JEE Main 2021 · 20 July · morning · Single correct · +4 / −1
The entropy of any system is given by
S = α ²β ln [ (μ kR)/(Jβ ²) + 3 ] where α and β are the constants. μ, J, k and R are no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively.
[Take S = dQ/T]
Choose the incorrect option from the following :
- (a) α and J have the same dimensions.
- (b) S and α have different dimensions
- (c) S, β, k and μR have the same dimensions
- (d) α and k have the same dimensions
Answer and solution
Answer (d) α and k have the same dimensions
Solution
Physics sir · 4 steps
Whatever sits inside a log must be dimensionless. Start there and every dimension falls into place.
- Inside the log. A logarithm only accepts a pure number, so (μ k R)/Jβ² must be dimensionless. Here J is a ratio of joules to calories, so J itself is dimensionless.
- Dimension of beta. Both k and μ R have the dimensions of entropy, [ML²T⁻²K⁻¹], the same as S = dQ/T. So β² must carry the dimensions of S².
[β]² = [k][μ R]/[J] = [S]² ⇒ [β] = [S] = [ML²T⁻²K⁻¹]
- Dimension of alpha. The log is a pure number, so [S] = [α]²[β]. Since [β] = [S], this needs [α]² = 1, so α is dimensionless, just like J.
- So. Options (a), (b) and (c) are all true. α is dimensionless but k has the dimensions of entropy, so (d) is the incorrect statement.
The idea to keep. Arguments of log, exp and sin are always pure numbers. Use that first to pin down the unknown constants, then match the dimensions on both sides of the equation.
Answer from the NTA final answer key. Solution by PrepWiser.