Question

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

	Column I		Column II
(A)	Velocity $v_0$ is	(p)	$3$
(B)	Velocity at point $B$ is	(q)	$\sqrt{qL}$
(C)	Velocity of point $C$ is	(r)	$\sqrt{5gL}$
(D)	Ratio of kinetic energy at $B$ and $C$ is	(s)	$\sqrt{3gL}$

• 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)

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

There are two external forces on the bob: gravity $\left(mg\right($ and tension $\left(T\right)$ in the string. The tension does no work as displacement is always perpendicular to the string. Total mechanical energy $\left(E\right)$ of the system is conserved.
If we take potential energy of the system to be zero at the lowest point $A$, then
At $A,E=\frac{1}{2}m{v}_0^2....\left(i\right)$
From Newton's second law
$T_A-mg=\frac{m{v}_0^2}{L}....\left(ii\right)$
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 $C,E=\frac{1}{2}m{v}_C^2+mg\left(2L\right).....\left(iii\right)$
From Newton's second law
$mg=\frac{m{v}_C^2}{L}....\left(iv\right)$
From (iv), $v_C=\sqrt{gL};C-q$
From (iii), $E=\frac{1}{2}m\left(gL\right)+2mgL=\frac{5}{2}mgL.....\left(v\right)$
Using (i), $\frac{1}{2}m{v}_0^5=\frac{5}{2}mgL$,
$v_0=\sqrt{5gL};\therefore A-r....\left(vi\right)$
At $B,E=\frac{1}{2}m{v}_B^2+mg\left(L\right)$
or $\frac{1}{2}m{v}_B^2=E-mg\left(L\right)=\frac{5}{2}mg-mgL$ (Using (v))
$\frac{1}{2}m{v}_{B}2=\frac{3}{2}mgL$
$\therefore v_B=\sqrt{3gL};\therefore B-s$
The ratio of kinetic energies at $B$ and $C$ is
$\therefore \frac{K_B}{K_C}=\frac{\frac{1}{2}m{v}_B^2}{\frac{1}{2}m{v}_C^2}=\frac{3gL}{gL}=\frac{3}{1};\therefore D-p$.