- 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
