______ m of air column ,on an average ,will exert as much pressure as 1 cm column of mercury.

100

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105

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Solution

The correct option is D 105
We may write,
Height of air column $\times$ Density of air $=$ Hight of mercury column $\times$ Density of mercury
Height of air column x $1.29 k g / m^{3} = 0.01 m \times 13600 k g / m^{3}$
Height of air column = $\frac{136}{1.29} = 105 m$ (approx)

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______ m of air column ,on an average ,will exert as much pressure as 1 cm column of mercury.

The correct option is D 105We may write, Height of air column × Density of air = Hight of mercury column × Density of mercuryHeight of air column x 1.29kg/m3=0.01m×13600kg/m3Height of air column =1361.29=105m (approx)

The correct option is D 105
We may write,
Height of air column $\times$ Density of air $=$ Hight of mercury column $\times$ Density of mercury
Height of air column x $1.29 k g / m^{3} = 0.01 m \times 13600 k g / m^{3}$
Height of air column = $\frac{136}{1.29} = 105 m$ (approx)