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The density `rho` of a liquid varies with depth h from the free surface as `rho=kh`. A small body of density `rho_(1)` is released from the surface of liquid. The body willA. come to a momentary rest at a depth `2rho_(1)//k` from the free surfaceB. execute S.H.M. about a point at a depth `rho_(1)//k` from the surfaceC. execute S.H.M. of amplitude `rho_(1)//k`D. all of these |
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Answer» Correct Answer - D Let V be the volume of body which is released on the surface of liquic. When body reaches at depth x from the surface of liquid, then upward thrust on body due to liquid `=V(kx)g` Weight of body `=Vrho_(1)g` At equilibrium , `V(kx)g=Vrho_(1)g or x=rho_(1)//k` If the body comes to rest at depth h, then work done on body by the upward thrust dur to liquid `=` change in potential energy of body `:. -int_(0)^(h)V(kx)g dx=-Vrho_(1) gh or h =(2rho_(1))/(k)` Now, the body will move upwards towards equilibrium position due to displacemtn x from equilibrium position . Hence the body will execute S.H.M. Choice (d) is right. |
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