Question

A glass full of water upto a height of 10 cm has a bottom of are 10 cm^(2), top of area 30cm^(2) and volume 1 litre. (a) Find the force exerted by the water on the bottom. (b) Find the resultant force exerted by the sides of the glass on the water. (c ) If the glass is convered by a jar and the air inside the jar is completely pumped out, what will be the answer to parts (a) and (b). (d) If a glass of different shape is used, provided the height, the bottom area, the top area and the volume are unchanged, will the answer to parts (a) and (b) change. Take g=10m//s^(3), density of water=10^(3) kg//m^(3) and atmospheric pressure =1.01xx10^(5)N//m^(2).

Answer 1

The net downwards force is 1 N and upward force is 211 N

Explanation:

(a) Force exerted by the water on the bottom

F1 = (p0+ρgh)A1 --------(1)

Here, P0 = atmospheric pressure = 1.01 × 10^5 N/m^2

ρ = density of water = 10^3 k/gm^3

g = 10 m/s^2, h = 10 cm = 0.1 m

and A1 = area of base =10 cm^2 = 10^−3 m^2

Substituting in Equation (1) and we get

F1 = (1.01 × 10^5 + 10^3 × 10 × 0.1) × 10^−3

or, F1 = 10^2 N (downwards)

(b) Force exerted by atmosphere on water

F2 = (p0)A2

Here,A2= area of top =30 cm^2 = 3×10^−3 m^2

= 303 N (downwards)

Force exerted by bottom on the water

F3 = − F1 or F3 = 102 N(upwards)

Weight of water W = (volume)(density)(g) = 10^3 x 10^3 (10)

= 10 N (downwards)

Let F be the force = net downward force

or, F + F3 = F2 + W

F = F2 + W − F3 = 303 + 10 − 102

F = 211 N (upwards)

(c ) If the air inside the jar is completely pumped out.

F1 = (ρgh) A1 (asp 0 = 0)

= 10^3 (10) (0.1)10^3 = 1 N (downwards)

In this case:

F2=0

and F3 = 1 N (upwards)

F=F2 + W − F3 = 0 + 10 − 1 = 9 N (upwards)

(d) No, the answer will remain the same. Because the answer depend upon p0ρ,g,h,A1 and A2.