Question
Physics Question on Electromagnetic waves
The potential energy of a particle varies with distance x from a fixed origin as ∨=(Axx+B), where A and B are constants. The dimensions of AB are
A
(A) [ML5/2T-2]
B
(B) [ML2T-2]
C
(C) [M3/2L3/2T-2]
D
(D) [ML7/2T-2]
Answer
(D) [ML7/2T-2]
Explanation
Solution
Explanation:
Given, V=Axx+BDimensions of V= dimensions of potential energy =[ML2T−2]From Eq. (i), Dimensions of B= dimensions of x=[M0LT0] Dimensions of A= dimensions of V× dimensions of (x+B) dimensions of x =[ML2T−2][M0LT0][M0L1/2T0]=[ML5/2T−2]Hence, dimensions of AB =[ML5/2T−2][M0LT0]=[ML7/2T−2]