Are these Physics questions free to attempt?

Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

Do I need an account to practise Physics MCQs?

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

Are these previous year NEET questions?

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.

Physics NEET Questions — Free MCQ Practice

● ● ● ● ●

ABCD is a quadrilateral. Forces $\vec{BA}, \vec{BC}, \vec{CD} & \vec{DA}$ act at a point. Their resultant is

A.

$2 \vec{AB}$

B.

$2 \vec{DA}$

C.

zero vector

D.

$2 \vec{BA}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

The resultant of the forces $\vec{P}$ and $\vec{Q}$ is $\vec{R}$ . If $\vec{Q}$ is double then the resultant also doubles in magnitude. Find the angle between $\vec{P}$ and $\vec{Q}$ .

A.

$\cos θ = \frac{Q}{2 P}$

B.

$\cos θ = \frac{- 4 Q}{3 P}$

C.

$\cos θ = \frac{- 2 Q}{3 P}$

D.

$\cos θ = \frac{- 3 P}{4 Q}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(D)

$Resultant of \vec{P} & \vec{Q} is \vec{R} R^{2} = P^{2} + Q^{2} + 2 PQ cosθ . . . . . (i) If Q is doubled, resultant is 2 R {(2 R)}^{2} = P^{2} + {(2 Q)}^{2} + 2 P (2 Q) cosθ . . . . . . (ii) From (i) & (ii) 4 (P^{2} + Q^{2} + 2 PQ cosθ) = P^{2} + 4 Q^{2} + 4 PQ cosθ 4 P^{2} + 8 PQ cosθ = P^{2} + 4 QP cosθ 4 Q cosθ = - 3 P cosθ = \frac{- 3 P}{4 Q}$

● ● ● ● ●

The maximum and minimum magnitude of the resultant of two vectors are 17 units and 7 units respectively. Then the magnitude of resultant of the vectors when they act perpendicular to each other is:

A.

14

B.

16

C.

18

D.

13

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

When two vectors are perpendicular, their resultant magnitude is given by the Pythagorean theorem: âˆš(a^2 + b^2), where a and b are the magnitudes of the vectors. If the maximum and minimum resultants are 17 and 7 units, then âˆš(17^2 + 7^2) = 13 units.

● ● ● ● ●

A vector $\vec{A}$ makes an angle of $20 °$ and $\vec{B}$ makes an angle of $110 °$ with the X-axis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the magnitude of the resultant.

A.

$1$

B.

$5 \sqrt{2}$

C.

$5$

D.

$7$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Angle between

$\vec{A} a n d \vec{B} = 90 ° R = \sqrt{A^{2} + B^{2}} = \sqrt{{(3)}^{2} + {(4)}^{2}} = 5 m$

● ● ● ● ●

A particle is moving westward with a velocity ${\vec{v}}_{1} = 5 m / s .$ Its velocity changed to ${\vec{v}}_{2} = 5 m / s$ northward. The change in velocity vector $(∆ \vec{V} = {\vec{v}}_{2} - {\vec{v}}_{1})$ is:

A.

$5 \sqrt{2} m / s$ towards north east

B.

$5 m / s$ towards north west

C.

zero

D.

$5 \sqrt{2} m / s$ towards north west

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

Consider east as positive x-axis, north as positive y-axis and vertically upward direction as z-axis. A helicopter first rises up to an altitude of 100 m than flies straight in north 500 m and then suddenly takes a turn towards east and travels 1000 m east. What is position vector of helicopter. (Take starting point as origin)

A.

$1000 \hat{i} - 500 \hat{j} + 100 \hat{k}$

B.

$1000 \hat{i} + 500 \hat{j} - 100 \hat{k}$

C.

$1000 \hat{i} + 500 \hat{j} + 100 \hat{k}$

D.

$- 1000 \hat{i} + 500 \hat{j} + 100 \hat{k}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Position of helicopter= $100 \hat{k} + 500 \hat{j} + 1000 \hat{i}$

= $1000 \hat{i} + 500 \hat{j} + 100 \hat{k}$

● ● ● ● ●

A force $\vec{F} = 6 \hat{i} - 8 \hat{j} + 10 \hat{k}$ Newton produces acceleration $1 m / s^{2}$ in a body. The mass of the body is (in kg)

A.

