Assume air in the atmosphere is replaced by water of uniform density such that the pressure is unchanged on the Earth's surface. i.e. 105 Pa. If a square plate of thickness 2 mm is kept on the surface of Earth and the effort required to lift it slowly is compared for both the cases (air and water), then choose the correct option?
Assume air in the atmosphere is replaced by water of uniform density such that the pressure is unchanged on the Earth's surface. i.e. 105 Pa. If a square plate of thickness 2 mm is kept on the surface of Earth and the effort required to lift it slowly is compared for both the cases (air and water), then choose the correct option?
The correct option is A Effort in air will be more as compared to water.
Consider the following representation of aforementioned situation. Let the pressure due to fluid (air/water) at a depth z be p and area of plate be Δs
Let us list all the forces acting upon this plate,
1.F1 vertically downwards due to fluid above it = P Δ s
2.F2 vertically upwards due to fluid below it = (P + dP) Δs
3.Weight = Mg vertically downwards
Now net force by the fluid =(F2−F1)
=(P+dP)Δs
=dPΔs
As you already know, dP = ρgdz Right??
butρ water>ρair
⇒dP water>dPair
Or in other words, the increase in pressure is more in case of water than air as one descends through a same depth Therefore the vertically upwards force will be more in case of water as a fluid. Hence, the apparent weight of this plate reduces more when kept in water than air. Therefore more effort is required when the plate is in air.Hence (a)
Effort in air will be more as compared to water.
Consider the following representation of aforementioned situation. Let the pressure due to fluid (air/water) at a depth z be p and area of plate be Δs
Let us list all the forces acting upon this plate,
1.F1 vertically downwards due to fluid above it = P Δ s
2.F2 vertically upwards due to fluid below it = (P + dP) Δs
3.Weight = Mg vertically downwards
Now net force by the fluid =(F2−F1)
=(P+dP)Δs
=dPΔs
As you already know, dP = ρgdz Right??
butρ water>ρair
⇒dP water>dPair
Or in other words, the increase in pressure is more in case of water than air as one descends through a same depth Therefore the vertically upwards force will be more in case of water as a fluid. Hence, the apparent weight of this plate reduces more when kept in water than air. Therefore more effort is required when the plate is in air.Hence (a)
