Explanation

(c) On pourring water on left side, mercury rises $x$ cm (say) from its previous level in the right limb of U-tube crating a differences of levels of mercury by$\ 2x$ cm. Equating pressure at A and B, we get
$p_{A} = p_{\mathbf{B}}$
$\therefore$ $11.2 \times 10^{- 2} \times \rho_{water} \times g = 2x \times \rho_{Hg} \times g$
$11.2 \times 10^{- 2} \times 1000\ {kgm}^{- 3} = 2x \times 13600\ {kgm}^{- 3}$
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$x = \frac{11.2 \times 10^{- 2} \times 1000}{2 \times 13600}m = 0.41\ cm$