# PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

## PSEB 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.

2 cubes each of volume 64 cm^{3} are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Let side of cube = x cm

Volume of cube = 64 cm^{3}

[volume of cube = (side)^{3}]

When cubes are joined end to end and cuboid is formed

whose Length = 2x cm = 2(4) = 8 cm

Width = x cm = 4 cm

Height = x cm = 4 cm

Surface area of cuboid = 2[LB + Bh + hL]

= 2 [8 × 4 + 4 × 4 + 4 × 8]

= 2 [32 + 16 + 32]

= 2 [80]

∴ Surface area of cuboid = 160 cm^{2}.

Question 2.

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere ¡s 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Solution:

Diameter of hemisphere = Diameter of cylinder

= 14 cm

2R = 14 cm

Radius of hemisphere (R) = 7 cm

Total height of vessel = 13 cm

∴ Height of cylinder = (13 – 7) = 6 cm

Inner surface area of vessel = inner surface area of cylinder + Inner surface area of hemisphere

= 2πRH + 2πR^{2}

= 2πR [H + R]

= 2 × 22/7 × 7(16 + 7)

= 44 × 13 = 572 cm^{2}

Hence, Inner surface area of vessel = 572 cm^{2}

Question 3.

A toy is ¡n the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution:

Radius of cone = Radius of hemisphere (R) = 3.5 cm

Total height of toy = 15.5 cm

∴ Height of cone (H) = (15.5 – 3.5) = 12 cm

Question 4.

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have?

Find the surface area of the solid.

Solution:

Side of cubical box = 7 cm

Diameter of hemisphere = Side of cubical box = 7 cm

2R = 7

R = 7/2 cm

Question 5.

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of (he hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution:

Let each side of cube = a

∴ Diameter of hemisphere = Side of cube

2R = a

R = a/2

Question 6.

A medicine capsule is ¡n the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Question 7.

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of 500 per (Note that the base of the tent will not be

covered with canvas.)

Solution:

Diameter of cone = Diameter of cylinder

2R = 4

R = 2 m

Question 8.

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

Solution:

Diameter of cylinder (D) = 1.4 cm = Diameter of cone

∴ Radius of cylinder = Radius of cone (R) = 0.7 cm

Height of cylinder (H) = 2.4 cm

Hence, Total surface area remaining solid to nearest cm^{2} = 18 cm^{2}.

Question 9.

A wooden article was made by scooping out a hemisphere form each end of a solid cylinder, as shown in Fig.

If the height of the cylinder is 10 cm, and its base ¡s of radius 3.5 cm, find the total surface area of the article.

Solution:

Height of cylinder (H) = 10 cm

Radius of cylinder = Radius of hemisphere (R) = 3.5 cm

Surface area of article = curved surface area of cylinder + 2 curved surface area of hemisphere

**Follow on Facebook page –** **Click Here**

**Google News join in – Click Here**

**Read More Asia News – Click Here**

**Read More Sports News – Click Here**

**Read More Crypto News – Click Here**