14 × 50 = 2200
curved surface area of the cone is 2200 cm^{2}.
Q7. A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Ans: Height (h)= 12 cm
Radius (r) = 5 cm, and
Slant height (l) = 13 cm
Volume of cone, V = (1 / 3) πr^{2}h
V = (1 / 3) × π × 5^{2} × 12 = 100π
Volume of the cone so formed is 100π cm^{3}.
Q8. If the triangle ABC in the Question 7 is revolved about the side 5cm, then find the volume of the solids so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Ans:
A rightangled ΔABC is revolved about its side 5cm, a cone will be formed of radius as 12 cm, height as 5 cm, and slant height as 13 cm.
Volume of cone = (1/3) πr^{2}h; where r is the radius and h be the height of cone
= (1/3) × π × 12 × 12 × 5
= 240 π
The volume of the cones of formed is 240π cm^{3}.
So, required ratio = (result of question 7) / (result of question 8) = (100π) / (240π) = 5 / 12 = 5 : 12.
Q9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas. (Assume π = 22 / 7)
Ans: Radius (r) of heap = (10.5 / 2) m = 5.25
Height (h) of heap = 3m
Volume of heap = (1 / 3)πr^{2}h
= (1 / 3) × (22 / 7) × 5.25 × 5.25 × 3
= 86.625
The volume of the heap of wheat is 86.625 m^{3}. Again,
Area of canvas required = CSA of cone = πrl, where l =
After substituting the values, we have
= (22 / 7) × 5.25 × 6.05
= 99.825
Therefore, the area of the canvas is 99.825 m^{2}.
Q1. Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m (Assume π = 22 / 7)
Ans: (i) Radius of sphere, r = 7 cm
Using, Volume of sphere = (4 / 3) πr^{3}
= (4 / 3) × (22 / 7) × 7^{3} = 4312 / 3
Hence, volume of the sphere is 4312 / 3 cm^{3}
(ii) Radius of sphere, r = 0.63 m
Using, volume of sphere = (4 / 3) πr^{3}
= (4 / 3) × (22 / 7) × 0.63^{3} = 1.0478
Hence, volume of the sphere is 1.05 m^{3} (approx).
Q2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m (Assume π = 22 / 7)
Ans: (i) Diameter = 28 cm
Radius, r = 28 / 2 cm = 14cm
Volume of the solid spherical ball = (4 / 3) πr^{3}
Volume of the ball = (4 / 3) × (22 / 7) × 14^{3} = 34496 / 3
Hence, volume of the ball is 34496 / 3 cm^{3}
(ii) Diameter = 0.21 m
Radius of the ball = 0.21 / 2 m = 0.105 m
Volume of the ball = (4 / 3)πr^{3}
Volume of the ball = (4 / 3) × (22 / 7) × 0.105^{3}
Hence, volume of the ball = 0.004851 m^{3}
Q3. The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm^{3}? (Assume π = 22 / 7)
Ans: Given,
Diameter of a metallic ball = 4.2 cm
Radius(r) of the metallic ball, r = 4.2 / 2 cm = 2.1 cm
Volume formula = 4 / 3 πr^{3}
Volume of the metallic ball = (4 / 3) × (22 / 7) × 2.1^{3}
Volume of the metallic ball = 38.808 cm^{3}
Now, using relationship between, density, mass and volume,
Density = Mass / Volume
Mass = Density × volume
= (8.9 × 38.808) g
= 345.3912 g
Mass of the ball is 345.39 g (approx).
Q4. The diameter of the moon is approximately onefourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Ans: Let the diameter of earth be “d”. Therefore, the radius of earth will be will be d/2
Diameter of moon will be d / 4 and the radius of moon will be d/8
Find the volume of the moon:
Volume of the moon = (4 / 3) πr^{3} = (4 / 3) π (d / 8)^{3} = 4 / 3π(d^{3} / 512)
Find the volume of the earth:
Volume of the earth = (4 / 3) πr^{3} = (4 / 3) π (d / 2)^{3} = 4 / 3π(d^{3} / 8)
Fraction of the volume of the earth is the volume of the moon
Volume of moon is of the 1 / 64 volume of earth.
Q5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22 / 7)
Ans: Diameter of hemispherical bowl = 10.5 cm
Radius of hemispherical bowl, r = 10.5 / 2 cm = 5.25 cm
Formula for volume of the hemispherical bowl = (2 / 3) πr^{3}
Volume of the hemispherical bowl = (2 / 3) × (22 / 7) × 5.25^{3} = 303.1875
Volume of the hemispherical bowl is 303.1875 cm^{3}
Capacity of the bowl = (303.1875) / 1000 L = 0.303 litres(approx.)
