Work, Energy and Power: NEET previous year questions

The useful power is work done against gravity divided by time, $P = \dfrac{mgh}{t}$. Substituting, $P = \dfrac{1000 \times 9.8 \times 20}{10} = 19600\ \text{W} = 19.6\ \text{kW}$ (option (3)). The 19.6 W answer drops a factor of 1000, and the 39.2 values double-count g or the height.

The condition $\mu_k(L) = 0$ fixes $\alpha = \mu_0/L$, so $\mu_k(x) = \mu_0\left(1 - \frac{x}{L}\right)$ falls linearly from $\mu_0$ to zero. Because the force varies with position, the work must be integrated: $|W| = \int_0^L \mu_k(x) Mg\,dx = \mu_0 Mg\left[L - \frac{L}{2}\right] = \frac{1}{2}\mu_0 MgL$. Comparing with $n\mu_0 MgL$ gives $n = 1/2$, which is simply the average of the initial and final friction coefficients since the variation is linear. NCERT Class 11, Work, Energy and Power (work done by a variable force).

In a head-on ELASTIC collision between equal masses the velocities are simply exchanged, so $A$ stops dead and $B$ leaves the collision at the full 10 $\text{ms}^{-1}$. Energy conservation on the swing then gives $h = \frac{v^2}{2g} = \frac{100}{20} = 5\ \text{m}$. The bob rises only 5 m, less than the 10 m string length, so it stays below the horizontal through the point of suspension and the string never goes slack. NCERT Class 11, Work, Energy and Power (collisions).

Why the other options are wrong:

- Option (1) (8 m) and option (2) (7 m) are higher than 5 m. Even if $B$ took all the kinetic energy of $A$, it could rise only $\frac{v^2}{2g} = 5\ \text{m}$. A greater height needs energy the collision never supplied.
- Option (4) (2.5 m) gives $B$ only half the kinetic energy of $A$, as if the energy were shared equally between the two masses. That halves the height. In an elastic collision of equal masses, all of it passes to $B$.