A mercury pellet of length 5 cm is trapped in a horizontal tube of small cross-section area. Length of tube enclosed by the mercury pellet is 20 cm when it is accelerated with a = g /2 in the directio

Question

A mercury pellet of length 5 cm is trapped in a horizontal tube of small cross-section area. Length of tube enclosed by the mercury pellet is 20 cm when it is accelerated with a = g /2 in the direction shown. The length of tube enclosed by the pellet if it is accelerated in the opposite direction with the same acceleration is : (Atmospheric pressure is 72.5 cm of Hg)

TestHub · Accepted Answer

Accelerating as shown: P 1 ​ = P 0 ​ + g a ​ × 5 = 72.5 + 2.5 = 75 cm of Hg, x 1 ​ = 20 cm. Reversed: P 2 ​ = 72.5 − 2.5 = 70 cm of Hg. Boyle's law: P 1 ​ x 1 ​ = P 2 ​ x 2 ​ ⇒ x 2 ​ = 70 75 × 20 ​ = 21.4 cm.

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