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An open U-tube contains mercury. When 6.8 cm of water is poured into one of the arms of the tube, how high does the mercury rise in the other arm from its initial level $(ρ_{H g} = 13.6 {gm/cm}^{3})$

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$0.41 c m$

$0.35 c m$

$0.25 c m$

$0.82 c m$

answer is C.

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Detailed Solution

$\frac{m g}{A} = h ρ_{H g} g \Rightarrow \frac{A . \frac{l}{2} ρ_{w} g}{A} = h ρ_{H g} g \Rightarrow h = \frac{l ρ_{w}}{2 ρ_{H g}} = \frac{6.8 \times 1}{2 \times 13.6} = \frac{1}{4} = 0.25 c m$

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