Q.34. A string A of length 0.314 m of Young's modulus is connected to another string B of length an

Q: What is the correct answer for JEE Main Physics Q34?

The correct answer is B. Sol. Let $L_A = 0.314 \text{ m}$, $Y_A = 2 \times 10^{10} \text{ N/m}^2$. Let $L_B = L_A = 0.314 \text{ m}$ (assuming 'length' means same length as A), $Y_B = 2Y_A = 4 \times 10^{10} \text{ N/m}^2$. Radius $r = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$. Area $A = \pi r^2 = \pi (0.2 \times 10^{-3})^2 = 4\pi \times 10^{-8} \text{ m}^2$. Force $F = mg = 0.8 \times 10 = 8 \text{ N}$. Elongation $

Q: Q.34. A string A of length 0.314 m of Young's modulus is connected to another string B of length and Young's modulus both twice of those of A. This series combination of strings is then suspended…

The correct answer is B. Sol. Let $L_A = 0.314 \text{ m}$, $Y_A = 2 \times 10^{10} \text{ N/m}^2$. Let $L_B = L_A = 0.314 \text{ m}$ (assuming 'length' means same length as A), $Y_B = 2Y_A = 4 \times 10^{10} \text{ N/m}^2$. Radius $r = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$. Area $A = \pi r^2 = \pi (0.2 \times 10^{-3})^2 = 4\pi \times 10^{-8} \text{ m}^2$. Force $F = mg = 0.8 \times 10 = 8 \text{ N}$. Elongation $

Question

Q.34. A string A of length 0.314 m of Young's modulus  is connected to another string B of length and Young's modulus both twice of those of A. This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass 0.8 kg. The net change in length of the combination is ________ mm. (radius of both the strings is 0.2 mm and acceleration due to gravity = ) (Mass of both strings is to be neglected as compared to the mass of load)

FirstInTest · Accepted Answer

Sol. Let $L_A = 0.314 	ext{ m}$, $Y_A = 2 	imes 10^{10} 	ext{ N/m}^2$. Let $L_B = L_A = 0.314 	ext{ m}$ (assuming 'length' means same length as A), $Y_B = 2Y_A = 4 	imes 10^{10} 	ext{ N/m}^2$. Radius $r = 0.2 	ext{ mm} = 0.2 	imes 10^{-3} 	ext{ m}$. Area $A = \pi r^2 = \pi (0.2 	imes 10^{-3})^2 = 4\pi 	imes 10^{-8} 	ext{ m}^2$. Force $F = mg = 0.8 	imes 10 = 8 	ext{ N}$. Elongation $\Delta L = \frac{FL}{AY}$. For string A: $\Delta L_A = \frac{FL_A}{A Y_A} = \frac{8 	imes 0.314}{4\pi 	imes 10^{-8} 	imes 2 	imes 10^{10}} = \frac{8 	imes 0.314}{8\pi 	imes 10^2} = \frac{0.314}{\pi 	imes 100} = \frac{0.314}{314} = 0.001 	ext{ m} = 1 	ext{ mm}$. For string B: $\Delta L_B = \frac{FL_B}{A Y_B} = \frac{8 	imes 0.314}{4\pi 	imes 10^{-8} 	imes 4 	imes 10^{10}} = \frac{8 	imes 0.314}{16\pi 	imes 10^2} = \frac{0.314}{2\pi 	imes 100} = \frac{0.314}{628} = 0.0005 	ext{ m} = 0.5 	ext{ mm}$. Total change in length = $\Delta L_A + \Delta L_B = 1 + 0.5 = 1.5 	ext{ mm}$. The solution provided gives 2 mm. Let's recheck the calculation. If $L_B$ is also $0.314 	ext{ m}$. $\Delta L_A = 1 	ext{ mm}$. $\Delta L_B = \frac{FL_B}{A Y_B} = \frac{8 	imes 0.314}{4\pi 	imes 10^{-8} 	imes (2 	imes 2 	imes 10^{10})} = \frac{8 	imes 0.314}{16\pi 	imes 10^2} = \frac{0.314}{2\pi 	imes 100} = 0.5 	ext{ mm}$. Total = 1.5 mm. If the length of string B is twice that of A, i.e., $L_B = 2L_A = 2 	imes 0.314 = 0.628 	ext{ m}$. Then $\Delta L_B = \frac{8 	imes 0.628}{4\pi 	imes 10^{-8} 	imes 4 	imes 10^{10}} = \frac{8 	imes 0.628}{16\pi 	imes 10^2} = \frac{0.628}{2\pi 	imes 100} = \frac{0.628}{628} = 0.001 	ext{ m} = 1 	ext{ mm}$. In this case, total change = $1 + 1 = 2 	ext{ mm}$. The problem states 'length and Young's modulus both twice of those of A'. This implies $L_B = 2L_A$. So, the total change is 2 mm.

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