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JEE Main 20257 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2025 · 7 April · Morning Shift · Chemistry · Chemical Kinetics

A person's wound was exposed to some bacteria and then bacteria...

A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay (r) was found to be proportional with the square of the existing number of bacteria at any instance. Which of the following set of graphs correctly represents the 'before' and 'after' situation of the application of the medicine?[Given : N=\mathrm{N}= No. of bacteria, t=\mathrm{t}= time, bacterial growth follows Ist I^{\text{st }} order kinetics.]A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The woundA person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The woundA person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The woundA person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The woundA person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound

Options

  1. AAssertion: Change in colour of acidic solution of potassium dichromate by breath is used to test drunken drivers.
  2. BAssertion: Change in colour of acidic solution of potassium dichromate by breath is used to test drunken drivers.
  3. CAssertion: Change in colour of acidic solution of potassium dichromate by breath is used to test drunken drivers.
  4. DAssertion: Change in colour of acidic solution of potassium dichromate by breath is used to test drunken drivers.

Correct answer

B. Assertion: Change in colour of acidic solution of potassium dichromate by breath is used to test drunken drivers.

Step-by-step solution

*Before applying medicine
dAdt=K[A]\frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{K}[\mathrm{A}] (First order growth) (Rate law)
A A0=N N0=eKt\frac{\mathrm{A}}{\mathrm{~A}_0}=\frac{\mathrm{N}}{\mathrm{~N}_0}=\mathrm{e}^{\mathrm{Kt}}
*After applying medicine
A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound
r=−dAdt=K[A]2\mathrm{r}=-\frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{K}[\mathrm{A}]^2 \quad (Rate law)
y=Kx2y=\mathrm{Kx}^2 Parabola
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