NCERT Solutions for Class 9th: Ch 13 Surface Areas and Volumes Maths
Page No: 225Exercise 13.4
1. Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
Answer
(i) Radius of the sphere (r) = 10.5 cm
Surface area = 4πr2
= (4 × 22/7 × 10.5 × 10.5) cm2
= 1386 cm2
(ii) Radius of the sphere (r) = 5.6 cm
Surface area = 4πr2
= (4 × 22/7 × 5.6 × 5.6) cm2
= 394.24 cm2
(iii) Radius of the sphere (r) = 14 cm
Surface area = 4πr2
= (4 × 22/7 × 14 × 14) cm2
= 2464 cm2
2. Find the surface area of a sphere of diameter:
(i) 14 cm (ii) 21 cm (iii) 3.5 m
Answer
(i) r = 14/2 cm = 7cm
Surface area = 4πr2
= (4 × 22/7 × 7 × 7)cm2
=616cm2
(ii) r = 21/2 cm = 10.5 cm
Surface area = 4πr2
= (4 × 22/7 × 10.5 × 10.5) cm2
= 1386 cm2
(iii) r = 3.5/2 m = 1.75 m
Surface area = 4πr2
= (4 × 22/7 × 1.75 × 1.75) m2
= 38.5 m2
3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Answer
r = 10 cm
Total surface area of hemisphere = 3πr2
= (3 × 3.14 × 10 ×10) cm2
= 942 cm2
4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Answer
Let r be the initial radius and R be the increased radius of balloons.
r = 7cm and R = 14cm
Ratio of the surface area =4πr2/4πR2
= r2/R2
= (7×7)/(14×14) = 1/4
Thus, the ratio of surface areas = 1 : 4
5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm2.
Answer
Radius of the bowl (r) = 10.5/2 cm = 5.25 cm
Curved surface area of the hemispherical bowl = 2πr2
= (2 × 22/7 × 5.25 × 5.25) cm2
= 173.25 cm2
Rate of tin - plating is = ₹16 per 100 cm2
Therefor, cost of 1 cm2 = ₹16/100
Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100
= ₹27.72
6. Find the radius of a sphere whose surface area is 154 cm2.
Answer
Let r be the radius of the sphere.
Surface area = 154 cm2
⇒ 4πr2 = 154
⇒ 4 × 22/7 × r2 = 154
⇒ r2 = 154/(4 × 22/7)
⇒ r2 = 49/4
⇒ r = 7/2 = 3.5 cm
7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Answer
Let the diameter of earth be r and that of the moon will be r/4
Radius of the earth = r/2
Radius of the moon = r/8
Ratio of their surface area = 4π(r/8)2/4π(r/2)2
= (1/64)/(1/4)
= 4/64 = 1/16
Thus, the ratio of their surface areas is 1:16
8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Answer
Inner radius of the bowl (r) = 5 cm
Thickness of the steel = 0.25 cm
∴ outer radius (R) = (r + 0.25) cm
= (5 + 0.25) cm = 5.25 cm
Outer curved surface = 2πR2
= (2 × 22/7 × 5.25 × 5.25) cm2
= 173.25 cm2
9. A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
(i) The surface area of the sphere with raius r = 4πr2
(ii) The right circular cylinder just encloses a sphere of radius r.
∴ the radius of the cylinder = r and its height = 2r
∴ Curved surface of cylinder =2πrh
= 2π × r × 2r
= 4πr2
(iii) Ratio of the areas = 4πr2:4πr2 = 1:1
Page No: 228
Exercise 13.5
1. A matchbox measures 4cm × 2.5cm × 1.5cm. What will be the volume of a packet containing 12 such boxes?
Answer
Dimension of matchbox = 4cm × 2.5cm × 1.5cm
l = 4 cm, b = 2.5 cm and h = 1.5 cm
Volume of one matchbox = (l × b × h)
= (4 × 2.5 × 1.5) cm3 = 15 cm3
Volume of a packet containing 12 such boxes = (12 × 15) cm3 = 180 cm3
2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1m3 = 1000 l)
Answer
Dimensions of water tank = 6m × 5m × 4.5m
l = 6m , b = 5m and h = 4.5m
Therefore Volume of the tank =ℓbh m3
=(6×5×4.5)m3=135 m3
Therefore , the tank can hold = 135 × 1000 litres [Since 1m3=1000litres]
= 135000 litres of water.
3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?
Answer
Length = 10 m , Breadth = 8 m and Volume = 380 m3
Volume of cuboid = Length × Breadth × Height
⇒ Height = Volume of cuboid/(Length × Breadth)
= 380/(10×8) m
= 4.75m
4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m3.
Answer
l = 8 m, b = 6 m and h = 3 m
Volume of the pit = lbh m3
= (8×6×3) m3
= 144 m3
Rate of digging = ₹30 per m3
Total cost of digging the pit = ₹(144 × 30) = ₹4320
5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Answer
length = 2.5 m, depth = 10 m and volume = 50000 litres
1m3 = 1000 litres
∴ 50000 litres = 50000/1000 m3 = 50 m3
Breadth = Volume of cuboid/(Length×Depth)
= 50/(2.5×10) m
= 2 m
6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last?
Answer
Dimension of tank = 20m × 15m × 6m
l = 20 m , b = 15 m and h = 6 m
Capacity of the tank = lbh m3
= (20×15×6) m3
= 1800 m3
Water requirement per person per day =150 litres
Water required for 4000 person per day = (4000×150) l
= (4000×150)/1000
= 600 m3
Number of days the water will last = Capacity of tank Total water required per day
=(1800/600) = 3
The water will last for 3 days.
7. A godown measures 40m × 25m × 15m. Find the maximum number of wooden crates each measuring 1.5m × 1.25m × 0.5m that can be stored in the godown.
Answer
Dimension of godown = 40 m × 25 m × 15 m
Volume of the godown = (40 × 25 × 15) m3 = 10000 m3
Dimension of crates = 1.5m × 1.25m × 0.5m
Volume of 1 crates = (1.5 × 1.25 × 0.5) m3 = 0.9375 m3
Number of crates that can be stored =Volume of the godown/Volume of 1 crate
= 10000/0.9375 = 10666.66 = 10666
8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Answer
Edge of the cube = 12 cm.
Volume of the cube = (edge)3 cm3
= (12 × 12 × 12) cm3
= 1728 cm3
Number of smaller cube = 8
Volume of the 1 smaller cube =1728/8 cm3 = 216 cm3
Side of the smaller cube = a
a3 = 216
⇒ a = 6 cm
Surface area of the cube = 6 (side)2
Ratio of their surface area = (6 × 12 × 12)/(6 × 6 × 6)
= 4/1 = 4:1
9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Answer
Depth of river (h) = 3 m
Width of river (b) = 40 m
Rate of flow of water (l) = 2 km per hour = (2000/60) m per minute
= 100/3 m per minute
Volume of water flowing into the sea in a minute = lbh m3
= (100/3 × 40 × 3) m3
= 4000 m3
Page No: 230
Exercise 13.6
1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm 3 = 1l)
Answer
(i) Radius of the sphere (r) = 10.5 cm
Go to Part I (Exercise 3.1 to 3.3)
Go to Part III (Exercise 3.7 to 3.8)
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