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PHYSICS MCQs
Initially potential energy = `1/2 kx^2`
`rArr u = 1/2 kx^2`
or ` 2u = kx^2`
`rArr k = (2u)/(x^2)`
when it is stretched to `nx` cm
`:.` `P.E. = 1/2 kx_1^2 = 1/2 xx (2u)/(x^2) xx n^2 x^2 = n^2 u`
`:. ` potential energy stored in the spring = `n^2 u`
In the direction of effective gravitational a acceleration , potential energy of body decreases when balloon rises, buoyancy force is greater then weight of balloon . It indicates that effected gravitational acceleration is in upward direction. Hence , (B) is correct.
The spring loses elastic potential energy which appears as energy of molecules of acid . Hence , (C) is correct
`because U(x) = a/x^12 - b/x^6`
At the stable equilibrium ` (dU)/(Dx) = 0`
`- (12a)/(x^13) + (6b)/(x^7) = 0
`x = ((2a)/(b))^(1/6)`
Here `F = - (Delta U)/(Delta x)`
= `- (-mgh)/(x)`
=` (mgh)/(x)`
The work done by force of friction in a closed loop is not zero . Hence , frictional force is non-conservative in nature .
`vec (F) * d vec(r) = ( x hat(i) + y hat(j)) * ( dx hat(i) + dy hat(j))`
= `xdx + ydy`
Since , work done by this force depends upon path followed by the particle between two point . Hence , the force must be non-conservative.