# Ncert Solutions For Class 10 Maths Ex 13.3

## Ncert Solutions For Class 10 Maths Chapter 13 Ex 13.3

1: A metallic sphere having radius of 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Calculate the height of the cylinder.

Ans.- Here, Radius of sphere is 4.2 cm, radius of cylinder is 6 cm

As we know that volume of sphere

=43πr3$\frac{4}{3}\pi r^{3}$

=43π×4.23$\frac{4}{3}\pi\times 4.2^{3}$

Similarly volume of cylinder

=πr2h=π×62×h$\pi r^{2}h=\pi \times 6^{2}\times h$

Since volume of cylinder = Volume of sphere

Hence, height of cylinder.

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

Ans.- Here, Radii of spheres = 6 cm, 8 cm, 10 cm

As we know that volume of sphere=43πr3$\frac{4}{3}\pi r^{3}$

So total volume of three spheres is

=43π(63+83+103)$\frac{4}{3}\pi\left ( 6^{3}+8^{3}+10^{3} \right )$

=43π(616+512+1000)$\frac{4}{3}\pi\left (616+512+1000 \right )$

=43π×1728$\frac{4}{3}\pi\times 1728$

Hence radius of biggest sphere is

=17283=12cm$\sqrt[3]{1728}=12cm$.

3: A 20 m deep well having diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Calculate the height of the platform.

Ans.- Here, Radius of well = 3.5 m, depth of well = 20 m

So the dimension of rectangular platform =22m×14m$22m\times 14m$

Now the volume of earth dug out

=πr2h$\pi r^{2}h$

=π×3.52×20=770m3$\pi\times 3.5^{2}\times 20=770m^{3}$

Area of top of platform = Area of Rectangle – Area of Circle

(as the circular portion of mouth of well is open)

=22×14π×3.52$22\times 14-\pi \times 3.5^{2}$

=30838.5=269.5m2$308-38.5=269.5m^{2}$

So as we know, height=volume/area

=770269.5=2.85m$\frac{770}{269.5}=2.85m$

4: A well is dug with diameter 3 m and a depth of 14 m. The earth taken out of it has been spread in a same quantity surrounding the well forming a shape of circular ring of width 4 m to form an embankment. Calculate the height of the embankment.

Ans.- Here, Radius of well = 1.5 m, depth of well = 14 m, width of embankment = 4 m

Now the volume of earth dug out,

=πr2h$\pi r^{2}h$

=π×1.52×14=31.5πm3$\pi \times 1.5^{2}\times 14=31.5\pi m^{3}$

Area of top of platform = (Area of bigger circle – Area of smaller circle)

So as we know, height=volume/area

=31.5π28π=1.125m$\frac{31.5\pi }{28\pi }=1.125m$

5: A vessel shaped like a right circular cylinder with diameter 12 cm and height 15 cm is full of ice cream. The ice cream needs to be filled into cones having height 12 cm and diameter 6 cm, forming a hemispherical shape on the top. Calculate the number of such cones which can be filled with ice cream.

Ans.- Here, Radius of cylinder = 6 cm, height of cylinder = 15 cm

Similarly, Radius of cone = 3 cm, height of cone = 12 cm

Given, Radius of hemispherical top on ice cream = 3 cm

Now as we know that, volume of cylinder

= πr2h$\pi r^{2}h$

= π×62×15=540πcm3$\pi \times 6^{2}\times 15=540\pi cm^{3}$

Volume of cone is

=13×π×32×12$\frac{1}{3}\times \pi \times 3^{2}\times 12$

=36πcm3$36\pi cm^{3}$

Now, volume of hemisphere is

=23πr3$\frac{2}{3}\pi r^{3}$

= 23×π×33=18πcm3$\frac{2}{3}\times \pi \times 3^{3}=18\pi cm^{3}$

So volume of ice-cream will be

= (36+18)π=54πcm3$\left ( 36+18 \right )\pi =54\pi cm^{3}$

Hence the number of ice creams = Volume of cylinder/Volume of ice cream

540π54π=10$\frac{540\pi }{54\pi }=10$

6: How many silver coins, 1.75 cm in diameter and of thickness 2 mm, can be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Ans.- Here, Radius of coin = 0.875 cm, height = 0.2 cm

Dimensions of cuboid = 5.5 cm x 10 cm x 3.5 cm

As we know that the volume of coin is

=πr2h$\pi r^{2}h$

= π×0.8752×0.2$\pi \times 0.875^{2}\times 0.2$= 0.48125cm3$0.48125cm^{3}$

Similarly, volume of cuboid is= 5.5×10×3.5=192.5cm3$5.5\times 10\times 3.5=192.5cm^{3}$

Therefore Number of coins

= 192.50.48125=400$\frac{192.5}{0.48125}=400$

7: A cylindrical bucket with 32 cm high and having 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, calculate the radius and slant height of the heap.

Ans.- Given, Radius of cylinder = 18 cm, height = 32 cm

Height of cone = 24 cm

As we know that the volume of cylinder is

= πr2h$\pi r^{2}h$

= π×182×32$\pi\times 18^{2}\times 32$

As, Volume of cone = Volume of cylinder

So, volume of cone is

= 13πr2×24$\frac{1}{3}\pi r^{2}\times 24$

Hence, the radius of cone can be calculated as follows:

r2=3×π×182×32π×24$r^{2}=\frac{3\times \pi \times 18^{2}\times 32 }{\pi \times 24}$

Or, r2=182×22$r^{2}=18^{2}\times 2^{2}$

Or, r = 36cm

So now the slant height of conical heap can be calculated as follows:

l = h2+r2$\sqrt{h^{2}+r^{2}}$

= 242+362$\sqrt{24^{2}+36^{2}}$

= 576+1296=1872$\sqrt{576+1296}=\sqrt{1872}$

= 3613cm$36\sqrt{13}cm$

8: Water in a canal with 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?

Ans.- Given, Depth = 1.5 m, width = 6 m, height of standing water = 0.08 m

In 30 minutes, length of water column = 5 km = 5000 m

Volume of water in 30 minutes = 1.5 x 6 x 5000 = 45000 cubic m

So as we know that,

Area = Volume/Height

= 450000.08=562500m2$\frac{45000}{0.08}=562500m^{2}$

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

Ans.- Given,  Radius of pipe = 10 cm = 0.1 m, length = 3000 m/h

Radius of tank = 5 m, depth = 2 m

So the volume of water in 1 hr. through pipe is

= πr2h$\pi r^{2}h$

= π×0.12×3000$\pi \times 0.1^{2}\times 3000$= 30πm3$30\pi m^{3}$

Similarly, the volume of tank is

= πr2h$\pi r^{2}h$

= π×52×2=50πm3$\pi \times 5^{2}\times 2=50\pi m^{3}$

Hence the time taken to fill the tank is = Volume of tank/Volume of water in 1 hr.

= 50π30π$\frac{50\pi }{30\pi }$=1 hr. 40min