PYQ NEET- Mechanical Properties Of Solids L-3

Question: Two wires are made of the same material and have the same volume. The first wire has cross-sectional area $A$ and the second wire has cross-sectional area 3A. If the length of the first wire is increased by $\Delta I$ on applying a force $F$, how much force is needed to stretch the second wire by the same amount?#

A) $4 F$

B) $6 F$

C) $9 F$

D) $F$

Answer: $9 F$#

Sol:#

According to the question,

For wire 1

Area of cross-section $=A_1$

Force applied $=F_1$

Increase in length $=\Delta$ l

From the relation of Young’s modulus of elasticity,

$$ Y=\frac{F l}{A \Delta l} $$

Substituting the values for wire 1 in the above relation, we get $$ \Rightarrow \quad Y_1=\frac{F_1 l_1}{A_1 \Delta l} $$

For wire 2

Area of cross-section $=A_2$

Force applied $=F_2$

Increase in length $=\Delta l$

Similarly,

$$ Y_2=\frac{F_2 I_2}{A_2 \Delta l} $$

$\because \quad$ Volume, $V=A I$

or

$$ I=\frac{V}{A} $$

Substituting the value of $/$ in Eqs. (i) and (ii), we get

$$ Y_1=\frac{F_1 V}{A_1^2 \Delta l} \text { and } Y_2=\frac{F_2 V}{A_2^2 \Delta l} $$

As it is given that the wires are made up of same material, i.e. $Y_1=Y_2$

$$ \Rightarrow \quad \frac{F_1 V}{A_1^2 \Delta l}=\frac{F_2 V}{A_2^2 \Delta l} $$

$\begin{aligned} & \Rightarrow \quad \frac{F_1}{F_2}=\frac{A_1^2}{A_2^2}=\frac{A^2}{9 A^2} \ & \left(\because A_1=A \text { and } A_2=3 A\right) \ & =\frac{1}{9} \ & \text { or } \ & F_2=9 F_1=9 F\left(\text { given, } F_1=F\right) \ & \end{aligned}$