Units & Dimensions

JEE Main level, single correct, medium

78 of 149

Question 78 of 149

The equation of a circle is given by x2+y2=a2{x}^{2}+{y}^{2}={a}^{2}, where aa is the radius. If the equation is modified to change the origin other than (0,0)(0,0), then find out the correct dimensions of AA and BB in a new equation :
(x−At)2+(y−tB)2=a2{\left(x-At\right)}^{2}+{\left(y-\dfrac{t}{B}\right)}^{2}={a}^{2}
The dimensions of tt is given as [T−1]\left[{T}^{-1}\right]

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Answer: (B) A=[LT],B=[L–1T–1]A=\left[LT\right],B=[{L}^{–1}T{}^{–1}]

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,
[At]=[x]⇒A×1T=L\left[At\right]=\left[x\right] \Rightarrow A\times \dfrac{1}{T}=L
∴[A]=T1L1∴\left[A\right]={T}^{1}{L}^{1}
For the second term,
[tB]=[y]\left[\dfrac{t}{B}\right]=\left[y\right]
⇒1T[B]=L\Rightarrow \dfrac{1}{T\left[B\right]}=L
⇒[B]=T−1L−1\Rightarrow \left[B\right]={T}^{-1}{L}^{-1}

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