Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

Step-by-Step Solution

Young's Modulus Formula: Young's modulus ( $Y$ ) is a property of the material and is defined as stress divided by strain: $Y = \frac{F/A}{\Delta l/L} \implies F = \frac{YA\Delta l}{L}$ .
Constraint (Same Volume): The volume ( $V$ ) of a wire is $V = A \times L$ . Since both wires have the same volume, $A_1 L_1 = A_2 L_2$ . Given $A_1 = A$ and $A_2 = 3A$ , we have: $A \cdot L_1 = 3A \cdot L_2 \implies L_2 = \frac{L_1}{3}$
Comparison of Forces: We need to find $F_2$ given that the extension $\Delta l$ and material ( $Y$ ) are the same for both. $F_1 = \frac{Y A_1 \Delta l}{L_1} = F$ $F_2 = \frac{Y A_2 \Delta l}{L_2}$
Substitution: Substitute $A_2 = 3A$ and $L_2 = L_1/3$ into the equation for $F_2$ : $F_2 = \frac{Y (3A) \Delta l}{(L_1/3)} = 3 \times 3 \times \frac{Y A \Delta l}{L_1} = 9 \left( \frac{Y A \Delta l}{L_1} \right)$ $F_2 = 9 F$

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