A mass of 20 kg is hanging with support of two strings of same li… - Vidyadip

MCQ

A mass of $20\ kg$ is hanging with support of two strings of same linear mass density. Now pulses are generated in both strings at same time near the joint at mass. Ratio of time, taken by a pulse travel through string $1$ to that taken by pulse on string $2$ is

A
$\frac{4}{3}$
B
$\frac{{\sqrt 4 }}{{\sqrt 3 }}$
✓
$\frac{{4\sqrt 4 }}{{3\sqrt 3 }}$
D
$\frac{{3\sqrt 4 }}{{4\sqrt 3 }}$

✓

Answer

Correct option: C.

$\frac{{4\sqrt 4 }}{{3\sqrt 3 }}$

c
$\frac{4 T_{1}}{5}=\frac{3 T_{2}}{5} \Rightarrow T_{1}=\frac{3}{4} T_{2}$

$v=\sqrt{\frac{T}{\mu}} \Rightarrow \frac{v_{1}}{v_{2}}=\sqrt{\frac{3}{4}}$

${l_1} = 10 \times \frac{4}{5} = 8\,m$

${l_2} = 10 \times \frac{3}{5} = 6\,m$

$\frac{t_{1}}{t_{2}}=\frac{v_{1} / v_{1}}{v_{2} / v_{2}}=\frac{8}{6} \sqrt{\frac{4}{3}}=\frac{4 \sqrt{4}}{3 \sqrt{3}}$

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