A hollow spherical shell is made of a metal of density 21 g/cm3. Its internal and external radius are 16 cm and 17 cm, respectively. What is the weight (in kg) of the shell ? 
(Take $$\pi = \frac{22}{7}$$​)

Correct option is B

Given:

Density of metal: 21 g/cm³

Internal radius of the spherical shell: 16 cm

External radius of the spherical shell: 17 cm

π =$$\frac{22}{7}$$​​

Formula Used:
The volume of the hollow spherical shell is calculated as the difference between the

volume of the outer sphere and the inner sphere:

Volume of the shell =$$\frac{4}{3} \pi \left( R^3 - r^3 \right)$$​

Where:

R is the external radius

r is the internal radius

Solution:

Volume of the shell = $$\frac{4}{3} \times \frac{22}{7} \times \left( 17^3 - 16^3 \right)$$​

Volume of the shell =$$\frac{4}{3} \times \frac{22}{7} \times (4913 - 4096)$$​​

Volume of the shell $$= \frac{4}{3} \times \frac{22}{7} \times 817$$​​

Volume of the shell =$$\frac{4 \times 17974}{21}$$​​

Volume of the shell=$$\frac{71896}{21} \approx 3423.61\, \text{cm}^3$$​

Mass =$$\text{Density} \times \text{Volume} = 21 \times 3423.61=$$​ 71896 g

Convert the mass into kilograms:

Mass =$$\frac{71896}{1000} = 71.896 \, \text{kg}$$​

The weight of the shell is approximately 71.896 kg.