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NEET NEET Questions — Free MCQ Practice

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A body is placed on a horizontal plank executing S.H.M. along vertical direction. Its amplitude of oscillation is $3.92 \times 10^{â€“ 3}m$. What should be the minimum periodic time so that the body does not loose contact with the plank?

A. 0.1256 s B. 0.1356 s C. 0.1456 s D. 0.1556

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Explanation

The body will not loose contact with the surface, $ if mg = m \omega ^2 r = { m4 \pi ^2 \over T^2 } . r $ (where r is amplitude ) $ \therefore T = 2 \pi \sqrt { r \over g} $

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If the kinetic energy of a particle executing S.H.M. is given by K = K Cos2Ã¹t, then the displacement of the particle is given by â€¦â€¦â€¦.

A. $ \left( { K_o \over m \omega^2 }\right)^{1/2} Sin \omega t $ B. $ \left( { 2K_o \over m \omega^2 }\right)^{1/2} Sin \omega t $ C. $ \left( { 2 \omega^2 \over m K_o }\right)^{1/2} Sin \omega t $ D. $ \left( { 2K_o \over m \omega }\right)^{1/2} Sin \omega t $

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Explanation

$ maximum kinetic energy K_o = { 1 \over 2 } m \omega^2 A^2 $ $ \therefore A = \left( { 2 K_o \over m \omega^2 } \right) ^{1 /2 } $ $ \therefore Equation for displacement is ; y = A sin \omega t = \left( { 2 K_o \over m \omega^2 } \right) ^{1 /2} sin \omega t $

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The equation for displacement of two identical particles performing S.H.M. is given by $x_1 =4Sin(20t+p/6)$ and $x_2 =10SinÃ¹t$. For what value of Ã¹ will both particles have same energy?

A. 4 units B. 8 units C. 16 units D. 20 units

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Explanation

$ E = { 1 \over 2 } m \omega^2 A^2 \Rightarrow E \alpha \omega^2 A^2 $ $ \therefore E \alpha ( A \omega ) ^2 \Rightarrow ( \omega_1 A_1 ) ^2 = ( \omega_2 A_2 ) ^2 $

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A spring having length l and spring constant k is divided into two parts having lengths $l_1$ and $l_2$. If $l_1 = nl_2$, the force constant of the spring having length $l_2$ is

A. k ( 1 + n ) B. $ k \left( { 1 + n \over n } \right) $ C. k D. $ { k \over (n+1) } $

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Explanation

$ k_2 be the spring constant of the spring having length l_2$ $ Now , l_1 + l_2 = l $ $ n l_2 + l_2 = l $

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When a mass m is suspended from the free end of a massless spring having force constant k, its oscillates with frequency f. Now if the spring is divided into two equal parts and a mass 2m is suspended from the end of anyone of them, it will oscillate with a frequency equal to â€¦â€¦â€¦â€¦.

A. f B. 2f C. $ { f \over \sqrt 2 } $ D. $ \sqrt 2 f $

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Explanation

$ f = { 1 \over 2 \pi } \sqrt { k \over m } and f' = { 1 \over 2 \pi } \sqrt { 2k \over 2m } $ $ { k' = 2 k } $ $ \therefore f' = f $

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A body of mass 1 kg suspended from the free end of a spring having force constant $400 Nm^{-1}$ is executing S.H.M. When the total energy of the system is 2 joule, the maximum acceleration is

â€¦â€¦â€¦$ms^{ â€“ 2}$ .

A. $ 8 ms^{-2} $ B. $ 10 ms^{-2} $ C. $ 40 ms^{-2} $ D. $ 40 cms^{-2} $

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Explanation

Energy stoved =Work done $ \therefore E = { 1 \over 2 } k A^2 $ Now maximum acceleration $ a_{max} = \omega^2 A $

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A spring is attached to the center of a frictionless horizontal turn table and at the other end a body of mass 2 kg is attached. The length of the spring is 35 cm. Now when the turn table is rotated with an angular speed of $10 rad s^{â€“ 1}$ , the length of the spring becomes 40 cm then the force constant of the spring is..... N/m.

A. $ 1.2 \times 10^3 $ B. $ 1.6 \times 10^3 $ C. $ 2.2 \times 10^3 $ D. $ 2.6 \times 10^3 $

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Explanation

Radius of the rotational motion r =0.4 m When the turn table rotates, the restoring force developed in the spring = centrifugal force $ \therefore F_{restore} = m \omega^2 r = 2 ( 10 ) ^2 \times 0.4 = 80 N $ Now increase in length of spring = 40-35 = 5 cm $ \therefore Force constant k - { F \over x } = { 80 \over 0.05 } = 1.6 \times 10^3 N/m $

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A simple pendulum is executing S.H.M. around point O between the end points B and C with a periodic time of 6 s. If the distance between B and C is 20 cm then in what time will the bob move from C to D? Point D is at the mid-point of C and O.

A. 1s B. 2 s C. 3 s D. 4 s

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Explanation

Here T=6 s Amplitude OB = OC = 1/2 BC = 10cm $ \therefore OD =5 cm$ $ Now displacement x = A sin ( wt + \phi) ...(1) $ $ where A = 10 gm , \omega = { 2 \pi \over T } = { \pi \over 3 } rad $ Now if at t = 0 , oscillator is at C i.e at t =0 , x = A $\therefore A= Asin(\omega \times 0 + \phi) (OR) A=Asin \phi$ $ \Rightarrow sin \phi = 1 $ $ \Rightarrow \phi ={\pi \over 2} $ putting this in eqn (1) $ x = A sin ( \omega t + { \pi \over 2 } ) = A cos \omega t = 10 cos \omega t $ $ \therefore for x = 5 cm $ $ 5 = 10 cos \omega t \Rightarrow cos \omega t = { 1 \over 2 } $ $ \therefore \omega t = { \pi \over 3 } $ $ \therefore t = 1 S $

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A small spherical steel ball is placed at a distance slightly away from the center of a concave mirror having radius of curvature 250 cm. If the ball is released, it will now move on the curved surface. What will be the periodic time of this motion? Ignore frictional force and take $g = 10 m / s^2$

A. $ \pi / 4 s $ B. ps C. $ \pi / 2 s $ D. 2 ps

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Explanation

Force responsible for oscillation in $ F = mg sin \theta = mg \theta $ $ \{ \theta is small \} $ $ = mg .{ x \over R} $ Comparing this with $ F = - kx $ $ k = { mg \over R } $

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A simple pendulum having length l issus pended at the roof of a train moving with constant accelerationâ€˜aâ€™ along horizontal direction. The periodic time of this pendulum

A. $ T = 2 \pi \sqrt { l \over g } $ B. $ T = 2 \pi \sqrt { l \over g+a } $ C. $ T = 2 \pi \sqrt { l \over g - a } $ D. $ T = 2 \pi \sqrt { l \over g^2 + a^2 } $

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Explanation

Here 2 acceleration vectors g. and a are acting along mutually prependicular direction . $ \therefore effective acceleratioin l^n g_{eff} = \sqrt { g^2 + a^2 } $ $ \therefore T = 2 \pi \sqrt { l \over g _{eff} } $

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