The length of the cruiser is 200m. If the cruiser travels at a speed of $\left( {\dfrac{{\sqrt 3 }}{2}} \right)c$ past a planet. Calculate the length of the cruiser measured by the inhabitants of the planet.
(A) 0
(B) Between 0 and 200m
(C) 200m
(D) Greater than 200m
(E) None of the above since it is impossible to reach the described speed.

Hint
The cruiser is moving with a speed comparable to that of light. So from Einstein’s special theory of relativity, the inhabitants of the planet will find the length of the cruiser contracted than the original length and hence it will look shorter.

Formula Used In this solution, we will be using the following formula,
$L' = \dfrac{L}{\gamma }$
where $L'$ is the new length,
$L$ is the original length and $\gamma $ is the conversion factor given by $\gamma = \dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{v}{c}} \right)}^2}} }}$

Complete step by step answer
From Einstein’s special theory of relativity, when an observer in a frame which is at rest sees a body moving at a speed which is close to the speed of light, then the observer finds there is a contraction in the length of that body. Due to this, the length of the body looks much shorter than it is.
In this question, the length of the cruiser as observed in a rest frame is given by 200m. It is then said to be moving at a speed which is close to the speed of light which is given by, $\dfrac{{\sqrt 3 }}{2}c$. So when the cruiser passes by the planet, the inhabitants of the planet, who are supposed to be at rest watch it pass by with a speed close to the speed of light. So according to Einstein’s theory of relativity, length contraction will take place for the cruiser. So the cruiser will look shorter than it is to the inhabitants of the planet.
Therefore the length of the cruiser will be less than 200m.
Hence the length of the cruiser will be between 0 to 200m. So the correct answer will be option B.

Note
We can calculate the amount by which the cruiser will be shorter by using the relativistic factor, $\gamma = \dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{v}{c}} \right)}^2}} }}$
Here we can put $v = \dfrac{{\sqrt 3 }}{2}c$
Therefore, we get
$\gamma = \dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}} }}$ which gives a value of,
$\gamma = \dfrac{1}{{\sqrt {1 - \dfrac{3}{4}} }}$
So doing the calculation in the denominator we get the LCM as 4
$\gamma = \dfrac{1}{{\sqrt {\dfrac{{4 - 3}}{4}} }}$
Therefore we get,
$\gamma = \dfrac{1}{{\sqrt {\dfrac{1}{4}} }}$
So this gives us
$\gamma = \dfrac{1}{{\dfrac{1}{2}}}$
$ \Rightarrow \gamma = 2$
So the contracted length will be given by $L' = \dfrac{L}{\gamma } = \dfrac{{200}}{2} = 100m$.