JEE Exam > JEE Questions > A granite rod of 60 cm length is clamped at i...
Start Learning for Free
A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?
- a)5 kHz
- b)2.5 kHz
- c)10 kHz
- d)7.5 kHz
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A granite rod of 60 cm length is clamped at its middle point and is se...
Since rod is clamped at centre. So centre it behaves as node & end it behave as antinode.
So,
So,
Most Upvoted Answer
A granite rod of 60 cm length is clamped at its middle point and is se...
Given data:
- Length of the granite rod (L) = 60 cm = 0.6 m
- Density of granite (ρ) = 2.7 x 10^3 kg/m^3
- Young's modulus of granite (Y) = 9.27 x 10^10 Pa
To find: Fundamental frequency of the longitudinal vibrations
Formula for the fundamental frequency of longitudinal vibrations in a rod (fixed at both ends):
f = (1/2L) * √(Y/ρ)
Calculation:
1. Convert the length of the rod to meters:
L = 0.6 m
2. Substitute the values of L, Y, and ρ into the formula and calculate the frequency:
f = (1/2 * 0.6) * √(9.27 x 10^10 / 2.7 x 10^3)
= (0.3) * √(9.27 x 10^10 / 2.7 x 10^3)
= (0.3) * √(9.27 x 10^10 / 2.7 x 10^3)
= (0.3) * √(9.27 x 10^7 / 2.7)
= (0.3) * √(34288888.889)
≈ (0.3) * 5859.7
≈ 1757.91 Hz
≈ 1.76 kHz
Therefore, the fundamental frequency of the longitudinal vibrations of the granite rod is approximately 1.76 kHz.
Hence, the correct answer is option 'A' - 5 kHz.
- Length of the granite rod (L) = 60 cm = 0.6 m
- Density of granite (ρ) = 2.7 x 10^3 kg/m^3
- Young's modulus of granite (Y) = 9.27 x 10^10 Pa
To find: Fundamental frequency of the longitudinal vibrations
Formula for the fundamental frequency of longitudinal vibrations in a rod (fixed at both ends):
f = (1/2L) * √(Y/ρ)
Calculation:
1. Convert the length of the rod to meters:
L = 0.6 m
2. Substitute the values of L, Y, and ρ into the formula and calculate the frequency:
f = (1/2 * 0.6) * √(9.27 x 10^10 / 2.7 x 10^3)
= (0.3) * √(9.27 x 10^10 / 2.7 x 10^3)
= (0.3) * √(9.27 x 10^10 / 2.7 x 10^3)
= (0.3) * √(9.27 x 10^7 / 2.7)
= (0.3) * √(34288888.889)
≈ (0.3) * 5859.7
≈ 1757.91 Hz
≈ 1.76 kHz
Therefore, the fundamental frequency of the longitudinal vibrations of the granite rod is approximately 1.76 kHz.
Hence, the correct answer is option 'A' - 5 kHz.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer?
Question Description
A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer?.
A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 103 kg/m3 and its young's modulus is 9.27 x 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?a)5 kHzb)2.5 kHzc)10 kHzd)7.5 kHzCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup