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Physics 2 Marks Questions

PART - 2 CH - 14 Waves · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 24 questions.

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Question 12 Marks
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is $1.7 km s ^{-1}$ ? The operating frequency of the scanner is 4.2 MHz .
Answer
Given that :
Speed of sound $v =1.7 km / s$
$
v=1.7 \times 10^3 m / s
$
Frequency $n =4.2 MHz =4.2 \times 10^6 Hz$
Wave length $(\lambda)=$ ?
We know that wavelength $\lambda$
$
\begin{array}{l}
=\frac{v}{n}=\frac{1.7 \times 10^3}{4.2 \times 10^6} \\
=0.405 \times 10^{-3} m=0.405 mm
\end{array}
$
Wave length$\quad$$(\lambda)=0.41 mm$.
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Question 22 Marks
A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is $340 ms^{-1}$ and in water $1486 ms^{-1}$.
Answer
Given that :
Frequency $n =1000 kHz =10^6 Hz$
Speed of sound in air $v_a=340 m / s$
Speed of sound in water $v _{ W }=1486 m / s$
(a) Wave length of reflected sound
$
\begin{array}{l}
\lambda_a=\frac{v_a}{n} \\
\lambda_a=\frac{340}{10^6}=3.4 \times 10^{-4} m
\end{array}
$
(b) Wave length of transmitted sound
$
\lambda_w=\frac{v_w}{n}
$
Put the value
$
\begin{aligned}
\lambda_w & =\frac{1486}{10^6}=1.486 \times 10^{-3} m \\
& \simeq 1.49 \times 10^{-3} m
\end{aligned}
$
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Question 32 Marks
Explain why bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”.
Answer
When bats fly they produce ultrasonic sound of very high frequency (i.e. less/short wave length). Here ultrasonic sound is get reflect from an obstacle. Their ears are so sensitive and trained that they received information only about the distance of the obstacle and they also know about the nature of the reflecting surface.
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Question 42 Marks
Explain why in a sound wave, a displacement node is a pressure antinode and vice versa.
Answer
The node is the point in a sound wave where amplitude of cycle means dispalcement is zero. Since here we gets a compression and refractions due to this reason pressure will maximum. It called pressure antinode. While antinode is that point where amplitude will maximum means displacement will maximum. If pressure is minimum therefore this pressure point is node. So displacement node and pressure antinode point is same.
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Question 52 Marks
Two sitar strings $A$ and $B$ playing the note '$G a$' are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string $A$ is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz. what is the frequency of $B$ ?
Answer
We know that frequency $n \propto \sqrt{T}$.
Here, T is tension. In any wire if tension is decrease the frequency will decrease. So it can say that frequency of $A n_A$ and frequency of $B$ is $n_B$.
Given : $\quad n_{ A }-n_{ B } =6 Hz $
$n_{ A } =324 Hz $
$324-n_{ B } =6 $
$324-6 =n_{ B }$
or $\quad n_{B}=318 Hz$
Reduce tension of A is
$\Delta n =3 Hz $
$\therefore \text { Frequency of } A \text { is } 324 -3=321 $
$n_{B} =318 Hz$
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Question 62 Marks
Given below are some functions of $x$ and $t$ to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all :
(a) $y=2 \cos (3 x) \sin 10 t$
(b) $y=2 \sqrt{x-v t}$
(c) $y=3 \sin (5 x-0.5 t)+4 \cos (5 x-0.5 t)$
(d) $y=\cos x \sin t+\cos 2 x \sin 2 t$
Answer
(a) Given equation is function of $x$ and $t$ (multiplicative). So it represent to stationary wave.
(b) Given equation is not any periodic function. So it doesn't represent to any wave equation.
(c) In given function value of sine and cosine is same. So it represent to equation of progressive periodic wave.
(d) Given equation is sum of two function $\cos x$. $\sin t$ and $\cos 2 x . \sin 2 t$. In which each stationary wave represented. Therefore given equation is represented to superposition of stationary wave.
