- Question 1 (JEE Main 2019):

The answer is (C).

The formula for Young’s modulus is $Y = \frac{F/A}{l/L}$. So, $\frac{l}{L} = \frac{F}{AY}$.

- Question 2 (JEE Main 2018):

The answer is (C).

The maximum speed of the block is $\frac{\sqrt{2kF}}{m}$.

This can be derived from the conservation of energy equation:

$KE + PE = constant$

At the maximum speed, the kinetic energy is at its maximum and the potential energy is at its minimum. So, $KE = PE$.

The kinetic energy is $\frac{1}{2}mv^2$ and the potential energy is $\frac{1}{2}kx^2$. So, $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$.

Solving for $v$, we get $v = \frac{\sqrt{2kx}}{m}$.

Substituting in the value of