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A block of weight w  produces an extension of 9 cm, when it is hang by an elastic spring of 60 cm length and is cut into two parts, one of length 40 cm and the other of length 20 cm. The same load w  hangs in equilibrium supported by both parts as shown in figure. The extension in cm now is

Option: 1

9


Option: 2

6


Option: 3

3


Option: 4

2


Answers (1)

best_answer

force constant,   k \propto \frac{1}{l}
let  k  be the force constant of original spring.                                             

Then,

k_{40}=\frac{60}{40} \,k=\frac{3}{2} \,k
and
k_{20}=\frac{60}{20} \,k=3\, k

k_{\text {net }}=k_{40}+k_{20}=\frac{9 k}{2}
Now,
\Delta l =\frac{F}{k} \propto \frac{1}{k}
\therefore \Delta l^{\prime} =\left(\frac{k}{9 k / 2}\right)(9 \mathrm{~cm}) \\
\Delta l^{\prime} =\frac{2}{9} \times 9=2 \mathrm{~cm}

Posted by

Ajit Kumar Dubey

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