# PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

## PSEB 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Question 1.

A metallic sphere of radius 4.2 cm is melted and recast into the shape of cylinder of radius 6 cm. Find the height of the cylinder.

Solution:

Question 2.

Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution:

Radius of first sphere (r_{1}) = 6 cm

Radius of second sphere (r_{2}) = 8 cm

Radius of third sphere (r_{3}) = 10 cm

Let radius of new sphere formed be R cm

On recasting volume remain same

Volume of all three spheres = Volume of big sphere

Question 3.

A 20 m deep well with-diameter 7 m is dug and the earth from digging is evenly spread out to form of platform 22 m by 14 m. Find the height of the platform.

Solution:

Diameter of well = 7 m

Radius of well (cylinder) R = 7/2 m

Height of well (H) = 20 m

Length of Platform (L) =22 m

Width of Platform (B) = 14 m

Let height of Platform be H m

Volume of earth dug out from well = Volume of platform formed

Question 4.

A well of diameter 3 m is dug 14 m deep. The earth taken out of ¡t has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution:

Depth of well (h) = 14 m

Radius of well (r) = 3/2 m

Embankment is in the shape of hollow cylinder whose inner radius is same as radius of well and width of embañkment 4 m

Timer radius of embankment = Radius of well(r) = 3/2 m

Outer radius of embankment (R) = (3/2 + 4) m

R = 11/2 = 5.5 m

Volume of earth dug out = Volume of embankment (so formed)

πR^{2}h = volume of outer cylinder – volume of inner cylinder

πr^{2}h = πR^{2}H – πr^{2}H

= πH[R^{2} – r^{2}]

Hence, Height of embankment H = 1.125 m.

Question 5.

A container shaped like a right circular cylinder having 4iameter 12 cm and height 15 cm Is full of ice-cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be tilled with ice-cream.

Dinieter of cvlrnder (D) = 12 cm

.. Radius of cylinder (R) = 6 cm

Height of cylinder (H) = 15 cm

Diameter of cone = 6 cm

Radius of cone (r) = 3 cm

Radius of hemisphere (r) = 3 cm

Height of cone (h) = 12 cm

Let us suppose number of cones used to fill the ice-cream = n

Volume of ice cream in container = n [Volume of ice cream in one cone]

n = 10

Hence, Number of cones formed = 10.

Question 6.

How many silver coins 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions

5.5 cm × 10 cm × 3.5 cm.

Solution:

Silver coin is in the form of cylinder

Diameter of silver coin = 1.75 cm

∴ Radius of silver coin (r) = 1.75/2 cm

Thickness of silver coin = Height of cylinder (H) = 2 mm

Question 7.

A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution:

Question 8.

Water in a canal 6m wide and 1.5m deep is flowing with a speed of 10 km/k How much area will it irrigate in 30 minutes, If 8 cm of standing water is needed?

Solution:

Width of canal = 6 m

Depth of water in canal = 1.5 m

Velocity at which water is flowing = 10km/hr

Volume of water discharge in one hour = (Area of cross section) velocity

= (6 × 1.5m2) × 10 km

= 6 × 1.5 × 10 × 6 × 1.5 × 10 × 1000 × 1000 m^{3}

∴ Volume of water discharge in 1/2 hour = 1/2 × 6×15/10 × 1000 = 45000 m^{3}

Let us suppose area to be irrigate = (x) m^{2}

According to question, 8 cm standing water is required in field

∴ Volume of water discharge by canal in 1/2 hours = Volume of water in field 45000 m^{3} = (Area of field) × Height of water

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