# Ncert Solutions For Class 10 Maths Ex 13.2

## Ncert Solutions For Class 10 Maths Chapter 13 Ex 13.2

1: A solid cone placed on a hemisphere with both their radii whose value is equal to 1 cm and the height of the cone is equal to its radius. Calculate the volume of the solid in terms of π.

Ans.-  radius = 1 cm, height = 1 cm

Volume of hemisphere

=23πr3$=\frac{2}{3}\pi r^{3}$ =23π×13$=\frac{2}{3}\pi\times 1^{3}$ =23πcm2$=\frac{2}{3}\pi cm^{2}$

Volume of cone

=13πr2h$=\frac{1}{3}\pi r^{2}h$ =13π×12×1=13πcm3$=\frac{1}{3}\pi\times 1^{2}\times 1=\frac{1}{3}\pi cm^{3}$

Total volume

=23π+13π=πcm3$=\frac{2}{3}\pi+\frac{1}{3}\pi =\pi cm^{3}$

2: Harish, who is student, he was asked to make a model of shape similar to a cylinder which contains two cones connected at its two ends by using a metal sheet. The diameter and the height of the model are 3 cm and 12 cm respectively. If both the cone has a height of 2 cm, calculate the volume of air present in the model that Harish made. (Assume that both the inner and outer dimension of the model is almost same).

Ans.- Height of cylinder = 12 – 4 = 8 cm, radius = 1.5 cm, height of cone = 2 cm

Volume of cylinder

=πr2h$=\pi r^{2}h$ =π×1.52×8=18πcm3$=\pi \times 1.5^{2}\times 8=18\pi cm^{3}$

Volume of cone

=13πr2h$=\frac{1}{3}\pi r^{2}h$ =13π×1.52×2$=\frac{1}{3}\pi\times 1.5^{2}\times 2$ =1.5πcm3$=1.5\pi cm^{3}$

Total volume

=1.5π+1.5π+18π$=1.5\pi +1.5\pi +18\pi$ =21π=66cm3$=21\pi =66cm^{3}$

3: A sweet, which contains sugar syrup up to about 30% of its volume. Calculate approximately how much syrup will be available in 45 sweets, each shaped like a cylinder having two hemispherical ends with length 5 cm and 2.8 cm diameter.

Ans.-  Length of cylinder = 5 – 2.8 = 2.2 cm, radius = 1.4 cm

Volume of cylinder

=πr2h$=\pi r^{2}h$ =π×1.42×2.2$=\pi \times 1.4^{2}\times 2.2$ =4.312πcm3$=4.312\pi cm^{3}$

Volume of two hemispheres

=43πr3$\frac{4}{3}\pi r^{3}$

=43π×1.43$=\frac{4}{3}\pi\times 1.4^{3}$ =10.9763πcm3$=\frac{10.976}{3}\pi cm^{3}$

Total volume

=4.312π+10.9763π$=4.312\pi +\frac{10.976}{3}\pi$

Volume of syrup = 30% of total volume

=π(4.312+10.9763)×30100$=\pi \left ( 4.312+\frac{10.976}{3} \right )\times \frac{30}{100}$ =23.9123×30100×227=7.515cm3$=\frac{23.912}{3}\times \frac{30}{100}\times \frac{22}{7}=7.515cm^{3}$

Volume of syrup in 45 sweets = 45 x 7.515 = 338.184 cm3$cm^{3}$

4: A flower pot that is made of wood having the shape of a cuboid with four conical depressions to hold flowers. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Calculate the volume of wood in the entire pot.

Ans.- Dimensions of cuboid is 15 cm x 10 cm x 3.5 cm, radius of cone is 0.5 cm, depth of cone is 1.4 cm

As we know Volume of cuboid = length x width x height

So, 15×10×3.5=525cm3$15\times 10\times 3.5=525cm^{3}$

Now volume of cone =13πr2h$=\frac{1}{3}\pi r^{2}h$

=13×227×0.52×1.4=1130cm3$=\frac{1}{3}\times \frac{22}{7}\times 0.5^{2}\times 1.4=\frac{11}{30}cm^{3}$

therefore$\ therefore$ Hence volume of wood =Volume of cuboid – 6 x volume of cone

=5256×1130$=525-6\times \frac{11}{30}$ =525115=522.8cm3$=525-\frac{11}{5}=522.8cm^{3}$

5: A container is in the shape of an inverted cone. The height and the radius at the top (which is open) of the container are 8 cm and 5 cm respectively. The container is filled with water up to the upper edge. When lead shots, each of which is a sphere of radius 0.5 cm are dropped inside the vessel, quarter quantity of the water flows out. Calculate the number of lead shots dropped in the vessel.

