| Two vessels \[A\]and \[B\]of different shapes have the same base area and are filled with water upto the same height h figure. The force exerted by water on the base is \[{{F}_{A}}\] for vessel \[A\]and \[{{F}_{B}}\] for vessel\[B\]. The respective weights of the vessels are \[{{W}_{A}}\] and \[{{W}_{B}}\]. Then, |
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A) \[{{F}_{A}}>{{F}_{B}};{{W}_{A}}>{{W}_{B}}\]
B) \[{{F}_{A}}={{F}_{B}};{{W}_{A}}>{{W}_{B}}\]
C) \[{{F}_{A}}={{F}_{B}};{{W}_{A}}<{{W}_{B}}\]
D) \[{{F}_{A}}>{{F}_{B}};{{W}_{A}}={{W}_{B}}\]
Correct Answer: B
Solution :
Since, \[h\] is the same for both vessels, pressure\[=h\rho g\]is the same. Now, force = pressure \[\times \] base area. Since, the base area is the same, the force exerted by water on the base is the same for both vessels. Thus, \[{{F}_{A}}={{F}_{B}}\]. The liquid exerts pressure and hence, force, not only on the base but also on the sides of the vessel. This force is normal to the sides of the vessel. For vessel\[B\], this force has no component in the downward direction. But, for vessel\[A\], this force (which is normal to the sides) has a non-zero component vertically downwards. Hence, vessel \[A\]gives a higher reading than vessel \[V\] when they are weighed on a weighing scale. Thus,\[{{W}_{A}}>{{W}_{B}}.\]
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