Q121 / 160 · TS EAMCET 2005 · Physics

Names of units of some physical quantities are given in List-I and their dimensions formulae are given in List-Il. Match the correct pairs in the lists $\begin{array}{ll|ll}<br />\hline & \text{ List-I } & \text{ List-II } \\<br />\hline A. & \mathrm{Pa}-\mathrm{s} & (i) & {\left[\mathrm{L}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]} \\<br />\hline B. & \mathrm{NmK}^{-1} & (ii) & {\left[\mathrm{MLT}^{-3} \mathrm{~K}^{-1}\right]} \\<br />\hline C. & \mathrm{J} \mathrm{kg}^{-1} \mathrm{~K}^{-1} & (iii) & {\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]} \\<br />\hline D. & \mathrm{Wm}^{-1} \mathrm{~K}^{-1} & (iv) & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]} \\<br />\hline<br />\end{array}$ $\begin{array}{llll}A & B & C & D\end{array}$

  • A
    (iv) (iii) (i) (ii)
  • B
    (iii) (ii) (iv) (i)
  • C
    (iii) (i) (iv) (ii)
  • Correct answer
    D
    (iii) (iv) (i) (ii)

Solution

Dimensions of Pa-s is $\begin{aligned} & =\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \cdot[\mathrm{T}] \\ & =\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] \end{aligned}$ Dimensions of $\mathrm{NmK}^{-1}$ is $\begin{aligned} & =\left[\mathrm{MLT}^{-2}\right][\mathrm{L}][\mathrm{K}]^{-1} \\ & =\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right] \end{aligned}$ Dimensions of $\mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ $\begin{aligned} & =\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right][\mathrm{M}]^{-1}[\mathrm{~K}]^{-1} \\ & =\left[\mathrm{L}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right] \end{aligned}$ Dimensions of $\mathrm{Wm}^{-1} \mathrm{~K}^{-1}$ $\begin{aligned} & =\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right][\mathrm{L}]^{-1}[\mathrm{~K}]^{-1} \\ & =\left[\mathrm{MLT}^{-3} \mathrm{~K}^{-1}\right] \end{aligned}$ So, the correct matching is (iii), (iv), (i), (ii), hence the answer is (d).

Hence, the correct answer is option: 'D'.

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