Assertion (A): | A glass tube partially filled with water represents an open organ pipe. |
Reason (R): | The open end corresponds to an antinode and the end in contact with water, to a node. |
1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
2. | Both (A) and (R) are True and (R) is not the correct explanation of (A). |
3. | (A) is True but (R) is False. |
4. | (A) is False but (R) is True. |
1. | \(\dfrac{2\pi\lambda}{a}\) | 2. | \(\dfrac{2\pi a}{\lambda}\) |
3. | \(\dfrac{\lambda}{a}\) | 4. | \(\dfrac{a}{\lambda}\) |
1. | \(3:1\) | 2. | \(1:2\) |
3. | \(2:1\) | 4. | \(1:3\) |
1. | \(8:9\) | 2. | \(9:7\) |
3. | \(9:8\) | 4. | \(7:9\) |
A string of length \(l\) is fixed at both ends and is vibrating in second harmonic. The amplitude at antinode is \(2\) mm. The amplitude of a particle at a distance \(l/8\) from the fixed end is:
1. \(2\sqrt2~\text{mm}\)
2. \(4~\text{mm}\)
3. \(\sqrt2~\text{mm}\)
4. \(2\sqrt3~\text{mm}\)
1. | \(420\) Hz | 2. | \(440\) Hz |
3. | \(484\) Hz | 4. | \(512\) Hz |
In a guitar, two strings \(A\) and \(B\) made of same material are slightly out of tune and produce beats of frequency \(6~\text{Hz}\). When tension in \(B\) is slightly decreased, the beat frequency increases to \(7~\text{Hz}\). If the frequency of \(A\) is \(530~\text{Hz}\), the original frequency of \(B\) will be:
1. | \(524~\text{Hz}\) | 2. | \(536~\text{Hz}\) |
3. | \(537~\text{Hz}\) | 4. | \(523~\text{Hz}\) |
The length of the string of a musical instrument is \(90\) cm and has a fundamental frequency of \(120\) Hz. Where should it be pressed to produce a fundamental frequency of \(180\) Hz?
1. | \(75\) cm | 2. | \(60\) cm |
3. | \(45\) cm | 4. | \(80\) cm |
1. | \(500\) m/s | 2. | \(156\) m/s |
3. | \(344\) m/s | 4. | \(172\) m/s |