Question

The rise of liquid into capillary tube is $h_1$. If the apparatus is taken in a lift moving up with acceleration, the height is $h_2$, then:

• A. $h_1=h_2$
• B. $h_1>h_2$
• C. $h_2>h_1$
• D. $h_2=0$

Answer 1

Answer: (B)

$h_1>h_2$

If lift is moving up than with $acc{l}^n$ a effective $acc{l}^n$ due to gravity becomes $(g+a)$

$h_1=\frac{2Tcos\theta }{r\rho g}$

$h_2=\frac{2Tcos\theta }{r\rho (g+a)}$

$\therefore h_2<h_1$