Question

Two wires \(A\) and \(B\) of same cross-section are connected end to end. When same tension is created in both wires, the elongation in \(B\) wire is twice the elongation in \(A\) wire. If \(L_A\) and \(L_B\) are the initial lengths of the wires \(A\) and \(B\) respectively, then \((\text{Young's modulus of material of wire }A=2\times 10^{11}\,\text{N m}^{-2}\text{ and Young's modulus of material of wire }B=1.1\times 10^{11}\,\text{N m}^{-2})\)

• A. \(\dfrac{L_A}{L_B}=\dfrac{10}{11}\)
• B. \(\dfrac{L_A}{L_B}=\dfrac{4}{5}\)
• C. \(\dfrac{L_A}{L_B}=\dfrac{9}{11}\)
• D. \(\dfrac{L_A}{L_B}=\dfrac{3}{7}\)

Hint:

For wires under the same tension and having the same cross-sectional area, \[ \Delta L\propto \frac{L}{Y} \] So, compare elongations using the ratio of length to Young's modulus.

Answer 1

The Correct Option is A

Step 1: Write the elongation formula for each wire.
The elongation of a wire under tension $F$ is: \[ \Delta L = \frac{FL}{AY} \] where $L$ = original length, $A$ = cross-sectional area, $Y$ = Young's modulus. Since both wires have the same cross-section $A$ and the same tension $F$: \[ \Delta L_A = \frac{FL_A}{AY_A}, \quad \Delta L_B = \frac{FL_B}{AY_B} \]
Step 2: Apply the given condition $\Delta L_B = 2\Delta L_A$.
\[ \frac{FL_B}{AY_B} = 2 \cdot \frac{FL_A}{AY_A} \] Cancelling $F$ and $A$ from both sides: \[ \frac{L_B}{Y_B} = \frac{2L_A}{Y_A} \]
Step 3: Rearrange to find the ratio $L_A / L_B$.
\[ \frac{L_A}{L_B} = \frac{Y_A}{2Y_B} \]
Step 4: Substitute the given values of Young's moduli.
$Y_A = 2 \times 10^{11}$ N/m^2, $Y_B = 1.1 \times 10^{11}$ N/m^2: \[ \frac{L_A}{L_B} = \frac{2 \times 10^{11}}{2 \times 1.1 \times 10^{11}} = \frac{2}{2.2} = \frac{10}{11} \]
Step 5: Verify the result.
$\frac{Y_A}{2Y_B} = \frac{2}{2 \times 1.1} = \frac{2}{2.2} = \frac{20}{22} = \frac{10}{11}$. Confirmed.
Step 6: State the final answer.
The ratio of initial lengths of wires A and B is $L_A : L_B = 10 : 11$. \[ \boxed{\dfrac{L_A}{L_B} = \dfrac{10}{11}} \]