Question

A uniform meter scale of mass is suspended by two vertical strings attached to its two ends. a body of mass 2m is placed on 75 cm mark. the tension in the two strings are in the ratio

• A. 1:2
• B. 1:3
• C. 1:4
• D. 3:4

Answer 1

The correct option is A 1:2

$T_1=\frac{mg}{2}+2mg\left(\frac{3}{4}\right)$

$=\frac{mg}{2}+\frac{3mg}{2}=2mg$

$T_2=\frac{mg}{2}+\frac{mg}{2}$ OR $(\frac{mg}{2}+2mg(\frac{1}{4}\left)\right)$

$=mg$

$\therefore \frac{T_1}{T_2}=\frac{T_{1}R}{T_{2}R}=\frac{T_1}{T_2}=\frac{1}{2}$

OR $1:2$

Hence, option $\left(A\right)$ is correct answer.