JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Analyzing +4 | -1

A particle of mass

{m}

is moving along the

{x}

- axis under the potential

V(x)=\frac{k x^2}{2}+\frac{\lambda}{x^{\prime}}

where

{k}

and

{x}^{\prime}

are positive constants of appropriate dimensions. The particle is slightly displaced from its equilibrium position. The particle oscillates with the angular frequency

({w})

given by

View answer and explanation

Correct option: C

\sqrt{\frac{k}{m}}

Option C is correct because the angular frequency is calculated from the second derivative of the potential energy at the equilibrium point.

More Questions from WORK-ENERGY-POWER

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A ball throwing cannon, lifts ball and throws with

72 \mathrm{~km} / \mathrm{hr}

Speed. If balls are raised upto

10 \mathrm{~m}

. then find Input power of cannon if it lifts 8 balls per minute and its efficiency is

80 \%

. Mass of each ball is

1 \mathrm{~kg}

. Assertion:When a particle moves in a circle with a uniform speed its acceleration is constant but the velocity changes.
Reason: Angular displacement is not an axial vector. A body of mass

m

thrown vertically upwards attains a maximum height

h

. As what height will its kinetic energy be

75 \%

of its initial value? Assertion: Centripetal force does no work.

Reason: Force and displacement are perpendicular to each other. Statement I: Both stretched and compressed springs possess potential energy.

Statement II: Work done against restoring force is stored as potential energy. A bullet of mass

m

moving at a speed

v

hits a ball of mass

M

kept at rest. A small part having mass

m^{\prime}

breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed

v_1

in the direction of the bullet. Find the velocity of the bullet after the collision.