NEET Physics Questions: Motion in a Lane

Pratcice NEET questions from all chapters from huge question bank for free. All MCQs are based on NCERT syllabus. To practice from a specific subject and chapter, select a subject below. Login to practice in a structured way with explanations, bookmarks, lists, notes etc. Click here to Login or Sign up for free.

Physics Chemistry Botany Zoology

Please Login or Sign up to use advanced filters.

Question bank:

A particle is moving in a circle of radius R with constant speed. It covers an angle $ \theta $ in some time interval. Find displacement in this interval of time




$ \triangle S = \sqrt { R^2 + R^2 - 2R^2 cos \theta } $ $ \triangle S= \sqrt { 2R^2 - 2R^2 Cos \theta } = \sqrt { 2R^2 (1 - Cos \theta ) }$ $ = 2 R sin { \theta \over 2 } $
Angle of projection, maximum height and time to reach the maximum height of a particle are , H and tm respectively. Find the true relation.




To find the time to reach the maximum height, we use the kinematic equation for a projectile. The time to reach the maximum height (tm) is given by: \\[ t_m = rac{u ext{sin} heta}{g} \\] The maximum height (H) is given by: \\[ H = rac{u^2 ext{sin}^2 heta}{2g} \\] Rearranging to solve for tm, we get: \\[ t_m = rac{u ext{sin} heta}{g} = rac{ ext{sin} heta}{g} imes rac{u^2 ext{sin} heta}{2u} \\] \\[ t_m = rac{2H}{g} imes rac{1}{u} \\] Simplifying, we get: \\[ t_m = rac{2H}{g} = rac{2H}{g} \\] Therefore, the correct option is o2: \\[ t_m = rac{ ext{2H}}{g} \\]
An object moves in x - y plane. Equations for displacement in x and y direction are x = 3sin2t and y = 3cos2t Speed of the particle is




The equations given for displacement are x = 3sin(2t) and y = 3cos(2t). These represent parametric equations of a circle with radius 3. The speed of a particle moving in a circle with constant radius and angular speed is constant. Thus, the speed is constant and non-zero. Hence, option o2 is correct.
To introduce a vector quantity ....




A vector quantity is defined by having both a magnitude and a direction. For example, velocity is a vector because it specifies how fast something is moving (magnitude) and in which direction. Therefore, to introduce a vector quantity, it needs both magnitude and direction.
Which from the following is a scalar ?




Electric current is a scalar quantity. Unlike vectors, scalar quantities have only magnitude and no direction. Electric current is described by its magnitude, which is the rate of flow of charge through a conductor, but it does not have a specific direction in the sense that a vector does.
  • 🎉 Independence Day Offer: Get 25% off with code FREEDOM25! 🎉
  • 📘 Chapter progress tracker with auto color coding
  • 🔍 Weak‑area identification & detailed analytics
  • ✏️ Daily DPPs, mini & custom tests
  • ✨ Offline progress recorder & many more
$ \vec P and \vec Q $ are equal vectors what from the followings is true.




If two vectors $ \\vec{P} and \\vec{Q} $ are equal, it means they have both the same magnitude and the same direction. Therefore, $ \\vec{P} and \\vec{Q} $ are parallel to each other.
$ \vec P = \vec Q $ is true , if




For two vectors \( \vec{P} \) and \( \vec{Q} \) to be equal, they must have both equal magnitudes and the same direction. This is because vectors are defined by both their magnitude and direction. Therefore, \( \vec{P} = \vec{Q} \) is true if and only if their magnitudes are equal and they are in the same direction.
$ \vec A and \vec B $ are in opposite direction so they are




When two vectors \( \vec{A} \) and \( \vec{B} \) are in opposite directions, they are referred to as antiparallel vectors. Antiparallel vectors have the same magnitude but opposite directions. This is different from parallel vectors, which have the same direction.
$ \vec C = \vec A + \vec B $ and A = B = C . Find the angle between $ \vec A and \vec B $




$$ \vec C = \vec A + \vec B $$ $$ C^2 = A^2 + B^2 + 2AB cos \theta $$
The resultant of two forces of magnitude 2N and 3N can never be.




The resultant of two forces can be found using the triangle law of vector addition. The magnitude of the resultant force \( \vec{R} \) of two forces of magnitudes 2N and 3N can range from the absolute difference of the two forces to the sum of the two forces. Hence, \( |2N - 3N| \leq R \leq 2N + 3N \), which simplifies to \( 1N \leq R \leq 5N \). Therefore, a resultant force of 0.5N is not possible.
The sum of $ \vec P and \vec Q $ is at right agnles to their difference then




Given that the sum of vectors $\vec{P}$ and $\vec{Q}$ is at right angles to their difference, we have: $\vec{P} + \vec{Q} \perp \vec{P} - \vec{Q}$. This implies that their dot product is zero: $(\vec{P} + \vec{Q}) \cdot (\vec{P} - \vec{Q}) = 0$. Expanding this, we get: $\vec{P} \cdot \vec{P} - \vec{P} \cdot \vec{Q} + \vec{Q} \cdot \vec{P} - \vec{Q} \cdot \vec{Q} = 0$. Simplifying, we obtain: $P^2 - Q^2 = 0$, or $P^2 = Q^2$. Hence, $P = Q$, which corresponds to $A = B$.
Out of the following pairs of forces, the resultant of which can not be 18N




To determine which pair of forces cannot have a resultant of 18N, we use the triangle inequality theorem for vectors. For two vectors with magnitudes $a$ and $b$: The resultant $R$ satisfies $|a - b| \leq R \leq a + b$. For the given options: 1. $|11N - 7N| \leq 18N \leq 11N + 7N$ ⟹ $4N \leq 18N \leq 18N$ (Possible) 2. $|11N - 8N| \leq 18N \leq 11N + 8N$ ⟹ $3N \leq 18N \leq 19N$ (Possible) 3. $|11N - 29N| \leq 18N \leq 11N + 29N$ ⟹ $18N \leq 18N \leq 40N$ (Possible) 4. $|11N - 5N| \leq 18N \leq 11N + 5N$ ⟹ $6N \leq 18N \leq 16N$ (Not Possible) Thus, the pair 11N and 5N cannot have a resultant of 18N.
$ \vec A = 2 \hat i + 2 \hat j - \hat k $ $ \vec B = 2 \hat i - \hat j - 2 \hat k $ Find $ 3 \vec A - 2 \vec B $




No explanation available.
Linear momentajm of a particle is $ (3 \hat i + 2 \hat j - \hat k ) kgms^{-1} $. Find its magnitude




No explanation available.
$ \vec A \times \vec B = \vec C $ Then $ \vec C $ is perpendicular to




The cross product $ \\vec{A} \\times \\vec{B} = \\vec{C} $ results in a vector \\vec{C} that is perpendicular to both \\vec{A} and \\vec{B} regardless of the angle between them. This is a fundamental property of the cross product. Hence, \\vec{C} is perpendicular to \\vec{A} and \\vec{B} whatever the angle between them.