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A 100 m long rod of density 10.0 x 104 kg/m3 and having Young’s modules Y = 1011 Pa, is clamped at one end. It is hammered at the other free end. The longitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time, the pulse take to go back to initial point.a)0.1 secb)0.2 secc)0.3 secd)2 secCorrect answer is option 'B'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A 100 m long rod of density 10.0 x 104 kg/m3 and having Young’s modules Y = 1011 Pa, is clamped at one end. It is hammered at the other free end. The longitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time, the pulse take to go back to initial point.a)0.1 secb)0.2 secc)0.3 secd)2 secCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 100 m long rod of density 10.0 x 104 kg/m3 and having Young’s modules Y = 1011 Pa, is clamped at one end. It is hammered at the other free end. The longitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time, the pulse take to go back to initial point.a)0.1 secb)0.2 secc)0.3 secd)2 secCorrect answer is option 'B'. Can you explain this answer?.

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Here you can find the meaning of A 100 m long rod of density 10.0 x 104 kg/m3 and having Young’s modules Y = 1011 Pa, is clamped at one end. It is hammered at the other free end. The longitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time, the pulse take to go back to initial point.a)0.1 secb)0.2 secc)0.3 secd)2 secCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A 100 m long rod of density 10.0 x 104 kg/m3 and having Young’s modules Y = 1011 Pa, is clamped at one end. It is hammered at the other free end. The longitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time, the pulse take to go back to initial point.a)0.1 secb)0.2 secc)0.3 secd)2 secCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A 100 m long rod of density 10.0 x 104 kg/m3 and having Young’s modules Y = 1011 Pa, is clamped at one end. It is hammered at the other free end. The longitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time, the pulse take to go back to initial point.a)0.1 secb)0.2 secc)0.3 secd)2 secCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A 100 m long rod of density 10.0 x 104 kg/m3 and having Young’s modules Y = 1011 Pa, is clamped at one end. It is hammered at the other free end. The longitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time, the pulse take to go back to initial point.a)0.1 secb)0.2 secc)0.3 secd)2 secCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A 100 m long rod of density 10.0 x 104 kg/m3 and having Young’s modules Y = 1011 Pa, is clamped at one end. It is hammered at the other free end. The longitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time, the pulse take to go back to initial point.a)0.1 secb)0.2 secc)0.3 secd)2 secCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 11 tests.