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Motion In A Plane

Question
CBSEENPH11026212

The potential energy of particle varies with distance x fixed a origin as v = Axx + B; where A and B are constants. The dimensions of AB are

  • [ M L5/2 T-2 ]

  • [ M L2 T-2 ]

  • [ M3/2 L3/2 T-2 ]

  • [M L7/2 T-2 ]

Solution

D.

[M L7/2 T-2 ]

Given;- v = A xx + B             .... (i)

Dimensions of v = dimensions of potential energy

                         = [ M1 L2 T-2 ]

From equqtion (i) 

Dimensions of B = Dimensions of x = [Mo L1 To ]

∴ Dimensions of A =dimension of v × dimensions of x + Bdimensions of x

                            = M1 L2 T-2 Mo L1 ToM1 L12 To

                             = [ M1 L5/2 T-2 ]

Hence dimensions of AB 

                            = [ M1 L5/2 T-2 ] [ Mo L1 To ]

                            = [ M L7/2 T-2 ]