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JEE Main 201910 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2019 · 10 April · Morning Shift · Chemistry · Chemical Kinetics

A bacterial infection in an internal wound grows as N ^ ' (t) = N _...

A bacterial infection in an internal wound grows as N′(t)=N0exp(t){N}^{'}\left(t\right) = {N}_{0}exp ( t ) , where the time tt is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as dNdt=−5N2\dfrac{dN}{dt} = - 5{N}^{2} . What will be the plot of N0N\dfrac{{N}_{0}}{N} vs tt after 11 hour?A bacterial infection in an internal wound grows as N ^ ' (t) = N _ 0 exp ( t ) , where the time t is in hours. A doseA bacterial infection in an internal wound grows as N ^ ' (t) = N _ 0 exp ( t ) , where the time t is in hours. A doseA bacterial infection in an internal wound grows as N ^ ' (t) = N _ 0 exp ( t ) , where the time t is in hours. A doseA bacterial infection in an internal wound grows as N ^ ' (t) = N _ 0 exp ( t ) , where the time t is in hours. A dose

Options

  1. AFour particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of
  2. BFour particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of
  3. CFour particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of
  4. DFour particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of

Correct answer

A. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of

Step-by-step solution

From 0 to 1 hour, N′=N0et{N}^{'} = {N}_{0}{e}^{t}
From 1 hour onwards dNdt=−5N2\dfrac{dN}{dt} = - 5{N}^{2}
So at t=1t = 1 hour, N′=eN0{N}^{'} = e{N}_{0}
dNdt=−5N2\dfrac{dN}{dt} = - 5{N}^{2}
∫eN0NN−2dN=−5∫1tdt{∫}_{e{N}_{0}}^{N}{N}^{- 2}dN = - 5{∫}_{1}^{t}dt
1N−1eN0=5(t−1)\dfrac{1}{N} - \dfrac{1}{e{N}_{0}} = 5\left(t - 1\right)
N0N−1e=5N0(t−1)\dfrac{{N}_{0}}{N} - \dfrac{1}{e} = 5{N}_{0} ( t - 1 )
N0N=5N0(t−1)+1e\dfrac{{N}_{0}}{N} = 5{N}_{0} ( t - 1 ) + \dfrac{1}{e}
N0N=5N0t+(1e−5N0)\dfrac{{N}_{0}}{N} = 5{N}_{0}t + \left(\dfrac{1}{e} - 5{N}_{0}\right)
Which is following y=mx+Cy = mx + C
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