c
(c)If the rise of level in the right limb be \( x cm.\) the fall of level of mercury in left limb be \(4x\) \( cm \) because the area of cross section of right limb is \(4\) times as that of left limb.
\(\therefore \) Level of water in left limb is \((36 + 4x) cm.\)
Now equating pressure at interface of \(Hg \) and water \((at A' B')\)
\((36 + 4x) \times 1 \times g = 5x \times 13.6 \times g\)
By solving we get \(x = 0.56 cm.\)
