","comment":{"@type":"Comment","text":"We can achieve the necessary aeroplane speed by applying the principle of length contraction based on special theory of relativity."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$0.13\\,c$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$0.34\\,c$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$0.43\\,c$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$0.31\\,c$$","position":2,"answerExplanation":{"@type":"Comment","text":" $$\\text{ From the theory of special relativity and the concept of length contraction, we have}\\\\ L=\\dfrac{{{L}_{0}}}{\\gamma }\\\\ \\Rightarrow L={{L}_{0}}\\sqrt{1-\\dfrac{{{v}^{2}}}{{{c}^{2}}}} \\\\ \\Rightarrow 38=40\\sqrt{1-\\dfrac{{{v}^{2}}}{{{c}^{2}}}} \\\\ \\Rightarrow 1-\\dfrac{{{v}^{2}}}{{{c}^{2}}}={{\\left( \\dfrac{19}{20} \\right)}^{2}} \\\\ \\Rightarrow \\dfrac{{{v}^{2}}}{{{c}^{2}}}=1-0.9025 \\\\ \\Rightarrow \\dfrac{{{v}^{2}}}{{{c}^{2}}}=0.9075 \\\\ \\therefore v=0.31c$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Relativity Quiz 1","text":" A kid is playing with a plastic scale of length $$50\\text{ cm}$$. He projects it in the air in such a way that the scale moves parallel to its length. The length of the scale when its mass is three-fourth of its rest mass in metres is ","comment":{"@type":"Comment","text":" The final length at a given moment is calculated using the special theory of relativity's principle of mass shift and length contraction."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$0.50$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$0.56$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" None of these","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$0.66$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$\\text{ From the theory of special relativity, we have}\\\\ m={{m}_{0}}\\gamma \\\\ \\text{ Given that rest mass}\\, m=\\dfrac{3}{4}{{m}_{0}}\\\\ \\text{Clearly, relativistic factor}\\, \\gamma =\\dfrac{3}{4}\\\\ \\text{Now, we know from the concept of length contraction that}\\\\ L=\\dfrac{{{L}_{0}}}{\\gamma }\\\\ \\text{Substituting}\\,{{L}_{0}}=50\\text{ cm}\\,\\text{ and}\\, \\gamma =\\dfrac{3}{4}\\,\\text{ in the above expression, we have}\\\\ L=50\\times \\dfrac{4}{3}\\\\ \\therefore L=66.66\\text{ cm = 0}\\text{.66 m}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Relativity Quiz 1","text":" The relativistic factor for a particle of mass $$8\\times {{10}^{-20}}\\text{ kg}$$ is $$6.4$$.Determine the particle’s relativistic kinetic energy. ","comment":{"@type":"Comment","text":"By directly substituting the given parameters in the expression for relativistic kinetic energy, the required kinetic energy can be easily determined."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$34.2\\,{{c}^{2}}\\times {{10}^{-20}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$43.2\\,{{c}^{2}}\\times {{10}^{-19}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$34.2\\,{{c}^{2}}\\times {{10}^{-19}}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$43.2\\,{{c}^{2}}\\times {{10}^{-20}}$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$\\text{ From the special theory of relativity,}\\\\ \\text{ we have the expression for relativistic kinetic energy as}\\\\ K{{E}_{rel}}=(\\gamma -1)m{{c}^{2}}\\\\ \\text{From the question, we are provided that}\\\\ m=8\\times {{10}^{-20}}\\text{ kg and} \\gamma =6.4\\\\ \\text{Clearly,}\\\\ K{{E}_{rel}}=(6.4-1)\\times 8\\times {{10}^{-20}}\\times {{c}^{2}} \\\\ \\therefore K{{E}_{rel}}=43.2{{c}^{2}}\\times {{10}^{-20}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Relativity Quiz 1","text":"The concept of relativity applies to both accelerated as well as unaccelerated motion. _ relativity handles with _ motion whereas _ relativity handles with _ motion only. In special relativity, a parameter called gamma factor_ plays the major role. This is also called the _ factor. It is given by _ where $$v$$ is the _ between inertial frames of reference and _ is the speed of light in vacuum. ","comment":{"@type":"Comment","text":" The given concept explains the theory of special relativity and can easily be filled without much confusion. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Special; accelerated; general; unaccelerated; $$\\gamma $$; Lambda; $$\\dfrac{1}{\\sqrt{1-\\dfrac{{{c}^{2}}}{{{v}^{2}}}}}$$; velocity; $$c$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" General; accelerated; special; unaccelerated$$\\gamma $$; Lambda; $$\\dfrac{1}{\\sqrt{1-\\dfrac{{{v}^{2}}}{{{c}^{2}}}}}$$; relative velocity; $$c$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" Special; accelerated; general; unaccelerated; $$\\gamma $$; Lorentz; $$\\dfrac{1}{\\sqrt{1-\\dfrac{{{c}^{2}}}{{{v}^{2}}}}}$$; velocity; $$c$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" General; accelerated; special; unaccelerated; $$\\gamma $$; Lorentz; $$\\dfrac{1}{\\sqrt{1-\\dfrac{{{v}^{2}}}{{{c}^{2}}}}}$$; relative velocity; $$c$$","position":3,"answerExplanation":{"@type":"Comment","text":" The concept of relativity applies to both accelerated as well as unaccelerated motion. General relativity handles with accelerated motion whereas special relativity handles with unaccelerated motion only. In special relativity, a parameter called gamma factor $$(\\gamma )$$ plays a major role. This is also called the Lorentz factor. It is given by $$\\dfrac{1}{\\sqrt{1-\\dfrac{{{v}^{2}}}{{{c}^{2}}}}}$$ where $$v$$ is the relative velocity between inertial frames of reference and $$c$$ is the speed of light in vacuum.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Relativity Quiz 1","text":" An artificial celestial body sends information to the earth at a frequency $$1.5\\,GHz$$, when it is moving away from the Earth at a speed of $$0.35\\,c$$. Evaluate the frequency at which the information is received on the earth.","comment":{"@type":"Comment","text":"Remember the expression for observed frequency in relation to the relativistic doppler effect principle."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2.04\\,GHz$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$3.04\\,GHz$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$4.04\\,GHz$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1.04\\,GHz$$","position":1,"answerExplanation":{"@type":"Comment","text":"$$\\text{ Using theory of special relativity in the concept of doppler effect, we have}\\\\ {{f}_{obs}}={{f}_{s}}\\sqrt{\\dfrac{1-\\dfrac{u}{c}}{1+\\dfrac{u}{c}}} \\\\ \\Rightarrow {{f}_{obs}}=1.5\\sqrt{\\dfrac{1-\\dfrac{0.35c}{c}}{1+\\dfrac{0.35c}{c}}} \\\\ \\Rightarrow {{f}_{obs}}=1.5\\sqrt{\\dfrac{0.65}{1.35}} \\\\ \\Rightarrow {{f}_{obs}}=1.5\\times 0.692\\\\ \\therefore {{f}_{obs}}=1.04\\,GHz$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}