Two metallic wires P and Q have same volume and are made up of same material. If their area of cross sections are in the ratio 4 : 1 and force F 1 is applied to P , an extension of ∆ l is produced. Th

Question

Two metallic wires P and Q have same volume and are made up of same material. If their area of cross sections are in the ratio 4 : 1 and force F 1 is applied to P , an extension of ∆ l is produced. The force which is required to produce same extension in Q is F 2 . The value of F 1 F 2 is ______.

TestHub · Accepted Answer

The formula for Young's modulus is given by

$\begin{array}{rcl} Y & = & \frac{\frac{F}{A}}{\frac{∆ l}{l}} \\ = & \frac{F l}{A ∆ l} \end{array}$

From above equation, it follows that

$Δ l = \frac{F l}{A Y} . . . (1)$

Again, volume of the wire is given by

$V = A l \Rightarrow l = \frac{V}{A}$

Hence, from equation (1),

$∆ l = \frac{F V}{A^{2} Y} . . . (2)$

As, $Y$ and $V$ is the same for both the wires,

$∆ l \propto \frac{F}{A^{2}}$

Hence, for two wires, it can be written that

$\frac{∆ l_{1}}{∆ l_{2}} = \frac{F_{1}}{A_{1}^{2}} \times \frac{A_{2}^{2}}{F_{2}} . . . (3)$

As $∆ l_{1} = ∆ l_{2}$ , from equation (3), it follows that

$F_{1} A_{2}^{2} = F_{2} A_{1}^{2} \Rightarrow \frac{F_{1}}{F_{2}} = \frac{A_{1}^{2}}{A_{2}^{2}} = {(\frac{4}{1})}^{2} = 16$

Physics - Elasticity Question with Solution | TestHub

Doubts & Discussion