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A block of mass 1 kg is released on wedge from position M as shown in figure (Neglect friction everywhere). The force exerted by vertical wall W on wedge, when the block is at position N is `(15)/(2)sqrt(P)` newton, then what is the value of P (Take g = 10 `m//s^(2)`)? |
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Answer» Correct Answer - 3 `rhogh=(1)/(2)rhoV_(1)^(2)` `Deltap+rhogh=(1)/(2)rhoV_(2)^(2)` and `V_(2)=2V_(1)` `thereforeDeltaP+rhogh=2rhoV_(1)^(2)=4rhoghimpliesDeltap=3rhogh=3xx10^(5)` Pascal = 3 atm |
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