Question

Water rises to a height $h_1$ in a capillary tube in a stationary lift. If the lift moves up with uniform acceleration, it rises to a height $h_2$, then acceleration of the lift is :

• A. $\left[\frac{h_2-h_1}{h_2}\right]g$
• B. $\left[\frac{h_2-h_1}{h_1}\right]g$
• C. $\left[\frac{h_1-h_2}{h_1}\right]g$
• D. $\left[\frac{h_1-h_2}{h_2}\right]g$

Answer 1

The correct option is D $\left[\frac{h_1-h_2}{h_2}\right]g$

Let the lift be going up with an acceleration of a.
$h_1=\frac{k}{g},h_2=\frac{k}{g_{eff}}=\frac{k}{g+a}$
$h_{1}g=h_2(g+a)$
$\frac{(h_1-h_2)g}{h_2}=a$