Question

A sonometer wire of $70{\rm{ cm}}$ length fixed at one end has a solid mass M, hanging from its end to produce tension in it. The wire produces a certain frequency. When the same mass ‘NI’ hangs in water it is found that the found that length of the wire has to be changed by $5{\rm{ cm}}$, on order to produce the same frequency then the density of the material of mass ‘M’ is
A. $\dfrac{{169}}{{27}}{\rm{ }}{{{\rm{gm}}} {\left/ {\vphantom {{{\rm{gm}}} {{\rm{cc}}}}} \right.} {{\rm{cc}}}}$
B. $1{\rm{ }}{{{\rm{gm}}} {\left/ {\vphantom {{{\rm{gm}}} {{\rm{cc}}}}} \right. } {{\rm{cc}}}}$
C. $5{\rm{ }}{{{\rm{gm}}} {\left/ {\vphantom {{{\rm{gm}}} {{\rm{cc}}}}} \right. } {{\rm{cc}}}}$
D. $\dfrac{{199}}{{27}}{\rm{ }}{{{\rm{gm}}} {\left/ {\vphantom {{{\rm{gm}}} {{\rm{cc}}}}} \right. } {{\rm{cc}}}}$

Answer 1

We will use the expression for the fundamental frequency of sonometer wire, which gives us the relation between the length of the wire, tension in the wire, and mass per unit length. We will compare the individual expression of frequencies to calculate the density of the material of mass.

Complete step by step answer:
Given:
The initial length of the wire is $l = 70{\rm{ cm}}$.
The change in length of wire is $\Delta l = 5{\rm{ cm}}$.

We have to find the density of the material of mass so that the final frequency produced in water is the same as that of its initial value.

We can write the expression for fundamental frequency for the given sonometer’s wire as below:
${f_1} = \dfrac{1}{{2l}}\sqrt {\dfrac{{{T_1}}}{\mu }}$……(1)
Here f is the initial frequency of sonometer wire, l is the initial length, ${T_1}$ is tension in the wire due to mass M and $\mu$ is the mass per unit length, which is the same in initial and final conditions.

Using the concept of the equilibrium, we can find that the weight of mass ‘M’ is equal to the tension produced in the wire, so we can write:
${T_1} = Mg$
Here ${T_1}$ is the initial tension in the sonometer wire; M is the mass attached to the wire, and g is the acceleration due to gravity.

We will substitute Mg for ${T_1}$ in equation (1) to get the value of the initial frequency of sonometer wire.${f_1} = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg}}{\mu }}$……(2)

The expression for the final frequency of wire can be written as below:
${f_2} = \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{{{T_2}}}{\mu }}$……(3)
Here ${T_2}$ is the final tension in the wire.

We can write the expression for tension in sonometer wire when it is in water, which is equal to the difference of weight of mass M and buoyant force applied by water.
${T_2} = \left( {M - \dfrac{M}{\rho }} \right)g$
Here $\rho$ is the density of the material of mass M.

We will substitute $\left( {M - \dfrac{M}{\rho }} \right)g$ for ${T_2}$ in equation (3).
${f_2} = \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{{\left( {M - \dfrac{M}{\rho }} \right)g}}{\mu }}$……(4)

It is given that the initial and final frequency is the same, so we can equate equation (2) and equation (4).
$\dfrac{1}{{2l}}\sqrt {\dfrac{{Mg}}{\mu }} = \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{{\left( {M - \dfrac{M}{\rho }} \right)g}}{\mu }} \\\ \implies \dfrac{M}{{\left( {M - \dfrac{M}{\rho }} \right)}} = {\left( {\dfrac{l}{{{l_2}}}} \right)^2}\\\ \implies \dfrac{1}{{\left( {1 - \dfrac{1}{\rho }} \right)}} = {\left( {\dfrac{l}{{{l_2}}}} \right)^2}$……(5)

We can write the final value of wire length as below:
${l_2} = l - \Delta l$

We will substitute 70 cm for l and 5 cm for $\Delta l$ in the above expression.

{l_2} = 70{\rm{ cm}} - 5{\rm{ cm}}\\\ = 65{\rm{ cm}} \end{array}$$ Now we will substitute 70 cm for l and $$65{\rm{ cm}}$$ for $${l_2}$$ in equation (5). $$\begin{array}{c} \dfrac{1}{{\left( {1 - \dfrac{1}{\rho }} \right)}} = {\left( {\dfrac{{70}}{{65}}} \right)^2}\\\ \rho = 7.294{\rm{ }}{{{\rm{gm}}} {\left/ {\vphantom {{{\rm{gm}}} {{\rm{cc}}}}} \right. } {{\rm{cc}}}}\\\ = 7.3{\rm{ }}{{{\rm{gm}}} {\left/ {\vphantom {{{\rm{gm}}} {{\rm{cc}}}}} \right. } {{\rm{cc}}}} \end{array}$$ Therefore, the density of the mass ‘M’ is $$7.294{\rm{ }}{{{\rm{gm}}} {\left/ {\vphantom {{{\rm{gm}}} {{\rm{cc}}}}} \right. } {{\rm{cc}}}}$$. **So, the correct answer is “Option D”.** **Note:** From the equation of equilibrium for final tension of the wire, we find that buoyant force is acting vertically upwards, and the weight of mass M is acting downwards, so we have taken tension in the wire as the difference of buoyant force and weight of mass M.