JEE Main 2023 · 30 January · morning · Single correct · +4 / −1
The charge flowing in a conductor changes with time as Q(t)=α t-β t²+γ t³. Where α, β and γ are constants. Minimum value of current is :
- (a) β-α²/(3 γ)
- (b) α-(3 β²)/γ
- (c) α-β²/(3 γ)
- (d) α-γ²/(3 β)
Answer and solution
Answer (c) α-β²/(3 γ)
Solution
Physics sir · 3 steps
Current is dQ/dt, and the lowest current comes where dI/dt is zero, so you differentiate twice.
- Current from charge. Current is the rate of flow of charge, so I = dQ/dt = α - 2β t + 3γ t².
- Find the turning point. Set dI/dt = -2β + 6γ t = 0, which gives t = β/3γ. Since d²I/dt² = 6γ > 0 (taking γ positive), this point is a minimum.
Iₘᵢₙ = α - 2β(β/3γ) + 3γ(β/3γ)² = α - 2β²/3γ + β²/3γ
- So. Iₘᵢₙ = α - β²/3γ, which is option (c).
The idea to keep. For the maximum or minimum of any quantity, set its first derivative to zero, and use the sign of the second derivative to tell which one it is. A quick dimension check also removes impossible options.
Answer from the NTA final answer key. Solution by PrepWiser.