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The potential energy of a particle varie...

The potential energy of a particle varies with distance `x` as `U=(Ax^(1//2))/(x^(2)+B)` where `a` and `B` are constants. The dimensional formula for `AXB` is :

A

`M^(1)L^(7//2)T^(-2)`

B

`M^(1)L^(11//2)T^(-2)`

C

`M^(1)L^(5//2)T^(-2)`

D

`M^(1)L^(9//2)T^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
B
Here `x^(2)` has the dimensions of `L^(2)`
`:.B=[L^(2)]`
Also `ML^(2)T^(-2)=(AL^(1/2))/(L^(2))`
or `A=ML^(7/2)T^(-2)`
`:.AxxB=ML^(11/2)T^(-2)`
Hence correct choice is `(b)`.
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