Physics MCQs for NEET — Practice Questions with Answers

The superposing waves are represented by the following equations : $y_1=5\sin 2\pi (10t-0.1x)$, $y_2=10\sin 2\pi (20t-0.2x)$ Ratio of intensities $\frac{I_{\max }}{I_{\min }}$ will be :

The displacement of a particle is given by $x=3\sin \left(5\pi t\right)+4\cos \left(5\pi t\right)$. The amplitude of the particle is :

The two interfering waves have intensities in the ratio 9 : 4. The ratio of intensities of maxima and minima in the interference pattern will be :

If the ratio of amplitude of two waves is 4 : 3. Then the ratio of maximum and minimum intensity will be :

Equation of motion in the same direction is given by $y_1=A\sin (\omega t-kx)$, $y_2=A\sin (\omega t-kx-\theta )$. The amplitude of the medium particle will be 

The displacement of the interfering light waves are $y_1=4\sin \omega t$ and $y_2=3\sin \left(\omega t+\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)$. What is the amplitude of the resultant wave :

Two waves are represented by $y_1=a\sin \left(\omega t+\frac{{\displaystyle \pi }}{{\displaystyle 6}}\right)$ and $y_2=a\cos \omega t$. What will be their resultant amplitude :

The amplitude of a wave represented by displacement equation $y=\frac{1}{\sqrt{a}}\sin \omega t\pm \frac{1}{\sqrt{b}}\cos \omega t$ will be

A tuning fork sounded together with a tuning fork of frequency 256 Hz emits two beats. On loading the tuning fork of frequency 256 Hz with wax, the number of beats heard are 1 per second. The frequency of the other tuning fork is :

If two tuning forks A and B are sounded together, they produce 4 beats per second. A is then slightly loaded with wax, they produce 2 beats when sounded again. The frequency of A is 256. The frequency of B will be :