Q197 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
The speed of the transverse wave on the stretched string under tension $T_1$ is $v_1$. If the tension in the string is increased by $T_1$, the speed of the transverse wave is $v_2$. If $\mu$ is the linear density of the string, then $v_2 = xv_1$. The value of $x$ is
- Correct answerA$\sqrt{2}$
- B$2$
- C$\dfrac{1}{2}$
- D$\dfrac{1}{\sqrt{2}}$
Solution
The speed of a transverse wave on a stretched string is given by $v = \sqrt{\dfrac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.
For the initial state, the tension is $T_1$, so the speed is:
$v_1 = \sqrt{\dfrac{T_1}{\mu}}$
The tension is increased by $T_1$, so the new tension is $T_2 = T_1 + T_1 = 2T_1$.
The new speed is:
$v_2 = \sqrt{\dfrac{2T_1}{\mu}} = \sqrt{2} \sqrt{\dfrac{T_1}{\mu}} = \sqrt{2} v_1$
Comparing this with $v_2 = x v_1$, we get $x = \sqrt{2}$.
Answer: $\sqrt{2}$
For the initial state, the tension is $T_1$, so the speed is:
$v_1 = \sqrt{\dfrac{T_1}{\mu}}$
The tension is increased by $T_1$, so the new tension is $T_2 = T_1 + T_1 = 2T_1$.
The new speed is:
$v_2 = \sqrt{\dfrac{2T_1}{\mu}} = \sqrt{2} \sqrt{\dfrac{T_1}{\mu}} = \sqrt{2} v_1$
Comparing this with $v_2 = x v_1$, we get $x = \sqrt{2}$.
Answer: $\sqrt{2}$
Hence, the correct answer is option: 'A'.