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Which statement is true ?
The cross product $ \\vec{A} \\times \\vec{B} $ is anticommutative, which means that $ \\vec{A} \\times \\vec{B} = - \\vec{B} \\times \\vec{A} $. This is a fundamental property of the cross product in vector algebra. Therefore, the correct statement is \vec{A} \times \vec{B} = - \vec{B} \times \vec{A}.
Two vectors A and B are such that lA+Bl=lA-Bl then find the angle between A and B
For two vectors \( \vec{A} \) and \( \vec{B} \), if \( |\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \), then the vectors are perpendicular to each other. This is because the magnitudes of the sum and difference of the vectors are equal only when the angle between them is 90°. Therefore, the angle between \( \vec{A} \) and \( \vec{B} \) is 90°.
$ \vec A = P \hat i - 2 P \hat j - \hat k and \vec B = - 3 \hat i + 2 \hat j + - 14 \hat k $ are perependicular to each other . Then p =
Find the unit vector in direction $ \hat i + 2 \hat j - 3 \hat k $
To find the unit vector in the direction of \( \hat{i} + 2 \hat{j} - 3 \hat{k} \), we first find the magnitude of the vector: \( \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \). The unit vector is then given by dividing each component by the magnitude: \( \frac{1}{\sqrt{14}} ( \hat{i} + 2 \hat{j} - 3 \hat{k} ) \). Therefore, the correct answer is \( \frac{1}{\sqrt{14}} ( \hat{i} + 2 \hat{j} - 3 \hat{k} ) \).
Find a unit vector perpendicular to both $ \vec A and \vec B $
To find a unit vector that is perpendicular to both vectors \( \vec{A} \) and \( \vec{B} \), we use the cross product \( \vec{A} \times \vec{B} \). The magnitude of the cross product is given by \( AB \sin \theta \). Therefore, the unit vector perpendicular to both \( \vec{A} \) and \( \vec{B} \) is \( \frac{ \vec{A} \times \vec{B} }{ AB \sin \theta } \).
$ \vec A and \vec B $ are two vectors $ \hat U_A = \hat U_B $ Now find the true option
$ \vec A + \vec B + \vec C = \hat j $
x and y co-ordinates of a particle moving in x-y plane at some instant are $ x = 2 t^2 $ and $ y = {3 \over 2 } t^2 $ . Calculate y co-ordinate when its x coordinate is 8 cm
$ x = 2t^2 = 8 $ $ \therefore t = 2s $ $ y = { 3 \over 2 } t^2 = {3 \over2 } (2)^2 = 6m $
A particle in xy plane is governed by $ x = A cos \omega t, y = A (1 – sin \omega t)$ . A and $ \omega $ are constants. What is the speed of the particle.
$$ x = A coswt $$ $$ y = A ( i - sin wt ) $$ $$ V \alpha = { dx \over dv } = - A \omega sin wt $$ $$Vy = - A \omega cos wt $$ $$ V = \sqrt { Vx^2 + Vy^2 } $$
Angle of projection of a projectile with horizonal line is $ \theta $ at time t = 0, After what time the angle will be again $ \theta $ ?
The time of flight for a projectile is given by \( \frac{2V_0 \sin \theta}{g} \). Since the angle of projection \( \theta \) will repeat at half the time of flight, the time when the angle will be \( \theta \) again is \( \frac{2V_0 \sin \theta}{g} \).
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.