1
MHT CET 2023 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

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

A
$$\frac{1}{3} \mathrm{~cm}$$
B
$$\frac{2}{3} \mathrm{~cm}$$
C
$$\frac{3}{2} \mathrm{~cm}$$
D
$$0.2 \mathrm{~cm}$$
2
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-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)

A
$$\mathrm{n}=\mathrm{n}_0\left[\frac{\mathrm{V}+\mathrm{V}_{\mathrm{L}}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right]$$
B
$$\mathrm{n}=\mathrm{n}_0\left[\frac{\mathrm{V}-\mathrm{V}_{\mathrm{L}}}{\mathrm{V}+\mathrm{V}_{\mathrm{s}}}\right]$$
C
$$\mathrm{n}=\mathrm{n}_0\left[\frac{\mathrm{V}-\mathrm{V}_{\mathrm{L}}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right]$$
D
$$\mathrm{n}=\mathrm{n}_0\left[\frac{\mathrm{V}+\mathrm{V}_{\mathrm{L}}}{\mathrm{V}+\mathrm{V}_{\mathrm{s}}}\right]$$
3
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Equation of simple harmonic progressive wave is given by $$y=\frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$$ then the resultant amplitude of the wave is $$\left(\cos 90^{\circ}=0\right)$$

A
$$\frac{a \pm b}{a b}$$
B
$$\frac{\sqrt{\mathrm{a}} \pm \sqrt{\mathrm{b}}}{\mathrm{ab}}$$
C
$$\frac{\sqrt{\mathrm{a}} \pm \sqrt{\mathrm{b}}}{\sqrt{\mathrm{ab}}}$$
D
$$\sqrt{\frac{a+b}{a b}}$$
4
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

When a string of length '$$l$$' is divided into three segments of length $$l_1, l_2$$ and $$l_3$$. The fundamental frequencies of three segments are $$\mathrm{n}_1, \mathrm{n}_2$$ and $$\mathrm{n}_3$$ respectively. The original fundamental frequency '$$n$$' of the string is

A
$$\mathrm{n}=\mathrm{n}_1+\mathrm{n}_2+\mathrm{n}_3$$
B
$$\sqrt{\mathrm{n}}=\sqrt{\mathrm{n}_1}+\sqrt{\mathrm{n}_2}+\sqrt{\mathrm{n}_3}$$
C
$$\frac{1}{\mathrm{n}}=\frac{1}{\mathrm{n}_1}+\frac{1}{\mathrm{n}_2}+\frac{1}{\mathrm{n}_3}$$
D
$$\frac{1}{\sqrt{\mathrm{n}}}=\frac{1}{\sqrt{\mathrm{n}_1}}+\frac{1}{\sqrt{\mathrm{n}_2}}+\frac{1}{\sqrt{\mathrm{n}_3}}$$
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