At the top of a mountain, a thermometer reads $7^{o} C$ and a barometer reads 70 cm of Hg. At the foot of a mountain, they read $27^{o} C$ and 76 cm of Hg respectively. The ratio of the density of air at the top to that at the bottom of the mountain is:

0.885

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0.987

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0.75

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Solution

The correct option is B 0.987
As we know,
$P V = \frac{w}{M} R T$
$∴ d = \frac{w}{V} = \frac{P M}{R T}$
$\frac{d_{1}}{d_{2}} = \frac{70 \times M \times R \times 300}{R \times 280 \times 76 \times M} = 0.987$

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At the top of a mountain a thermometer reads $7^{\circ} C$ and a barometer reads 70 cm of Hg. At the bottom of the mountain these read $27^{\circ} C$ and 76 cm of Hg respectively. Comparison of density of air at the top with that of bottom is

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At the top of a mountain, a thermometer reads 7oC and a barometer reads 70 cm of Hg. At the foot of a mountain, they read 27oC and 76 cm of Hg respectively. The ratio of the density of air at the top to that at the bottom of the mountain is:

The correct option is B 0.987As we know,PV=wMRT∴d=wV=PMRTd1d2=70×M×R×300R×280×76×M=0.987

At the top of a mountain, a thermometer reads $7^{o} C$ and a barometer reads 70 cm of Hg. At the foot of a mountain, they read $27^{o} C$ and 76 cm of Hg respectively. The ratio of the density of air at the top to that at the bottom of the mountain is:

The correct option is B 0.987
As we know,
$P V = \frac{w}{M} R T$
$∴ d = \frac{w}{V} = \frac{P M}{R T}$
$\frac{d_{1}}{d_{2}} = \frac{70 \times M \times R \times 300}{R \times 280 \times 76 \times M} = 0.987$