Question

The frequency of a sonometer wire is $100\,Hz$ . When the weights producing the tensions are completely immersed in water, the frequency becomes $80\,Hz$ and on immersing the weights in a certain liquid, the frequency becomes $60\,Hz$ . The specific gravity of the liquid is
(A) $1.42$
(B) $1.77$
(C) $1.82$
(D) $1.21$

Answer 1

From the given data in the questions, compare the frequency in air and that of the water. From the obtained value calculate the ratio of the density of the water and the medium and that of the liquid and the medium. Divide both the values to obtain the specific gravity.

Formula used:
The formula of the specific gravity is given by
$s = \dfrac{{{d_l}}}{{{d_w}}}$
Where $s$ is the specific gravity of the liquid, ${d_l}$ is the density of the liquid and ${d_w}$ is the density of the water which is kept as the reference.

Complete step by step solution:
The frequency of the wire, $F = 100\,Hz$
The frequency of the wire in water, ${F_w} = 80\,Hz$
The frequency of the wire in the liquid, ${F_l} = 60\,Hz$
It is known that the force is directly proportional to the square root of the acceleration due to gravity.
$F\alpha \sqrt g$
From the given data it is known that ${F_w} = 0.8{F_{}}$
Squaring both sides and substituting the value of the frequency of the air in the above step.
${0.8^2} = \dfrac{{g'}}{g}$
The value of the $\dfrac{g}{{g'}}$ is also written as $1 - \dfrac{{{d_w}}}{{{d_m}}}$.
$1 - 0.64 = \dfrac{{{d_w}}}{{{d_m}}}$
By simplifying the above step,
$\dfrac{{{d_w}}}{{{d_m}}} = 0.36$ ----------------(1)
The above value is also written as $1 - \dfrac{{{d_l}}}{{{d_m}}}$.
$1 - \dfrac{{{d_l}}}{{{d_m}}} = 0.36$
By further simplification,
$\dfrac{{{d_l}}}{{{d_m}}} = 0.64$ ---------------(2)
From the equation (1) and (2),
$\dfrac{{{d_l}}}{{{d_w}}} = \dfrac{{0.64}}{{0.36}}$
By the formula of the specific gravity,
$s = 1.77$
Hence the specific gravity of the liquid is obtained as $1.77$

Note: The value of the specific gravity of the wire can provide us information whether the wire floats or sinks in the liquid. The obtained answer for specific gravity of the liquid is $1.77$ , it is greater than that of the water( specific gravity of water is $1$ ).