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JEE Main 202329 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
29 January
Shift
Evening Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2023 · 29 January · Evening Shift · Physics · Units and Dimensions

The equation of a circle is given by x ^ 2 + y ^ 2 = a ^ 2 , where...

The equation of a circle is given by x2+y2=a2{x}^{2} + {y}^{2} = {a}^{2}, where aistheradius.Iftheequationismodifiedtochangetheoriginotherthana is the radius. If the equation is modified to change the origin other than( 0, 0 ),thenfindoutthecorrectdimensionsof, then find out the correct dimensions ofA and Binanewequation:</p><p>in a new equation :</p><p>{\left(x - At\right)}^{2} + {\left(y - \dfrac{t}{B}\right)}^{2} = {a}^{2}</p><p>Thedimensionsof</p><p>The dimensions oftisgivenas[T−1]is given as \left[{T}^{- 1}\right]

Options

  1. AA=[L–1T],B=[LT–1]A = [ {L}^{– 1}T ] , B = [ {LT}^{– 1} ]
  2. BA=[LT],B=[L–1T–1]A = \left[LT\right], B = [ {L}^{– 1}T{}^{– 1} ]
  3. CA=[L–1T–1],B=[LT–1]A = [ {L}^{– 1}T{}^{– 1} ] , B = [ {LT}^{– 1} ]
  4. DA=[L–1T–1],B=[LT]A = [ {L}^{– 1}{T}^{– 1} ] , B = [ LT ]

Correct answer

B. A=[LT],B=[L–1T–1]A = \left[LT\right], B = [ {L}^{– 1}T{}^{– 1} ]

Step-by-step solution

Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. Then,

(x−At)2+(y−tB)2=a2{\left(x - At\right)}^{2} + {\left(y - \dfrac{t}{B}\right)}^{2} = {a}^{2}

For first term,

\left[At\right] = \left[x\right] ⇒ A \times \dfrac{1}{T} = L</p><p></p><p>∴ \left[A\right] = {T}^{1}{L}^{1}</p><p>Forthesecondterm,</p><p>[tB]=[y]</p><p>For the second term,</p><p>\left[\dfrac{t}{B}\right] = \left[y\right]

⇒1T[B]=L⇒ \dfrac{1}{T\left[B\right]} = L

⇒[B]=T−1L−1⇒ \left[B\right] = {T}^{- 1}{L}^{- 1}

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