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

A, B, C, D

Explanation

(a) Force exerted by the water on the bottom
`F_(1)=(p_(0)+rho gh)A_(1)` …(i)
Here, `p_(0)= `atmospheric pressure `=1.01xx10^(5)N//m^(2)`
`rho=`density of water `=10^(3)k//gm^(3)`
`g=10m//s^(2), h=10cm=0.1 m`
and `A_(1) =` area of base `=10 cm^(2)=10^(-3)m^(2)`
Subsituting in Eq. (i) and we get,
`F_(1)=(1.01xx10^(5)+10^(3)xx10xx0.1)xx10^(-3)`
or, `F_(1)=102 N` (downwards)
(b) Force exerted by atmosphere on water
`F_(2)=(p_(0))A_(2)`
Here,` A_(2)=` area of top `=30cm^(2)=3xx10^(-3)m^(2)`
`=303N` (downwards).
Force exerted by bottom on the water
`F_(3)=-F_(1)` or `F_(3)=102 N`(upwards)
weight of water `W=("volume")("density")(g)=10^(3)10^(3)(10)`
`=10N` (downwards)
Let` F` be the force= net downward froce
or, `F+F_(3)=F_(2)+W`
`:. F=F_(2)+W-F_(3)=303+10-102`
or,` F=211N` (upwards)
(c ) If the air inside the jar is completely pumped out,
` F_(1)=(rho gh)A_(1) (as p_(0)=0)`
`=10^(3)(10)(0.1)10^(3)=1N` (downwards)
In this case,
`F_(2)=0`
and `F_(3)=1N` (upwards)
`:. F=F_(2)+W-F_(3)=0+10-1=9 N` (upwards)
(d) No, the answer will remain the same. Because the answer depend upon `p_(0) rho,g,h,A_(1)` and `A_(2)`.