MHT CET · Physics · Kinetic Theory of Gases
An open U-tube contains mercury. When \(11.2 \mathrm{~cm}\) 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 unit?
- A \(0.56 \mathrm{~cm}\)
- B \(1.35 \mathrm{~cm}\)
- C \(0.41 \mathrm{~cm}\)
- D \(2.32 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(C) \(0.41 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
On pouring water on left side, mercury rises \(x\) \(\mathrm{cm}\) (say) from its previous level in the right limb of U-tube creating a difference of levels of mercury by \(2 x \mathrm{~cm} .\) Equating pressure at \(A\) and \(B\), we get
\(
\begin{array}{l}
\quad p_{A}=p_{B} \\
\therefore 11.2 \times 10^{-2} \times \rho_{\text {water }} \times g=2 x \times \rho_{\mathrm{Hg}} \times g \\
11.2 \times 10^{-2} \times 1000 \mathrm{~kg} / \mathrm{m}^{3} \\
=2 x \times 13600 \mathrm{~kg} / \mathrm{m}^{3} \\
x=\frac{11.2 \times 10^{-2} \times 1000}{2 \times 13600} \mathrm{~m} \\
=0.41 \mathrm{~cm}
\end{array}
\)
\(
\begin{array}{l}
\quad p_{A}=p_{B} \\
\therefore 11.2 \times 10^{-2} \times \rho_{\text {water }} \times g=2 x \times \rho_{\mathrm{Hg}} \times g \\
11.2 \times 10^{-2} \times 1000 \mathrm{~kg} / \mathrm{m}^{3} \\
=2 x \times 13600 \mathrm{~kg} / \mathrm{m}^{3} \\
x=\frac{11.2 \times 10^{-2} \times 1000}{2 \times 13600} \mathrm{~m} \\
=0.41 \mathrm{~cm}
\end{array}
\)
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