A hollow cylinder ( $\rho =2.2\times {10}^{-8}\Omega -m$) of length 3 m has inner and outer diameters are 2 mm and 4 mm respectively. The resistance of the cylinder is

The resistance of a hollow cylinder can be calculated using the formula:

$R=\frac{\rho l}{A}$

where $\rho$ is the resistivity of the material, l is the length of the cylinder, and A is the cross-sectional area of the cylinder.

The cross-sectional area of a hollow cylinder is given by:

$A=\pi (r_2^2-r_1^2)$

where $r_1$ and $r_2$ are the inner and outer radii of the cylinder, respectively.

Substituting the given values, we get:

$r_1=1\text{ mm}=0.001\text{ m}$

$r_2=2\text{ mm}=0.002\text{ m}$

$l=3\text{ m}$

$\rho =2.2\times {10}^{-8}\text{ Ω m}$

Therefore, the cross-sectional area is:

$A=\pi \left(\right(0.002\text{ m}{)}^2-(0.001\text{ m}{)}^2)=3\times \pi \times {10}^{-6}{\text{ m}}^2$

Substituting these values in the formula for resistance, we get:

$R=\frac{(2.2\times {10}^{-8}\text{ Ω m})·\left(3\text{ m}\right)\times 7}{3\times 22\times {10}^{-6}{\text{ m}}^2}\approx 7\times {10}^{-3}\text{ Ω}$

Hence, the correct answer is $7\times {10}^{-3}\Omega$.