Question

solve this problem from
hc verma :-


juniors don't answer plz !​

Answer 1

Given,

Area of base is same for all vessels.

Equal volumes of a liquid are poured in the vessels.

To find: The force on the base

Solution:

Liquid pressure can be given as:

P_L = ρgh ...(i)

where, ρ = density of liquid, g = acceleration due to gravity

and, h = height

And, Pressure can also be given as:

P = Force/ Area ...(ii)

Putting (ii) in (i):

F / A = ρgh

or, F = ρgh * A

Therefore, Force on the base will be Aρgh.

Since, the density, area, acceleration is constant for all vessel so Force will be directly proportional to height of liquid filled in vessel.

If we pour the liquid of same quanitity in each vessel, the triangular shaped vessel will have liquid to a greater height which is followed by square vessel and then trapezium type vessel.

Therefore, Force will be maximum in 3rd vessel.

Order will be: 3rd vessel (Triangular) > 2nd vessel (square) > 1st vessel.

Answer 2

Given,

Area of base is same for all vessels.

Equal volumes of a liquid are poured in the vessels.

To find: The force on the base

Solution:

Liquid pressure can be given as:

P_L = ρgh ...(i)

where, ρ = density of liquid, g = acceleration due to gravity

and, h = height

And, Pressure can also be given as:

P = Force/ Area ...(ii)

Putting (ii) in (i):

F / A = ρgh

or, F = ρgh * A

Therefore, Force on the base will be Aρgh.

Since, the density, area, acceleration is constant for all vessel so Force will be directly proportional to height of liquid filled in vessel.

If we pour the liquid of same quanitity in each vessel, the triangular shaped vessel will have liquid to a greater height which is followed by square vessel and then trapezium type vessel.

Therefore, Force will be maximum in 3rd vessel.

Order will be: 3rd vessel (Triangular) > 2nd vessel (square) > 1st vessel.