A clock is set right at 9 a.m. The clock gains 10 min. in 24 hrs. What will be the correct time, when clock shows 2 p.m. on the following day?

The clock is set right at 9 a.m. on the first day.
The faulty clock shows 2 p.m. on the following day.
The time elapsed from 9 a.m. on day 1 to 9 a.m. on day 2 is 24 hours.
The time elapsed from 9 a.m. on day 2 to 2 p.m. on day 2 is $2 \text{ p.m.} - 9 \text{ a.m.} = 5$ hours.
The total time elapsed on the faulty clock is $24 \text{ hours} + 5 \text{ hours} = 29 \text{ hours}$.

The clock gains 10 minutes in 24 hours. This means that for every 24 hours of correct time, the faulty clock shows $24 \text{ hours} + 10 \text{ minutes}$.
Convert 10 minutes to hours: $10 \text{ min} = \frac{10}{60} \text{ hours} = \frac{1}{6} \text{ hours}$.
So, when the correct clock shows 24 hours, the faulty clock shows $24 + \frac{1}{6} = \frac{144+1}{6} = \frac{145}{6}$ hours.

Let $T_F$ be the time shown by the faulty clock and $T_C$ be the time shown by the correct clock.
The ratio of faulty time to correct time is constant:

We know that the faulty clock has shown $T_F = 29$ hours. We need to find the corresponding correct time $T_C$.

To solve for $T_C$:

Since $145 = 5 \times 29$, we can simplify:

Convert the fractional part to minutes:

So, the correct time elapsed is 28 hours and 48 minutes.

The clock was set right at 9 a.m. on day 1.
Add 28 hours and 48 minutes to 9 a.m. (day 1):
9 a.m. (day 1) + 24 hours = 9 a.m. (day 2)
9 a.m. (day 2) + 4 hours = 1 p.m. (day 2)
So, 9 a.m. (day 1) + 28 hours = 1 p.m. (day 2).
Now add the remaining 48 minutes:
1 p.m. (day 2) + 48 minutes = 1:48 p.m. (day 2).

The correct time is 1:48 p.m. on the following day. This corresponds to option (C).

The final answer is $\text{C) 48 min past 1 / 1 ਵਜੇ ਕੇ 48 ਮਿੰਟ}$.