Heat Lost in a Stretched Wire
Question: A 50 m long light wire having cross-sectional area \(6.25 \times 10^{-4}\,\text{m}^2\) and Young's modulus \(E = 10^{10}\,\text{N/m}^2\) is subjected to a load of 100 kg. Calculate the heat lost (in J) when the wire is stretched by the load.
(Note: The load creates a tension \(T = mg\) in the wire and the stored elastic energy, which may be lost as heat in an irreversible process, is given by the energy stored in the wire.)
- (1) \(4\,\text{J}\)
- (2) \(2\,\text{J}\)
- (3) \(6\,\text{J}\)
- (4) \(8\,\text{J}\)
Select the Correct Option
Step‑by‑Step Explanation
Step 1: Determine the Tension in the Wire
The load is 100 kg, so its weight is:
\[
T = mg = 100 \times 9.8 \approx 980\,\text{N}.
\]
Step 2: Calculate the Elastic Potential Energy Stored in the Wire
For a stretched wire (or string), the elastic potential energy stored is given by:
\[
U = \frac{T^2 L}{2AE},
\]
where:
- \(T\) is the tension,
- \(L = 50\,\text{m}\) is the length of the wire,
- \(A = 6.25 \times 10^{-4}\,\text{m}^2\) is the cross-sectional area, and
- \(E = 10^{10}\,\text{N/m}^2\) is Young's modulus.
Step 3: Substitute the Values
First, calculate \(T^2\):
\[
T^2 = (980)^2 \approx 960{,}400\,\text{N}^2.
\]
Next, plug in all values:
\[
U = \frac{960{,}400 \times 50}{2 \times (6.25 \times 10^{-4}) \times 10^{10}}.
\]
Step 4: Simplify the Expression
Compute the denominator:
\[
2 \times (6.25 \times 10^{-4}) \times 10^{10} = 2 \times 6.25 \times 10^{6} = 12.5 \times 10^{6} = 1.25 \times 10^{7}.
\]
Now the energy:
\[
U = \frac{960{,}400 \times 50}{1.25 \times 10^{7}}.
\]
Multiply the numerator:
\[
960{,}400 \times 50 = 48{,}020{,}000.
\]
Then,
\[
U = \frac{48{,}020{,}000}{12{,}500{,}000} \approx 3.84\,\text{J}.
\]
Rounding to a neat value, this is approximately \(4\,\text{J}\).
Final Answer: The heat lost is approximately \(4\,\text{J}\), which corresponds to option (1).
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