The potential energy of a particle varies with distance x as U=Ax1/2x2+B, where A and B are constants. The dimensional formula for A x B is:
M1L7/2T−2
M1L11/4T−2
M1L5/2T−2
M1L9/2T−2
A
M1L7/2T−2
B
M1L5/2T−2
C
M1L11/4T−2
D
M1L9/2T−2
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Solution
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Given, U=Ax1/2x2+B
x2+B should have some dimension ⇒L2=dim(B)
dim(A)=dim(U×(x2+B)x1/2)=[ML2T−2][L2]L1/2
=ML7/2T−2
dim(A×B)=[ML7/2T−2][L2]=[ML11/2T−2]
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