Question

A tube of length L is filked completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizortal plane about one of its ends with.a uniform angular velocity o.The force exerted by the liquid at the other end is Lo 2 2.2 (3) MLo 2 19. A block B is pushed momentarily along a

answer is 1...pls explain well

Answer 1

Here consider a small length 'dr' at a distance 'r' from the axis of rotation. Then centrifugal fugal force on liquid of dr length is

dF= dm r w^{​​​​​​2}

^{where dm= mass of liquid in dr length}

^{= (M/L)dr}

^{so for total length L}

^{integration dF= integration dm r w^2}

^{F = integration (M/L)dr r w^2}

^{= (M/L)w^2 $\int$r dr Here limits are 0 to L}

^{= (M/L)w^2 (r^2/2) limits 0to L}

^{= (M/L)w^2( L^2/2)}

^{F = ( Mw^2 L )/2}

^{do option 1 is correct}