Question

A bob of mass $3\text{kg}$ is attached with a light string of length $R$. If it is given an initial velocity $u=\sqrt{9gR}$ at the bottom most point, the tension in the string at the topmost point is (Take $g=10{\text{m/s}}^2$)

• A. $80\text{N}$
• B. $0\text{N}$
• C. $120\text{N}$
• D. $100\text{N}$

Answer 1

The correct option is C $120\text{N}$

FBD of the bob:


Let the tension at the topmost point ($B$) be $T$ and $v$ be the velocity at top most point.
$\therefore$ From FBD, $T+mg=\frac{m{v}^2}{R}\dots \dots \left(1\right)$

Applying energy conservation between $A$ and $B$ :
Taking $A$ as a reference
$U_A+K_A=U_B+K_B$
$0+\frac{m{u}^2}{2}=\frac{m{v}^2}{2}+mgh$
$\Rightarrow v^2=u^2-2gh$
$v=\sqrt{u^2-2gh}$
$=\sqrt{9gR-2g\left(2R\right)}$
[$\because u=\sqrt{9gR}\text{is given}$ and $\because h=2R$]
$\therefore v=\sqrt{5gR}$

On putting the value of $v$ in equation (1), we get
$T=\frac{m{v}^2}{R}-mg$
$\Rightarrow T=\frac{m\left(5gR\right)}{R}-mg$
$\Rightarrow T=4mg$
$=4\times 3\times 10$
$=120\text{N}$