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) \( 13600 cm \)
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) \( 13600 cm \)
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Given ρliquid = 0.8 gram/cm3
ρHg = 13.6 gram/cm3
hHg = 76 cm
hliquid = ?
(Patm)liquid = (Patm)Hg
ρliquidghliquid = ρHgghHg
hliquid = \(\frac{\rho_{Hg}.h_{Hg}}{\rho_{liquid}}\)
⇒ \(\frac{13.6\times76}{0.8}\)
⇒ 1292 cm.
option (c) is correct.
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