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(B) The speed of light in vacuum is denoted by $c$.
According to the problem, a new unit of length $(\alpha)$ is defined such that $1 \alpha = c (	e

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$400$

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(B) The speed of light in vacuum is denoted by $c$.
According to the problem, a new unit of length $(\alpha)$ is defined such that $1 \alpha = c (	ext{speed of light})$.
This implies that in this new system of units, the speed of light is $1 \alpha/	ext{s}$.
The time taken $t$ is $6 	ext{ min } 40 	ext{ s}$.
Converting time into seconds: $t = (6 	imes 60) 	ext{ s} + 40 	ext{ s} = 360 	ext{ s} + 40 	ext{ s} = 400 	ext{ s}$.
The distance $d$ is given by the formula $d = 	ext{speed} 	imes 	ext{time}$.
Substituting the values: $d = (1 \alpha/	ext{s}) 	imes (400 	ext{ s}) = 400 \alpha$.

List $I$	List $II$
$A$. Torque	$I$. $[M^1 L^1 T^{-2} A^{-2}]$
$B$. Magnetic field	$II$. $[L^2 A^1]$
$C$. Magnetic moment	$III$. $[M^1 T^{-2} A^{-1}]$
$D$. Permeability of free space	$IV$. $[M^1 L^2 T^{-2}]$

$A$ new unit $(\alpha)$ of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of $\alpha$ units if light takes $6 \text{ min } 40 \text{ s}$ to cover this distance (in $\text{ } \alpha$)?

Similar Questions

Match List $I$ with List $II$:

List $I$ List $II$

$A$. Torque $I$. $[M^1 L^1 T^{-2} A^{-2}]$

$B$. Magnetic field $II$. $[L^2 A^1]$

$C$. Magnetic moment $III$. $[M^1 T^{-2} A^{-1}]$

$D$. Permeability of free space $IV$. $[M^1 L^2 T^{-2}]$

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