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Question 72 Marks
One end of a long string of linear mass density $8.0 \times 10 kg . m ^{-1}$ is connected to an electrically operated tunning fork of frequency 256 Hz . The other end of the string is passed over a stationary pulley and tied to the scale of a balance on which 90 kg weight are hanging. The pulley end absorb all the incoming energy due 90 which the amplitude of the waves reflected from this end is negligible. At $t=0$ the transverse displacement at the left end of the string at $x=0$ is zero $(y=0)$ and it is moving along the positive direction. The amplitude of the wave is 5.0 cm write the transverse displacement $y$ describe this wave on the string as a function of $X$ and $t$.
Answer
Given,
$\begin{aligned}M & =90 kg \\T & =Mg \\& =90 \times 9.8=882.0 N\end{aligned}$
Linear mass density $m=8 \times 10^{-3} kg / m$
Speed of transverse wave in string
$\begin{aligned}v & =\sqrt{\frac{T}{m}} \\v & =\sqrt{\frac{882}{8 \times 10^{-3}}}=3.32 \times 10^2 \\& =332 m / s
\end{aligned}$
The equation of progressive wave in positive x-direction
$y=a \sin (\omega t-k x)\ldots\ldots (1)$
Where, $\omega=2 \pi n$ and $a=5.0 cm$ or $a=0.05 m$
$\begin{aligned}n & =256 Hz \\\therefore \omega & =2 \pi \times 256=2 \times 3.14 \times 256 \\& =1607.68 \text { Radian per second } \\& \simeq 1.61 \times 10^3 \text { Radian per second }\end{aligned}$
Propagation constant
$\begin{aligned} k & =\frac{\omega}{ v }=\frac{2 \pi \times 256}{3.32 \times 10^2} \\ & =\frac{1607.68}{3.32 \times 10^2}=4.84 \text { per meter }\end{aligned}$
Since wave is propagating along positive x-direction then equation of wave.
$\begin{array}{l}\quad \quad y=a \sin (\omega t-k x) \\\text {or}\quad y=0.05 \sin \left(1.6 \times 10^3 t-4.84 x\right)\end{array}$
Hence, x and y in meter and is in second.
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Question 82 Marks
A disturbance pulse takes 0.2 second to travel across a long wire and return. Tension is created in the wire by continuously hanging a weight 100 times the total weight of the wire on a pulley. Find the total length of the wire. Here g = 9.8 m/s.
Answer
For the velocity of non-progressive wave.
$ \begin{aligned} v & =\sqrt{\frac{T}{m}} \\ v & =\sqrt{\frac{m \times l \times 100 \times g}{m}} \\ & {[\because \text { Total weight of wire }=(m l) g] } \\ \therefore v & =\sqrt{100 l g}\ldots\ldots (1) \end{aligned} $
Given : $\quad t=0.2=\frac{2 l}{ v }=\frac{2 l}{\sqrt{100 l g}}$
Put value from equation (1)
Taking square both side
$ \Rightarrow \quad(0.2)^2=\frac{4 l^2}{100 l g} $
$\Rightarrow \quad 0.04=\frac{4 l^2}{100 l g}$
$\Rightarrow \quad 0.04 \times 100=\frac{4 l}{g}$
$\begin{array}{ll}\Rightarrow & 4=\frac{4 l}{g} \\ \Rightarrow & l=g=9.8 meter\end{array}$
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Question 92 Marks
Explain the principle of superposition
Answer
Principle of superposition : When two waves propogate simultaneously in a medium the displacement of each particle of the medium at any instant is equal to vector sum of the separate displacement of both. This principle is called superposition principle.
If at any moment the displacement by both the wive $y_1$ and $y_2$ then resultant displacement will be :
$
y=y_1+y_2
$
The frequency and amplitude of the resultant wave obtained as a result of superposition depend the following things:
(i) Frequency of wave (ii) amplitude of wave (iii) phase difference between wave (iv) direction of wave propagation.
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Question 102 Marks
A disturbance pulse takes 0.2 second to travel across a long wire and return. Tension is created in the wire by continuously hanging a weight 100 times the total weight of the wire on a pulley. Find the total length of the wire. Here g = 9.8 m/s.
Answer
For the velocity of non-progressive wave.
