Question

# The frequency of a sonometer wire is 300 Hz. The frequency becomes half when the mass producing the tension in the wire is completely immersed in water and on immersing the mass in a different liquid, the frequency become one-third. The relative density of the liquid is (upto 2 decimals )

A
1.18
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B
1.19
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Solution

## Let ρl be the density of liquid and ρw be the density of water. As we know, frequency f∝√T [where T is tension] ⇒f1f2=√T1T2...(i) $$\begin{array}{cll}\rho \operatorname{Vg} & \rho \operatorname{Vg} & \rho \operatorname{Vg} \\ \text { fig(I) } & \text { fig(II) } & \text { fig(III) }\end{array}$$ From figure (I), (II) & (III) T1=ρVg, [where ρ is the density of mass] T2=ρVg−ρwVg and T3=ρVg−ρlVg So, putting all these values in equation (i) we get, f1f2=√ρVgρVg−ρwVg ⇒300150=√ρρ−ρw ⇒4=ρρ−ρw⇒4ρ−4ρw=ρ ⇒ρ=43ρw .....(2) Again, f1f3=√T1T3=√ρVgρVg−ρlVg ⇒300100=√ρρ−ρl⇒9=ρρ−ρl ⇒9ρ−9ρl=ρ ⇒ρ=98ρl .....(3) From equation (2) & (3), 43ρw=98ρl ⇒ρlρw=43×89=1.185≈1.18 [∵ρlρw=Relative density]

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