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NEET
Exam
NEET
Subject
Physics
Chapter
Units and Dimensions

NEET · Physics · Units and Dimensions

A physical quantity X is defined by the relation X = (a b^2)/(c +...

A physical quantity XX is defined by the relation X=ab2c+d2X = \dfrac{a b^2}{c + d^2}. It is given that XX has the dimensions of pressure, aa has the dimensions of force, and cc has the dimensions of length. The dimensional formula of the ratio bd\dfrac{b}{d} is:

Options

  1. A[L−3/2]\left[\mathrm{L}^{-3/2}\right]
  2. B[M0L0T0]\left[\mathrm{M}^0 \mathrm{L}^0 \mathrm{T}^0\right]
  3. C[L]\left[\mathrm{L}\right]
  4. D[L−1]\left[\mathrm{L}^{-1}\right]

Correct answer

D. [L−1]\left[\mathrm{L}^{-1}\right]

Step-by-step solution

According to the principle of homogeneity of dimensions, quantities added together must have the same dimensions. From the denominator (c+d2)(c + d^2):
[d2]=[c]=[L][d^2] = [c] = [\mathrm{L}]
[d]=[L1/2][d] = [\mathrm{L}^{1/2}]

The dimensional formula of the entire equation is:
[X]=[a][b2][c][X] = \dfrac{[a][b^2]}{[c]}

We are given that XX is pressure and aa is force. Substituting their dimensional formulas:
[ML−1T−2]=[MLT−2][b2][L][\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}] = \dfrac{[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}][b^2]}{[\mathrm{L}]}

Simplifying the right side:
[ML−1T−2]=[MT−2][b2][\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}] = [\mathrm{M} \mathrm{T}^{-2}][b^2]

Solving for [b2][b^2]:
[b2]=[ML−1T−2][MT−2]=[L−1][b^2] = \dfrac{[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}]}{[\mathrm{M} \mathrm{T}^{-2}]} = [\mathrm{L}^{-1}]
[b]=[L−1/2][b] = [\mathrm{L}^{-1/2}]

The dimensional formula of the ratio bd\dfrac{b}{d} is:
\left[\dfrac{b}{d}\right] = \dfrac{[\mathrm{L}^{-1/2}]}{[\mathrm{L}^{1/2}]} = [\mathrm{L}^{-1}]<br><br>Answer:[L−1]<br><br>Answer: \left[\mathrm{L}^{-1}\right]
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