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A block of weight $W$ produces an extension of $9 \mathrm{~cm}$ , when it is hang by an elastic spring of $60 \mathrm{~cm}$ length.

It is cut into two parts, one of length $40 \mathrm{~cm}$ and the other of length $20 \mathrm{~cm}$ . The same load $W$ hangs in equilibrium supported by both parts as shown in figure.

The extension in $cm$ is now __
Option: 1
$9$

Option: 2
$6$

Option: 3
$3$

Option: 4
$2$

Answers (1)

best_answer

Force constant,

$K \alpha \frac{1}{l}$

Let $K$ be the force constant of original spring.

Then

$K_{40}=\frac{60}{40} K=\frac{3}{2} K$
and
$\begin{aligned} K_{20}&=\frac{60}{20} K=3 K \\ \therefore K_{\text {net }}&=K_{40}+K_{20}=\frac{9 K}{2} \end{aligned}$
Now,
$\begin{aligned} \Delta l&=\frac{F}{K} \alpha \frac{1}{K} \\ \therefore \Delta l^{\prime}&=\left(\frac{K}{9 K / 2}\right)\times (9 \mathrm{~cm}) \\ \Delta l^{\prime}&=\frac{2}{9} \times 9=2 \mathrm{~cm} \end{aligned}$

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