**KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3.

## Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3

(Take π \(=\frac{22}{7}\) unless stated otherwise)

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:

Let the height of cylinder be ‘h’ cm.

Radius of cylinder, r = 6 cm.

i) Volume of metallic sphere, V = \(\frac{4}{3} \pi r^{3}\)

ii) Volume of Cylinder, V = \(\pi r^{2} h\)

\(\frac{22}{7}\) × 36 × h = 310.46

∴ Height of cylinder is 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 the given spheres are

r_{1} = 6 cm, r_{2} = 8 cm and r_{3} = 10 cm

⇒ Volume of the given spheres are

= \(\frac{4}{3} \times \frac{22}{7}\) × [1728] cm^{3}

Let the radius of the new big sphere be R. Volume of the new sphere

= \(\frac{4}{3}\) × π × R_{3} = \(\frac{4}{3} \times \frac{22}{7}\) × R_{3}

Since, the two volumes must be equal.

∴ \(\frac{4}{3} \times \frac{22}{7} \times R^{3}=\frac{4}{3} \times \frac{22}{7} \times 1728\)

⇒ R_{3} = 1728 ⇒ R = 12 cm

Thus, the required radius of the resulting sphere is 12 cm.

Question 3.

A 20 m deep well with a 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.

Solution:

Radius of well, ‘r’ = 3.5 m.

Height of well, h = 20 m.

(i) Volume of soil dug out of well, V = \(\pi r^{2} h\) (∵ cylinder sphere)

= \(\frac{22}{7} \times(3.5)^{2} \times 20\)

= \(=\frac{22}{7} \times \frac{35}{10} \times \frac{35}{10} \times 20\)

= 770m^{3}

(ii) Length of platform, l = 22 m.

Breadth, b = 14 m.

Height, h =?

Volume of Platform, V= l × b × h (∵ Cuboid)

= 22 × 14 × h

∴ Volume of Platform = Volume of soil dug out of well.

∴ Height of platform is 2.5 m

Question 4.

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

Solution:

Diameter of cylindrical well = 3 m

⇒ Radius of the cylindrical well = \(\frac{3}{2}\) m = 1.5 m

Depth of well (h) = 14 m

∴ Volume of cylindrical well

Let the height of the embankment = H m.

Internal radius of the embankment (r) = 1.5 m.

External radius of the embankment (R)

= (4 + 1.5) m = 5.5 m.

∴ Volume of the embankment

= πR^{2}H – πr^{2}H = πH [R^{2} – r^{2}]

= πH (R + r) (R – r)

= \(\frac{22}{7}\) × H (5.5 + 1.5)(5.5 – 1.5)

= \(\frac{22}{7}\) × H × 7 × 4m^{3}

Since, Volume of the embankment=Volume of the cylindrical well

⇒ \(\frac{22}{7}\) × H × 7 × 4 = 99

⇒ H = 99 × \(\frac{7}{22} \times \frac{1}{7} \times \frac{1}{4} m=\frac{9}{8} m\) = 1.125 m

So, the required height of the embankment = 1.125 m.

Question 5.

A container shaped like a right cylinder diameter 12height 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 cones which can be filled with ice cream.

Solution:

Diameter of right circular cylinder is 12 cm ➝ r = 6 cm.

Height: 15 cm. ➝ h = 15 cm.

Diameter of Cone, 6 cm. ➝ r = 3 cm.

Height 12 cm. ➝ h = 12 cm.

(i) Volume of the ice cream which has cone-shaped and hemisphere.

(ii) Volume of cylindrical vessel, V = \(\pi r^{2} h\)

Let number of cones be ‘n’,

n × 54π = π × 36 × 15

54n = 36 × 15

\(n=\frac{36 \times 15}{54}=\frac{540}{54}\)

∴ n = 10

Ice cream can be filled in 10 cones.

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:

For a circular coin:

Diameter = 1.75 cm

⇒ Radius (r) = \(\frac{175}{200}\) cm

Thickness (h) = 2mm = \(\frac{2}{10}\) cm

∴ Volume of one coin = πr^{2}h = \(\frac{22}{7} \times\left(\frac{175}{200}\right)^{2} \times \frac{2}{10} \mathrm{cm}^{3}\)

For a cuboid:

Length (l) = 10 cm,

Breadth (b) = 5.5 cm

and height (h) = 3.5 cm

∴ Volume = l × b × h = 10 × \(\frac{55}{10} \times \frac{35}{10}\) cm^{3}

Thus, the required number of coins = 400.

Question 7.

A cylindrical bucket, 32 cm high and with a 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:

Height of cylindrical bucket, h = 32 cm.

Radius, r = 18 cm.

Height of conical heap of sand = 24 cm.

Radius, r =?

Slant height, l =?

Volume of the conical heap of sand = Volume of Cylinder.

Radius of Cone, r = 36 cm.

Slant height of cone, l = \(\sqrt{\mathrm{h}^{2}+\mathrm{r}^{2}}\)

∴ Radius of cone, r = 36 cm.

Slant height of cone, h = \(12 \sqrt{13}\) 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:

Breadth of a canal, b = 6 m.

Height, h = 1.5 m

Speed of water =10 km/hr.

Quantity of water = 8 cm.

Quantity of water flowing in 30 minutes?

Volume of water flowing in 60 minutes,

= (10 × 1000) × 6 × 1.5 cm^{3}.

( ∵ Speed = 10 km/hr.; 10 × 1000 m/hr.)

Volume of water flowing in 30 minutes,

= \(10000 \times 9 \times \frac{30}{60} \mathrm{m}^{3}\)

∴ Let the area of land be ‘x’ sq.m.

x= 562500 sq.m.

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 of the pipe = 20 cm

⇒ Radius of the pipe (r) = \(\frac{20}{2}\) cm = 10 cm

Since, the water flows through the pipe at 3 km/hr.

∴ Length of water column per hour(h) = 3 km

= 3 × 1000 m = 3000 × 100 cm = 300000 cm.

Length of water column per hour(h) = 3 km

Volume of water flown in one hour = πr^{2}h

= π × 10^{2} × 300000 cm^{3} = π × 30000000 cm^{2}

Now, for the cylindrical tank :

Diameter = 10 m

⇒ Radius (R) = \(\frac{10}{2}\) m = 5 × 100 cm = 500 cm

Height (H) = 2 m = 2 × 100 cm = 200 cm

∴ Volume of the cylindrical tank = πR^{2}H

= π × (500)^{2} × 200 cm^{3}

Now, time required to fill the tank

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3, drop a comment below and we will get back to you at the earliest.