JEE Mains · Physics · STD 11 - 14. waves and sound
Two strings (A, B) having linear densities \(\mu_{ A }=2 \times 10^{-4} kg / m\) and \(\mu_{ B }=4 \times 10^{-4} kg / m\) and lengths \(L _{ A }=2.5 m\) and \(L _{ B }=1.5 m\) respectively are joined. Free ends of A and B are tied to two rigid supports C and D , respectively creating a tension of 500 N in the wire. Two identical pulses, sent from \(C\) and \(D\) ends, take time \(t_1\) and \(t_2\), respectively, to reach the joint. The ratio \(t_1 / t_2\) is :
- A 1.08
- B 1.9
- C 1.67
- D 1.18
Answer & Solution
Correct Answer
(D) 1.18
Step-by-step Solution
Detailed explanation
Given \(L _{ A }=2.5 m\), \(L _{ B }=1.5 m\), \(T =500 N\) \(v_A=\sqrt{\frac{T}{\mu_A}}=\sqrt{\frac{500}{2 \times 10^{-4}}}=5 \sqrt{10} \times 10^2 m / s\) \(v _{ B }=\sqrt{\frac{ T }{\mu_{ B }}}=\sqrt{\frac{500}{4 \times 10^{-4}}}=5 \sqrt{5} \times 10^2 m / s\)…
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