Question

A cylindrical vessel filled with water is released on an inclined surface of inclination angle $\theta$ as shown in figure. The coefficient of friction between surface with vessel is $\mu (<tan\theta )$. Then the constant angle made by the surface of water with the incline will be

• A. $ta{n}^{-1}\mu$
• B. $\theta -ta{n}^{-1}\mu$
• C. $\theta +ta{n}^{-1}\mu$
• D. $co{t}^{-1}\mu$

Answer 1

The correct option is A $ta{n}^{-1}\mu$

Figure shows forces acting on a 'particle' on the surface, with respect to vessel.


Here $mgsin\theta$ & $\mu mgcos$ $\theta$ are pseudo forces.
$tan\varphi =\mu$
$\therefore \varphi =ta{n}^{-1}\mu$
Here $\varphi$ is angle between normal to the inclined surface and the resultant force.
As free surface is ${\perp }^{lar}$ to the resultant force acting on it, same angle will be formed between the surface of water and the inclined surface.