TS EAMCET · Physics · Mechanical Properties of Solids
The Young's modulus and Poisson's ratio of a material are respectively Y and \(\sigma\). The force required to decrease the area of cross-section of a wire made of this material by \(\Delta \mathrm{A}\) is
- A \(\frac{\mathrm{Y} \Delta \mathrm{A}}{4 \sigma}\)
- B \(\frac{2 \mathrm{Y} \Delta \mathrm{A}}{\sigma}\)
- C \(\frac{\mathrm{Y} \Delta \mathrm{A}}{2 \sigma}\)
- D \(\frac{\mathrm{Y} \Delta \mathrm{A}}{\sigma}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{Y} \Delta \mathrm{A}}{2 \sigma}\)
Step-by-step Solution
Detailed explanation
\(\frac{\Delta \mathrm{A}}{\mathrm{A}} = 2 \frac{\Delta \mathrm{r}}{\mathrm{r}}\) \(\frac{\Delta \mathrm{r}}{\mathrm{r}} = \sigma \frac{\Delta \mathrm{L}}{\mathrm{L}}\) \(\frac{\Delta \mathrm{L}}{\mathrm{L}} = \frac{\mathrm{F}}{\mathrm{YA}}\)…
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