Question

a physical quantity is given by h=fv×v/l, find the dimension of h

Answer 1

Answer:

LT^-3

Explanation:

F=[MLT^-2]

V=[LT^-1]

L=[ML^2T^-1]

FV^2/L = [MLT^-2] [LT^-1]^2/[ML^2T^-1]

=MLT^-2L^2T^-2/ML^2T^-1

=ML^3T^-4/ML^2T^-1

=LT^-3

Answer 2

Given

h = $\frac{fv^{2} }{L}$

Where , f denotes Force.

              v denotes velocity.

              L denotes Angular momentum.

Therefore dimensions of h is calculated as:

h =$\frac{[MLT^{-2} ][LT^{-1}] ^{2} }{[ML^{2}T^{-1} ]}$

Simplifying the above equation, we get:

h = $[M^{0} LT^{-3} ]$

Therefore, the dimension of h is $[M^{0} LT^{-3} ]$

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