Two wires are fixed in a sonometer. Their tensions are in the rat… - Vidyadip

MCQ

Two wires are fixed in a sonometer. Their tensions are in the ratio $8 : 1$. The lengths are in the ratio $36:35.$ The diameters are in the ratio $4 : 1$. Densities of the materials are in the ratio $1 : 2$. If the lower frequency in the setting is $360 Hz.$ the beat frequency when the two wires are sounded together is

A
$5$
B
$8$
C
$6$
✓
$10$

✓

Answer

Correct option: D.

$10$

d
(d) Frequency in a stretched string is given by

$n = \frac{1}{{2l}}\sqrt {\frac{T}{{\pi {r^2}\rho }}} = \frac{1}{l}\sqrt {\frac{T}{{\pi {d^2}\rho }}} $ ($d =$ Diameter of string)

==> $\frac{{{n_1}}}{{{n_2}}} = \frac{{{l_2}}}{{{l_1}}}\sqrt {\frac{{{T_1}}}{{{T_2}}} \times {{\left( {\frac{{{d_2}}}{{{d_1}}}} \right)}^2} \times \left( {\frac{{{\rho _2}}}{{{\rho _1}}}} \right)} $

$ = \frac{{35}}{{36}}\sqrt {\frac{8}{1} \times {{\left( {\frac{1}{4}} \right)}^2} \times \frac{2}{1}} = \frac{{35}}{{36}}$$ \Rightarrow {n_2} = \frac{{36}}{{35}} \times 360 = 370$

Hence beat frequency = ${n_2} - {n_1} = 10$

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