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The range of water flowing out of a small hole made at a depth `10 m` below water surface in a large tank is `R`. Find the extra pressure (in atm) applied on the water surface so that range becomes `2R`. Take `1atm=10^(5)Pa`. |
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Answer» Correct Answer - 3 Initial velocity of efflux: `v_1=sqrt(2gh)` For range to become double, velocity of efflux should be doubled: `v_2=2_(v_1)=2sqrt(2gh)` Let extra pressur applied be p, then `(p_0+p_0+pgh)-p_0=1/2pv_2^2` `implies p+pgh=1/2p8gh` `impliesp+3rhogh=3xx10^3xx10xx10xx10=3xx10^5Pa=3atm` |
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