Question

A particle is suspended from a string of length $R$ and it is given a velocity $u=3\sqrt{gR}$ at the bottom keeping the other end fixed. Identify the correct statement from the options given below:

• A. $v_B<v_C$, $T_B>T_C$
• B. $v_B>v_C$, $T_B>T_C$
• C. $v_B>v_C$, $T_B<T_C$
• D. $v_B<v_C$, $T_B<T_C$

Answer 1

The correct option is B $v_B>v_C$, $T_B>T_C$

Given :$v_A=u=3\sqrt{gR}\text{m}/s$
Applying energy conservation between A and B:
$\frac{1}{2}m{v}_A^2=\frac{1}{2}m{v}_B^2+mgR$
$v_B^2=v_A^2-2gR=(3\sqrt{gR}{)}^2-2gR=7gR$
$\Rightarrow v_B=\sqrt{7gR}\text{m/s}$
Simillarly applying energy conservation between A and C:
$\frac{1}{2}m{v}_A^2=\frac{1}{2}m{v}_C^2+mg\left(2R\right)$
$v_C^2=v_A^2-4gR=9gR-4gR=5gR$

$\Rightarrow v_C=\sqrt{5gR}$
Applying Newton's $2^{nd}$ law,

At B:
$T_B=\frac{m{v}_B^2}{R}=7mg$

Simillarly, at C:
$mg+T_C=\frac{m{v}_C^2}{R}$
$\Rightarrow T_C=\frac{m{v}_C^2}{R}-mg$
$\therefore T_c=4mg$

$\therefore v_B>v_C$ and $T_B>T_C$

Hence option B is the correct answer