a
(a)Weight of cylinder = upthrust due to both liquids
\(V \times D \times g = \left( {\frac{A}{5}\, \times \frac{3}{4}L} \right) \times d \times g + \left( {\frac{A}{5} \times \frac{L}{4}} \right) \times 2d \times g\)
==>\(\left( {\frac{A}{5} \times L} \right)\, \times D \times g = \frac{{A \times L \times d \times g}}{4}\)==>\(\frac{D}{5} = \frac{d}{4}\) \(\therefore \;D = \frac{5}{4}d\)