An expression for a dimensionless quantity P is given by \(P = \frac{α }{β }{\log _e}\left( {\frac{{kt}}{{β x}}} \right);\) where α and β are constants, x is distance; k is the Boltzmann constant and t is the temperature. Then the dimensions of α will be :

  1. [ML-1 T0]
  2. [M L0 T-2]
  3. [M L T-2]
  4. [M L2 T-2]

Answer (Detailed Solution Below)

Option 3 : [M L T-2]
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JEE Main 04 April 2024 Shift 1
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Detailed Solution

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CONCEPT:

  • Boltzmann's constant is the proportionality constant that relates to the average relative kinetic energy of the particles in a gas with the thermodynamics temperature of the gas. Its value is written as;

           k = 1.380649 \(\times\) 10-23 m2 kg s-2 K-1 

  • Temperature is defined as  the degree and intensity of the heat  and its dimensional formula is written as;

            t = [M0L0T0K1]

  • Distance is defined as the total movement of an object and  its dimensional formula is written as;

​           x =  [M0L1T0]

CALCULATION:

Given:

\(P = \frac{α }{β }{\log _e}\left( {\frac{{kt}}{{β x}}} \right)\)      -----(1)

Here, we have α and β constants, x is distance; k is the Boltzmann constant and t is the temperature.

The dimensional formula of Boltzmann constant, k = [M1L2T-2K-1]

The dimensional formula of P = [M0L0T0]

The dimensional formula of distance,x = [M0L1T0]

α and β are constants.

and the dimensional formula of temperature, t = [M0L0T0K1]

Now, let us put these values in equation (1) we have;

\(P = \frac{α }{β }{\log _e}\left( {\frac{{kt}}{{β x}}} \right)\)

⇒ [M0L0T0] = \(\frac{α}{β}\) loge \((\frac{[M^1L^2T^{-2}K^{-1}][M^0L^0T^0K^1]}{β [M^0L^1T^0]})\)

⇒  [M0L0T0] = \(\frac{α}{β}\) loge \((\frac{[M^1L^2T^{-2}K^{0}]}{β [M^0L^1T^0]})\)

⇒  [M0L0T0] = \(\frac{α}{β}\) loge \((\frac{[M^1L^1T^{-2}K^{0}]}{β })\)

⇒ \(\frac{β}{α}\) [M0L0T0] = loge \((\frac{[M^1L^1T^{-2}K^{0}]}{β })\)

Taking the exponential on both sides we get;

 loge \((\frac{[M^1L^1T^{-2}K^{0}]}{β })\) =  \(\frac{β}{α}\) [M0L0T0

⇒ \(\frac{[M^1L^1T^{-2}K^{0}]}{β }\) = \(e^ {\frac{β}{α} [M^0L^0T^0]}\)

\(e^ {\frac{β}{α} [M^0L^0T^0]}\) should be dimensionless therefore we have;

\(\frac{β}{α}\) =1

α = β      ------(2)

and \(\frac{[M^1L^1T^{-2}K^{0}]}{β }\) =1

⇒ β  = [MLT-2]

Now, from equation (2) we have;

→ α = [MLT-2]

Hence, option 3) is the correct answer.

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