Units & Dimensions

JEE Main level, single correct, medium

28 of 149

Question 28 of 149

Match the LIST-I with LIST-II
\(\begin{array}{||c|l||c||l||} \hline & \textbf{List-I} & & \textbf{List-II} \\ \hline A. & \text{Spring constant} & I. & \mathrm{M L^{2} T^{-2} K^{-1}} \\ \hline B. & \text{Thermal conductivity} & II. & \mathrm{M L^{0} T^{-2}} \\ \hline C. & \text{Boltzmann constant} & III. & \mathrm{M L^{2} T^{-3} A^{-2}} \\ \hline D. & \text{Inductive reactance} & IV. & \mathrm{M L T^{-3} K^{-1}} \\ \hline \end{array}\)
Choose the correct answer from the options given below:

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Answer: (A) A-II, B-IV, C-I, D-III

A. Spring constant: F=kx⇒[k]=MLT−2/L=ML0T−2F = kx \Rightarrow [k] = MLT^{-2}/L = ML^0T^{-2} →\rightarrow II
B. Thermal conductivity: Q/t=KA(dT/dx)⇒[K]=ML2T−3/(L2⋅K/L)=MLT−3K−1Q/t = KA(dT/dx) \Rightarrow [K] = ML^2T^{-3}/(L^2 \cdot K/L) = MLT^{-3}K^{-1} →\rightarrow IV
C. Boltzmann constant: E=kBT⇒[kB]=ML2T−2K−1E = k_BT \Rightarrow [k_B] = ML^2T^{-2}K^{-1} →\rightarrow I
D. Inductive reactance: same dimension as resistance =ML2T−3A−2= ML^2T^{-3}A^{-2} →\rightarrow III
Matching: A-II, B-IV, C-I, D-III.

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