[ 2 Marks Questions ] - Physics STD 11 Questions - Vidyadip

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[ 2 Marks Questions ]

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Question 12 Marks

The displacement $y$ of a wave travelling in the $x$-direction $;$ s given by $y=10^{-4} \sin (600 t-2 x+$ $\pi / 3$ )
Where $x$ is expressed in metres and $t$ is seconds. What is the speed of the wave motion (in $ms ^{-1}$ )?

Answer

$y=10^{-4} \sin \left(600 t-2 x+\frac{\pi}{3}\right) \text { and } y=a \sin \left(\omega t-k x+\frac{\pi}{3}\right) $
$ \therefore \omega=600 rad s ^{-1}, k=2 rad ^{-1} m$
$v=\frac{\omega}{k}=\frac{600}{2}=300 ms ^{-1} \text {} $

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Question 22 Marks

A police car with a siren of frequency 8 kHz ¡s moving with uniform velocity 36 km/h towards a tall building which reflect the sound waves. The speed of sound In air is 320m/s. What is the frequency of the siren heard by the car driver?

Answer

(a) Frequency received by the building.
$
f^{\prime}=\left(\frac{v}{v-v_c}\right)
$
The wall (source) reflect this frequency, So frequency heard by the car driver is
$
\begin{aligned}
f^{\prime \prime} & =\left(\frac{v+v_c}{v}\right) f^{\prime}=\left(\frac{v+v_c}{v}\right)\left(\frac{v}{v-v_c}\right) f \\
& =\left(\frac{v+v_c}{v-v_c}\right) f=\left(\frac{320+10}{320-10}\right) \times 8 kHz =\frac{33}{31} \times 8=8.5 kHz
\end{aligned}
$

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Question 32 Marks

A train moves towards a stationary observer with speed 34 mIs. The train sounds a whistle and its frequency registered by the observer is f1. If the train’s speed is reduced to 17 m/s, the frequency registered f2. If the speed of sound is 340 m/s, then find the ratio f1/f2

Answer

For the stationary observer $f_1=\frac{v}{v-v_s} \times f$
$
\begin{aligned}
\therefore \quad f_1 & =\frac{340}{340-34} \times f \text { and } f_2=\frac{340}{340-17} \times f \\
\frac{f_2}{f_1} & =\frac{340-17}{340-34}=\frac{19}{18} \Rightarrow f_1: f_2=19: 18
\end{aligned}
$

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Question 42 Marks

Tube A has both ends open, while B has on end closed otherwise the two tubes are identical. What Is the ratio of fundamental frequency of the tubes A and B?

Answer

The fundamental frequency for tube A with both ends open is $f _{ A }=\frac{v}{2 L }$
The fundamental frequency for tube $B$ with one end closed is $f _{ B }=\frac{v}{4 L }$ $\frac{f_{ A }}{f_{ B }}=\frac{v / 2 L }{v / 4 L }=2 \Rightarrow f_{ A }: f_{ B }=2: 1$

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Question 52 Marks

An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. What Is the percentage increase in the apparent frequency?

Answer

Here observer moves towards the stationary source.
$ v_0=-v / 5, v_s=0$
$\text { Apparent frequency } \quad f^{\prime}=\frac{v-v_o}{v-v_s} \times f=\frac{v-v / 5}{v-0} \times f=\frac{6}{5} v=1.2 v $
The percentage increase in apparent frequency
$\frac{f^{\prime}-f}{f} \times 100=\frac{1.2 f-f}{f} \times 100=20 \%$

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Question 62 Marks

Estimate the speed of sound $in$ air at standard temperature and pressure by using
(i) Newton's formula and
(ii) Laplace formula. The mass of 1 mole of air $=29.0 \times 10^{-3} kg$. For air, $\gamma=1.4$

Answer

Density of air,
$
\begin{aligned}
\rho & =\frac{\text { Mass of } 1 \text { mole of air }}{\text { Volume of } 1 \text { mole of air }} \\
& =\frac{29.0 \times 10^{-3} kg }{22.9 \text { litre }}=\frac{29.0 \times 10^{-3} kg }{22.4 \times 10^{-3} m ^3}=1.29 kg m ^{-3}
\end{aligned}
$
Standard pressure,
$
P =1.01 \times 10^5 Pa
$
(i) According to Newton's formula, speed of sound in air at S.T.P is
$
v=\sqrt{\frac{ P }{\rho}}=\sqrt{\frac{1.01 \times 10^5}{1.29}}=280 ms ^{-1}
$
(ii) According to Newton's formula speedof sound in air at S.T.P is
$
v=\sqrt{\frac{\gamma P }{\rho}}=\sqrt{\frac{1.4 \times 1.01 \times 10^5}{1.29}}=331.5 ms ^{-1}
$

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Question 72 Marks

A steel wire $0.72 m$ long has a mass of $5.0 \times 10^{-3} kg$. If the wire is under a tension of $60 N$. What is the speed of transverse waves on the wire?

