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JEE Main 202225 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Evening Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2022 · 25 June · Evening Shift · Physics · Atomic Physics

Which of the following figure represents the variation of ( R R _ 0...

Which of the following figure represents the variation of ln⁡(RR0)\ln \left(\dfrac{R}{{R}_{0}}\right) with lnAlnA (if R=R = radius of a nucleus and A=A = its mass number)Which of the following figure represents the variation of ( R R _ 0 ) with lnA (if R = radius of a nucleus and A = itsWhich of the following figure represents the variation of ( R R _ 0 ) with lnA (if R = radius of a nucleus and A = itsWhich of the following figure represents the variation of ( R R _ 0 ) with lnA (if R = radius of a nucleus and A = itsWhich of the following figure represents the variation of ( R R _ 0 ) with lnA (if R = radius of a nucleus and A = its

Options

  1. ALet C _ 1 and C _ 2 be the centres of the circles x ^ 2 + y ^ 2 - 2x - 2y - 2 = 0 and x ^ 2 + y ^ 2 - 6x - 6y + 14 = 0
  2. BLet C _ 1 and C _ 2 be the centres of the circles x ^ 2 + y ^ 2 - 2x - 2y - 2 = 0 and x ^ 2 + y ^ 2 - 6x - 6y + 14 = 0
  3. CLet C _ 1 and C _ 2 be the centres of the circles x ^ 2 + y ^ 2 - 2x - 2y - 2 = 0 and x ^ 2 + y ^ 2 - 6x - 6y + 14 = 0
  4. DLet C _ 1 and C _ 2 be the centres of the circles x ^ 2 + y ^ 2 - 2x - 2y - 2 = 0 and x ^ 2 + y ^ 2 - 6x - 6y + 14 = 0

Correct answer

B. Let C _ 1 and C _ 2 be the centres of the circles x ^ 2 + y ^ 2 - 2x - 2y - 2 = 0 and x ^ 2 + y ^ 2 - 6x - 6y + 14 = 0

Step-by-step solution

We know the relation between radius and atomic mass number is R=R0A13R = {R}_{0}A\dfrac{1}{3} ⇒lnR=lnR0+13lnA⇒ lnR = ln{R}_{0} + \dfrac{1}{3}lnA

⇒ln⁡(RR0)=13lnA⇒ \ln \left(\dfrac{R}{{R}_{0}}\right) = \dfrac{1}{3}lnA
It is similar to y=mxy = mx which is equation of line passing through origin.

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