Question

A uniform metre stick of mass $250\text{gm}$ is suspended from the ceiling through two vertical strings of equal lengths. A small object of mass $50\text{gm}$ is placed on the stick at a distance of $70\text{cm}$ from the left end. Find the tension in both the strings. Take $g=10{\text{m/s}}^2$.

• A. $T_1=1.15\text{N},T_2=2.15\text{N}$
• B. $T_1=1.06\text{N},T_2=1.14\text{N}$
• C. $T_1=1.40\text{N},T_2=1.60\text{N}$
• D. $T_1=1.72\text{N},T_2=1.28\text{N}$

Answer 1

The correct option is C $T_1=1.40\text{N},T_2=1.60\text{N}$


Let $C$ be the centre of mass of the stick.
For translation equilibrium,
$T_1+T_2=(m_1+m_2)g$
$T_1+T_2=0.3g=3...\left(i\right)$

For rotational equilibrium, let the net torque be zero about centre $C$:
${\tau }_{\text{net}}=0...\left(ii\right)$
Taking anticlockwise sense of rotation for torque as $+ve$
$\&\tau =F\times r_{\perp }$
$\because$ Force $m_{2}g$ passes through reference point $C$, hence its torque is zero about point $C$.
Here, $AC=CB=50\text{cm}=0.50\text{m}$
$m_1=0.05\text{kg},m_2=0.25\text{kg}$
Substituting the torques with proper signs in Eq $\left(ii\right)$,
$\because \sum \tau =0$
$\Rightarrow -(T_1\times 0.5)+0-(m_{1}g\times 0.20)+(T_2\times 0.50)=0$
$\Rightarrow -(T_1\times 0.50)-(0.05\times 10\times 0.20)+(T_2\times 0.50)=0$
$\Rightarrow \frac{T_2-T_1}{2}=0.1$
$\Rightarrow T_2-T_1=0.20...\left(iii\right)$

Adding Eq $\left(i\right)\&\left(iii\right)$,
$2T_2=3.20$
$\therefore T_2=1.60\text{N}$
and $T_1=3-1.6=1.40\text{N}$