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NEET PHYSICSWAVEEasy

Question

If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:

A

1:21:2

B

1:11:1

C

2:1\sqrt{2}:1

D

1:21:\sqrt{2}

Step-by-Step Solution

The speed of a transverse wave on a stretched string is given by the formula v=Tμv = \sqrt{\frac{T}{\mu}}, where TT is the tension in the string and μ\mu is its linear mass density. Let the initial tension be T1=TT_1 = T, then the initial speed is v1=Tμv_1 = \sqrt{\frac{T}{\mu}}. If the tension is doubled, the new tension becomes T2=2TT_2 = 2T. The final speed is then v2=2Tμ=2Tμ=2v1v_2 = \sqrt{\frac{2T}{\mu}} = \sqrt{2} \sqrt{\frac{T}{\mu}} = \sqrt{2} v_1. The ratio of the initial speed to the final speed is v1v2=12\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}, which can be written as 1:21:\sqrt{2}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVEinitialtensionstretchedstringdoubled

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