An open U-tube contains mercury. When $11.2 c m$ 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 unit?

0.56 c m

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1.35 c m

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0.41 c m

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2.32 c m

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Solution

The correct option is C $0.41 c m$

On pouring water on left side, mercury rises $x c m$ (say) from its previous level in the right limb of U-tube creating a difference of levels of mercury by $2 x$ cm. Equating at $A$ and $B$ , we get
$P_{A} = P_{B}$
$∴$ $11.2 \times 10^{- 2} = ρ_{w a t e r} \times g = 2 x \times ρ H g \times g$
$11.2 \times 10^{- 2} \times 1000 k g / m^{3} = 2 x \times 13600 k g / m^{3}$
$x = \frac{11.2 \times 10^{- 2} \times 1000}{2 \times 13600} m$
$= 0.41 c m$

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An open U-tube contains mercury. When 11.2cm 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 unit?

An open U-tube contains mercury. When $11.2 c m$ 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 unit?