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

If time (t), velocity $(υ)$ , and angular momentum $(l)$ are taken as the fundamental units. Then the dimension of mass (m) in terms of t, $υ, a n d l$ is:

Option 1 -

$[t^{- 2} υ^{- 1} l^{1}]$

Option 2 -

$[t^{- 1} υ^{1} l^{- 2}]$

Option 3 -

$[t^{- 1} υ^{- 2} l^{1}]$

Option 4 -

$[t^{1} υ^{2} l^{- 1}]$

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Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

Let Mass α $[t^{a} υ^{b} l^{c}] \Rightarrow [M^{1} L^{0} T^{0}] \equiv T^{a} \times {[L T^{- 1}]}^{b} \times {[M L^{2} T^{- 1}]}^{c}$

=> a – b – c = 0 c = 1, and b + 2c = 0 Þ b = -2c = -2

=> a = b + c = 1 – 2 = -1

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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:

Raj Pandey

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

$\frac{k t}{β x}$ = 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

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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Using F = MA = $m \frac{V}{T}$ $\frac{}{}$

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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:

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The dimensional formula for energy E is [ML²T? ²].
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If time (t), velocity $(υ)$ , and angular momentum $(l)$ are taken as the fundamental units. Then the dimension of mass (m) in terms of t, $υ, a n d l$ is:

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If time (t), velocity ( υ ) , and angular momentum ( l ) are taken as the fundamental units. Then the dimension of mass (m) in terms of t, υ , a n d l is:

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If time (t), velocity $(υ)$ , and angular momentum $(l)$ are taken as the fundamental units. Then the dimension of mass (m) in terms of t, $υ, a n d l$ is: