QNEET-2024-Ph-146: NEET 2024 Physics - Class 12 NEET-2024-Ph-146 Solution: A uniform wire of diameter d carries a current of 100 mA when…

To solve this problem, we need to understand the relationship between the current, the drift velocity, and the cross-sectional area of the wire. The electric current I in a wire is given by I = n e A v d where: n is the number density of electrons, e is the charge of an electron, A is the cross-sectional area of the wire, and v d is the mean drift velocity of the electrons. Let's denote the current in the thicker wire as I 1 = 100   mA = 0.1   A and the current in the thinner wire as I 2 = 200   mA = 0.2   A . The diameter of the thicker wire is d , so its cross-sectional area, A 1 , is A 1 = � ( d 2 ) 2 = � d 2 4 For the thinner wire, the diameter is d 2 , so its cross-sectional area, A 2 , is A 2 = � ( d 4 ) 2 = � d 2 16 Using the relation for current in the thicker wire, we have I 1 = n e A 1 v = n e ( � d 2 4 ) v = 0.1   A Let the mean drift velocity in the thinner wire be denoted by v _ . For the thinner wire: I 2 = n e A 2 v _ = n e ( � d 2 16 ) v _ = 0.2   A Now, we can equate the expressions for current and solve for v _ : � d 2 16 n e v _ = 0.2   A v _ = 0.2   A n e ( � d 2 16 ) From the equation for the thicker wire: 0.1   A = � d 2 4 n e v v = 0.1   A n e ( � d 2 4 ) Dividing the two equations, we get v _ v = 0.2   A ( � d 2 16 ) n e _ ( � d 2 4 ) n e 0.1   A = 0.2   A _ 4 0.1   A _ 16 = 4 1 = 8 Thus, v _ = 8 v The mean drift velocity of electrons in the thinner wire is 8 times the mean drift velocity in the thicker wire. Therefore, the correct option is: Option B : 8 v