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Question 53 of 80

Q.A sonometer wire vibrates with frequency n1n_1 in air under a suitable load of specific gravity 'σ'. When the load is immersed in water, the frequency of vibration of the wire n2n_2 will be ______. (A) n1σ+1σn_1\sqrt{\dfrac{\sigma+1}{\sigma}}
(B) n1σ−1σn_1\sqrt{\dfrac{\sigma-1}{\sigma}}
(C) n1σσ+1n_1\sqrt{\dfrac{\sigma}{\sigma+1}}
(D) n1σσ−1n_1\sqrt{\dfrac{\sigma}{\sigma-1}}

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2016MCQ· 1mImportance★★★★★
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Frequency of a stretched wire depends on the tension; find how the tension (net weight) changes when the load is immersed in water (buoyancy reduces the net downward pull).

The frequency of vibration of a stretched sonometer wire under tension TT is

n=12LTm⇒n∝Tn=\frac{1}{2L}\sqrt{\frac{T}{m}}\quad\Rightarrow\quad n\propto\sqrt{T}

(all else — length LL, linear mass density mm — being unchanged).

In air, the tension in the wire equals the weight of the suspended load:

T1=VsdgT_1=Vsdg

where VV is the volume of the load, sdsd (i.e. σρw\sigma\rho_w, with ρw\rho_w the density of water) is the density of the load material expressed via specific gravity σ\sigma, so T1=VσρwgT_1=V\sigma\rho_w g.

In water, the load additionally experiences an upward buoyant force equal to the weight of water it displaces, VρwgV\rho_w g. The net tension becomes …

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