Question

A glass of water upto a height of 10 cm has a bottom of area 10 cm^(2) , top of area 30 cm^(2) and volume 1 litre. The downward force exerted by water on the bottom is…(Taking g=10 m//s^(2) ,density of water =10^(3) kg//m^(3) , atmosphereic pressure =1.01xx10^(5) N//m^(2))

Answer 1

Force exerted by the water on the bottom

F1=(p0+ρgh)A1 …(i)

Here, p0=atmospheric pressure =1.01×105N/m2

ρ=density of water =103k/gm3

g=10m/s2,h=10cm=0.1m

and A1= area of base =10cm2=10−3m2

Subsituting in Eq. (i) and we get,

F1=(1.01×105+103×10×0.1)×10−3

or, F1=102N (downwards)

(b) Force exerted by atmosphere on water

F2=(p0)A2

Here,A2= area of top =30cm2=3×10−3m2

=303N (downwards).

Force exerted by bottom on the water

F3=−F1 or F3=102N(upwards)

weight of water W=(volume)(density)(g)=103103(10)

=10N (downwards)

LetF be the force= net downward froce

or, F+F3=F2+W

∴F=F2+W−F3=303+10−102

or,F=211N (upwards)

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

F1=(ρgh)A1(asp0=0)

=103(10)(0.1)103=1N (downwards)

In this case,

F2=0

and F3=1N (upwards)

∴F=F2+W−F3=0+10−1=9N (upwards)

(d) No, the answer will remain the same. Because