Question

On dipping a capillary of radius '$r$' in water, water rises upto a height $H$ and potential energy of water is $u_1$. If a capillary of radius $2r$ is dipped in water, then the potential energy is $u_2$. The ratio $\frac{u_1}{u_2}$ is :-

• A. $2:1$
• B. $1:2$
• C. $4:1$
• D. $1:1$

Answer 1

Answer: (D)

$1:1$

The weight of liquid of height 'h' in tube

$=mg$

= (volume $\times$ dencity) g

$=(\frac{\pi }{4}{d}^{2}h\times \rho g)$

$=\frac{\pi }{4}{d}^{2}h\rho g$

Potential energy $\left(u\right)=mgh$

$=\left(\frac{\pi }{4}{d}^{2}h\rho g\right)h$

$u=\frac{\pi }{4}\rho g\left(d^2{h}^2\right)$

$u\alpha (dh{)}^2$

& we know

$h=\frac{4Tws\theta }{\rho gd}$

$dh=\frac{4Tws\theta }{\rho g}$ = constant

So, u = constant $\Rightarrow u_1:i_2=1:1$