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Q.
What is the relative humidity on a day, when partial pressure of water vapour is $0.012 \times 10^{5} Pa$ and temperature is $12^{\circ}C$ ? The vapour pressure of water at this temperature is $0.016 \times 10^{5}$ Pa
Given that standard vapour pressure at $12^{\circ} C$
$=0.016 \times 10^{5}\, Pa$
The partial pressure of water vapour at $12^{\circ} C$ is
$=0.012 \times 10^{5} Pa$
$\therefore R . H. =\frac{\text { Vapour pressure of air }}{\text { SVP at the same temperature }}$
$=\frac{0.012 \times 10^{5}}{0.016 \times 10^{5}}$
$=0.75=75 \%$