Waves

JEE Main level, single correct, medium

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Question 4 of 142

Two strings (A,B)(A, B) having linear densities μA=2×10−4 kg/m\mu_{A}=2 \times 10^{-4} \mathrm{~kg} / \mathrm{m} and, μB=4×10−4 kg/m\mu_{B}=4 \times 10^{-4} \mathrm{~kg} / \mathrm{m} and lengths LA=2.5 mL_{A}=2.5 \mathrm{~m} and LB=1.5 mL_{B}=1.5 \mathrm{~m} respectively are joined. Free ends of AA and BB are tied to two rigid supports CC and DD, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from CC and DD ends, take time t1t_{1} and t2t_{2}, respectively, to reach the joint. The ratio t1/t2t_{1} / t_{2} is :

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Answer: (D) 1.18

Wave velocity in string: v=Tμv = \sqrt{\frac{T}{\mu}}
Time to reach joint:
t=Lv=LμTt = \frac{L}{v} = L\sqrt{\frac{\mu}{T}}
t1=LAμATt_1 = L_A\sqrt{\frac{\mu_A}{T}} and t2=LBμBTt_2 = L_B\sqrt{\frac{\mu_B}{T}}
t1t2=LALB×μAμB=2.51.5×2×10−44×10−4\frac{t_1}{t_2} = \frac{L_A}{L_B} \times \sqrt{\frac{\mu_A}{\mu_B}} = \frac{2.5}{1.5} \times \sqrt{\frac{2 \times 10^{-4}}{4 \times 10^{-4}}}
=53×12=532=526≈1.18= \frac{5}{3} \times \sqrt{\frac{1}{2}} = \frac{5}{3\sqrt{2}} = \frac{5\sqrt{2}}{6} \approx 1.18

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