FORCE ACTING on the BASE             \[F=P\times A=hdgA\]\[=0.4\times 900\times 10\times 2\times {{10}^{-3}}=7.2N\] "> FORCE ACTING on the BASE             \[F=P\times A=hdgA\]\[=0.4\times 900\times 10\times 2\times {{10}^{-3}}=7.2N\] ">
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A uniformly tapering vessel is filled with a liquid of density 900 kg/m3. The force that acts on the base of the vessel due to the liquid is \[(g=10\,m{{s}^{-2}})\]

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) 10 months ago

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FORCE ACTING on the BASE             \[F=P\times A=hdgA\]\[=0.4\times 900\times 10\times 2\times {{10}^{-3}}=7.2N\]
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