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10-1.Thermometry, Thermal Expansion and Calorimetry
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The relative humidity on a day, when partial pressure of water vapour is $0.012 \times {10^5}\,Pa$ at $12°C$ is ......... $\%$ (take vapour pressure of water at this temperature as $0.016 \times {10^5}\,Pa$)
A
$70$
B
$40$
C
$75$
D
$25$
(AIIMS-1998)
Solution
(c) Partial pressure of water vapour ${P_W} = 0.012 \times {10^5}\,Pa$,
Vapour pressure of water ${P_V} = 0.016 \times {10^5}\,Pa$.
The relative humidity at a given temperature is given by $ = \frac{{{\rm{Partial\,\,pressure \,\,of\,\, water\,\, vapour}}}}{{{\rm{Vapour\,\, pressure\,\, of\,\, water}}}}$
$ = \frac{{0.012 \times {{10}^5}}}{{0.016 \times {{10}^5}}} = 0.75 = 75\% $
Standard 11
Physics
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