Question

Solve this:
Q.33. 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
1) $\left[\frac{h_2-h_1}{h_2}\right]g$
2) $\left[\frac{h_2-h_1}{h_1}\right]g$
3) $\left[\frac{h_1-h_2}{h_1}\right]g$
4) $\left[\frac{h_1-h_2}{h_2}\right]g$

Answer 1

Dear Student,
$$\begin{gathered} Accordingtoquestion \\ {h}_1=\frac{2S\cos \left(\theta \right)}{r\rho g}----1 \\ and \\ {h}_2=\frac{2S\cos \left(\theta \right)}{r\rho (g+a)}-----2 \\ divideequation1by2weget \\ \frac{h_1}{h_2}=\frac{g+a}{g} \\ or \\ a=\left(\frac{h_1-h_2}{h_2}\right)g \end{gathered}$$
Regard