Question

If a barometer uses a liquid of density \( 0.8 gram / cm ^{3} \), instead of Mercury height of liquid column would be.... a) \( 76 cm \) b) \( 1.013 \times 10^{5} cm \) c) \( 1292 cm \) d) \( 13600 cm \)

Answer 1

Given ρ_liquid = 0.8 gram/cm³

ρ_Hg = 13.6 gram/cm³

h_Hg = 76 cm

h_liquid = ?

(P_atm)_liquid = (P_atm)_Hg

ρ_liquidgh_liquid = ρ_Hggh_Hg

h_liquid = \(\frac{\rho_{Hg}.h_{Hg}}{\rho_{liquid}}\)

⇒ \(\frac{13.6\times76}{0.8}\)

⇒ 1292 cm.

option (c) is correct.