Question

Find the extension in the spring connecting pulley and ceiling in the equilibrium condition. The block is of mass ${}^{\prime }{m}^{\prime }$ and all springs are of spring constant $k$.
Consider spring, string and pulley to be very light.

• A. $\frac{mg}{k}$
• B. $\frac{mg}{2k}$
• C. $\frac{2mg}{k}$
• D. $\frac{mg}{4k}$

Answer 1

The correct option is C $\frac{2mg}{k}$

At equilibrium the spring force will balance the weight of mass ${}^{\prime }{m}^{\prime }$.

From F.B.D of mass $m$


$\Rightarrow kx=mg$
$\text{Or}x=\frac{mg}{k}$

F.B.D of spring connecting the string and mass ${}^{\prime }{m}^{\prime }$,


Since spring is massless the tension in entire spring will be equal throughout i.e $T=kx$
$\Rightarrow T=kx=mg...\left(i\right)$

Let the extension in spring connecting the pulley and ceiling be $x^{\prime }$.


Balancing the forces on massless pulley we get,

$\Rightarrow k{x}^{\prime }=2T$
or, $x^{\prime }=\frac{2T}{k}$

Substituting value of tension from Eq.$\left(i\right)$
$\Rightarrow x^{\prime }=\frac{2mg}{k}$

$\begin{array}{c}Ans.C\end{array}$

$\begin{array}{c}\text{Why this question}?\\ \text{Concept: In problems involving string and}\\ \text{sprig connected through pulley, always}\\ \text{remember that "tension in a string}\\ \text{connected with spring will be same}\\ \text{everywhere, if it is a single string".}\end{array}$