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A large open tank is filled with water upto a height H. A small hole is made at the base of the tank. It takes `T_1` time to decrease the height of water to H/n(ngt1) and it takes `T_(2)` time to take out the remaining water. If `T_(1)=T_(2)`, then the value of n isA. 2B. 3C. 4D. `2sqrt(2)` |
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Answer» `-A(dy)/(dt)=asqrt(2gy)` `(2A)/(asqrt(2)g)(sqrt(H)-sqrt(H/n))=T_(1)` `(2A)/(asqrt(2)g)(sqrt(H/n)-0)=T_(2)` `T_(1)=T_(2)` n=4. |
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