Question

A straight tube of length L contains incompressible liquid of mass M and the closed tube is whirled in horizontal plane about one of the ends. If ωis the uniform angular velocity, the force exerted by the liquid on the other end is

• A. $\frac { M L \omega ^ { 2 } } { 2 }$
• B. 2MLω²
• C. $\frac { M L \omega ^ { 2 } } { 4 }$
• D. MLω²

Answer 1

Answer: (A)

$\frac { M L \omega ^ { 2 } } { 2 }$

Answer 2

Let there be a small element of length dl at a distance L from the end of rotational axis.

Mass of the element dl = $\frac { \mathrm { M } } { \mathrm { L } }$ dl

Small radial force on this element = $\left( \frac { \mathrm { M } } { \mathrm { L } } \mathrm { dl } \right)$ lω²

∴ Total force = $\int _ { 0 } ^ { \mathrm { L } } \left( \frac { \mathrm { M } } { \mathrm { L } } \mathrm { dl } \right) \omega ^ { 2 }$

= $\frac { \mathrm { M } } { \mathrm { L } } \omega ^ { 2 } \int _ { 0 } ^ { \mathrm { L } } 1 \mathrm { dl } = \frac { \mathrm { ML } \omega ^ { 2 } } { 2 }$