1.

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 :

Answer»

`M^(1)L^(7//2)T^(-2)`
`M^(1)L^(11//2)T^(-2)`
`M^(1)L^(5//2)T^(-2)`
`M^(1)L^(9//2)T^(-2)`

SOLUTION :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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