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A rod flies with constant velocity past a mark which is stationary in the reference frame $K$. In the frame $K$ it takes $\Delta t = 20\;ns$for the rod to fly past the mark. In the reference frame fixed to the rod the mark moves past the rod for $\Delta t' = 25\;ns$. Find the proper length of the rod.

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Last updated date: 15th Jul 2024
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Answer
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Hint: The rod moves with constant velocity and the image of the rod took $20\;ns$ to move past the reference frame. And the reference frame took $25\;ns$ to move past the rod. Hence, by using the length contraction formula, the original length of the rod can be calculated. The length contraction formula relates the original length and the observed length of the rod.

Useful formula:
The length contraction formula,
$l = {l_0}\sqrt {1 - {{\left( {\dfrac{v}{c}} \right)}^2}} $
Where, $l$ is the observed length of the rod, ${l_0}$ is the proper length of the rod, $v$ is the velocity of the rod and $c$ is the speed of the light.

Given data:
$\Delta t = 20\;ns$
$\Delta t' = 25\;ns$
The speed of light, $c = 3 \times {10^8}\;m{s^{ - 1}}$

Complete step by step solution:
In the frame $K$, the length observed of the rod $l$ related with $\Delta t$ is given by,
$l = v\Delta t$
Where, $v$ is the velocity of the rod.

In the reference frame fixed to the rod, the proper length of the rod ${l_0}$ related with $\Delta t'$ is given by,
${l_0} = v\Delta t'$

By the length contraction formula,
$l = {l_0}\sqrt {1 - {{\left( {\dfrac{v}{c}} \right)}^2}} $
Taking the term inside the square root to left hand side, we get
${l_0} = \dfrac{l}{{\sqrt {1 - {{\left( {\dfrac{v}{c}} \right)}^2}} }}$
Substituting the values of $l$ and ${l_0}$ in above equation, we get
$
  v\Delta t' = \dfrac{{v\Delta t}}{{\sqrt {1 - {{\left( {\dfrac{v}{c}} \right)}^2}} }} \\
  \sqrt {1 - {{\left( {\dfrac{v}{c}} \right)}^2}} = \dfrac{{v\Delta t}}{{v\Delta t'}} \\
 $
Cancel the term $v$ in right hand side at numerator and denominator,
$\sqrt {1 - {{\left( {\dfrac{v}{c}} \right)}^2}} = \dfrac{{\Delta t}}{{\Delta t'}}$
Squaring on both sides, we get
${\left( {\sqrt {1 - {{\left( {\dfrac{v}{c}} \right)}^2}} } \right)^2} = {\left( {\dfrac{{\Delta t}}{{\Delta t'}}} \right)^2}$
By cancelling square and square root, we get
$
  1 - {\left( {\dfrac{v}{c}} \right)^2} = {\left( {\dfrac{{\Delta t}}{{\Delta t'}}} \right)^2} \\
  1 - {\left( {\dfrac{{\Delta t}}{{\Delta t'}}} \right)^2} = {\left( {\dfrac{v}{c}} \right)^2} \\
 $
Taking square root on both sides, we get
$\sqrt {1 - {{\left( {\dfrac{{\Delta t}}{{\Delta t'}}} \right)}^2}} = \sqrt {{{\left( {\dfrac{v}{c}} \right)}^2}} $
By cancelling square and square root, we get
$
  \sqrt {1 - {{\left( {\dfrac{{\Delta t}}{{\Delta t'}}} \right)}^2}} = \left( {\dfrac{v}{c}} \right) \\
  c \times \sqrt {1 - \left( {\dfrac{{\Delta {t^2}}}{{\Delta {{t'}^2}}}} \right)} = v \\
 $
By taking LCM inside the square root,
$c \times \sqrt {\left( {\dfrac{{\Delta {{t'}^2} - \Delta {t^2}}}{{\Delta {{t'}^2}}}} \right)} = v$
Taking out the term $\Delta t'$ by cancelling the square and square root, we get
$
  \dfrac{c}{{\Delta t'}} \times \sqrt {\left( {\Delta {{t'}^2} - \Delta {t^2}} \right)} = v \\
  c \times \sqrt {\left( {\Delta {{t'}^2} - \Delta {t^2}} \right)} = v\Delta t' \\
 $
Since, ${l_0} = v\Delta t'$
Then substitute ${l_0}$ in the place of $v\Delta t'$ in the above equation, we get
$c \times \sqrt {\left( {\Delta {{t'}^2} - \Delta {t^2}} \right)} = {l_0}$
Substitute the given values in above equation, we get
${l_0} = 3 \times {10^8}\;m{s^{ - 1}} \times \sqrt {\left( {{{\left( {25\;ns} \right)}^2} - {{\left( {20\;ns} \right)}^2}} \right)} $
Since, the value of nano second is $ns = {10^{ - 9}}\;s$
Substitute the value in the above equation, we get
$
  {l_0} = 3 \times {10^8}\;m{s^{ - 1}} \times \sqrt {\left( {{{\left( {25 \times {{10}^{ - 9}}\;s} \right)}^2} - {{\left( {20 \times {{10}^{ - 9}}\;s} \right)}^2}} \right)} \\
  {l_0} = 3 \times {10^8} \times \sqrt {\left( {\left( {6.25 \times {{10}^{ - 16}}} \right) - \left( {4 \times {{10}^{ - 16}}} \right)} \right)} \\
 $
Taking arithmetic operation inside the square root,
${l_0} = 3 \times {10^8} \times \sqrt {2.25 \times {{10}^{ - 16}}} $
Taking square root to the term inside the root, we get
$
  {l_0} = 3 \times {10^8} \times 1.5 \times {10^{ - 8}} \\
  {l_0} = 4.5\;m \\
 $

Hence, the proper length of the rod is ${l_0} = 4.5\;m$

Note: The proper length of the rod must be larger value than the observed length of the rod in the reference plane. This is because of the principle of relativity. It states the relation between the original size of the object and to the size which is observed in the reference plane.