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Question: The frequency of a sonometer wire is f, but when the weights producing the tensions are completely i...

The frequency of a sonometer wire is f, but when the weights producing the tensions are completely immersed in water the frequency becomes f/2 and on immersing the weights in a certain liquid the frequency becomes f/3. The specific gravity of the liquid is :
A. 43\dfrac{4}{3}
B. 169\dfrac{16}{9}
C. 1512\dfrac{15}{12}
D. 3227\dfrac{32}{27}

Explanation

Solution

The formula for frequency of sonometer is being used. The frequency of the sonometer when immersed in water is different. When it is immersed in liquid other than the water the frequency is different. Formula for frequency is different. Formula for water has also been used.
Formula used:
Frequency of sonometer f=12lMgmf=\dfrac{1}{2l}\sqrt{\dfrac{Mg}{m}}
Frequency of sonometer when immersed in medium fw=12lMgVρwgm{{f}_{w}}=\dfrac{1}{2l}\sqrt{\dfrac{Mg-V{{\rho }_{w}}g}{m}}

Complete answer:
Let us consider a sonometer having frequency and be the length of the sonometer and is the mass per unit length of the string. So frequency can be given as
f=12lMgmf=\dfrac{1}{2l}\sqrt{\dfrac{Mg}{m}}
When a mass produces tension in the string and is completely immersed in water its frequency becomes f2\dfrac{f}{2}. When mass is immersed in water the frequency of the sonometer will be
fw=12lMgVρwgm{{f}_{w}}=\dfrac{1}{2l}\sqrt{\dfrac{Mg-V{{\rho }_{w}}g}{m}}
VρgV\rho g Is the buoyancy force
When mass is completely immersed completely in a certain liquid its frequency becomes f3\dfrac{f}{3}.
Frequency of water fw=f2=14lMgm{{f}_{w}}=\dfrac{f}{2}=\dfrac{1}{4l}\sqrt{\dfrac{Mg}{m}}
fw=12lMgVρwgm{{f}_{w}}=\dfrac{1}{2l}\sqrt{\dfrac{Mg-V{{\rho }_{w}}g}{m}}
The density of water ρw=1{{\rho }_{w}}=1
Squaring both the sides, we get
M4=MV\dfrac{M}{4}=M-V
V=MM4=3M4V=M-\dfrac{M}{4}=\dfrac{3M}{4}………..(1)
When the mass is immersed in an unknown liquid the frequency of sonometer is

& {{f}_{l}}=\dfrac{1}{2l}\sqrt{\dfrac{Mg-V{{\rho }_{l}}g}{m}} \\\ & =\dfrac{1}{6l}\sqrt{\dfrac{Mg}{m}} \end{aligned}$$ From equation (1) and using the value of $$V$$ $$\begin{aligned} & \dfrac{M}{9}=M-V{{\rho }_{l}} \\\ & \dfrac{M}{9}=M-\dfrac{3M}{4}{{\rho }_{l}} \\\ & {{\rho }_{l}}=\dfrac{32}{27} \end{aligned}$$ Therefore specific gravity of liquid is $$\dfrac{32}{27}$$. **Hence option D is correct.** **Additional information:** Sonometer is a device which describes the relation between frequency of sound which is produced when the string is plucked, the tension, length and mass per unit length of the string. **Note:** When the string is stretched the frequency of vibration varies. The frequency is directly proportional to the square root of its tension while the resonating length and mass per unit length remains constant.