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A curved glass vessel full 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 L. (i) Find the force exerted by the water on the bottom. (ii) Find the resultant force exerted by the sides of the glass on the water. (iii) If the glass vessel is covered by a jar and the air inside the jar is completely pumped out, what will be the answers to parts (a) and (b) (iv) If a glass vessel of different shape is used provided the height, the bottom area and the volume are unchanged, will the answers to parts (a) and (b) change. (Take, g=10 ms^(-2) , density of water =10^(3) kgm^(-2) and atmospheric pressure =1.01xx10^(5)Nm^(-2)) |
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Answer» Solution :(i) Force exerted by the water on the BOTTOM `F_(1)=(rho_(0)+rhogh)A_(1)` Here, `rho_(0)`= atmospheric pressure `=1.01xx10^(5) Nm^(-2)` `rho= " density of water " =10^(3) kgm^(-3)` `g=10 ms^(-2), h=10 cm =0.1 m` and `A_(1) = " AREA of BASE " =10 cm^(2)=10^(-3) m^(2)` ![]() Substituting these values in Eq. (i), we get `F_(1)=(1.01xx10^(5)+10^(3)xx10xx0.1)xx10^(-3)` or `F_(1)=102 N` (downwards) (ii) Force exerted by atmosphere on water `F_(2)=(rho_(0))A_(2)` Here, `A_(2)= " area of top " =30 cm^(2)=3xx10^(-3) m^(2)` ` therefore F_(2)=(1.01xx10^(5))(3xx10^(-3))` =303 N (dwonwards) 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)` =10 N (downwards) Let F the force exerted by side walls on the water (upwards). Then for equilibrium of water (FBD) Net upward force=net downwardforce or ` F+F_(3)=F_(2)+w` `therefore "" F=F_(2)+w-F_(3)=303+10-102` or "" F=211 N (upwards) (iii) If the air inside the jar is completely pumped out, `F_(1)=(rhogh)A_(1) "" (as rho_(0)=0)` `=(10^(3))(10)(0.1)(10^(-3))` =1 N (downwards) In this case, `F_(2)=0` and "" `F_(3)=1 N` (upwards) `therefore "" F=F_(2)+w-F_(3)` `=0+10-1` =9 N (upwards) (iv) No, the answer will remain the same. Because the answers depend upon `rho_(0),rho,g, h,A_(a) " and " A_(2)` |
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