Answer

Here $T =60 N$, Mass $=5.0 \times 10^{-3} Kg$, Length $=0.72 m$ mass per unit length, $\quad m=\frac{5.0 \times 10^{-3} kg }{0.72 m }=6.9 \times 10^{-3} kg m ^{-1}$
The speed of the transverse wave on the wire,
$
v=\sqrt{\frac{ T }{m}}=\sqrt{\frac{60 N }{6.9 \times 10^{-3} kg m ^{-1}}} \Rightarrow v=93 ms ^{-1}
$

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Question 82 Marks

The second overtone of an open organ pipe has the same frequency as the 1st overtone of a closed pipe L metre long. Then what will be the length of the open pipe.

Answer

$2^{\text {nd }} \text { overtone of an open organ pipe }=\frac{3 v}{2 L_O}$
$1^{\text {st }} \text { overtone of an closed organ pipe }=\frac{3 v}{4 L_C}$
$\frac{3 v}{3 L _{ O }}=\frac{3 v}{4 L _{ C }}$
$\frac{1}{ L _{ O }}=\frac{1}{2 L _{ C }}$
$L _{ O }=2 L _{ C }$
The Length of the open pipe is two times of the length of the closed pipe.

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Question 92 Marks

Two cars moving in opposite directions approach each other with speed of $22 ms ^{-1}$ and $16.5 ms ^{-}$ ${ }^1$ respectively. The driver of the first car blows a horn having a frequency $400 Hz$. To find the frequency heard by the driver of the second car.

Answer

$\begin{aligned} f_{ A } & =f\left[\frac{v+v_o}{v-v_s}\right]=400\left[\frac{340+16.5}{340-22}\right] \\ & =400\left[\frac{356.5}{318}\right]=400 \times 1.1210 \\ f_{ A } & =448.4 Hz \end{aligned}$

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Question 102 Marks

The fundamental frequency in an open organ pipe is equal to the 3rd harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm. What is the length of the open organ pipe.

Answer

For closed organ pipe, 3 rd harmonics $=\frac{3 v}{4 l}$
For open organ pipe, fundamental frequency, $=\frac{v}{2 l^{\prime}}$
Given,
$
\frac{3 v}{4 l}=\frac{v}{2 l^{\prime}} \text {}
$
$
\begin{aligned}
l^{\prime} & =\frac{4 l}{3 \times 2}=\frac{4 l}{6} \\
l=20 cm \quad l^{\prime} & =\frac{4 \times 20}{6} \\
l^{\prime} & =13.33 cm
\end{aligned}
$

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Question 112 Marks

Define reverberation time?

Answer

It is defined as the time which sound takes to fall in intensity to one millionth $\left(10^{-6}\right)$ part of its original intensity after it was stopped.

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Question 122 Marks

What is musical scale?

Answer

A series of notes whose fundamental frequencies have definite ratios and which produce a pleasing effect on the ear when sounded in succession constitute a musical scale.

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Question 132 Marks

What is meant by reverberation?

Answer

The persistence of audible sound after the source has ceased to emit sound is called reverberation.

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Question 142 Marks

What is meant by threshold of heating?

Answer

The lowest intensity of sound that can be perceived by the human ear is called threshold of hearing. For a sound of frequency $10 kHz$, the threshold of hearing is $10^{-12} Wm ^{-2}$

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Question 152 Marks

What are stationary waves?

Answer

When two identical waves of same amplitude and frequency travelling in opposite directionals with the same speed along the same path superpose each other, the resultant wave does not travel in the either direction and is called stationary or standing waves.

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Question 162 Marks

Define wave velocity or phase velocity?

Answer

The distance covered by a wave in the direction of its propagation per unit time is called the wave velocity.

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Question 172 Marks

What do you mean by phase of a wave?

Answer

The phase of a harmonic is a quantity that gives complete information of the wave at any time and at any position.

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Question 182 Marks

What is a plane progressive harmonic wave?

Answer

If during the propagation of a wave through a medium the particles of the medium vibrate simple harmonically about their mean positions, than the wave is said to be plane progressive harmonic wave.

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Question 192 Marks

What is a progressive wave?

Answer

A wave that travels from one point of the medium to another is called a progressive wave.

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Question 202 Marks

Define the term wave motion?

Answer

Wave motion is a kind of disturbances which travels through a medium due to repeated vibrations of the particles of the medium about their mean positions, the disturbance being handed over from one particle to the next.

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[ 2 Marks Questions ] - Physics STD 11 Questions - Vidyadip