Top 50 Most Repeated LAWS OF MOTION PYQs | JEE MAINS
A curated collection of the most important questions from LAWS OF MOTION, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from LAWS OF MOTION, fully solved with step-by-step concepts to prepare for JEE MAINS.
For banking, $\tan\theta = \frac{v^2}{rg}$ balances centripetal force without friction....
Read Full Step-by-Step Solution →Using $T = \frac{2m_1m_2g}{m_1+m_2}$, $T = \frac{2\cdot2\cdot3\cdot10}{5} = 24\,\text{N}$....
Read Full Step-by-Step Solution →The normal reaction N = mg = 5 × 9.8 = 49 N. Maximum static friction = μ_s N = 0.4 × 49 = 19.6 N. The block begins to move when the applied force just...
Read Full Step-by-Step Solution →Convert the speed to metres per second: $80\ \text{km h}^{-1}=22.22\ \text{m s}^{-1}$. Using $\tan\theta = \frac{v^{2}}{r g}$ with $g=9.8\ \text{m s}^...
Read Full Step-by-Step Solution →The component of weight down the incline is $mg\sin\theta = 2 \times 10 \times \sin 30^\circ = 10\,\text{N}$. Maximum static friction is $\mu mg\cos\t...
Read Full Step-by-Step Solution →Applying Newton's second law to each mass and eliminating the tension gives a = (M – m)g/(M + m) = (3 – 2)·10 / (3 + 2) = 2 m/s²....
Read Full Step-by-Step Solution →The block experiences an acceleration \(a = F/m = 10/2 = 5\,\text{m/s}^2\). After \(3\,\text{s}\) the velocity becomes \(v = a t = 5 \times 3 = 15\,\t...
Read Full Step-by-Step Solution →Using $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$, $a = 2.5\,\text{m/s}^2$....
Read Full Step-by-Step Solution →The component of the block's weight down the plane is \(5g\sin30° = 24.5\,N\). The hanging mass exerts \(3g = 29.4\,N\) downward. Net force = 4.9 N, t...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Laws of Motion....
Read Full Step-by-Step Solution →For banking, $\tan\theta = \frac{v^2}{rg}$ balances centripetal force without friction....
Read Full Step-by-Step Solution →Using $T = \frac{2m_1m_2g}{m_1+m_2}$, $T = \frac{2\cdot2\cdot3\cdot10}{5} = 24\,\text{N}$....
Read Full Step-by-Step Solution →The normal reaction N = mg = 5 × 9.8 = 49 N. Maximum static friction = μ_s N = 0.4 × 49 = 19.6 N. The block begins to move when the applied force just...
Read Full Step-by-Step Solution →Convert the speed to metres per second: $80\ \text{km h}^{-1}=22.22\ \text{m s}^{-1}$. Using $\tan\theta = \frac{v^{2}}{r g}$ with $g=9.8\ \text{m s}^...
Read Full Step-by-Step Solution →The component of weight down the incline is $mg\sin\theta = 2 \times 10 \times \sin 30^\circ = 10\,\text{N}$. Maximum static friction is $\mu mg\cos\t...
Read Full Step-by-Step Solution →Applying Newton's second law to each mass and eliminating the tension gives a = (M – m)g/(M + m) = (3 – 2)·10 / (3 + 2) = 2 m/s²....
Read Full Step-by-Step Solution →The block experiences an acceleration \(a = F/m = 10/2 = 5\,\text{m/s}^2\). After \(3\,\text{s}\) the velocity becomes \(v = a t = 5 \times 3 = 15\,\t...
Read Full Step-by-Step Solution →Using $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$, $a = 2.5\,\text{m/s}^2$....
Read Full Step-by-Step Solution →The component of the block's weight down the plane is \(5g\sin30° = 24.5\,N\). The hanging mass exerts \(3g = 29.4\,N\) downward. Net force = 4.9 N, t...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Laws of Motion....
Read Full Step-by-Step Solution →For banking, $\tan\theta = \frac{v^2}{rg}$ balances centripetal force without friction....
Read Full Step-by-Step Solution →Using $T = \frac{2m_1m_2g}{m_1+m_2}$, $T = \frac{2\cdot2\cdot3\cdot10}{5} = 24\,\text{N}$....
Read Full Step-by-Step Solution →The normal reaction N = mg = 5 × 9.8 = 49 N. Maximum static friction = μ_s N = 0.4 × 49 = 19.6 N. The block begins to move when the applied force just...
Read Full Step-by-Step Solution →Convert the speed to metres per second: $80\ \text{km h}^{-1}=22.22\ \text{m s}^{-1}$. Using $\tan\theta = \frac{v^{2}}{r g}$ with $g=9.8\ \text{m s}^...
