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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-$II$
$(1)$ Planck's constant	$(i)$ $[ML^{-1} T^{-2}]$
$(2)$ Gravitational constant	(ii) $[ML^{-1} T^{-1}]$
$(3)$ Bulk modulus	(iii) $[ML^2 T^{-1}]$
$(4)$ Coefficient of viscosity	(iv) $[M^{-1} L^3 T^{-2}]$

$A$ 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 $\text{ } \alpha$)?

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