Question

The force \(F\) acting on a particle in terms of its distance \(x\) from a fixed point is given by \[ F=\frac{A}{B+x^{1.5}}. \] If the dimensional formula of \(AB\) is \[ [M^aL^bT^c], \] then the value of \(a+b+c\) is

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

Only quantities having the same dimensions can be added or subtracted. First determine the dimensions of the denominator, then use \[ \boxed{[A]=[F]\times[B].} \]

Answer 1

The Correct Option is B

Step 1: Find the dimensions of \(B\). Since \[ B+x^{1.5} \] is a sum, both terms must have the same dimensions. Therefore, \[ [B]=[x^{1.5}] =L^{3/2}. \]

Step 2: Find the dimensions of \(A\). Given, \[ F=\frac{A}{B+x^{1.5}}. \] Hence, \[ [A]=[F][B]. \] Now, \[ [F]=MLT^{-2}. \] Therefore, \[ [A] = (MLT^{-2})(L^{3/2}) = ML^{5/2}T^{-2}. \]

Step 3: Find the dimensions of \(AB\). \[ [AB] = [A][B] = \left(ML^{5/2}T^{-2}\right) \left(L^{3/2}\right) = ML^4T^{-2}. \] Thus, \[ a=1,\qquad b=4,\qquad c=-2. \] Hence, \[ a+b+c = 1+4-2 = 3. \] Therefore, \[ \boxed{3}. \] Thus, \[ \boxed{(B)} \] is the correct answer.