Question

A bob of mass $m$ is suspended by a light string of length $L$ . It is imparted a horizontal velocity $v_0$ at the lowest point A such that it completes a circle in the vertical plane. Match Column I with Column II.

• A. $A - p$ , $B - q$ , $C - s$ , $D - r$
• B. $A - q$ , $B - r$ , $C - p$ , $D - s$
• C. $A - r$ , $B - s$ , $C - q$ , $D - p$
• D. $A - s$ , $B - p$ , $C - r$ , $D - q$

Answer 1

Answer: (C)

$A - r$ , $B - s$ , $C - q$ , $D - p$

Answer 2

There are two external forces on the bob : gravity $(mg)$ and tension $(T)$ in the string. The tension does no work as displacement is always perpendicular to the string. Total mechanical energy $(E)$ of the system is conserved. If we take potential energy of the system to be zero at the lowest point $A$ , then At $A$ , $E = \frac{1}{2}mv^{2}_{0} \quad\cdots\left(i\right)$ From Newtons second law $T_{A} - mg = \frac{mv^{2}_{0}}{L}\quad\cdots\left(ii\right)$ where $T_{A}$ is the tension in the string at $A$ . At the highest point $C$ , to complete the circle tension in string will be minimum (zero). At $C$ , $E = \frac{1}{2}mv^{2}_{C}+mg\left(2L\right)\quad\cdots\left(iii\right)$ From Newtons second law $mg = \frac{mv_{C}^{2}}{L}\quad\quad\left(iv\right)$ From $\left(iv\right), v_{C} =\sqrt{gL}$ ; $C-q$ From $\left(iii\right)$ , $E = \frac{1}{2}m\left(gL\right)+2mgL= \frac{5}{2}mgL \quad\left(v\right)$ Using $\left(i\right)$ , $\frac{1}{2}mv^{2}_{0} = \frac{5}{2}mgL$ , $v_{0} = \sqrt{5\,gL}$ ; $\therefore A-r\quad\cdots\left(vi\right)$ At $B$ , $E =\frac{1}{2}mv^{2}_{B}+mg\left(L\right)$ or $\frac{1}{2}mv^{2}_{B} = E-mg\left(L\right)$ $\frac{5}{2}mgL-mgL\quad$ (Using $\left(v\right)$ ) $\frac{1}{2} mv^{2}_{B} = \frac{3}{2} mgL$ $\therefore v_{B} = \sqrt{3\,gL}$ ; $\therefore B-s$ The ratio of kinetic energies at $B$ and $C$ is $\therefore \frac{K_{B}}{K_{C}} = \frac{\frac{1}{2}mv^{2}_{B}}{\frac{1}{2}mv^{2}_{C}}$ $= \frac{3gL}{gL} = \frac{3}{1}$ ; $\therefore D-p$