A barometer has become faulty because some air has leaked into the space above the mercury column. The height of the mercury column is $70 c m$ and the length of the space above the mercury is $10 c m$ . The barometer space is pushed further down into the mercury trough. The length of space above mercury reduces to $6 c m$ and the height of mercury column is found to decrease to $67 c m$ . This information shows that the true atmospheric pressure must be:

74.5 c m

of Hg

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75.0 c m

of Hg

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75.5 c m

of Hg

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76.0 c m

of Hg

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Solution

The correct option is A $74.5 c m$ of Hg
$V_{1}$ is volume of air trapped
$V_{1}$ = A $\times$ 10
$V_{2}$ is volume of air trapped in 2
$V_{2}$ = A $\times$ 6
& $P_{1} = (P_{0} - 70)$ un of Hg
$P_{2} = (P_{0} - 67)$ un of Hg
So from Boyle's law
$P_{1} V_{1} = P_{2} V_{1}$
$(P_{0} - 70) \times 10 \times A = (P_{0} - 67) \times 6 \times A$
${10 P}_{0} - 700 = 6 P_{0} - 6 \times 67$
${4 P}_{0} = 700 - 6 \times 67$
$P_{0} = \frac{298}{4} = 74.5 u n . o f H g$

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The correct option is A 74.5cm of HgV1 is volume of air trapped V1 = A × 10V2 is volume of air trapped in 2 V2 = A × 6& P1=(P0−70) un of Hg P2=(P0−67) un of HgSo from Boyle's law P1V1=P2V1 (P0−70)×10×A=(P0−67)×6×A 10P0−700=6P0−6×67 4P0=700−6×67P0=2984=74.5un.ofHg