1
MHT CET 2021 20th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

A particle is suspended from a vertical spring which is executing S.H.M. of frequency $$5 \mathrm{~Hz}$$. The spring is unstretched at the highest point of oscillation. Maximum speed of the particle is $$(\mathrm{g} =10 \mathrm{~m} / \mathrm{s}^2)$$

A
$$\frac{1}{\pi} \mathrm{m} / \mathrm{s}$$
B
$$\frac{1}{4 \pi} \mathrm{m} / \mathrm{s}$$
C
$$\frac{1}{2 \pi} \mathrm{m} / \mathrm{s}$$
D
$$\pi~ \mathrm{m} / \mathrm{s}$$
2
MHT CET 2021 20th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

A body performs S.H.M. under the action of force '$$\mathrm{F}_1$$' with period '$$\mathrm{T}_1$$' second. If the force is changed to '$$\mathrm{F}_2$$' it performs S.H.M. with period '$$\mathrm{T_2}$$' second. If both forces '$$\mathrm{F_1}$$' and '$$\mathrm{F_2}$$' act simultaneously in the same direction on the body, the period in second will be

A
$$\frac{T_1+T_2}{T_1 T_2}$$
B
$$\frac{T_1^2+T_2^2}{T_1 T_2}$$
C
$$\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}$$
D
$$\frac{T_1 T_2}{T_1+T_2}$$
3
MHT CET 2021 20th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

A mass '$$\mathrm{m}_1$$' is suspended from a spring of negligible mass. A spring is pulled slightly in downward direction and released, mass performs S.H.M. of period '$$\mathrm{T}_1$$'. If the mass is increased by '$$\mathrm{m}_2$$', the time period becomes '$$\mathrm{T}_2$$'. The ratio $$\frac{\mathrm{m}_2}{\mathrm{~m}_1}$$ is

A
$$\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2}$$
B
$$\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{~T}_1}$$
C
$$\frac{\mathrm{T}_2^2-\mathrm{T}_1^2}{\mathrm{~T}_1^2}$$
D
$$\frac{T_1^2-T_2^2}{T_1^2}$$
4
MHT CET 2021 20th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

Two particles $$\mathrm{P}$$ and $$\mathrm{Q}$$ performs S.H.M. of same amplitude and frequency along the same straight line. At a particular instant, maximum distance between two particles is $$\sqrt{2}$$ a. The initial phase difference between them is

$$\left[\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\right]$$

A
$$\frac{\pi}{6}$$
B
$$\frac{\pi}{2}$$
C
zero
D
$$\frac{\pi}{3}$$
MHT CET Subjects
Physics
Mechanics
Optics
Electromagnetism
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
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Coordinate Geometry
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