NEET-XII-Physics
47: The Special Theory of Relativity
- #5An aeroplane travels over a rectangular field 100 m × 50 m, parallel to its length. What should be the speed of the plane so that the field becomes square in the plane frame?Ans : The rectangular field appears to be a square when the length becomes equal to the breadth, i.e. 50 m.
Contracted length, L' = 50
Original length, L = 100
Let the speed of the aeroplane be v.
Velocity of light, c = 3 × 108 m/s
We know,
`` L\text{'}=L\sqrt{1-{v}^{2}/{c}^{2}}``
`` \Rightarrow 50=100\sqrt{1-{v}^{2}/{c}^{2}}``
`` \Rightarrow {\left(1/2\right)}^{2}=1-{v}^{2}/{c}^{2}``
`` \Rightarrow \frac{1}{4}=1-\frac{{v}^{2}}{{c}^{2}}``
`` \Rightarrow \frac{1}{4}=\frac{{c}^{2}-{v}^{2}}{{c}^{2}}``
`` \Rightarrow \frac{{c}^{2}}{4}={c}^{2}-{v}^{2}``
`` \Rightarrow {v}^{2}={c}^{2}-\frac{{c}^{2}}{4}``
`` \Rightarrow {v}^{2}=\frac{3{c}^{2}}{4}``
`` \Rightarrow v=\frac{\sqrt{3}c}{2}=0.866c``
Hence, the speed of the aeroplane should be equal to 0.866c, so that the field becomes square in the plane frame.
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