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 to a point other than $(0,0)$,find the correct dimensions of $A$ and $B$ in the new equation: $(x-At)^2+(y-\frac{t}{B})^2=a^2$. The dimensions of $t$ are given as $[T^{-1}]$.
- A$A=[L^{-1}T], B=[LT^{-1}]$
- B$A=[LT], B=[L^{-1}T^{-1}]$
- C$A=[L^{-1}T^{-1}], B=[LT^{-1}]$
- D$A=[L^{-1}T^{-1}], B=[LT]$
Explore More
Similar Questions
$A$ quantity $f$ is given by $f = \sqrt{\frac{hc^5}{G}}$,where $c$ is the speed of light,$G$ is the universal gravitational constant,and $h$ is Planck's constant. The dimension of $f$ is that of:
Planck's constant $(h)$,speed of light in vacuum $(c)$,and Newton's gravitational constant $(G)$ are three fundamental constants. Which of the following combinations of these has the dimension of length?
DifficultNEET 2016
View SolutionThe physical quantity which has the dimensional formula ${M^1}{T^{ - 3}}$ is
Difficult
View SolutionIf $C$ is the velocity of light,$h$ is Planck's constant,and $G$ is the gravitational constant,and these are taken as fundamental quantities,then the dimensional formula of mass is:
Frequency $(n)$ is a function of density $(\rho)$,length $(a)$,and surface tension $(T)$. Then its value is:
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo