The potential energy U of a particle varies with distance x from a fixed origin as U = (Asqrt(x))/(x^(2)+B)where A and B are dimensional constants. The dimensional formula for AB is

The potential energy U of a particle varies with distance x from a fix

1.	The potential energy U of a particle varies with distance x from a fixed origin as U = (Asqrt(x))/(x^(2)+B)where A and B are dimensional constants. The dimensional formula for AB 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 :In the expression `U= (Asqrt(x))/(x^(2)+B)` B must have the dimensions of `A= [(UX^(2))/(sqrt(x))]= ([ML^(2)T^(-2)][L^(2)])/([L^(1//2)])= [ML^(7//2)T^(-2)]` `:.`the DIMENSIONAL FORMULA for `AB = [ML^(7//2)T^(-2)][L^(2)]= [M^(1)L^(11//2)T^(-2)]`