MHT CET · Physics · Mechanical Properties of Solids
A \(5 \mathrm{~m}\) long aluminium wire \(\left(Y=7 \times 10^{10} \mathrm{Nm}^{-2}\right.\) ) of diameter \(3 \mathrm{~mm}\) supports a \(40 \mathrm{~kg}\) mass. In order to have the same elongation in the copper wire \(\left(Y=12 \times 10^{10} \mathrm{Nm}^{-2}\right)\) of the same
length under the same weight, the diameter should now be (in \(\mathrm{mm}\) )
- A \(1.75\)
- B \(1.5\)
- C \(2.5\)
- D \(5.0\)
Answer & Solution
Correct Answer
(C) \(2.5\)
Step-by-step Solution
Detailed explanation
\(l=\frac{F L}{\pi r^{2} Y}\)
\(\Rightarrow r^{2} \propto \frac{1}{Y}(F, L\) and \(l\) are constant) \(\frac{r_{2}}{r_{1}}=\left[\frac{Y_{1}}{Y_{2}}\right]^{1 / 2}=\left[\frac{7 \times 10^{10}}{12 \times 10^{10}}\right]^{1 / 2}\)
\(\Rightarrow \quad r_{2}=1.5 \times\left(\frac{7}{12}\right)^{1 / 2}=1.145 \mathrm{~mm}\)
\(\therefore\) diameter \(=2.29 \mathrm{~mm}\).
\(\Rightarrow r^{2} \propto \frac{1}{Y}(F, L\) and \(l\) are constant) \(\frac{r_{2}}{r_{1}}=\left[\frac{Y_{1}}{Y_{2}}\right]^{1 / 2}=\left[\frac{7 \times 10^{10}}{12 \times 10^{10}}\right]^{1 / 2}\)
\(\Rightarrow \quad r_{2}=1.5 \times\left(\frac{7}{12}\right)^{1 / 2}=1.145 \mathrm{~mm}\)
\(\therefore\) diameter \(=2.29 \mathrm{~mm}\).
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