KVPY SX SB Exam 20141 Question Paper

$\mathit{\sigma }={\mathit{h}}^{\mathit{\alpha }}{\mathit{k}}_{\mathit{B}}^{\mathit{\beta }}{\mathit{c}}^{\mathit{\gamma }}$
σ → Steffan boltzmann constant
h → Planck’s constant
kB → Boltzmann constant
c → speed of sight
According to stefan’s law
$\frac{\mathit{Q}}{\mathit{A}\mathit{t}}=\mathit{\sigma }{\mathit{T}}^4$
$\mathit{\sigma }=\frac{\mathit{Q}}{\mathit{A}\mathit{T}}\times \frac{1}{{\mathit{T}}^4}$
$\left[\mathit{\sigma }\right]=\frac{\left[{\mathit{M}}^3{\mathit{L}}^2{\mathit{T}}^{-2}\right]}{\left[{\mathit{L}}^2\right]\left[\mathit{T}\right]\left[{\mathit{K}}^4\right]}=\left[{\mathit{M}}^1{\mathit{T}}^{\mathit{–}3}{\mathit{K}}^{\mathit{–}4}\right]$
&E = hν
⇒ $\mathit{h}=\frac{\mathit{E}}{\mathit{\nu }}\Rightarrow \left[\mathit{h}\right]=\frac{\left[{\mathit{M}}^3{\mathit{L}}^2{\mathit{T}}^{-2}\right]}{\left[{\mathit{T}}^{-1}\right]}$
$\left[\mathit{h}\right]=\left[{\mathit{M}}^1{\mathit{L}}^2{\mathit{T}}^{\mathit{–}1}\right]$
$\left[\mathit{c}\right]=\left[\mathit{L}{\mathit{T}}^{\mathit{–}1}\right]$
$\mathit{E}=\frac{3}{2}{\mathit{k}}_{\mathit{B}}\mathit{T}$
⇒ $\left[{\mathit{k}}_{\mathit{B}}\right]=\frac{\left[{\mathit{M}}^1{\mathit{L}}^2{\mathit{T}}^{-2}\right]}{\left[\mathit{K}\right]}$
$\left[{\mathit{k}}_{\mathit{B}}\right]=\left[{\mathit{M}}^1{\mathit{L}}^2{\mathit{T}}^{\mathit{–}2}{\mathit{K}}^{\mathit{–}1}\right]$
According to homogeneity principle of dimension
$\left[{\mathit{M}}^1{\mathit{T}}^{\mathit{–}3}{\mathit{K}}^{\mathit{–}4}\right]={\left[{\mathit{M}}^1{\mathit{L}}^2{\mathit{T}}^{\mathit{–}1}\right]}^{\mathit{\alpha }}{\left[{\mathit{M}}^1{\mathit{L}}^2{\mathit{T}}^{\mathit{–}2}{\mathit{K}}^{\mathit{–}1}\right]}^{\mathit{\beta }}$ ${\left[{\mathit{L}}^1{\mathit{T}}^{\mathit{–}1}\right]}^{\mathit{\gamma }}$
on comparing powers 00000000 of M, L, T, K on both sides
⇒ α + β = 1 ….(1)
2α + 2β + γ = 0 ….(2)
– α – 2β – γ = – 3 ….(3)
– β = – 4 ….(4)
⇒ β = 4
α = – 3
Putting values is equation (2)
2 (– 3) + 2 (4) + γ = 0
γ = – 2
Ans :
α = – 3
β = 4
γ = – 2