A capillary tube of radius $R$ is immersed in water
and water rises in it to a height $H$. Mass of water
in the capillary tube is $M$. If the radius of the tube
is doubled, mass of water that will rise in the
capillary tube will now be :
- M
- 2 M
- M/2
- 4 M
The Correct Option is A
Solution and Explanation
$A =$ area of cross-section,
$h =$ height, $\rho=$ density
$\Rightarrow m =\left(\pi r ^{2}\right) h \rho=\pi r ^{2} \times \frac{2 \sigma \cos \theta}{ r \rho g}$
$\Rightarrow m \propto r$
$\therefore$ Doubling radius
$\Rightarrow$ mass will also get doubled
$m '=2 M$
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Concepts Used:
Gauss Law
Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.
Gauss Law:
According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.
For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.
Gauss Law Formula:
As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;
Q = ϕ ϵ0
The Gauss law formula is expressed by;
ϕ = Q/ϵ0
Where,
Q = total charge within the given surface,
ε0 = the electric constant.