When a stretched wire and a tuning fork are sounded together, `5` beats per second are produced, when length of wire is `95 cm` or `100 cm`, frequency of the fork is

Options

A. `90 Hz`
B. `195 Hz`
C. `100 Hz`
D. `105 hz`

Answer

B

Explanation

Let the frequency of tuning fork be `n`.
Let at length `l_1=95cm` the frequency in `n_1` and at `l_2=100cm` is frequency is `n_2(ltn_1)`
Given `n_1-n=5` .(i)
and `n-n_2=5` …(ii)
Adding Eqs. (i) and (ii) we get
`n_1-n_2=10` .(iii)
for stretched wire `nprop(1)/(l)`
Also, `(n_1)/(n_2)=(l_2)/(l_1)=(100)/(95)`
`n_1=(100)/(95)n_2`
From Eq. (iii), `(100)/(95)n_2-n_2=10impliesn_2=190`
From Eq. (ii), `n=n_2+5=190+5=195Hz`.