Read Full Step-by-Step Solution →The component of weight down the incline is $mg\sin\theta = 2 \times 10 \times \sin 30^\circ = 10\,\text{N}$. Maximum static friction is $\mu mg\cos\t...
Read Full Step-by-Step Solution →Applying Newton's second law to each mass and eliminating the tension gives a = (M – m)g/(M + m) = (3 – 2)·10 / (3 + 2) = 2 m/s²....
Read Full Step-by-Step Solution →The block experiences an acceleration \(a = F/m = 10/2 = 5\,\text{m/s}^2\). After \(3\,\text{s}\) the velocity becomes \(v = a t = 5 \times 3 = 15\,\t...
Read Full Step-by-Step Solution →Using $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$, $a = 2.5\,\text{m/s}^2$....
Read Full Step-by-Step Solution →The component of the block's weight down the plane is \(5g\sin30° = 24.5\,N\). The hanging mass exerts \(3g = 29.4\,N\) downward. Net force = 4.9 N, t...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Laws of Motion....
Read Full Step-by-Step Solution →For banking, $\tan\theta = \frac{v^2}{rg}$ balances centripetal force without friction....
Read Full Step-by-Step Solution →Using $T = \frac{2m_1m_2g}{m_1+m_2}$, $T = \frac{2\cdot2\cdot3\cdot10}{5} = 24\,\text{N}$....
Read Full Step-by-Step Solution →The normal reaction N = mg = 5 × 9.8 = 49 N. Maximum static friction = μ_s N = 0.4 × 49 = 19.6 N. The block begins to move when the applied force just...
Read Full Step-by-Step Solution →Convert the speed to metres per second: $80\ \text{km h}^{-1}=22.22\ \text{m s}^{-1}$. Using $\tan\theta = \frac{v^{2}}{r g}$ with $g=9.8\ \text{m s}^...
Read Full Step-by-Step Solution →The component of weight down the incline is $mg\sin\theta = 2 \times 10 \times \sin 30^\circ = 10\,\text{N}$. Maximum static friction is $\mu mg\cos\t...
Read Full Step-by-Step Solution →Applying Newton's second law to each mass and eliminating the tension gives a = (M – m)g/(M + m) = (3 – 2)·10 / (3 + 2) = 2 m/s²....
Read Full Step-by-Step Solution →The block experiences an acceleration \(a = F/m = 10/2 = 5\,\text{m/s}^2\). After \(3\,\text{s}\) the velocity becomes \(v = a t = 5 \times 3 = 15\,\t...
Read Full Step-by-Step Solution →Using $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$, $a = 2.5\,\text{m/s}^2$....
Read Full Step-by-Step Solution →The component of the block's weight down the plane is \(5g\sin30° = 24.5\,N\). The hanging mass exerts \(3g = 29.4\,N\) downward. Net force = 4.9 N, t...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Laws of Motion....
Read Full Step-by-Step Solution →For banking, $\tan\theta = \frac{v^2}{rg}$ balances centripetal force without friction....
Read Full Step-by-Step Solution →Using $T = \frac{2m_1m_2g}{m_1+m_2}$, $T = \frac{2\cdot2\cdot3\cdot10}{5} = 24\,\text{N}$....
Read Full Step-by-Step Solution →The normal reaction N = mg = 5 × 9.8 = 49 N. Maximum static friction = μ_s N = 0.4 × 49 = 19.6 N. The block begins to move when the applied force just...
Read Full Step-by-Step Solution →Convert the speed to metres per second: $80\ \text{km h}^{-1}=22.22\ \text{m s}^{-1}$. Using $\tan\theta = \frac{v^{2}}{r g}$ with $g=9.8\ \text{m s}^...
Read Full Step-by-Step Solution →The component of weight down the incline is $mg\sin\theta = 2 \times 10 \times \sin 30^\circ = 10\,\text{N}$. Maximum static friction is $\mu mg\cos\t...
Read Full Step-by-Step Solution →Applying Newton's second law to each mass and eliminating the tension gives a = (M – m)g/(M + m) = (3 – 2)·10 / (3 + 2) = 2 m/s²....
Read Full Step-by-Step Solution →The block experiences an acceleration \(a = F/m = 10/2 = 5\,\text{m/s}^2\). After \(3\,\text{s}\) the velocity becomes \(v = a t = 5 \times 3 = 15\,\t...
Read Full Step-by-Step Solution →Using $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$, $a = 2.5\,\text{m/s}^2$....
Read Full Step-by-Step Solution →The component of the block's weight down the plane is \(5g\sin30° = 24.5\,N\). The hanging mass exerts \(3g = 29.4\,N\) downward. Net force = 4.9 N, t...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Laws of Motion....
Read Full Step-by-Step Solution →