Physics MCQs for NEET — Practice Questions with Answers

For hearing beats, difference of frequencies should be approximately 10 Hz.

No of beats, x = $\Delta n=\frac{30}{3}=10Hz$

⇒ Also $\Delta n=v\left[\frac{{\displaystyle 1}}{{\displaystyle {\lambda }_1}}-\frac{{\displaystyle 1}}{{\displaystyle {\lambda }_2}}\right]=v\left[\frac{{\displaystyle 1}}{{\displaystyle 5}}-\frac{{\displaystyle 1}}{{\displaystyle 6}}\right]$=10⇒ v = 300 m/s

No. of beats = frequency difference = $\frac{4}{0.25}=16$

$n_x=300Hz, n_y=?$

x = beat frequency = 4 Hz, which is decreasing (4 → 2)

after increasing the tension of the string y.

Also tension of wire y increasing so $n_y\text{}\uparrow$ $(\because n\propto \sqrt{T})$

Hence $n_x-n_y\text{}\downarrow =x\text{}\downarrow$ → Correct

$n_y\text{}\uparrow -n_x=x\text{}\downarrow$ → Wrong

⇒ $n_y=n_x-x=300-4=296Hz$

From the given equations of progressive waves ${\omega }_1=500\pi$ and ${\omega }_2=506\pi$

∴ $n_1=250$ and $n_2=253$

So beat frequency $=n_2-n_1=253-250=3$ beats per sec

∴ Number of beats per min = 180.

Frequency $=\frac{360}{60}\times 60=360Hz.$

In a stationary wave, nodes are points of zero displacement, and antinodes are points of maximum displacement. The distance between a node and an adjacent antinode is one-quarter of the wavelength (λ/4). Therefore, the correct answer is (λ/4).

Comparing given equation with standard equation

$y=2a\sin \frac{2\pi x}{\lambda }\cos \frac{2\pi vt}{\lambda }$ gives us $\frac{2\pi }{\lambda }=\frac{\pi }{15}\Rightarrow \lambda =30$

Distance between nearest node and antinodes = $\frac{\lambda }{4}=\frac{30}{4}=7.5$

On comparing the given equation with the standard equation $y=2a\sin \frac{2\pi x}{\lambda }\cos \frac{2\pi vt}{\lambda }$

We get $\frac{2\pi }{\lambda }=\frac{\pi }{20}\Rightarrow \lambda =40$

The separation between two consecutive nodes = $\frac{\lambda }{2}=\frac{40}{2}=20cm$

Since the point x = 0 is a node and reflection is taking place from point x = 0. This means that reflection must be taking place from the fixed end and hence the reflected ray must suffer an additional phase change of π or a path change of $\frac{\lambda }{2}$.

So, if $y_{\text{incident}}=a\cos (kx-\omega t)$

⇒ $y_{\text{reflected}}=a\cos (-kx-\omega t+\pi )=-a\cos (\omega t+kx)$