Two strings (A, B) having linear densities and and lengths respectively are joined. Free ends of A and B are tied to tw

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Solution

Q.
Two strings (A, B) having linear densities $μ_{A} = 2 \times 10^{- 4} kg/m$ and $μ_{B} = 4 \times 10^{- 4} kg/m$ 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 [2026]

1 1.18

2 1.90

3 1.08

4 1.67

Ans.
(1)

$Given L_{A} = 2.5 m, L_{B} = 1.5 m, T = 500 N$

$v_{A} = \sqrt{\frac{T}{μ_{A}}} = \sqrt{\frac{500}{2 \times 10^{- 4}}} = 5 \sqrt{10} \times 10^{2} m/s$

$v_{B} = \sqrt{\frac{T}{μ_{B}}} = \sqrt{\frac{500}{4 \times 10^{- 4}}} = 5 \sqrt{5} \times 10^{2} m/s$

$t_{1} = \frac{L_{A}}{v_{A}} = \frac{2.5}{5 \sqrt{10}} \times 10^{- 2} s$

$t_{2} = \frac{L_{B}}{v_{B}} = \frac{1.5}{5 \sqrt{5}} \times 10^{- 2} s$

$∴$ $\frac{t_{1}}{t_{2}} = \frac{2.5}{5 \sqrt{10}} \times \frac{5 \sqrt{5}}{1.5} = \frac{5}{3} \times \frac{1}{\sqrt{2}} = \frac{1.66}{1.41} = 1.18$

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