AP EAMCET · PHYSICS · Mechanical Properties of Solids
When a \(8 \mathrm{~m}\) long wire is stretched by a load of \(10 \mathrm{~kg}-\mathrm{wt}\), it is elongated by \(1.5 \mathrm{~mm}\). The energy stored in the wire in this process is \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
- A \(7.5 \mathrm{~J}\)
- B \(0.05 \mathrm{~J}\)
- C \(5 \mathrm{~J}\)
- D \(0.075 \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(0.075 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Given that, weight of load, \(w=10 \mathrm{~kg}-\mathrm{wt}\) \[ F=w=10 \times 10 \mathrm{~N}=100 \mathrm{~N} \] Elongation in wire, \(\Delta l=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m}\) Length of wire, \(l=8 \mathrm{~m}\) We know that, elastic potential energy stored in…
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