A massless conical flask filled with a liquid is kept on a table in a vacuum. The force exerted by the liquid on the base of the flask is $W_{1}$ .The force exerted by the flask on the table is $W_{2}$ .

W_{1} = W_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

W_{1} > W_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

W_{1} < W_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

The force exerted by the liquid on the walls of the flask is

(W_{1} - W_{2})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
B $W_{1} > W_{2}$
D The force exerted by the liquid on the walls of the flask is $(W_{1} - W_{2})$ .
Pressure acts equally in all directions, and a liquid exerts forces on these walls of its container normal to the walls. Hence, the liquid exerting forces on the sloping walls of the conical flask will have a net upward component. As the force exerted by the liquid on the base of the flask is $W_{1}$ ,we get the force exerted by the liquid on the walls of the flask as $W_{1} - W_{2}$

Suggest Corrections

Similar questions

W_{1}

W_{2}

θ_{1}, θ_{2}

θ_{3}

w_{1}, w_{2}

w_{3}

θ_{1} > θ_{2} > θ_{3}

w_{1}, w_{2}

w_{3}

w_{1}, w_{2}, a n d w_{3}

Join BYJU'S Learning Program