Given below are the

a. Multiple choice questions

b. True and False

c. Calculation problems

A solid metallic hemisphere of radius 8 cm is melted and recasted into a right circular cone of base radius 6 cm. Determine the height of the cone.

(a) 20 cm

(b) 28.44 cm

(c) 26.44 cm

(d) None of the above

Radius of the hemisphere = 8 cm

Volume of the hemisphere =$ \frac{2}{3}\pi R^{3}$

Radius of the right circular cone = r=6 cm

Let the height of the cone = h cm

Volume of the right circular cone

= $ \frac {1}{3}\pi R^{2}h$

Now as per question

Volume of the hemisphere = volume of the right circular cone

$ \frac {2}{3}\pi R^{3}=\frac{1}{3}\pi R^{2}h$

Substituting the values

h = 28. 44 cm

Hence (b) is the answer

A cone, a hemi-sphere and a cylinder stand on equal bases and have the same height. Find the ratio of their volumes as well

(a) 3:2:1

(b) 2:4:8

(c) 1:2:3

(d) 3:9:27

The volume of cone is given by formula

V =$ \frac {1}{3}\pi R^{2}h$

Volume of the hemisphere =$ \frac {2}{3}\pi R^{3}$

Volume of the cylinder =$ \pi R^{2}h$

Now Height(H)= Radius (R) as they have height

Volume of Cone=$ \frac {1}{3}\pi R^{3}$

Volume of the hemisphere =$ \frac {2}{3}\pi R^{3}$

Volume of the cylinder =$ \pi R^{3}$

Ratio will be 1:2:3

Hence the Answer (c)

Find the ratio of their total surface areas

(a) (√2+1):3:4

(b) 2:4:8

(c) 1:2:3

(d) 3:9:27

Surface area of cone is given by = $\pi R^{2} +\pi R \sqrt {R^2 + H^2} =(1+\sqrt{2})\pi R^{2}$ as ( H=R)

Surface area of hemisphere is given by =$3\pi r^{2}$

Surface area of cylinder is given by =$2\pi rh + 2\pi r^{2}= 4\pi r^{2}$ as ( H=R)

So ratio will be (√2+1):3:4

Hence the Answer (a)

(i)Two identical solid hemispheres of equal base radius

(ii)A solid cylinder of radius

(iii)A solid cone of radius

(iv)A solid ball is exactly fitted inside the cubical box of side

(i) True

(ii) false

(iii) True

(iv) False

Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9cm.

The edge of cube is 9 cm So this will be the altitude of the Largest cone.

The radius of Largest circular cone will be 9/2=4.5 cm.

The volume of cone is given by formula

V =$ \frac {1}{3}\pi R^{2}h$

where R is the radius and H the altitude of cone.

Substituting the values and taking $\pi$ as 22/7

we get

V=190.93 cm

A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the one is 2cm and diameter of the base is 4cm. If a right circular cylinder circumscribes the toy, find how much more space it will cover.

8π cm^{3}

A hemispherical tank full of water is emptied by a pipe at the rate of 3 litres per second. How much time will it take to make the tank half empty, if the tank is 3m in diameter?

(16.5 min)

A toy is in the shape of a right circular cylinder with a hemisphere at one end and a cone on the other. The radius and height of the cylindrical part are 5cm and 13cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is 30cm

770cm^{2}

The radii of the internal and the external surfaces of metallic spherical shells are 3cm and 5cm respectively. It is melted and recast into a solid right circular cylinder of height 10 cm. Find the diameter of the base of the cylinder

A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of < 5 per 100 sq. cm. [ Use π=22/7]

This Class 10 Maths Extra Questions for Surface Area and Volume with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.

**Notes**-
**Assignments** -
**NCERT Solutions**- NCERT Solution Surface Area and Volume Class10 Exercise 13.1
- NCERT Solution Surface Area and Volume Class10 Exercise 13.2
- NCERT Solution Surface Area and Volume Class10 Exercise 13.3

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