Two tuning forks \(P\) and \(Q\) when set vibrating gives 4 beats/ second. If the prong of fork \(P\) is filed, the beats are reduced to \(2 / \mathrm{s}\). What is the frequency of \(\mathrm{P}\), if that of \(\mathrm{Q}\) is \(250 \mathrm{~Hz}\) ? (A) \(246 \mathrm{~Hz}\) (B) \(250 \mathrm{~Hz}\) (C) \(254 \mathrm{~Hz}\) (D) \(252 \mathrm{~Hz}\)

The initial beat frequency is given as 4 beats/second. Let the frequencies of forks P and Q be \(f_P\) and \(f_Q\) (with \(f_Q = 250\) Hz) respectively. Then, the beat frequency is given by the absolute difference between their frequencies: \[ |f_P - f_Q| = 4 \]

After filing fork P, the beat frequency reduces to 2 beats/second. Let the new frequency of fork P after filing be \(f'_P\). Then, the modified beat frequency is given by: \[ |f'_P - f_Q| = 2 \]

We now have a system of two equations and two variables, \(f_P\) and \(f'_P\). We are given that the filing of fork P leads to a decrease in its frequency, so we can write this relation as: \[f'_P = f_P - x\] Where, \(x\) is the decrease in frequency due to filing. Now, we substitute this expression into our 2nd equation: \[| (f_P - x) - f_Q | = 2\] Plugging in the given values: \[| (f_P - x) - 250 | = 2\] We know from equation 1 that \(| f_P - 250 | = 4\), so there are two possible cases: 1. \(f_P - x = 252 \implies f_P = x + 252\) 2. \(f_P - x = 248 \implies f_P = x + 248\)

We need to satisfy both equations for the same values of \(f_P\) and \(x\). Let's analyze the two cases: 1. If \(f_P = x + 252\), this would mean that filing would increase the frequency of fork P. However, we know that filing decreases the frequency, so this is not a valid solution. 2. If \(f_P = x + 248\), this means the initial frequency of fork P is higher than the final frequency after filing, which is consistent with the given information. Substituting this into equation 1, we get: \[| (x + 248) - f_Q | = 4\] Plugging in the given values: \[| x + 248 - 250 | = 4\] \[|x - 2| = 4\] So, \(x = 6\) or \(x=-6\). Since the frequency is decreasing, we choose \(x=6\). Now, we can find the initial frequency of fork P: \[f_P = x + 248 = 6 + 248 = 254 Hz\] Thus, the frequency of fork P is \(254 Hz\), and the correct answer is (C) \(254 Hz\).