Question
The potential energy of particle varies with distance x fixed a origin as ; 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 = .... (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 =
= [ M1 L5/2 T-2 ]
Hence dimensions of AB
= [ M1 L5/2 T-2 ] [ Mo L1 To ]
= [ M L7/2 T-2 ]