Question

A mass 2kg is suspended as shown in figure. The system is in equilibrium in vertical plane. Assume pulleys to be massless. The force constant of the spring is 100 N/m. If the extension in the spring is given by $\frac{X}{10}m$. $X$ is $(g=10m/s^2)$

• A. 8
• B. 4
• C. 16
• D. 20

Answer 1

The correct option is A 8

From FBD of Block A,
$T=Mg$
From FBD of pulley B,
$2\left(2T\right)=Kx$
or,
$x=4Mg/K$=$\frac{4\times 2\times 10}{100}=\frac{8}{10}$m
Comparing with given data,
$X/10=8/10$
or, $X=8m$