A mercury manometer consists of two unequal arms of equal cross-section $1 mathrm{~cm}^2$ and lengths 100 cm and 50 cm. The two open ends are sealed with air in the tube at a pressure of 80 cm of mercury. Some amount of mercury is now introduced into the monometer through the stopcock connected to it. If the mercury rises in the shorter tube to a length of 10 cm in a steady state, find the length of the mercury column rising in the longer tube.

A mercury manometer consists of two unequal arms of equal cross-section and

A mercury manometer consists of two unequal arms of equal cross-section $1 \mathrm{~cm}^{2}$ and lengths $100 \mathrm{~cm}$ and $50 \mathrm{~cm}$ . The two open ends are sealed with air in the tube at a pressure of $80 \mathrm{~cm}$ of mercury. Some amount of mercury is now introduced in the monometer through the stopcock connected to it. If mercury resis in the shorter tube to a length $10 \mathrm{~cm}$ in steady state, find the length of the mercury column risen in the longer tube.
Option: 1
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Option: 2
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Option: 3
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Option: 4
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Answers (1)

Let $\mathrm{p}_{1}$ and $\mathrm{p}_{2}$ and be the pressures in centimetre of mercury in the two arms after introducing mercury in the tube. Suppose the mercury column rises in the second arm to $l_{0} \mathrm{~cm}$ .

Using $\mathrm{pV}=$ constant for the shorter arm,
$(80 \mathrm{~cm})(50 \mathrm{~cm})=\mathrm{p}_{1}(50 \mathrm{~cm}-10 \mathrm{~cm})$
$or \, \mathrm{p}_{1}=100 \mathrm{~cm}\quad \ldots(i)$

Using $\mathrm{pV}=$ constant for the longer arm,
$(80 \mathrm{~cm})(100 \mathrm{~cm})=\mathrm{p}_{2}\left(100-l_{0}\right) \mathrm{cm}\quad \ldots(ii)$

From the figure,
$\mathrm{p}_{1}=\mathrm{p}_{2}+\left(l_{0}-10\right) \mathrm{cm}$ .
Thus by (i),
$100 \mathrm{~cm}=\mathrm{p}_{2}+\left(l_{0}-10\right) \mathrm{cm}$ .
or
$\mathrm{p}_{2}=110 \mathrm{~cm}-l_{0} \mathrm{~cm}$
Putting in (ii),
$\begin{array}{ll} & \left(110-l_{0}\right)\left(100-l_{0}\right)=8000 \\ \text { or, } & l_{0}{ }^{2}-210 l_{0}+3000=0 \\ \text { or } & I_{0}=15.5 \end{array}$