Answer is (c) 5 c
\(\frac{1}{f_{air}}=\Big(\frac{1.5}{1}-1\Big)\) \(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(1)
\(\frac{1}{f_{liquid}}=\Big(\frac{1.5}{1.25}-1\Big)\)\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(2)
eq(2)/eq(1)
\(\frac{f_{air}}{f_{liquid}}\) = \(\frac{2}{5}\) , \(f_{liquid}=\frac{5}{2}f_{air}=\frac{5}{2}\times2\) = 5cm