Gujarat Board GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths 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 a cylinder of radius 6 cm. Find the height of the cylinder.

Solution:

In case of sphere, r = 4.2 cm

Volume of sphere = \(\frac {4}{3}\)πr^{3} = \(\frac {4}{3}\)π(4.2)^{3}

In case of cylinder, r = 6 cm and h = ?

Volume of cylinder = πr^{3}h = π x 6^{3} x h

Volume of cylinder = Volume of sphere

⇒ π x 6^{2} x h = \(\frac {4}{3}\)π (4.2)^{3}

h = \(\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6}\)

h = 4 x 1.4 x 0.7 x 0.7

h = 2.74 cm

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:

Radii of solid sphere are

r_{1} = 6 cm

r_{2} = 8 cm

and r_{3} = 10 cm

Let radius of resulting sphere be r.

Volume of three spheres = Volume of resulting sphere

\(\frac {4}{3}\)πr^{3}_{1} + \(\frac {4}{3}\)πr^{3}_{2} + \(\frac {4}{3}\)πr^{3}_{3} = \(\frac {4}{3}\)πr^{3}

\(\frac {4}{3}\)π [r^{3}_{1} + r^{3}_{2} + r^{3}_{3}] = \(\frac {4}{3}\)πr^{3}

⇒ r^{3}_{1} + r^{3}_{2} + r^{3}_{3} = r^{3}

6^{3} + 8^{3} + 10^{3} = r^{3}

r^{3} = 216 + 512 + 1000

r^{3} = 1728

r = \(\sqrt[3]{1728}\)

r = 12 cm

Question 3.

A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform. [NCERT 20121]

Solution:

In case of cylindrical well

Diameter = 7 m

r = \(\frac {7}{2}\)m

Depth of well h = 20 m

Volume of well or earth dug out = πr^{2}h

= \(\frac {22}{7}\) x \(\frac {7}{2}\) x \(\frac {7}{2}\) x 20 = 11 x 7 x 10 = 770 m^{3}

In case of cuboidal platform

l = 22 m

b = 14 m

Let height of platform be h m.

Then volume of earth spread out in the form of cuboidal plate form = Volume of earth dug out from the well

l x b x h = 770

22 x 14 x h = 770

h = \(\frac{770}{22 \times 14}\) = \(\frac{70}{2 \times 14}\)

h = 2.5 m

Question 4.

A well of diameter 3 m is dug 14 m deep. The earth taken out of it 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. [CBSE 20121

Solution:

In case of cylindrical well

Diameter = 3 m

Radius r_{1} = \(\frac {3}{2}\)m

Depth h = 14m

Volume of earth dug out from well

= πr^{2}_{1}h

= π (\(\frac {3}{2}\))^{2} x 14

In case of circular embankment

Radius of embankment be R

= r + 4

= \(\frac {3}{2}\) + 4 = \(\frac {11}{2}\) m

Let height of embankment be H.

Volume of embankment = Volume of earth dug out

πR^{2}H – πr^{2}H = πr^{2}_{1}h

π(R^{2} – r^{2})H = π(\(\frac {3}{2}\))^{2} x 14

H = \(\frac {9}{8}\)

H = 1.125 m

Hence, the height of the embankment is 1.125 m.

Question 5.

A container shaped like a right circular cylinder having a diameter 12 cm and a 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 filled with ice cream.

Solution:

In case of right circular cylinder, diameter = 12 cm

r_{1} = 6 cm

h_{1} = 15 cm

Volume of ice cream in right circular cylinder

= πr_{1}^{2}h_{1}

= π x 6^{2} x 15

In case of cone

Diameter = 6 cm

Radius r_{2} = 3 cm

Height h_{2} = 12 cm

Volume of cone = \(\frac {1}{3}\) πr_{2}^{2}h_{2}

= \(\frac {1}{3}\) π x 3_{2} x 12 = 36π cm^{3}

Volume of hemisphere = \(\frac {2}{3}\) πr^{3}

= \(\frac {2}{3}\)π(3)^{3} = 187πcm^{3}

Number of cones filled with ice cream

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 x 10 cm x 3.5 cm?