$\begin{aligned}v & =\sqrt{\frac{T}{m}} \\v & =\sqrt{\frac{m \times l \times 100 \times g}{m}} \\& {[\because \text { Total weight of wire }=(m l) g] } \\\therefore \quad v & =\sqrt{1001 g}\quad\quad\ldots\ldots (1) \\\text { Given: } \quad t & =0.2=\frac{2 l}{v}=\frac{2 l}{\sqrt{100 l g}}\end{aligned}$
Put value from equation (1)
Taking square both side
$\Rightarrow(0.2)^2=\frac{4 l^2}{1001 g}$
$\begin{array}{lr}\Rightarrow & 0.04=\frac{4 l^2}{100 l g} \\ \Rightarrow & 0.04 \times 100=\frac{4 l}{g}\end{array}$
$\begin{array}{ll}\Rightarrow & 4=\frac{4 l}{g} \\ \Rightarrow & l=g=9.8 meter\end{array}$
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Question 112 Marks
The equation of transverse vibration of a 3 meter long string tied at both end is $y=0.3 \sin \left(\frac{2 \pi x}{3}\right) \cos 120 \pi t$ meter. Find the total number of node and antinode formed by the wire.
Answer
Given equation of transverse wave
$y=0.3 \sin \left(\frac{2 \pi x}{3}\right) \cos 120 \pi t$
Compare the equation
$y=a \sin \frac{2 \pi}{\lambda} x \cos \omega t$
$\lambda=3$ the length of string is also 3 m .

Image
Hence, 3 nodes and 3 antinode are being formed on the string.
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Question 122 Marks
Earthquake generate wave within the earth unlike gasous. Earth can sense both transverse (s) and longitudnal ( $P$ ) waves. The speed of $S$ wave is about 4.0 $km S ^{-1}$ and speed of $P$ wave is about $8 km s ^{-1}$. A seismometer records the $P$ and $S$ wave of an earth quake. The first $P$ wave arrives 4 minute earlier than first $S$ wave. Assuming that the wave travels in straight line. Find out what is the distance to the place where the earthquake occur.
Answer
Given
Speed of transverse (s) waves $v _1=4 km / s$
Time to reach seismometer $=t_1$
Speed of longitudnal wave $( P ) v _2=8 km / s$
Time to reach seismometer $= t _2$
Since both waves originate at the same place and reach the seismometer both will cover same distance $l$ Hence,
$\begin{aligned} & & l & = v _1 t_1= v _2 t_2 \\ \Rightarrow & & 4 t_1 & =8 t_2 \Rightarrow t_1=2 t_2\end{aligned}$
Given : $\quad t_1=t_2+(4 \times 60)$
Put value in equation (1)
$\begin{array}{l}t_2=240 sec . \text { and } \\ t_1=2 t_2=480 sec .\end{array}$
Distance to the place where earthquake occur
$l= v _1 t_1=4 \times 480=1920 km$
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Question 132 Marks
To open ended pipes standing near each other are played together. Prove that the length of one pipe among them is $y cm$. When converted to $\frac{ v y}{2 l^2}$ approximately. Length of pipe $l$ and speed of sound in velocity $v$.
Answer
The length of one pipe is $l$ then the length of other pipe will be $l \times  y$.
The speed of sound in air is v.
Then $n _1- n _2=$ frequency of pulse
$\begin{array}{l}\Rightarrow \quad \frac{v}{2 l}-\frac{v}{2(l+Y)}=\frac{v[l+Y-l]}{2 l(l+Y)} \\\Rightarrow \quad \frac{vy}{2 l(l+Y)} \\\text { But for } l \gg Y=\frac{vy}{2 l \times l\left(1+\frac{y}{l}\right)} \\\frac{vy}{2 l^2(1+0)}=\frac{vy}{2 l^2} \because l \gg y\end{array}$
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Question 142 Marks
Two waves of wavelength 50 cm and 50.5 cm produce 6 pulse per second. Find the velocity of solution in gas
Answer
Let speed of sound $= v cm / s$
Frequency of first wave $n_1=\frac{ v }{\lambda_1}=\frac{ v }{50}$
$\therefore \quad n_1=\frac{v}{50}\ldots\ldots (1)$
Frequency of second wave
$n_2=\frac{ v }{\lambda_2}$
$ n_2=\frac{v}{50.5} $
Pulses per second $=n_1-n_2=6$
$ \begin{aligned}\therefore 6 & =\frac{v}{50}-\frac{v}{50.5} \\ 6 & =v\left[\frac{50.5-50}{50 \times 50.5}\right] \end{aligned} $
or $\quad 6 \times 50 \times 50.5=0.5 v$
$\begin{array}{l} v =\frac{6 \times 50 \times 50.5}{0.5} \\ v =\frac{6 \times 50 \times 505}{5}=30300 cm / s \\ v =303 m / \text { second }\end{array}$
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Question 152 Marks
The string of a tunning fork and a tuning fork are played simultaneously. At length of the wire is kept at 0.49 meter or 0.50 meter. 4 beats per second are heared. what is the frequency of tunning fork?
