If energy (E), momentum (p) and force (F) are chosen as fundamental units. The dimensions of mass in new system is

Options

A. `[E^(-1)p^(3)]`
B. `[E^(-1)p^(2)]`
C. `[E^(-2)p^(2)]`
D. `[E^(-1)p]`

Answer

B

Explanation

The dimensions of E, p and F in terms of M, L and T are `[E]= [ML^(2)T^(-2)], [p]= [MLT^(-1)], [F]= [MLT^(-2)]`
Let `m = kE^(a)p^(b)F^(c)` where k is a dimensionless constant.
`:. [M]= [ML^(2)T^(-2)]^(a)[MLT^(-1)]^(b)[MLT^(-2)]^(c)`
or `[ML^(0)T^(0)]= [M^(a+b+c)L^(2a+b+c)T^(-2a-b-2c)]`
Equating the powers of M, L and T we get
a + b +c= 1, 2a + b + c = 0, -2a - b- 2c =0
Solving, we get a=-1, b= 2 and c=0
Hence `[m] = [E^(-1)p^(2)]`