A column of air at 51∘C produces 4beats/s with a tuning fork. As the temperature of air column is decreased to 16∘C, the number of beats reduce to 1 beat/s . The frequency of tuning fork is :

Let Vf and Ve is the frequency of fork and air column respectively.Ve−Vf= beat frequencyV4L−Vf=ΔV(∵Ve=V4l) KT−Vf=ΔV Initially, K51+273−Vf=4 …(1)Finally, K16+273−Vf=1 …( 2)From (1) Vf=k324−4Substituting in (2)k289−k324+4=1 k(17−18)=−3 −k=−3 k=3 From (1) Vf=3×18−4=54−4 =50 HzOn solving, Vf=50 Hz

A column of air at 51∘C produces 4beats/s with a tuning fork. As the temperature of air column is decreased to 16∘C, the number of beats reduce to 1 beat/s . The frequency of tuning fork is :

Let Vf and Ve is the frequency of fork and air column respectively.Ve−Vf= beat frequencyV4L−Vf=ΔV(∵Ve=V4l) KT−Vf=ΔV Initially, K51+273−Vf=4 …(1)Finally, K16+273−Vf=1 …( 2)From (1) Vf=k324−4Substituting in (2)k289−k324+4=1 k(17−18)=−3 −k=−3 k=3 From (1) Vf=3×18−4=54−4 =50 HzOn solving, Vf=50 Hz

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A column of air at $51^{\circ} C$ produces 4beats/s with a tuning fork. As the temperature of air column is decreased to $16^{\circ} C$ , the number of beats reduce to 1 beat/s . The frequency of tuning fork is :

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25Hz

150Hz

50 Hz

75 Hz

answer is C.

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Detailed Solution

Let $V_{f}$ and $V_{e}$ is the frequency of fork and air column respectively.
$V_{e} - V_{f} =$ beat frequency
$\frac{V}{4 L} - V_{f} = Δ V (∵ V_{e} = \frac{V}{4 l})$
$K \sqrt{T} - V_{f} = Δ V$
Initially, $K \sqrt{51 + 273} - V_{f} = 4$ …(1)

Finally, $K \sqrt{16 + 273} - V_{f} = 1$ …( 2)
From (1) $V_{f} = k \sqrt{324} - 4$
Substituting in (2)
$k \sqrt{289} - k \sqrt{324} + 4 = 1$
$k (17 - 18) = - 3$
$- k = - 3$
$k = 3$
From (1) $V_{f} = 3 \times 18 - 4$
$= 54 - 4$
$= 50 Hz$
On solving, $V_{f} = 50 Hz$