NEET (UG)

Practice Test 1 — Work, Energy & Power

12 questions • 18 minutes • auto-graded with full solutions

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Section A — MCQ (Single Correct)

Question 1

A $5\ \text{N}$ force moves a body $3\ \text{m}$ in its own direction. Work done is:

Solution: $W = Fs = 5\times3 = 15\ \text{J}$.

Question 2

The kinetic energy of a $2\ \text{kg}$ body moving at $3\ \text{m/s}$ is:

Solution: $KE = \tfrac{1}{2}(2)(9) = 9\ \text{J}$.

Question 3

Work done by the centripetal force in uniform circular motion is:

Solution: It is perpendicular to velocity, so it does no work.

Question 4

A body falls freely from height $h$. At half the height, the ratio KE : PE is:

Solution: At $h/2$, half the PE has become KE, so KE = PE.

Question 5

A $1000\ \text{W}$ motor running for $1\ \text{minute}$ does work:

Solution: $W = Pt = 1000\times60 = 60000\ \text{J}$.

Question 6

If momentum is halved, kinetic energy becomes:

Solution: $KE \propto p^2$, so halving $p$ gives $\tfrac{1}{4}$.

Question 7

A bullet embeds in a wooden block. The collision is:

Solution: Sticking together ⇒ perfectly inelastic.

Question 8

Energy lost to friction reappears mainly as:

Solution: Friction converts mechanical energy chiefly into heat.

Section B — Assertion & Reason

Question 9

A: The work done by a conservative force around a closed path is zero.
R: For a conservative force, work depends only on the end positions, not the path.

Solution: Path-independence implies zero round-trip work — R explains A.

Question 10

A: Kinetic energy can never be negative.
R: It depends on the square of the speed.

Solution: $KE = \tfrac{1}{2}mv^2 \ge 0$ because $v^2 \ge 0$ — R explains A.

Question 11

A: In a perfectly inelastic collision, kinetic energy is conserved.
R: Momentum is conserved in all collisions.

Solution: KE is NOT conserved in inelastic collisions (A false); momentum is conserved in all collisions (R true).

Question 12

A: A car needs more engine power to climb a hill than to move on level ground at the same speed.
R: On a slope the engine force must also balance the component of gravity along the incline.

Solution: Extra force against gravity at the same speed means more power ($P = Fv$) — R explains A.