correct answer is: d) 2\large\frac{1}{3}

Explanation

A pump can fill the tank in 2 hours.
\therefore In 1 hour, it fills =\large\frac{1}{2} part of the tank.
A leak can empty the full tank in 14 hours.
\therefore In 1 hour, it empties =\large\frac{1}{14} part of the tank.
So that,
Presence of the leak the pump can fills =\large\left(\frac{1}{2}-\frac{1}{14}\right) part of the tank per hour.
\rightarrow\large\left(\frac{7-1}{14}\right) _{[L.C.M 14]}
\rightarrow\large\frac{6}{14}
\therefore Presence of the leak the pump can fills \large\frac{6}{14} part of the tank per hour.
Let, Full tank capacity ‘1’ .
Now,
To fill \large\frac{6}{14} part of tank the pump takes 1 hour
\therefore To fill 1 or full tank the pump takes =\left(1\div\large\frac{6}{14}\right) hour
\rightarrow\left(1\times\large\frac{14}{6}\right) hour
\rightarrow\large\frac{14}{6} or \large\frac{7}{3} or 2\large\frac{1}{3} hours
Ans: The pump takes 2\large\frac{1}{3} hours to fill the tank in presence of the leak.

Other ways to ask this same question are:
1. How long does it take for the pump to fill the tank when there’s also a leak that can empty the tank in 14 hours, if the pump alone can fill the tank in 2 hours?
2. What is the duration needed for the pump to fill the tank in the presence of a leak, given that the pump alone can fill it in 2 hours and the leak can empty it in 14 hours?
3. Find out how long it takes for the pump to fill the tank when there’s a leak that can empty it in 14 hours, if the pump can fill it in 2 hours normally.
4. Calculate the time required for the pump to fill the tank with a leak present, knowing that the pump alone takes 2 hours to fill it and the leak can empty it in 14 hours.