Waves Test

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Waves Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

The irreducible phase difference in any wave of 5000 A from a source of light is

A
$$\pi$$

B
$$12\pi$$

C
$$12\pi \times{10}^{6}$$

D
$$\pi \times {10}^{6}$$

Solution
Question 2
1 / -0

Two waves having intensity ratio 9 : 1 produce interference. The ratio of maximum to minimum intensity is:

A
10 : 8

B
9 : 1

C
4 : 1

D
2 : 1

Solution

Hint:

If two waves of intensity $${{I}_{1}}\,and\,{{I}_{2}}$$ interfere, then the ration of maximum to minimum intensity in interference is given as,

$$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \dfrac{\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}}}{\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}}} \right)}^{2}}$$

Step 1: Assume values of intensities.

Given,

Ratio of Intensities of two waves $${{I}_{1}}\,and\,{{I}_{2}}$$
$${{I}_{1}}:{{I}_{2}}=9:1$$.

$$\Rightarrow {{I}_{1}}=9x,\,{{I}_{2}}=1x$$
for any constant $$x$$.

Step 2: Calculate ratio of intensities.

The ratio of maximum to minimum intensity in interference is given as,

$$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \dfrac{\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}}}{\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}}} \right)}^{2}}$$

$$\Rightarrow \dfrac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \dfrac{\sqrt{9x}+\sqrt{1x}}{\sqrt{9x}-\sqrt{1x}} \right)}^{2}}$$

$$\Rightarrow \dfrac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \dfrac{4}{2} \right)}^{2}}$$
$$\Rightarrow \dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{4}{1}$$
$$\Rightarrow {{I}_{\max }}:{{I}_{\min }}=4:1$$

The required ratio is 4:1. Option C.
Question 3
1 / -0

A stretched sting of length L,fixed at both ends can sustain stationary waves of wavelength $$\lambda $$ Choose which of the following value of wavelength is not possible?

A
2 L

B
4 L

C
L

D
$$\dfrac { L }{ 2 } $$

Solution
Question 4
1 / -0

The ratio longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is

A
$$\dfrac{4}{3}$$

B
$$\dfrac{525}{376}$$

C
$$25$$

D
$$\dfrac{900}{11}$$

Solution

$$\textbf{Hint:}$$ Use hydrogen atom spectrum wavelength formula.
$$\textbf{Step 1:}$$The wave number is given by
$$\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$$
For shortest wavelength, $$n_1=infinite\, n_2=1$$
$$\dfrac{1}{\lambda _{min}}=R(\dfrac{1}{1^2}-\dfrac{1}{(infinite)^2})=R$$
For longest wavelength, $$n_1=6$$, $$n_2=5$$
$$\dfrac{1}{\lambda _{max}}=R(\dfrac{1}{5^2}-\dfrac{1}{6^2})=\dfrac{11R}{900}$$
$$\textbf{Step 2:}$$Ratio,
$$\dfrac{1/\lambda _{min}}{1/\lambda_{max}}=\dfrac{R}{11R/900}$$
$$\dfrac{\lambda _{max}}{\lambda_{min}}=\dfrac{900}{11}$$
$$\textbf{Option D is correct.}$$.
Question 5
1 / -0

Three waves of amplitude $$10\mu m,4\mu m$$ and $$7\mu m$$ arrival at a point with successive phase difference of $$\pi /2$$. The amplitude of the resultant wave is

A
$$2\mu m$$

B
$$7\mu m$$

C
$$5\mu m$$

D
$$1$$

Solution
Question 6
1 / -0

Equations $$Y_1=a$$ sin $$\omega t$$ and $$Y_2=a/2$$ sin $$\omega t + a/2$$ cos wt represent S.H.M. The ratio of the amplitude of first to that of second S.H.M. is

A
1

B
0.5

C
2

D
$$\sqrt 2$$

Solution
Question 7
1 / -0

Two particles execute $$SHM$$ of same amplitude of $$20cm$$ with same period along the same line about the same equilibrium position. The maximum distance between the two is $$20cm$$. Their phase difference in radians is :-

A
$$\cfrac{2\pi}{3}$$

B
$$\cfrac{\pi}{2}$$

C
$$\cfrac{\pi}{3}$$

D
$$\cfrac{\pi}{4}$$

Solution

we have,
$$\frac{{d\left| {{{\overline r }_{12}}} \right|}}{{dt}} = 0$$
Rel. velocity=0
vel. same in 2nd and 3rd quad.
$$\begin{array}{l} A\sin \theta =10 \\ \theta =\pi /6 \\ 2\theta =\pi /3 \end{array}$$
Phase diff b/w particle.
So,
Option $$C$$ is correct answer.
Question 8
1 / -0

Two waves are represented as $$y_1 = 2a sin (\omega t + \pi/6)$$ and $$y_2 = -2a cos \Big \lgroup \omega t - \frac{\pi}{6} \Big \rgroup$$. The phase difference between the two waves is

A
$$\dfrac{\pi}{3}$$

B
$$\dfrac{4\pi}{3}$$

C
$$\dfrac{3\pi}{3}$$

D
$$\dfrac{5\pi}{6}$$

Solution
Question 9
1 / -0

The time lag between two particles vibrating in a progressive wave separated by a distance $$20$$ m is $$0.02 s$$ . The wave velocity if the frequency of the wave is $$500 Hz$$ , is

A
$$1000 ms^{-1}$$

B
$$500 ms^{-1}$$

C
$$2000 ms^{-1}$$

D
$$250 ms^{-1}$$

Solution

Given,
distance, $$s=20m$$
$$t=0.02s$$
$$f=500Hz$$
Velocity, $$v=\dfrac{s}{t}$$
$$v=\dfrac{20}{0.02}=1000m/s$$
The correct option is A.
Question 10
1 / -0

A particle moves with a simple harmonic motion in a straight line. In the first second starting from rest, it travels a distance $$a$$ and in the next second, it travels a distance $$b$$ in the same direction. The amplitude of the motion is:

A
$$\cfrac {2a^2}{3b-a}$$

B
$$\cfrac {3a^2}{3a-b}$$

C
$$\cfrac {2a^2}{3a-b}$$

D
$$\cfrac {3a^2}{3b-a}$$

Solution

Let the acceleration be $$f, f=\omega^2 x$$
Therefore, distance of the particle from the centre at any time $$t$$ is given by
$$x=r \cos (\omega t)$$, where $$r$$ is the amplitude
when $$t=1\ s, x=r-a$$
$$\therefore \ (r-a)=r\cos \omega$$
$$\cos \omega =\dfrac {r-a}{r}$$
When $$t=2\ s, x=r-a-b$$,
therefore $$r-a-b=\cos 2\omega$$
$$\therefore \ r-a-b=r(2\cos^2 \omega -1)$$
Substituting the value of $$\cos \omega $$ from equation.
we get $$r-a-b=r \left[2\dfrac {(r-a)^2}{r^2}-1\right]$$
$$r-a-b=\dfrac {2(r-a)^2}{r}-r$$
$$\therefore \ r(3a-b)=2a^2 \ \Rightarrow \ r=\dfrac {2a^2}{3a-b}$$