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The potential energy of a particle is g...

The potential energy of a particle is given by the formula `U= 100 -5x+100x^2`, U and x are in SI units. If mas of the particle is 0.1 kg then magnitude of its acceleration .

A

at 0.05m from the origin is `50 m*s^(-2)`

B

at 0.5m from the mean position is `100 m*s^(-2)`

C

at 0.05 m from the origin is `150 m*s^(-2)`

D

at 0.05 m from the mean position is `200 m*s^(-2)`

Text Solution

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The correct Answer is:
A, B, C
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Knowledge Check

  • The potential energy of a particle varies with distance x from a fixed origin as U=(Asqrtx)/(x+B) where A and B are constants. The dimension of AB are

    A
    `ML^(5//2) T^-2`
    B
    `ML^2 T^-2`
    C
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    D
    `ML^(7//2) T^-2`

  • The potential energy of a particle varies with distance x from a fixed origin as U=(Asqrtx)/(x+B) where A and B are constants. The dimension of AB are

    A
    `ML^(5//2)T^-2`
    B
    `ML^2T^-2`
    C
    `M^(3//2)L^(3//2)T^-2`
    D
    `ML^(7//2)T^-2`

  • Displacement (x) and time (t) of a particle in motion are related as x = at + bt^(2) -ct^(3) where a,b, and c, are constants. Velocity of the particle when its acceleration becomes zero is

    A
    `a +(b^(2))/(c )`
    B
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    D
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