MHT CET 2023 9th May Evening Shift | Waves Question 43 | Physics

If the length of an open organ pipe is $$33.3 \mathrm{~cm}$$, then the frequency of fifth overtone is [Neglect end correction, velocity of sound $$=333 \mathrm{~m} / \mathrm{s}$$ ]

$$3500 \mathrm{~Hz}$$

$$3000 \mathrm{~Hz}$$

$$2500 \mathrm{~Hz}$$

$$2000 \mathrm{~Hz}$$

MHT CET 2023 9th May Evening Shift

MCQ (Single Correct Answer)

-0

If the end correction of an open pipe is $$0.8 \mathrm{~cm}$$, then the inner radius of that pipe is

$$\frac{1}{3} \mathrm{~cm}$$

$$\frac{2}{3} \mathrm{~cm}$$

$$\frac{3}{2} \mathrm{~cm}$$

$$0.2 \mathrm{~cm}$$

MHT CET 2023 9th May Morning Shift

MCQ (Single Correct Answer)

-0

When both source and listener are approaching each other the observed frequency of sound is given by $$\left(V_L\right.$$ and $$V_S$$ is the velocity of listener and source respectively, $$\mathrm{n}_0=$$ radiated frequency)

$$\mathrm{n}=\mathrm{n}_0\left[\frac{\mathrm{V}+\mathrm{V}_{\mathrm{L}}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right]$$

$$\mathrm{n}=\mathrm{n}_0\left[\frac{\mathrm{V}-\mathrm{V}_{\mathrm{L}}}{\mathrm{V}+\mathrm{V}_{\mathrm{s}}}\right]$$

$$\mathrm{n}=\mathrm{n}_0\left[\frac{\mathrm{V}-\mathrm{V}_{\mathrm{L}}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right]$$

$$\mathrm{n}=\mathrm{n}_0\left[\frac{\mathrm{V}+\mathrm{V}_{\mathrm{L}}}{\mathrm{V}+\mathrm{V}_{\mathrm{s}}}\right]$$

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