JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Analyzing +4 | -1

Statement I: The maximum speed at which a car can turn on level curve of radius

40 \mathrm{~m}

, is

11 \mathrm{~m} / \mathrm{s} ; \mu=0.3

.

Statement II:

v=\sqrt{\mu \mathrm{Rg}} \Rightarrow \mu=\frac{\mathrm{v}^2}{\mathrm{Rg}}=\frac{11 \times 11}{40 \times 10}=0.3

View answer and explanation

Correct option: A

If both Statement - I and Statement - II are true, and Statement - II is the correct explanation of Statement - I

Option A is correct. Statement I is true because the maximum speed is calculated using the formula v = √(μRg), and Statement II correctly explains how the coefficient of friction is derived from this formula.

More Questions from WORK-ENERGY-POWER

Practice All Chapter MCQs →

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.