Q121 / 160 · AP EAMCET 2024 (18 May Shift 1) · Physics

Match the following:
Figure

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

Solution

$\frac{Q}{t}=\frac{k A\left(\theta_1-\theta_2\right)}{1}$ $\Rightarrow\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right]=[\mathrm{k}][\mathrm{L}][\mathrm{K}]$ $\therefore \quad[\mathrm{k}]=\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-3} \mathrm{~K}^{-1}\right]$ $\mathrm{p}^{\mathrm{V}}=\mathrm{N} \mathrm{k}_{\mathrm{B}} \mathrm{T} \Rightarrow\left[\mathrm{k}_{\mathrm{B}}\right]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\mathrm{K}]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]$ $\begin{aligned} & \mathrm{Q}=\mathrm{mL} \Rightarrow[\mathrm{L}]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\mathrm{M}]}=\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right] \\ & \mathrm{Q}=\mathrm{ms} \Delta \theta \Rightarrow[\mathrm{S}]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\mathrm{M}][\mathrm{K}]}=\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]\end{aligned}$

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

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