NEET-XII-Physics
47: The Special Theory of Relativity
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- #3-c3 × 107 m s-1?Ans : Here,
v = 3 × 107 m/s
`` L\text{'}=1\times \sqrt{1-\frac{9\times {10}^{14}}{9\times {10}^{16}}}``
`` =1\sqrt{1-{10}^{-2}}``
`` =0.995\,\mathrm{\,m\,}``
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- Qstn #4A person standing on a platform finds that a train moving with velocity 0.6c takes one second to pass by him. FindAns : Given:
Velocity of train, v = 0.6c
Time taken, t = 1 s
- #4-athe length of the train as seen by the person andAns : Length observed by the observer, L= vt
⇒ L = 0.6 × 3 × 108
= 1.8 × 108 m
- #4-bthe rest length of the train.Ans : Let the rest length of train be L0. Then,
`` L={L}_{0}\sqrt{1-{v}^{2}/{c}^{2}}``
`` \Rightarrow 1.8\times {10}^{8}={L}_{0}\sqrt{1-\frac{{\left(0.6\right)}^{2}{c}^{2}}{{c}^{2}}}``
`` \Rightarrow 1.8\times {10}^{8}={L}_{0}\times 0.8``
`` \Rightarrow {L}_{0}=\frac{1.8\times {10}^{8}}{0.8}``
`` \Rightarrow {L}_{0}=2.25\times {10}^{8}\,\mathrm{\,m\,}``
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- Qstn #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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- Qstn #6The rest distance between Patna and Delhi is 1000 km. A nonstop train travels at 360 km h-1.Ans : Given:
Rest distance between Patna and Delhi, L0 = 1000 km = 106 m
Speed of train, v = 360 km/h`` =\frac{360\times 5}{18}=100\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
- #6-aWhat is the distance between Patna and Delhi in the train frame?Ans : Apparent distance between Patna and Delhi is given by
`` L\text{'}={L}_{0}\sqrt{1-{v}^{2}/{c}^{2}}``
`` L\text{'}={10}^{6}\sqrt{1-{\left(\frac{100}{3\times {10}^{8}}\right)}^{2}}``
`` L\text{'}={10}^{6}\sqrt{1-\frac{{10}^{4}}{9\times {10}^{16}}}``
`` L\text{'}={10}^{6}\sqrt{1-\frac{1}{9}\times {10}^{-12}}``
`` L\text{'}=56\times {10}^{-9}\,\mathrm{\,m\,}``
Change in length = 56 nm
So, the distance between Patna and Delhi in the train frame is 56 nm less than 1000 km.
- #6-bHow much time elapses in the train frame between Patna and Delhi?Ans : Actual time taken by the train is given by
t = `` \frac{{L}_{0}}{v}=\frac{1000\times {10}^{3}}{100}={10}^{4}\,\mathrm{\,s\,}=\frac{500}{3}\,\mathrm{\,min\,}``
Change in time, `` ∆t=\frac{∆L}{v}=\frac{L\text{'}}{v}=\frac{56\times {10}^{-9}}{100}=0.56\times {10}^{-9}\,\mathrm{\,s\,}=0.56\,\mathrm{\,ns\,}``
So, 0.56 nm less than `` \frac{500}{3}`` min elapse in the train frame between Patna and Delhi.
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- Qstn #7A person travels by a car at a speed of 180 km h-1. It takes exactly 10 hours by his wristwatch to go from the station A to the station B. (a) What is the rest distance between the two stations? (b) How much time is taken in the road frame by the car to go from the station A to the station B?digAnsr: bAns : Given:
Speed of car, v = 180 km/hr = 50 m/s
Time, t = 10 h
Let the rest distance be L0.
Apparent distance is given by
`` L\text{'}={L}_{0}\sqrt{1-{v}^{2}/{c}^{2}}``
Apparent distance, L' = 10 × 180 = 1800 km = 18 × 105 m
Now,
`` 18\times {10}^{5}={L}_{0}\sqrt{1-\frac{{\left(50\right)}^{2}}{{\left(3\times {10}^{8}\right)}^{2}}}``
`` {L}_{0}=\left(18\times {10}^{5}+25\times {10}^{-9}\right)\,\mathrm{\,m\,}``
Thus, the rest distance is 25 nm more than 1800 km.
(b) Let the time taken by the car to cover the distance in the road frame be t. Then,
`` t=\frac{1.8\times {10}^{5}+25\times {10}^{-9}}{50}``
`` =0.36\times {10}^{5}+5\times {10}^{-8}``
`` =10\,\mathrm{\,hours\,}+0.5\,\mathrm{\,ns\,}``
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- Qstn #8A person travels on a spaceship moving at a speed of 5c/13.Ans : Given: Speed of spaceship, `` u=\frac{5c}{13}``
- #8-aFind the time interval calculated by him between the consecutive birthday celebrations of his friend on the earth.Ans : Proper time, t = 1 yr
`` \,\mathrm{\,Time\,}intervalcalculatedbyhim=t\text{'}=\frac{t}{\sqrt{1-{u}^{2}/{c}^{2}}}``
`` =\frac{1\,\mathrm{\,yr\,}}{\sqrt{1-{\displaystyle \frac{25{c}^{2}}{169{c}^{2}}}}}``
`` =\frac{13}{12}\,\mathrm{\,yr\,}``
`` ``
Thus, the time interval calculated by him between the consecutive birthday celebrations of his friend on Earth is 13/12 years.
- #8-bFind the time interval calculated by the friend on the earth between the consecutive birthday celebrations of the traveller.Ans : The friend on Earth also considers the same speed, so the time interval calculated by him is also 13/12 years.
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- Qstn #9According to the station clocks, two babies are born at the same instant, one in Howrah and other in Delhi.Ans : The time interval recorded in a moving train, i.e. improper time is always greater than proper time (recorded in ground frame). The birth timings recorded by the station clocks is the proper time interval because it is the ground frame.
Also, the time recorded in the moving train is improper because it records time at two different places.
Hence, the proper time interval ∆T is less than that of improper, i.e. ∆T' = v∆T.
Thus,
- #9-aWho is elder in the frame of 2301 Up Rajdhani Express going from Howrah to Delhi?Ans : Delhi baby is elder in the frame of 2301 Up Rajdhani Express going from Howrah to Delhi.
- #9-bWho is elder in the frame of 2302 Dn Rajdhani Express going from Delhi to Howrah.Ans : Howrah baby is elder in the frame of 2302 Dn Rajdhani express going from Delhi to Howrah.
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