Question

When a certain weight is suspended from a long uniform wire, its length increases by $1cm$. If the same weight is suspended from a second wire of the same material and length but having a diameter half of the diameter of the first then the increase in length will be

(A) $0.5cm$
(B) $2cm$
(C) $4cm$
(D) $8cm$

Answer 1

Hint: Since the same material is used for both the wire hence its Young’s modulus will remain same therefore in order to find the change in length of the second wire equate the Young’s modulus of both the wire and you will get the required answer. Use the formula of Young’s modulus in terms of force and length.

Complete answer:
We know that, Young’s modulus is given by:
$Y = \dfrac{F}{A} \times \dfrac{1}{\Delta l}$
Where, F is force, $A$ is area and $l$ is length of the wire.

Since, the same material is used hence the Young’s modulus will be same:
${Y_1} = {Y_2}$

From the question;
$\Delta {l_1} = 1cm$ and $\Delta {l_2} = ?$
${A_1} = \pi {r^2}$ and ${A_2} = \pi {\left( {\dfrac{r}{2}} \right)^2}$

By putting the values in above equation, we get;
$\dfrac{F}{\Delta {l_1}A_1} = \dfrac{F}{\Delta {l_2}A_2}$
$\Delta {l_2} = \dfrac{\Delta {l_1}A_1}{A_2} = \dfrac{1 \times \pi {r^2}}{\pi \times \dfrac{{{r^2}}}{4}}$

By solving;
$\Delta {l_2} = 4cm$

Hence the correct answer is Option(C).

Note: Here in this case the material used for both the wire is same therefore their Young’s modulus will also be same but it is not the case all the time, when different material is used for both the wire young’s modulus will also be different. Therefore, the required answer will also be get changed.