Units And Dimensions Question 6
Question 6 - 29 January - Shift 2
The equation of a circle is given by $x^{2}+y^{2}=a^{2}$, where $a$ is the radius. If the equation is modified to change the origin other than $(0,0)$, then find out the correct dimensions of $A$ and $B$ in a new equation: $(x-A t)^{2}+(y-\frac{t}{B})^{2}=a^{2}$.
The dimensions of $t$ is given as $[T^{-1}]$.
(1) $A=[L^{-1} T], B=[LT^{-1}]$
(2) $A=(LT], B=[L^{-1} T^{-1}]$
(3) $A=[L^{-1} T^{-1}], B=[LT^{-1}]$
(4) $A=[L^{-1} T^{-1}], B=[LT]$
Show Answer
Answer: (2)
Solution:
Formula: Principle of homogeneity of dimensions
$(x-A t)^{2}+(y-\frac{t}{B})^{2}=a^{2}$
$[At]=A \times \frac{1}{T}=L$
$\therefore[A]=T^{1} L^{1}$
$\frac{t}{B}$ is in meters
$\therefore \frac{1}{T[B]}=L$
$\therefore \quad[B]=T^{-1} L^{-1}$
$\therefore$ Correct Ans. (2)
