Question

Same tension is applied to the following four wires made of the same material. The elongation is longest in

• A. Wire of length \(50 \, cm\) and diameter \(0.5 \, mm\)
• B. Wire of length \(200 \, cm\) and diameter \(2 \, mm\)
• C. Wire of length \(300 \, cm\) and diameter \(3 \, mm\)
• D. Wire of length \(100 \, cm\) and diameter \(1 \, mm\)

Hint:

For wires under the same force and material, \[ \Delta L \propto \frac{L}{d^2} \] A longer and thinner wire undergoes greater elongation.

Answer 1

The Correct Option is A

Step 1: Use the formula for elongation of a wire.
The elongation produced in a wire is given by
\[ \Delta L=\frac{FL}{AY} \] where,
\[ F=\text{applied force} \] \[ L=\text{length of wire} \] \[ A=\text{cross-sectional area} \] \[ Y=\text{Young's modulus} \] Since all wires are made of the same material and same tension is applied, \(F\) and \(Y\) are constant.
Therefore,
\[ \Delta L \propto \frac{L}{A} \]

Step 2: Express area in terms of diameter.
Cross-sectional area of a wire is
\[ A=\frac{\pi d^2}{4} \] Hence,
\[ \Delta L \propto \frac{L}{d^2} \]

Step 3: Calculate \(\dfrac{L}{d^2}\) for each option.
For option (1),
\[ \frac{L}{d^2} = \frac{50}{(0.5)^2} = \frac{50}{0.25} = 200 \] For option (2),
\[ \frac{L}{d^2} = \frac{200}{(2)^2} = \frac{200}{4} = 50 \] For option (3),
\[ \frac{L}{d^2} = \frac{300}{(3)^2} = \frac{300}{9} = 33.33 \] For option (4),
\[ \frac{L}{d^2} = \frac{100}{(1)^2} = 100 \]

Step 4: Compare the values.
The elongation is maximum for the largest value of \(\dfrac{L}{d^2}\).
Among all options, the maximum value is
\[ 200 \] which corresponds to option (1).

Step 5: Final conclusion.
Hence, the elongation is longest in
\[ \boxed{\text{Wire of length }50 \, cm\text{ and diameter }0.5 \, mm} \]