Correct Answer – A
(a) Velocity is muximum when K. E. is maximum for minimum P.E.
` (dV)/(dx) = 0 rArr x^(2) – x = 0 rArr x = +- 1`
`rArr Min P.E. = (1)/(4) – (1)/(2) = – (1)/(4) J`
`K. E _(max) + P.E_(min) = 2(Given)`
`:. K.E _(max) = 2 + (1)/(4) = (9)/(4) = (1) /(2) m nu_(max) ^(2)`
`rArr (1)/(2) xx 1 xx nu_(max)^(2) . = (9)/(4) rArr nu_(max) . = (3)/(sqrt(2))`

Correct Answer – C
Max. `K.E.=(1)/(2)m upsilon_(max)^(2)0`total mechanical energy
`:. (1)/(2)xx1xxupsilon_(max)^(2)=2`
`upsilon_(max)=sqrt(2xx2)=2m//s`

Correct Answer – C
Total mechanical energy of particle,
`E_(T) = 2J`
When kinetic energy is maximum, the potential energy should be minimum.
The potential of the particle is given by
`V (x) = (x^(4))/(4) -(x^(2))/(2)`
or `(dV)/(dx) =(4x^(3))/(4)-(2x)/(2) = x^(3) – x = x (x^(2) -1)`
For `V` to be minimum, `(dV)/(dx) = 0`
`:. x(x^(2) -1) = 0`, or `x = 0 pm 1`
At `x = 0, V (x) = 0`
At `x = “pm” 1, V (x) = – (1)/(4) J`
`:. (“Kinetic energy”)_(max) = E_(T) – V_(min)`.
or `(“Kinetic energy”)_(max) = 2 2-(-(1)/(4)) = (9)/(4) J`
or `(1)/(2) mv_(m)^(2) = (9)/(4)`
or `v_(m)^(2) = (9xx2)/(mxx4)`
or `v_(m)^(2) = (9xx2)/(1xx4) = (9)/(2)`
`rArr v_(m) = (3)/(sqrt(2)) m//s`.

Correct Answer – D
`V(x)=(x^4)/(4)-(x^2)/(2)`
`F=-(dV_(x))/(dx)=-[x^3-x]=0impliesx(x^2-1)=0`
`x=0`,`x=+-1`
`(d^2V_(x))/(dx)=3x^2-1`
At `x=+-1`,`(d^2V_(x))/(dx)=+ve`, i.e., at `x=+-1`, P.E. is minimum
`V_(min)=-(1)/(4)`
`E=K_(max)+V_(min)`
`implies2=(1)/(2)xx1xxv_(max)^2-(1)/(4)impliesv_(max)=(3)/(sqrt2)(m)//(s)`

Correct Answer – C
`U =(x^(4))/(4)-(x^(2))/(2),(dU)/(dx) =x^(3) -x = 0 rArrx =0, x = pm 1`
`(d^(2)U)/(dx^(2)) = 3x^(2) -1(d^(2)U)/(dx^(2)) = +ve` for `x =pm1`
`U(pm 1)=-(1)/(4)`
`K_(max) =Ulmin=T.E = 2J`
`K_(max) =(9)/(4)`
`K_(max) =(1)/(2) mv^(2) , v = (3)/(sqrt(2))`.