The system of 2 rods shown in fig is vibrating at the same frequency and forming a standing wave. The ratio of the number of antinodes in the two rods if radius of rod B is twice the radius of A is:

JEE Main 2019 April Solved Paper of Physics

a.1

b.2

c.3

d.4

Answer: (b)

\(f_{1} = \frac{n}{2l}\sqrt{\frac{T}{\rho A}}f_{2} = \frac{m}{2l}\sqrt{\frac{T}{\rho 4A}}\)

Given f1 = f2

\(\frac{f_{1}}{f_{2}} = \frac{n}{m}\times 2\) \(\frac{n}{m}\times 2\)

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