Physics MCQs for NEET — Practice Questions with Answers

$ I = I_0 cos ^2 \theta = { I_0 \over 2 } and { E^2 \over E_0^2 } = {1 \over 2} , { E \over E_0} = { 1 \over \sqrt 2 } $ $ \therefore { |E- E_0| \over E_0} = 0.29 = 29 \%$

$ from fig \theta p + 90 ^\circ +r = 180 ^\circ $ $ \therefore \theta_p + r = 90 ^\circ and \theta p -r = 22 ^\circ $ $ \therefore r = 34 ^ \circ $

$ here { I_1 \over I_2 } = \alpha , \therefore {E_1 \over E_2 } = \sqrt \alpha $ $ and { E_1 + E_2 \over E_1 - E_2 } = { \sqrt \alpha + 1 \over \sqrt \alpha - 1 } $ $ \therefore { I_{max} \over I_{min} } = { (\sqrt \alpha + 1)^2 \over ( \sqrt \alpha -1 ) ^2 } $ $ \therefore { I _{max} + I_{min} \over I_{max} - I_{min} } = { 2 ( \alpha + 1 ) \over 4 \sqrt \alpha }= { \alpha +1 \over 2 \sqrt \alpha } $

In Simple Harmonic Motion (SHM), the distance traveled by a particle in one complete cycle (periodic time \( T \)) is four times the amplitude \( A \). This is because the particle travels from one extreme to the other extreme, back to the initial extreme, and then to the starting point, covering a total distance of \( 4A \).

$ T = 2 \pi \sqrt { l \over g } $ When lift moves up with accleration g/3 the effective graritatianl acclenations in $ g^1 = g + { g \over 3} = {4g \obver 3 } $ $ \therefore new peliodic time T ' = 2 \pi \sqrt { l \over g } $

$ v_1 = { dy_1 \over dt } = 2 \times 10 cos ( 10t + \theta ) $ $ v_2 = -3 \times 10 sin t = 30 cos ( 10 + { \pi \over 2 } ) $ $ \therefore Phase difference = (10 t + \theta) - ( 10 + { \pi \over 2 } ) = \theta - { \pi \over 2 } $

For two springs with the same mass executing SHM, the maximum velocity \( v_{max} \) is given by \( v_{max} = A \omega \), where \( A \) is the amplitude and \( \omega \) is the angular frequency. Since \( \omega = \sqrt{\frac{k}{m}} \), and the maximum velocities are the same, we have \( A_1 \sqrt{\frac{k_1}{m}} = A_2 \sqrt{\frac{k_2}{m}} \). Simplifying, we get \( \frac{A_1}{A_2} = \sqrt{\frac{k_2}{k_1}} \). Therefore, the ratio of their amplitudes is \( \sqrt{\frac{k_2}{k_1}} \).

The velocity of the bob of a simple pendulum when it passes through its equilibrium position can be derived using the principle of conservation of mechanical energy. At the maximum displacement (angle θ), all the energy is potential, and at the equilibrium position, all the energy is kinetic. Therefore, we equate the potential energy at the maximum displacement to the kinetic energy at the equilibrium position. The formula for the velocity (v) is given by: $$ v = \\sqrt{2gl(1 - \\cos \\theta)} $$ where g is the acceleration due to gravity, l is the length of the pendulum, and θ is the angle of displacement.

$K_1 l _ 1 = K_2 l_2 = kl $ $ \therefore K_1 \left( l \over 4 \right) = K_2 \left( { 3 \over 4 } l \right) = kl $ $ force constant of spring having length {3 \over 4 } l \;in $ $ k_2 = { 4 \over 3 } k $

Amplitude of SHM given by $x = a sin \omega t+b cos \omega t $in $ A = \sqrt { a^2 + b^2 } = ( 3^2 + 4 ^2 ) ^ { 1/2 } = 5 m $