MCQ
Two closed pipe produce $10$ beats per second when emitting their fundamental nodes. If their length are in ratio of $25 : 26$. Then their fundamental frequency in $Hz$, are
- A$270, 280$
- B$260, 270$
- ✓$260, 250$
- D$260, 280$
Using ${n_1}, = \frac{v}{{4{l_1}}}$ and ${n_2} = \frac{v}{{4{l_2}}}$
==> $\frac{{{n_1}}}{{{n_2}}} = \frac{{{l_2}}}{{{l_1}}} = \frac{{26}}{{25}}$…..$(ii)$
After solving these equation ${n_1} = 260Hz$,${n_2} = 250$$Hz$
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(use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \mathrm{R}=$ radius of earth)
(Specific heat of water is $4.2 \mathrm{~kJ}^{-1} \mathrm{~kg}^{-1}$ and the density of water is $10.$) $0 \mathrm{k}^2 \mathrm{~m}^{-3}$ )