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BITSAT 2022 Paper

Section: Physics

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Question : 4 of 130

Marks: +1, -0

Transverse waves of the same frequency are generated in two steel wires

A

B

. The diameter of

A

is twice that of

B

and the tension in

A

is half that in

B

. The ratio of the velocities of the waves in

A

B

1:2
$1 : \sqrt{2}$
$1 : 2 \sqrt{2}$
$3 : 2 \sqrt{2}$

Solution: 👈: Video Solution

The velocity if transverse waves is given by

v = \sqrt{\frac{T}{m}}

where,

T =

Tension and m = mass per unit

length of the wire. If r is the radius if the wire and ρ its density then,

m = \pi r^2 \rho

\therefore v = \sqrt{\frac{T}{m}} = \sqrt{\frac{T}{\pi r^2 \rho}}

\therefore v_A = \frac{\sqrt{T_A}}{r_A \sqrt{\pi \rho}}

and

v_B = \frac{\sqrt{T_B}}{r_B \sqrt{\pi \rho}}

Now,

\frac{v_a}{v_B} = \sqrt{\frac{T_A}{T_B}} \cdot \frac{r_B}{r_A}

\because r_A = 2 r_B \text{ and } T_A = \frac{1}{2} T_B

Hence,

\frac{v_A}{v_B} = \frac{1}{2\sqrt{2}}

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