Nuclei of radioactive element {A} are being produced at rate {t}^{2} at time {t}. The element {A} has decay constant equal to \\lambda. If {N} is the number of active nuclei of element {A} at any time {t} then {{d} {N}}/{{dt}} was found to be minimum at t=t_{0} . N at time t=t_{0} is

Correct Answer is : 2t0−λt02λ2

{{t}_{0}-{xt}_{0}^{2}}/{{x}^{2}}

{{t}_{0}-2 \\lambda{t}^{2}}/{\\lambda^{2}}

{4 t_{0}-x t_{0}^{2}}/{\\lambda^{2}}

{2 {t}_{0}-\\lambda {t}_{0}^{2}}/{\\lambda^{2}}

JEE Advanced 2021 Full Test 4 Paper 1

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PART – I

Physics

SECTION – A (One Options Correct Type)
This section contains 06 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct.

Question : 4 of 54

Marks: +1, -0

Nuclei of radioactive element A are being produced at rate t2 at time t. The element A has decay constant equal to λ. If N is the number of active nuclei of element A at any time t then

was found to be minimum at t=t0.N at time t=t0 is

t0−xt02
x2
t0−2λt2
λ2
4t0−xt02
λ2
2t0−λt02
λ2

Validate

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