QJEEM23Ph823f: JEE Main 2023 Physics - Class 11 JEEM23Ph823f Solution: A metal block of mass is suspended from a rigid support through …

To find the mass $m$ of the metal block, we need to consider the tensile stress developed in the wire. The formula for tensile stress is:

$\text{Tensile Stress} = \frac{\text{Force}}{\text{Area}}$

The force acting on the wire is the weight of the metal block, which can be represented as $F = mg$ .

The cross-sectional area of the wire, given its diameter $d = 14 \, mm$ , can be calculated using the formula for the area of a circle:

$A = \pi (\frac{d}{2})^2 = \pi (\frac{14}{2})^2 \, mm^2$

Now, convert the area to $m^2$ :

$A = \pi (\frac{14 \times 10^{-3}}{2})^2 \, m^2$

We are given that the tensile stress developed in the wire is $7 \times 10^5 \, Nm^{-2}$ . Using the tensile stress formula, we can write:

$7 \times 10^5 \, Nm^{-2} = \frac{mg}{A}$

Now, solve for the mass $m$ :

$m = \frac{7 \times 10^5 \, Nm^{-2} \cdot A}{g}$

Substitute the values of A and g into the equation:

$m = \frac{7 \times 10^5 \, Nm^{-2} \cdot \pi (\frac{14 \times 10^{-3}}{2})^2 \, m^2}{9.8 \, ms^{-2}}$

After calculating, we get:

$m \approx 11 \, kg$

Therefore, the mass of the metal block is approximately $11 \, kg$ .