Match List-I with List-II: List-I (Electric field) List-II (Mobility) (A) Low electric field (I) Mobility decreases by $ \frac{1}{E} $ (B) Medium electric field (II) Mobility decreases by $ \frac{1}{\sqrt{E}} $ (C) High electric field (III) Mobility remains constant Choose the correct answer:

(A) – (III), (B) – (I), (C) – (II)

Match List-I with List-II: List-I (Electric field) List-II (Mobility) (A) Low electric field (I) Mobility decreases by $$ \frac{1}{E} $$ (B) Medium electric field (II) Mobility decreases by $$ \frac{1}{\sqrt{E}} $$ (C) High electric field (III) Mobility remains constant Choose the correct answer:}

(A) – (III), (B) – (I), (C) – (II)

(A) – (III), (B) – (II), (C) – (I)

(A) – (I), (B) – (III), (C) – (II)

(A) – (II), (B) – (III), (C) – (I)

(A) – (III), (B) – (I), (C) – (II)

List-I (Electric field)	List-II (Mobility)
(A) Low electric field	(I) Mobility decreases by \( \frac{1}{E} \)
(B) Medium electric field	(II) Mobility decreases by \( \frac{1}{\sqrt{E}} \)
(C) High electric field	(III) Mobility remains constant

Match List-I with List-II

List-I	List-II
(A) Faraday’s Law	(ii) $ \vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} $
(B) Conservation Law	(iii) $ \vec{\nabla} \times \vec{A} = 0 $
(C) Ohm’s Law	(iv) $ \vec{J} = \sigma \vec{E} $
(D) Gauss’s Law	(i) $ \frac{\rho}{\epsilon_0} $

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List-I	List-II
(A) Faraday’s Law	(ii) \( \vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \)
(B) Conservation Law	(iii) \( \vec{\nabla} \times \vec{A} = 0 \)
(C) Ohm’s Law	(iv) \( \vec{J} = \sigma \vec{E} \)
(D) Gauss’s Law	(i) \( \frac{\rho}{\epsilon_0} \)