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Question 1:

Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m

Solution 1:

i) Radius of the sphere(r) = 7 cm

Volume of the sphere = 4/3πr³

= (4/3 × 22/7 × 7 × 7 × 7) cm³

=(88×7×49)/3×7 cm³

=4312/3 cm³
= 1437 cm³

Volume of the sphere =1437 cm³

(ii) Given:

Radius of the sphere(r) = 0.63 m

ii) Volume of the sphere = 4/3πr³

= (4/3 × 22/7 × 0.63 × 0.63 × 0.63) m³

= 4×22×.09×.21× 0.63 m³

= 1.0478 m³

Volume of the sphere =1.0478 m³

Comments

Sushmit Datta

Question 2:

A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square meter, find the

(i) inside surface area of the dome,

(ii) volume of the air inside the dome.

Solution 2:

(i) 249.48 square metre

(ii) 39.69 cubic metre

Comments

Shubham Jakhar

Question 3:

Twenty seven 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 S and S'.

Solution 3:

(i) r' = 3r

(ii) 1/9

Comments

Shubham Jakhar

Question 4:

Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m

Solution 4:

(i) Radius of sphere = 7 cm

Volume of sphere =

Therefore, the volume of the sphere is 1437 cm³.

(ii) Radius of sphere = 0.63 m

Volume of sphere =

Therefore, the volume of the sphere is 1.05 m³ (approximately).

Comments

Tasneem

Question 5:

Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m

Solution 5:

(i) Radius (r) of ball =

Volume of ball =

Therefore, the volume of the sphere is cm³.

(ii)Radius (r) of ball = = 0.105 m

Volume of ball =

Therefore, the volume of the sphere is 0.004851 m³.

Comments

Tasneem

Question 6:

The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of
the metal is 8.9 g per $cm^{3}$ ?

Solution 6:

Radius (r) of metallic ball =

Volume of metallic ball =

Mass = Density × Volume

= (8.9 × 38.808) g

= 345.3912 g

Hence, the mass of the ball is 345.39 g (approximately)

Comments

Tasneem

Question 7:

The diameter of the moon is approximately one-fourth of the diameter of the earth.
What fraction of the volume of the earth is the volume of the moon?

Solution 7:

Let the diameter of earth be d. Therefore, the radius of earth will be .

Diameter of moon will be and the radius of moon will be .

Volume of moon =

Volume of earth =

Therefore, the volume of moon is of the volume of earth.

Comments

Tasneem

Question 8:

How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Solution 8:

Radius (r) of hemispherical bowl = = 5.25 cm

Volume of hemispherical bowl =

= 303.1875 cm³

Capacity of the bowl =

Therefore, the volume of the hemispherical bowl is 0.303 litre.

Comments

Tasneem

Question 9:

A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m,
then find the volume of the iron used to make the tank.

Solution 9:

Thickness of iron sheet: 1cm
Inner radius of tank (r): 1m or 100 cm.
Volume of iron used to build the tank= (1/2) x (Outer volume of tank - Inner volume of tank.)

Outer radius of tank (R) = (100+1) cm= 101cm
Outer volume of tank = $\frac{4}{3}\pi \times R^{3}$
Similarly, inner volume = $\frac{4}{3}\pi \times r^{3}$

Therefore, (Outer volume) - (inner volume) =

$\frac{4}{3}\times \frac{22}{7}\times \left ( R^{3}-r^{3} \right )=$ $\left ( \frac{88}{21} \right )\times \left [ \left ( 101-100 \right )\left \{ \left ( 101\times 101 \right )+\left ( 100\times 100 \right )\right \} \right ]$

$=\left ( \frac{88}{21} \right )\times 30301$
= 12575.62 cc.
Therefore, volume of iron used
= (1/2)x 126975.62 cc
= 63487.81 cc.

Comments

Sushmit Datta

Question 10:

Find the volume of a sphere whose surface area is 154 $cm^{2}$

Solution 10:

Let radius of sphere be r.

Surface area of sphere = 154 cm²

⇒ 4πr² = 154 cm²

Volume of sphere =

Therefore, the volume of the sphere is cm³.

Comments

Tasneem

Question 11:

A dome of a building is in the form of a hemisphere. From inside, it was white-washed
at the cost of Rs.4989.60. If the cost of white-washing is Rs. 20 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.

Solution 11:

(i) Total cost of white washing = Rs.4989.60
Rate= Rs. 20/ sq.metre.
Therefore, CSA of the inner side of dome =
= 249.48 sq.metre.

(ii) Let the inner radius of the hemispherical dome be r.

CSA of inner side of dome = 249.48 m²

2πr² = 249.48 m²

⇒ r = 6.3 m

Volume of air inside the dome = Volume of hemispherical dome

= 523.908 m³

= 523.9 m³ (approximately)

Therefore, the volume of air inside the dome is 523.9 m³.

Comments

Sushmit Datta

Question 12:

Twenty seven 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 S and S′.

Solution 12:

(i)Radius of 1 solid iron sphere = r

Volume of 1 solid iron sphere

Volume of 27 solid iron spheres

27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this new sphere be r'.

Volume of new solid iron sphere

(ii) Surface area of 1 solid iron sphere of radius r = 4πr²

Surface area of iron sphere of radius r' = 4π (r')²

= 4 π (3r)² = 36 πr²

Comments

Tasneem

Question 13:

A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much
medicine (in $mm^{3}$ ) is needed to fill this capsule?

Solution 13:

Diameter of sphere capsule(d)= 3.5mm
Therefore, radius of the sphere(r)= 1.75mm.
Volume of a sphere = 4/3π $r^{3}$
So,

Volume of spherical capsule

= 22.458 mm³

= 22.46 mm³ (approximately)

Therefore, the amount of medicine that would be needed to fill the capsule would be 22.46 $mm^{3}$

Comments

Sushmit Datta

All Exercises - Chapter 13 - Surface Areas and Volumes Class 9

Exercise 13.8

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Learn Exercise 13.8 with Free Lessons & Tips

All Exercises - Chapter 13 - Surface Areas and Volumes Class 9

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