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)`)?

Answer

3

Explanation

`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