JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 16)
A cylindrical vessel containing a liquid is
rotated about its axis so that the liquid rises at
its sides as shown in the figure. The radius of
vessel is 5 cm and the angular speed of
rotation is $$\omega $$ rad s–1. The difference in the
height, h (in cm) of liquid at the centre of
vessel and at the side will be :
_2nd_September_Morning_Slot_en_16_1.png)
$${{2{\omega ^2}} \over {25g}}$$
$${{5{\omega ^2}} \over {2g}}$$
$${{25{\omega ^2}} \over {2g}}$$
$${{2{\omega ^2}} \over {5g}}$$
Explanation
_2nd_September_Morning_Slot_en_16_2.png)
Applying pressure equation from A to B
P0 + $$\rho $$.$${{R{\omega ^2}} \over 2}$$.R - $$\rho $$gh = P0
$$ \Rightarrow $$ $${{\rho {R^2}{\omega ^2}} \over 2}$$ = $$\rho $$gh
$$ \Rightarrow $$ h = $${{{R^2}{\omega ^2}} \over 2}$$ = $${\left( 5 \right)^2}{{{\omega ^2}} \over {2g}}$$ = $${{25} \over 2}{{{\omega ^2}} \over g}$$
Comments (0)
