JEE Main 24 Feb 2021 Shift 1 Solved Paper

Given, work done, W=α⋅β2e−‌

x2

akT

where, k is Boltzmann constant,
T is temperature and x is displacement.
We know that, ‌

x2

αkT

is a dimensionless quantity.
∴‌‌[‌

x2

αkT

]=[M0L0T0]⇒[α]=‌

[x2]

[k][T]

⇒‌‌[α]=‌

[L2]

[k][T]

Since, dimensions of k are
[k]=[M1L2T−2K−1] ... (i)
Dimensions of temperature are
[T]=[K]
Substituting Eqs. (ii) and (iii) in Eq. (i), we get
[α]‌‌=‌

[L2]

[M1L2T−2K−1][K]

[α]‌‌=[M−1T2]
According to dimensional analysis,
[W]‌=[αβ2]
⇒‌‌[β2]‌=‌

[W]

[α]

⇒‌‌[β2]‌=‌

[M1L2T−2]

[M−1T2]

=[M2L2T−4]
⇒‌‌[β]‌=[MLT−2]