When a stretched wire and a tuning fork are sounded together, 5 beats per second are produced when the length of wire is 95cm or 100cm then the frequency of the fork is:
A) 90 Hz
B) 100 Hz
C) 105 Hz
D) 195 Hz

Hint: For this problem we will use the concept of beats:
For longer length frequency will be lower:
${f_2} = f - 5$ (${f_2}$ is the lower frequency, 5 is the beat we have assumed to let understand the formula, f is the frequency of the fork)
For shorter length frequency will be high:
${f_1} = f + 5$ ( ${f_1}$ is the high frequency)
Also the law of length;
$\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{{l_2}}}{{{l_1}}}$ (${l_1}$ and ${l_2}$ are the length of the wire respectively)

Complete step by step solution:
Let us define a few terms first:
Beats: The phenomenon of alternate waxing and waning of sound at regular intervals is called beats. The number of beats per second is called beat frequency. Beats are only heard when the difference of two frequencies is not more than 10 which is due to the persistence of hearing. One loud sound and one fade sound constitute a beat. The time from each loud sound to the next loud sound is called one beat period.
Let’s calculate the frequency of the tuning fork:
We have;
${f_2} = f - 5$..............(1)
${f_1} = f + 5$..............(2)
As per law of length:
$\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{{l_2}}}{{{l_1}}}$............(3)
On substituting the values of length and values of f1 and f2 from equation 1 and 2)
$ \Rightarrow \dfrac{{f + 5}}{{f - 5}} = \dfrac{{100}}{{95}}$..........(4)
On cross multiplying in equation (4)
$
   \Rightarrow 95f + 475 = 100f - 500 \\
   \Rightarrow 947 = 5f \\
   \Rightarrow f = 195Hz \\
 $ (Rearranged the terms of frequency and constant terms)

Option: (D) is correct.

Note: Beats have many applications such as to determine the unknown frequency as we have done in the question above, used in tuning musical instruments such as guitar, sitar etc, used in electronics in oscillators, used in mines for the presence of various dangerous gases .