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Ncert solutions

CHAPTERS

EXERCISES

1. Surface Areas and Volumes - Exercise 13.1

2. Surface Areas and Volumes - Exercise 13.2

3. Surface Areas and Volumes - Exercise 13.3

4. Surface Areas and Volumes - Exercise 13.4

5. Surface Areas and Volumes - Exercise 13.5

6. Surface Areas and Volumes - Exercise 13.6

7. Surface Areas and Volumes - Exercise 13.7

.A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then

find the volume of the iron used to make the tank. (Assume π = 22/7)

Inner Radius of the tank, (r ) = 1m

Outer Radius (R ) = 1.01m

The volume of the iron used in the tank = 2 / 3 \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right)

Put values,

Volume of the iron used in the hemispherical tank = 2/3 x 22/7 x \left(1.01^{3}-1^{3}\right)=0.06348

So volume of the iron used in the hemispherical tank is 0.06348 \mathrm{m}^{3}

.A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then

find the volume of the iron used to make the tank. (Assume π = 22/7)

Inner Radius of the tank, (r ) = 1m

Outer Radius (R ) = 1.01m

The volume of the iron used in the tank = 2 / 3 \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right)

Put values,

Volume of the iron used in the hemispherical tank = 2/3 x 22/7 x \left(1.01^{3}-1^{3}\right)=0.06348

So volume of the iron used in the hemispherical tank is 0.06348 \mathrm{m}^{3}

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