Ans.- Given, radius of cone is 5 cm, height of cone is 8 cm, radius of sphere is 0.5 cm

So volume of cone=13πr2h$\frac{1}{3}\pi r^{2}h$

13π×52×8$\frac{1}{3}\pi\times 5^{2}\times 8$ 2003πcm3$\frac{200}{3}\pi cm^{3}$

Similarly volume of lead shot=43πr3$\frac{4}{3}\pi r^{3}$

43π×0.53$\frac{4}{3}\pi \times 0.5^{3}$ 16πcm3$\frac{1}{6}\pi cm^{3}$

Now number of lead shots will be

=2003π×14÷16π$\frac{200}{3}\pi \times \frac{1}{4}\div \frac{1}{6}\pi$

=50π3×6π=100$\frac{50\pi }{3}\times \frac{6}{\pi }=100$

6: A solid metal pole consisting of a cylinder of height and base diameter as 220 cm and 24 cm respectively is surmounted by another cylinder of height and radius as 60 cm and 8 cm respectively. Calculate the mass of the pole, given that 1 cm3 of iron has approximately 8g mass.

Ans.- Here radius of bigger cylinder = 12 cm, height of bigger cylinder = 220 cm

Similarly, radius of smaller cylinder = 8 cm, height of smaller cylinder = 60 cm

As we know volume of bigger cylinder=πr2h$\pi r^{2}h$

=π×122×220$\pi\times 12^{2}\times 220$

=31680πcm3$31680\pi cm^{3}$

Now volume of small cylinder=πr2h$\pi r^{2}h$

=π×82×60$\pi \times 8^{2}\times 60$

=3840πcm3$3840\pi cm^{3}$

Therefore total volume=31680π+3840π$31680\pi +3840\pi$

=35520πcm3$35520\pi cm^{3}$

Hence, Mass= Density. Volume

=8×35520π=892262.4gm$8\times 35520\pi =892262.4gm$= 892.3kg$892.3kg$

7: A solid containing of a right circular cone of height and radius are 120 cm and 60 cm respectively standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of liquid such that it touches the bottom. Calculate the volume of liquid left in the cylinder, if the radius and height of the cylinder are 60 cm and 180 cm respectively.

Ans.- Here, Radius of cone = 60 cm, height of cone = 120 cm

Radius of hemisphere = 60 cm

Radius of cylinder = 60 cm, height of cylinder = 180 cm

So as we know that volume of cone is =13πr2h$\frac{1}{3}\pi r^{2}h$

=13π×602×120$\frac{1}{3}\pi \times 60^{2}\times 120$

=144000πcm3$144000\pi cm^{3}$

Volume of hemisphere

=43πr3$\frac{4}{3}\pi r^{3}$

23π×603=144000cm3$\frac{2}{3}\pi \times 60^{3}=144000cm^{3}$

Similarly, volume of solid=(144000+144000)π$\left ( 144000+144000 \right )\pi$

=288000πcm3$288000\pi cm^{3}$

Now volume of cylinder=πr2h$\pi r^{2}h$

=π×602×180=648000πcm3$\pi \times 60^{2}\times 180=648000\pi cm^{3}$

Hence the total volume of liquid left in the cylinder is

=(648000288000)π=360000π$\left ( 648000-288000 \right )\pi =360000\pi$

=1130400cm3$1130400cm^{3}$

8: A spherical glass vessel having a cylindrical neck of height 8 cm and diameter 2 cm; the radius of the spherical part is 4.25 cm. By determining the amount of water it holds, a child calculates its volume to be 345 cm3. Identify whether he is correct, taking the above as the inside measurements, and π = 3.14.

Ans.- Here, Radius of cylinder = 1 cm, height of cylinder = 8 cm, radius of sphere = 4.25 cm

As we know that volume of cylinder=πr2h$\pi r^{2}h$

=π×12×8$\pi \times 1^{2}\times 8$

=8πcm3$8\pi cm^{3}$

Similarly volume of sphere=43πr3$\frac{4}{3}\pi r^{3}$

=43π×(8.52)3$\frac{4}{3}\pi \times \left ( \frac{8.5}{2} \right )^{3}$

=6141256000πcm3$\frac{614125}{6000}\pi cm^{3}$

therefore$\ therefore$,the total volume will be

=(6141256000+8)π$\left (\frac{614125}{6000}+8 \right )\pi$

=(614125+480006000)π=346.51cm3$\left (\frac{614125+48000}{6000} \right )\pi =346.51cm^{3}$