Therefore, hemispherical bowl can hold 0.303 litres of milk.
Q6. A hemi spherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume π = 22/7)
Ans: Inner Radius of the tank, (r) = 1m
Outer Radius (R) = 1.01m
Volume of the iron used in the tank = (2 / 3) π (R^{3} – r^{3})
Put values,
Volume of the iron used in the hemispherical tank = (2 / 3) × (22 / 7) × (1.01^{3} – 1^{3}) = 0.06348
So, volume of the iron used in the hemispherical tank is 0.06348 m^{3}.
Q7. Find the volume of a sphere whose surface area is 154 cm^{2}. (Assume π = 22 / 7)
Ans: Let r be the radius of a sphere.
Surface area of sphere = 4πr^{2}
4πr^{2} = 154 cm^{2} (given)
r^{2} = (154 × 7) / (4 × 22) r = 7 / 2
Radius is 7 / 2 cm
Now, Volume of the sphere = (4 / 3) πr^{3}
Volume of the sphere = (4 / 3) x (22 / 7) x (7 / 2)^{3} = 179 x 2 / 3
Volume of the sphere is 179 x 2 / 3 cm^{3}
Q8. A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of Rs. 4989.60. If the cost of whitewashing isRs20 per square meter, find the
(i) Inside surface area of the dome
(ii) volume of the air inside the dome (Assume π = 22 / 7)
Ans: (i) Cost of whitewashing the dome from inside = Rs 4989.60
Cost of whitewashing 1m^{2 }area = Rs 20
CSA of the inner side of dome = 498.96 / 2 m^{2} = 249.48 m^{2}
(ii) Let the inner radius of the hemispherical dome be r.
CSA of inner side of dome = 249.48 m^{2} (from (i))
Formula to find CSA of a hemi sphere = 2πr^{2}
2πr^{2} = 249.48
2 × (22 / 7) × r^{2} = 249.48
r^{2} = (249.48 × 7) / (2 × 22)
r^{2} = 39.69
r = 6.3
So, radius is 6.3 m
Volume of air inside the dome = Volume of hemispherical dome
Using formula, volume of the hemisphere
= 2 / 3 πr^{3} = (2 / 3) × (22 / 7) × 6.3 × 6.3 × 6.3
= 523.908
= 523.9(approx.)
Volume of air inside the dome is 523.9 m^{3}.
Q9. Twentyseven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the
(i) radius r’ of the new sphere,
(ii) ratio of Sand S’.
Ans: Volume of the solid sphere = (4 / 3)πr^{3}
Volume of twenty seven solid sphere = 27 × (4 / 3)πr^{3} = 36 πr^{3}
(i) New solid iron sphere radius = r’
Volume of this new sphere = (4/3)π(r’)^{3}
(4 / 3)π(r’)^{3} = 36 πr^{3}
(r’)^{3} = 27r^{3}
r’= 3r
Radius of new sphere will be 3r (thrice the radius of original sphere)
(ii) Surface area of iron sphere of radius r, S = 4πr^{2}
Surface area of iron sphere of radius r’= 4π (r’)^{2}
Now
S / S’ = (4πr^{2}) / ( 4π (r’)^{2})
S / S’ = r^{2} / (3r’)^{2} = 1 / 9
The ratio of S and S’ is 1: 9.
Q10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm^{3}) is needed to fill this capsule? (Assume π = 22 / 7)
Ans: Diameter of capsule = 3.5 mm
Radius of capsule, say r = diameter / 2 = (3.5 / 2) mm = 1.75mm
Volume of spherical capsule = 4 / 3 πr^{3}
Volume of spherical capsule = (4 / 3) × (22 / 7) × (1.75)^{3} = 22.458
The volume of the spherical capsule is 22.46 mm^{3}.
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Short Answer Type Questions: Surface Areas & Volumes Doc  4 pages 
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Long Question Answer: Surface Areas And Volume Doc  6 pages 
1. What is the formula for calculating the surface area of a cube? 
2. How do you find the surface area of a cylinder? 
3. What is the difference between lateral surface area and total surface area? 
4. How do you calculate the volume of a cone? 
5. What is the surface area of a sphere? 
Short Answer Type Questions: Surface Areas & Volumes Doc  4 pages 
NCERT Exemplar Solutions: Surface Area & Volume Doc  18 pages 
Long Question Answer: Surface Areas And Volume Doc  6 pages 

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