Question

An open U-tube contains mercury. When $11.2 \,cm$ of water is poured into one of the arms of the tube, how high does the mercury rise in the other arm form its initial level?

• A. 0.56 cm
• B. 1.35 cm
• C. 0.41 cm
• D. 2.32 cm

Answer 1

Answer: (C)

Key Idea : At the same level in the two limbs of a U-tube, pressure is same.
On pouring water on left side, mercury rises $x cm$ (say) from its previous level in the right limb of $U$-tube creating a difference of levels of mercury by $2 x\, cm$. Equating pressures at $A$ and $B$, we get

$P_{A}=P_{B}$
$\therefore 11.2 \times 10^{-2} \times \rho_{\text {water }} \times g=2 x \times \rho_{\text {mercury }} \times g$
$\Rightarrow 11.2 \times 10^{-2} \times 1000\, kg / m ^{3}$
$=2 x \times 13600 \times\, kg / m ^{3}$
$\Rightarrow x=\frac{11.2 \times 10^{-2} \times 1000\, m }{2 \times 13600}$
$=0.41\, cm$