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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)`. |
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Answer» Correct Answer - A::B::C::D (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)`. |
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