JEE PYQ: Waves And Sound Question 21
Question 21 - 2020 (08 Jan Shift 2)
A transverse wave travels on a taut steel wire with a velocity of $v$ when tension in it is $2.06 \times 10^4$ N. When the tension is changed to $T$, the velocity changed to $v/2$. The value of $T$ is close to:
(1) $2.50 \times 10^4$ N
(2) $5.15 \times 10^3$ N
(3) $30.5 \times 10^4$ N
(4) $10.2 \times 10^2$ N
Show Answer
Answer: (2)
Solution
The velocity of a transverse wave in a stretched wire is given by
$v = \sqrt{\frac{T}{\mu}}$
Where, $T =$ Tension in the wire, $\mu =$ linear density of wire
$(\because V \propto T)$
$\therefore \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$
$\Rightarrow \frac{v}{v} \times 2 = \sqrt{\frac{2.06 \times 10^4}{T_2}}$
$\Rightarrow T_2 = \frac{2.06 \times 10^4}{4} = 0.515 \times 10^4$ N
$\Rightarrow T_2 = 5.15 \times 10^3$ N
BETA
