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Cosmic ray muons generated at a top of the earth atmosphere decay law.{ Given : T1/2of muons = 1.92 μ sec , velocity v = 0.96 c }An observer from the top of the mountain of height 2 km about mean sea level detect muons with their speed of light and count 1000 muons, the no. of muons (n) of the same speed detected by an observer at mean sea level at the same period of the time would be n, thena)T1/2 = 6.85 μ secb)T1/2 = 1.944 μ secc)n = 495d)n = 821Correct answer is option 'A,C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Cosmic ray muons generated at a top of the earth atmosphere decay law.{ Given : T1/2of muons = 1.92 μ sec , velocity v = 0.96 c }An observer from the top of the mountain of height 2 km about mean sea level detect muons with their speed of light and count 1000 muons, the no. of muons (n) of the same speed detected by an observer at mean sea level at the same period of the time would be n, thena)T1/2 = 6.85 μ secb)T1/2 = 1.944 μ secc)n = 495d)n = 821Correct answer is option 'A,C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Cosmic ray muons generated at a top of the earth atmosphere decay law.{ Given : T1/2of muons = 1.92 μ sec , velocity v = 0.96 c }An observer from the top of the mountain of height 2 km about mean sea level detect muons with their speed of light and count 1000 muons, the no. of muons (n) of the same speed detected by an observer at mean sea level at the same period of the time would be n, thena)T1/2 = 6.85 μ secb)T1/2 = 1.944 μ secc)n = 495d)n = 821Correct answer is option 'A,C'. Can you explain this answer?.

Solutions for Cosmic ray muons generated at a top of the earth atmosphere decay law.{ Given : T1/2of muons = 1.92 μ sec , velocity v = 0.96 c }An observer from the top of the mountain of height 2 km about mean sea level detect muons with their speed of light and count 1000 muons, the no. of muons (n) of the same speed detected by an observer at mean sea level at the same period of the time would be n, thena)T1/2 = 6.85 μ secb)T1/2 = 1.944 μ secc)n = 495d)n = 821Correct answer is option 'A,C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics. Download more important topics, notes, lectures and mock test series for Physics Exam by signing up for free.

Here you can find the meaning of Cosmic ray muons generated at a top of the earth atmosphere decay law.{ Given : T1/2of muons = 1.92 μ sec , velocity v = 0.96 c }An observer from the top of the mountain of height 2 km about mean sea level detect muons with their speed of light and count 1000 muons, the no. of muons (n) of the same speed detected by an observer at mean sea level at the same period of the time would be n, thena)T1/2 = 6.85 μ secb)T1/2 = 1.944 μ secc)n = 495d)n = 821Correct answer is option 'A,C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Cosmic ray muons generated at a top of the earth atmosphere decay law.{ Given : T1/2of muons = 1.92 μ sec , velocity v = 0.96 c }An observer from the top of the mountain of height 2 km about mean sea level detect muons with their speed of light and count 1000 muons, the no. of muons (n) of the same speed detected by an observer at mean sea level at the same period of the time would be n, thena)T1/2 = 6.85 μ secb)T1/2 = 1.944 μ secc)n = 495d)n = 821Correct answer is option 'A,C'. Can you explain this answer?, a detailed solution for Cosmic ray muons generated at a top of the earth atmosphere decay law.{ Given : T1/2of muons = 1.92 μ sec , velocity v = 0.96 c }An observer from the top of the mountain of height 2 km about mean sea level detect muons with their speed of light and count 1000 muons, the no. of muons (n) of the same speed detected by an observer at mean sea level at the same period of the time would be n, thena)T1/2 = 6.85 μ secb)T1/2 = 1.944 μ secc)n = 495d)n = 821Correct answer is option 'A,C'. Can you explain this answer? has been provided alongside types of Cosmic ray muons generated at a top of the earth atmosphere decay law.{ Given : T1/2of muons = 1.92 μ sec , velocity v = 0.96 c }An observer from the top of the mountain of height 2 km about mean sea level detect muons with their speed of light and count 1000 muons, the no. of muons (n) of the same speed detected by an observer at mean sea level at the same period of the time would be n, thena)T1/2 = 6.85 μ secb)T1/2 = 1.944 μ secc)n = 495d)n = 821Correct answer is option 'A,C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Cosmic ray muons generated at a top of the earth atmosphere decay law.{ Given : T1/2of muons = 1.92 μ sec , velocity v = 0.96 c }An observer from the top of the mountain of height 2 km about mean sea level detect muons with their speed of light and count 1000 muons, the no. of muons (n) of the same speed detected by an observer at mean sea level at the same period of the time would be n, thena)T1/2 = 6.85 μ secb)T1/2 = 1.944 μ secc)n = 495d)n = 821Correct answer is option 'A,C'. Can you explain this answer? tests, examples and also practice Physics tests.