Class 11th Physics Units and Measurement DIMENSIONAL ANALYSIS AND ITS APPLICATIONS

An expression for a dimensionless quantity P is given by $P = \frac{α}{β} l o g_{e} (\frac{k t}{β t});$ where a and b are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of a will be:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>[</mo> <mrow> <msup> <mrow> <mi>M</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msup> <msup> <mrow> <mi>L</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <msup> <mrow> <mi>T</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msup> </mrow> <mo>]</mo> </mrow> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>[</mo> <mrow> <mi>M</mi> <msup> <mrow> <mi>L</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msup> <msup> <mrow> <mi>T</mi> </mrow> <mrow> <mo>−</mo> <mn>2</mn> </mrow> </msup> </mrow> <mo>]</mo> </mrow> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>[</mo> <mrow> <mi>M</mi> <mi>L</mi> <msup> <mrow> <mi>T</mi> </mrow> <mrow> <mo>−</mo> <mn>2</mn> </mrow> </msup> </mrow> <mo>]</mo> </mrow> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>[</mo> <mrow> <mi>M</mi> <msup> <mrow> <mi>L</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msup> <mrow> <mi>T</mi> </mrow> <mrow> <mo>−</mo> <mn>2</mn> </mrow> </msup> </mrow> <mo>]</mo> </mrow> </mrow> </math> </span></p>

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Raj Pandey | Contributor-Level 9

4 months ago

Correct Option - 4

Detailed Solution:

$P = \frac{α}{β} l o g_{e} (\frac{k t}{β x})$

$\frac{k t}{β x}$ = Dimensionless

$β = \frac{k t}{x} = \frac{[M L^{2} T^{- 2} k^{- 1}] [k]}{[L]}$

$\frac{α}{β} =$ dimensionless

a = dimensionless of b

a = MLT^-2

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If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be:

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m = FTV¹

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The work done by a gas molecule in an isolated system is given by $W = α β^{2} e^{- \frac{x^{2}}{α k T}}$ , where x is the displacement, k is the Boltzmann constant and T is the temperature. a and b are constants. Then the dimensions of b will be:

Alok Kumar Singh

Since, $\frac{x^{2}}{α k T}$ should be dimensionless.

So, dimension of $α, [α] = \frac{L^{2}}{M L^{2} T^{- 2}} = M^{- 1} T^{2}$

Dimension of $α β^{2}$ should be that of W.

So, $[α β^{2}] = M L^{2} T^{- 2}$

$\Rightarrow [β^{2}] = \frac{M L^{2} T^{- 2}}{M^{- 1} T^{2}} = M^{2} L^{2} T^{- 4} \Rightarrow [β] = M L T^{- 2}$

If E and G respectively denotes energy and gravitational constant, then E/G has the dimensions of:

Alok Kumar Singh

The dimensional formula for energy E is [ML²T? ²].
The dimensional formula for the gravitational constant G is [M? ¹L³T? ²].
The ratio E/G has dimensions: [ML²T? ²] / [M? ¹L³T? ²] = [M²L? ¹T? ].

A quantity X is given by (IFv²MLL⁴) in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of:


x = (IFV²)/ (WL? )
I = [ML²]
F = [MLT? ²]
V² = [L² T? ²]
W = [ML² T? ²]
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X = [ML? ¹ T? ²]

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Physics Units and Measurement 2025

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