Answer
The frequency of the tunning fork for length of wire $l$ will be $==\frac{1}{2 l} \sqrt{\frac{T}{m}}$
Let the frequency of tunning fork = n. vibrating with tunning frok 4 beat/second will generate. Therefore, frequency of tunning frok will be (n+4) or (n-4). 0.49 meter lengths wire frequency will greater the 0.50 meter length's wire. There- fore the frequency of 0.49 meter wire will be n + 4 and the frequency of 0.50 meter wire will be n-4.
$\begin{aligned}\therefore \quad n+4 & =\frac{1}{2 \times 0.49} \sqrt{\frac{T}{m}}\ldots\ldots (1) \\n-4 & =\frac{1}{2 \times 0.50} \sqrt{\frac{T}{m}}\ldots\ldots (2)\end{aligned}$
Divide equation (1) and (2)
$\begin{array}{rlrl}& & \frac{n+4}{n-4} & =\frac{0.50}{0.49}=\frac{50}{49} \\\Rightarrow & & 49 n+196 & =50 n-200 \\\Rightarrow & 196+200 & =50 n-49 n \\\Rightarrow & & 396 & =n\end{array}$
Then frequency of tuning fork = 396 vibration/second
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Question 162 Marks
When a tunning fork of unknown frequency is played with a 252 Hz frequency tuner, 2 beats per second are heard. Then find the frequency of unknown tuning fork.
Answer
The possible frequency of the unknown running bork will be :
$=254 \pm 4=258$ or 250
According to second experiment, the possible frequency of the unknown tunning fork will be:
$=252 \pm 2=254$ or 250
But both the experiments the frequency of the unknown tunning fork will same.
Hence, the frequency of the unknown tunning fork will be 250 vibration/second.
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Question 172 Marks
The tension of a wire is 2.5 kg. When the load increase its frequency changes to 3: 2. Find the original stress.
Answer
 Let original stress is T.
$n=\frac{1}{2 l} \sqrt{\frac{T}{m}}\ldots\ldots (1)$
To increase 2.5 kilogram then frequency is equal to
$\frac{3}{2} n .$
$\frac{3}{2} n=\frac{1}{2 l} \sqrt{\frac{T+2.5}{m}}\ldots\ldots (2)$
Divide equation (1) and (2)
$\Rightarrow \frac{2}{3}=\sqrt{\frac{T}{T+2.5}}$
Taking square both side
$\frac{4}{9}=\frac{T}{T+2.5}$
$4 T+10=9 T$
$10=9 T-4 T$
$5T=10$
$T=\frac{10}{5}=2 kg \cdot w t$.
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Question 182 Marks
The speed of wave is 360 m/sec. And the frequency is 500 Hz. and the phase difference between adjacent particle is $60^{\circ} $. What will be the path difference between them?
Answer
$\begin{array}{l}\text {Given :}\quad v=360 m / s \\n=500 Hz \\\phi=60^{\circ}=\pi / 3 \text { Radian }\end{array}$
Let the path difference between adjacent particle $=x$
$\begin{aligned} \text { Phase difference } & =\frac{2 \pi}{\lambda} \times \text { Path difference } \\
\text { Path difference } & =\frac{\lambda}{2 \pi} \times \text { Phase difference } \\ & =\frac{ v / n}{2 \pi} \times \frac{\pi}{3} \\ & =\frac{360}{500} \times \frac{1}{6}=0.12 \text { meter }\end{aligned}$
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Question 192 Marks
Necessary condition for sustained interference write down.
