The equation of a circle is given by x2+y2=a2, 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-(t/B))2=a2. The dimensions of t is given as [ T -1].

Question

The equation of a circle is given by x2+y2=a2, 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-(t/B))2=a2. The dimensions of t is 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 ]

Tardigrade Admin · Accepted Answer

( x - At )2+( y -( t / B ))2= a 2 [ At ]= A × (1/ T )= L ∴ [ A ]= T 1 L 1 ( t / B ) is in meters ∴ (1/ T [ B ])= L 2 ∴ [ B ]= T -1 L -1

$A = [L^{- 1} T], B = [L T^{- 1}]$

$A = (L T], B = [L^{- 1} T^{- 1}]$

$A = [L^{- 1} T^{- 1}], B = [L T^{- 1}]$

$A = [L^{- 1} T^{- 1}], B = [L T]$

$(x - A t)^{2} + (y - \frac{t}{B})^{2} = a^{2}$
$[A t] = A \times \frac{1}{T} = L$
$∴ [A] = T^{1} L^{1}$
$\frac{t}{B}$ is in meters
$∴ \frac{1}{T [ B ]} = L^{2}$
$∴ [B] = T^{- 1} L^{- 1}$

Q. The equation of a circle is given by x2+y2=a2, 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−At)2+(y−Bt​)2=a2. The dimensions of t is given as [T−1].

A=[L−1T],B=[LT−1]

A=(LT],B=[L−1T−1]

A=[L−1T−1],B=[LT−1]

A=[L−1T−1],B=[LT]

(x−At)2+(y−Bt​)2=a2 [At]=A×T1​=L ∴[A]=T1L1 Bt​ is in meters ∴T[B]1​=L2 ∴[B]=T−1L−1

$A = [L^{- 1} T], B = [L T^{- 1}]$

$A = (L T], B = [L^{- 1} T^{- 1}]$

$A = [L^{- 1} T^{- 1}], B = [L T^{- 1}]$

$A = [L^{- 1} T^{- 1}], B = [L T]$

$(x - A t)^{2} + (y - \frac{t}{B})^{2} = a^{2}$
$[A t] = A \times \frac{1}{T} = L$
$∴ [A] = T^{1} L^{1}$
$\frac{t}{B}$ is in meters
$∴ \frac{1}{T [ B ]} = L^{2}$
$∴ [B] = T^{- 1} L^{- 1}$