Question

A mass $M$ is suspended as shown in figure. If the system is in equilibrium and $K$ is the spring constant, then the extension produced in the spring is given by (assume pulleys to be massless and strings inextensible.)

• A. $\frac{4Mg}{K}$
• B. $\frac{Mg}{K}$
• C. $\frac{2Mg}{K}$
• D. $\frac{3Mg}{K}$

Answer 1

The correct option is A $\frac{4Mg}{K}$


Let extension in spring be $X$,
From diagram, we get,
$T_1=Mg.......\left(1\right)$
$T_2=T_1+T_1=2T_1......\left(2\right)$
$KX=T_2+T_2=2T_2......\left(3\right)$
Therefore, from $\left(2\right)$ and $\left(3\right)$
$KX=4T_1$
Now, using $\left(1\right)$, we get
$KX=4Mg$
$\Rightarrow X=\frac{4Mg}{K}$
Hence option A is correct