The earth is considered as a sphere of radius r and density ρ, the gravitational acceleration at a point on the surface of the earth is given as g = krρ where k is a constant and g is the gravitational acceleration at the poles. The dimension of k is:

  1. M1 L1 T-2
  2. M1 L-1 T1
  3. M1 L-3 T2
  4. M-1 L3 T-2

Answer (Detailed Solution Below)

Option 4 : M-1 L3 T-2
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Detailed Solution

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CONCEPT:

Dimensions:

  • Dimensions of a physical quantity are the powers to which the fundamental units are raised to obtain one unit of that quantity.

Principle of homogeneity of dimensions:

  • According to this principle, a physical equation will be dimensionally correct if the dimensions of all the terms occurring on both sides of the equation are the same.
  • This principle is based on the fact that only the physical quantities of the same kind can be addedsubtracted, or compared.
  • Thus, velocity can be added to velocity but not to force.

EXPLANATION:

The dimension of g is given as,

⇒ [g] = [M0 L1 T-2]     -----(1)

The dimension of r is given as,

⇒ [r] = [M0 L1 T0]     -----(2)

The dimension of density ρ is given as,

⇒ [ρ] = [M1 L-3 T0]     -----(3)

Let the dimension of k be,

⇒ [k] = [Mx Ly Tz]     -----(4)

  • We know that only the physical quantities of the same kind can be addedsubtracted, or compared.
  • Therefore the dimension of krρ must be equal to the dimension of g.

So the dimension of krρ is given as,

⇒ [krρ] = [g]     -----(5)

By equation 1, equation 2, equation 3, equation 4 and equation 5,

⇒ [Mx Ly Tz]×[M0 L1 T0]×[M1 L-3 T0] = [M0 L1 T-2]

⇒ [Mx+1 Ly-2 Tz] = [M0 L1 T-2]     -----(6)

By comparing LHS and RHS of equation 6,

⇒ x + 1 = 0

⇒ x = -1

⇒ y - 2 = 1

⇒ y = 3

⇒ z = -2

So the dimension of k is given as,

⇒ [k] = [M-1 L3 T-2]

  • Hence, option 4 is correct.
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