NEET Practice Questions (MCQs) with Answers & Solutions

Optical fibers work on the principle of total internal reflection. When light enters the fiber at a certain angle, it gets totally internally reflected within the core of the fiber, allowing it to travel long distances with minimal loss. This ensures that the light signal is transmitted efficiently from one end of the fiber to the other.

$ Here , I_A = I_1 + I_2 + 2 \sqrt { I_1 I_2 } $ $ cos { \pi \over 2 } = I \times 4 I \times 2 \sqrt { I \times 4 I } \times cos 90 ^\circ $ $ I_A = 5I $ $ and I_B = 5I + 2 \sqrt { I \times 4 I } \times cos \pi = 5 I -4 I = I $ $ \therefore I_A - I_B = 4 I $

$ 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 \( \sqrt{\frac{k_2}{k_1}} \).