$6 \hat{i} - 8 \hat{j} + 10 \hat{k}$

B.

$100$

C.

$10 \sqrt{2}$

D.

$10$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$\vec{F} = m \vec{a} \vec{a} = \frac{\vec{F}}{m} = \frac{6 \hat{i} - 8 \hat{j} + 10 \hat{k}}{m} |\vec{a}| = \frac{|\vec{F}|}{m} = \frac{\sqrt{{(6)}^{2} + {(- 8)}^{2} + {(10)}^{2}}}{m} 1 = \frac{10 \sqrt{2}}{m} m = 10 \sqrt{2} k g$

● ● ● ● ●

$\vec{A} = \hat{i} + \hat{j} - \hat{k}; \vec{B} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$ angle between $\vec{A}$ and $\vec{B}$ is

A.

$120 °$

B.

$90 °$

C.

$60 °$

D.

$30 °$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Angle between two vectors = $\cos^{- 1} [\frac{\vec{A} . \vec{B}}{A B}]$

$= \cos^{- 1} [\frac{2 + 3 - 5}{\sqrt{3} \times \sqrt{38}}] = \cos^{- 1} (0) = 90 °$

● ● ● ● ●

Given the vectors

$\vec{A} = 2 \hat{i} + 3 \hat{j} - \hat{k} \vec{B} = 3 \hat{i} - 2 \hat{j} - 2 \hat{k} & \vec{C} = p \hat{i} + p \hat{j} + 2 p \hat{k}$

Find the angle between $(\vec{A} - \vec{B}) & \vec{C}$

A.

$θ = \cos^{- 1} (\frac{2}{\sqrt{3}})$

B.

$θ = \cos^{- 1} (\frac{\sqrt{3}}{2})$

C.

$θ = \cos^{- 1} (\frac{\sqrt{2}}{3})$

D.

none of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(C)

$\vec{A} - \vec{B} = (2 \hat{i} + 3 \hat{j} - \hat{k}) - (3 \hat{i} - 2 \hat{j} - 2 \hat{k}) = - \hat{i} + 5 \hat{j} + \hat{k} Let \vec{A} - \vec{B} = \vec{D} (let) ∴ \vec{D} = - \hat{i} + 5 \hat{j} + \hat{k} |\vec{D}| = \sqrt{{(- 1)}^{2} + {(5)}^{2} + {(1)}^{2}} = \sqrt{27} To find the angle between \vec{D} & \vec{C} |\vec{C}| = \sqrt{{(p)}^{2} + {(p)}^{2} + {(2 p)}^{2}} = P \sqrt{6} \vec{D} . \vec{C} = DC cosθ (- \hat{i} + 5 \hat{j} + \hat{k}) . (p \hat{i} + p \hat{j} + 2 p \hat{k}) = \sqrt{27} \times p \sqrt{6} cosθ - p + 5 p + 2 p = p \sqrt{27 \times 6} cosθ 6 = \sqrt{27 \times 6} cosθ cosθ = \sqrt{\frac{6}{27}} = \sqrt{\frac{2}{9}} cosθ = \frac{\sqrt{2}}{3} or θ = \cos^{- 1} \frac{\sqrt{2}}{3}$

● ● ● ● ●

If vector $\vec{P}, \vec{Q}$ and $\vec{R}$ have magnitude 5, 12 and 13 units and $\vec{P} + \vec{Q} = \vec{R}$ the angle between $\vec{Q}$ and $\vec{R}$ is -

A.

$\cos^{- 1} \frac{5}{12}$

B.

$\cos^{- 1} \frac{5}{13}$

C.

$\cos^{- 1} \frac{12}{13}$

D.

$\cos^{- 1} \frac{2}{13}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Physics MCQs for NEET — Practice Questions with Answers

Frequently Asked Questions