Question

If speed V, area A and force F are chosen as fundamental units, then the dimension of Young's modulus will be :

Options

• FA²V^-3

• FA²V^-1

• FA²V^-2

• FA^-1V⁰

Answer 1

FA^-1V⁰

Explanation:

Given: The fundamental units are, speed V, area A and force F.

To find: The dimensions of Young's modulus (Y) in the new system of fundamental units.

Dimensions of Young's modulus:

y = [ML^-1T^-2]               ... (i)

Dimensions for force :

[F] = [MLT^-2]               ... (ii)

Dimensions for area :

[A] =  [L²]                ... (iii)

Dimensions for velocity :

[V] = [LT^-1]            ... (iv)

Let the dimensions for Young's modulus in the new system of fundamental units be :

[Y] = [F]^a [A]^b[V]^c    ... (v)

Using equations (i), (ii), (iii), (iv) and (v) :

[ML^-1T^-2] = [MLT^-2]^a [L²]^b [LT^-1]^c

Compare the exponents.

a = 1

a + 2b + c = -1

-2a - c = -2

On solving the above set of equations we get:

a = 1, b = -1, c = 0

That gives,

[Y] = [F]¹ [A]^-1[V]⁰