Q132 / 160 · TS EAMCET 2024 (09 May Shift 1) · Physics
A steel wire of length 3 m and a copper wire of length 2.2 m are connected end to end. When the combination is stretched by a force, the net elongation is 1.05 mm . If the area of cross-section of each wire is $6 \mathrm{~mm}^2$, then the load applied is
(Young's module of steel and copper are respectively $2 \times 10^{11} \mathrm{Nm}^{-2}$ and $1.1 \times 10^{11} \mathrm{Nm}^{-2}$ )
- Correct answerA180 N
- B90 N
- C135 N
- D120 N
Solution
Young's modulus, $\mathrm{Y}=\frac{\mathrm{F}}{\mathrm{A}} \left\lvert\, \frac{\Delta \mathrm{l}}{\mathrm{I}_{\mathrm{o}}} \Rightarrow \Delta \mathrm{l}=\frac{\mathrm{F}}{\mathrm{Y}_{\mathrm{A}}} \mathrm{l}_{\mathrm{o}}\right.$
$\begin{aligned}
& \text { Given } \Delta 1_{\mathrm{s}}+\Delta 1_{\mathrm{c}}=1.05 \mathrm{~mm}=1.05 \times 10^{-3} \mathrm{~m} \\
& \therefore \frac{\mathrm{~F}}{\mathrm{Y}_{\mathrm{s}}} \frac{\ell_{\mathrm{s}}}{\mathrm{~A}_{\mathrm{s}}}+\frac{\mathrm{F}}{\mathrm{Y}_{\mathrm{c}}} \frac{\ell_{\mathrm{c}}}{\mathrm{~A}_{\mathrm{c}}}=1.05 \times 10^{-3} \\
& \Rightarrow \mathrm{~F}\left(\frac{3}{2 \times 10^{11} \times 6 \times 10^{-6}}+\frac{2.2}{1.1 \times 10^{12} \times 0 \times 10^{-6}}\right) \\
& =1.05 \times 10^{-3}
\end{aligned}$
Solving above equation we get load applied $\mathrm{F}=180 \mathrm{~N}$
Hence, the correct answer is option: 'A'.