Q. No.| 1. | Wavelength of the component waves is \(10~\text{cm}.\) |
| 2. | The separation between a node and the nearest antinode is \(2.5~\text{cm}.\) |
| 3. | Frequency of the component wave is \(0.25~\text{Hz}\). |
| 4. | All of these |
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1. \(\frac{1}{n}= \frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}\)
2. \(n= n_1\times n_2\times n_3\)
3. \(n= n_1+ n_2+ n_3\)
4. \(n= \frac{n_1+ n_2+ n_3}{3}\)
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| 1. | \(10^{4}~\text{m/s}\) |
| 2. | \(1~\text{m/s}\) |
| 3. | \(10^{-4}~\text{m/s}\) |
| 4. | Not derivable from the above data |
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1. \(1:2:3:~.......\)
2. \(1:3:5:7~.......\)
3. \(1:2:4:8~.......\)
4. \(1:3:9:~.......\)
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1. \(20~\text{cm}\)
2. \(10~\text{cm}\)
3. \(40~\text{cm}\)
4. \(30~\text{cm}\)
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| 1. | \(100~\text{Hz}\) | 2. | \(200~\text{Hz}\) |
| 3. | \(50~\text{Hz}\) | 4. | \(400~\text{Hz}\) |
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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 |
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| 1. | \(100~\text{cm}\) | 2. | \(150~\text{cm}\) |
| 3. | \(200~\text{cm}\) | 4. | \(66.7~\text{cm}\) |
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1. \(1.21~\mathring{A}\)
2. \(2.42~\mathring{A}\)
3. \(0.605~\mathring{A}\)
4. \(4.84~\mathring{A}\)
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A string of length \(3\) m and a linear mass density of \(0.0025\) kg/m is fixed at both ends. One of its resonance frequencies is \(252\) Hz. The next higher resonance frequency is \(336\) Hz. Then the fundamental frequency will be:
1. \(84~\text{Hz}\)
2. \(63~\text{Hz}\)
3. \(126~\text{Hz}\)
4. \(168~\text{Hz}\)
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