Solution:

Silver coins

Diameter = 1.75 cm

Radius r_{1} = \(\frac {1.75}{2}\) cm = \(\frac {1.75}{200}\) = \(\frac {7}{8}\)cm

Thickness of coin h_{1} = 2 mm = \(\frac {2}{10}\)cm = \(\frac {1}{5}\)cm

Volume of a coin = πr^{2}_{1}h_{1}

= π x (\(\frac {7}{8}\))^{2} x \(\frac {1}{5}\)

= \(\frac {22}{7}\) x \(\frac {49}{64}\) x \(\frac {1}{5}\)

= \(\frac{11 \times 7}{32 \times 5}\) cm^{2}

In case of cuboid

Length (l) = 5.5 cm

Breadth (b) = 10 cm

Height (h) = 3.5 cm

Volume of cuboid = l x b x h

= 5.5 x 10 x 3.5 cm^{3}

= 192.5 cm^{3
}

Hence, 400 silver coins must be method.

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 help of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution:

In case of cylindrical bucket

Radius (r) of bucket = 18 cm

Height h = 32 cm

Volume of sand filled in cylindrical bucket

= πr^{2}h

= π x 18^{2 }x 3^{2}

= 10368 π cm^{3}

In case of conical heap

Height (H) = 24 cm

Volume of conical heap = Volume of cylindrical bucket

\(\frac {1}{3}\)πR^{2}H = 10368π

\(\frac {1}{3}\) x R^{2} x 24 = 10368

R^{2} = \(\frac{10368 \times 3}{24}\) = 1296

R = \(\sqrt{1296}\)

R = 36 cm

Hence radius of conical heap of sand = 36 cm

I = \(\sqrt{\mathrm{R}^{2}+\mathrm{H}^{2}}\) = \(\sqrt{36^{2}+24^{2}}\)

= \(\sqrt{1296+576}\) = \(\sqrt{1872}\) = 12 \(\sqrt{13}\)

Hence, slant height of sand = 12\(\sqrt{13}\) cm.

Question 8.

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

Solution:

In case of canal

Width of canal (b) = 6 m

Depth of canal (h) = 1.5 m

Speed of flow of water = 10 km/h

= \(\frac{10 \times 1000 \mathrm{m}}{60 \mathrm{min}}\) = \(\frac{500}{3}\) m / minute

speed of water in 30 minute

= \(\frac{500}{3}\) x 30 = 5000 m

Volume of water flowed in 30 minutes

= l x b x h

= 6 x \(\frac{3}{2}\) x 5000 = 45000 m^{2}

Volume of water stand in a field to irrigate = Volume of water flowed in 30 minutes

⇒ Area of field x \(\frac{8}{100}\) = 45000

( ∴ height = 8 cm = \(\frac{8}{100}\) m)

Area of field = \(\frac{45000 x 100}{8}\) = 562500

1 hectare = 10000 m^{2}

Area of field irrigate = \(\frac{562500}{10000}\) = 56.25 hectares

Question 9.

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Solution:

Diameter in case of cylindrical pipe = 20 cm

Radius (r) = 10 cm = \(\frac{10}{100}\) m = 0.1 m

Speed of flow of water in pipe = 3 km / h

= \(\frac{3 \times 1000}{60 \times 60}\) = \(\frac{5}{6}\) m / s

Volume of water flowed per second

= πr^{2}h = π(0.1)^{2} x \(\frac{5}{6}\)

In case of tank

Diameter = 10 m

R = 5m

Depth H = 2m

Volume of tank = πr^{2}h

= π x 5^{2} x 2 = 50π

Time taken to fill the tank

= \(\frac{50 \pi}{\pi \times 0.1 \times 0.1 \times \frac{5}{6}}\)

= \(\frac{5000 \times 6}{1 \times 1 \times 5}\)

= \(\frac{6000}{60}\) minutes = 100 minutes

= \(\frac{100}{60}\) hour = 1 hour 40 minutes