Question

A uniform meter sticks of mass 200 g is suspended from the ceiling through two verticle strings of equal lengths fixed at the ends. A small object of 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tension in the two strings.

• A. $T_1=1.06N,T_2=1.14N$
• B. $T_1=1.03N,T_2=1.17N$
• C. $T_1=1.10N,T_2=1.10N$
• D. $T_1=1.00N,T_2=1.20N$

Answer 1

The correct option is A $T_1=1.06N,T_2=1.14N$

C is the centre of mass of the stick.

For translation equilibrium

$T_1+T_2=220g--------\left(1\right)$

Here,g is acceleration due to gravity $=1000cm/s^2$

For rotation equilibrium torque about centre of mass shall be zero or sum of clockwise torque is equal to sum anti clockwise torque.

So, AC=CB=50cm

$T_1\times 50+20g\times (70-50)=T_2\times 50$

$⟹T_2-T_1=8g---------\left(2\right)$

So,from (1) & (2) we get

$T_2=114g=1.14\times {10}^{5}dyne=1.14N$

$T_1=106g=1.06\times {10}^{5}dyne=1.06N$