Answer
(i) Both wave have same frequencies.
(ii) Phase difference doesn't depend on time means it will co-hearent.
(iii) Both wave propagate in same direction in straitght line.
(iv) For rectify interference both wave should have same amplitude but this is not compulsory.
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Question 202 Marks
How are the fundamental and harmonic frequency related? What do you understand about the har monic of close organ pipe stretched string and an argon pipe.
Answer
The lowest frequency tone produced from a sound source is the fundamental tone. Harmonic are perfect multiples of fundamental frequency. Odd multiples of fundamental frequencies is odd harmonic and even multiples of frequencies is even harmonic. Close argon pipe generate only odd harmonic stretched string or open organ pipe is even or odd harmonic.
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Question 212 Marks
The wave $y_1=a \sin (\omega t-k x)$ is superimposed by another wave $y_2=a \sin (\omega t+k x)$ then find the ampli-tude of the resultant wave and also explain what are the condition for the amplitude to be maximum and minimum?
Answer
According to the principle of superposition the amplitude of the resultant wave is
$y = y _1+ y _2$
$\therefore \quad y=a \sin (\omega t-k x)+a \sin (\omega t+k x)$
Therefore, $y =2 a \sin \omega t \cos kx$ or $y =(2 a \cos kx ) \sin \omega t$ so, resultant wave amplitude $A=(2 a \cos k x)$
For maximum amplitude
$\begin{aligned} \cos k x= \pm 1 \Rightarrow & k x= m \pi \\ & (\text { where } m =0,1,2, \ldots \ldots \ldots .)\end{aligned}$
$x=\frac{ m \pi}{ k }($ where $m =0,1,2, \ldots \ldots \ldots)$
For minimum amplitude
$\begin{aligned} \cos k x=0 \Rightarrow & k x=(2 m-1) \pi \\ & (\text { where } m =1,2,3, \ldots \ldots \ldots \ldots .)\end{aligned}$
$\begin{aligned} \therefore \quad & x=\frac{(2 m-1) \pi}{ k } \\ & (\text { where } m =1,2,3, \ldots \ldots \ldots \ldots)\end{aligned}$
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Question 222 Marks
If carbon dioxide $\left( CO _2\right)$ gas is filled in the resonance tube appartus insted of air, then what will be the difference in the length of the resonance tube if the tube tunning frok is used?
Answer
If $CO _2$ gas is filled in the resonant tube appartus instead of air, then the length of the column of resonant $CO _2$ will be less air column. It will be following :
Fundamental frequency $n =\frac{ v }{2 l}$
$n =\frac{ v }{2 l} \quad \Rightarrow l=\frac{ v }{2 n }$
$\begin{array}{ll}\Rightarrow & l \propto v \quad(\prime n =\text { constant }) \\ \text { But } & v =\sqrt{\frac{\gamma P }{ d }} \Rightarrow v \propto \frac{1}{\sqrt{d}}\end{array}$
Due to heavily weight of $CO _2$ sound velocity in air is less.
Therefore, the length of resonant air column is less in $CO _2$ will be.
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Question 232 Marks
Write the formula for the speed of transverse waves in a stretched string and explain the symbols used in it
Answer
If both the end of a string are tied and when the string is pulled and released from the middle in a perpendicular direction, then transverse progressive waves start moving toward its ends.
Speed of these waves $\text v =\sqrt{\frac{ T }{m}}=\sqrt{\frac{ Mg }{\pi r^2 d}}$
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Question 242 Marks
What is wave? Mention those two properties of the medium due to which transmission of mechanical waves is possible in the medium?
Answer
Wave : The disturbance or movement generated due to a vibrating source means the form of transmission of energy to any other point is called wave and the method of energy transmission is called wave motion. Propagation of waves in any medium is possible due to properties like elasticity of the medium and if a medium does not have these properties then waves can't propagate in it. Water waves, sound waves, spring waves and waves produced in a rope etc. are mechnical waves.
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