2 Marks Questions

Question 512 Marks

Explain why (or how):
Bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,

Answer

Bats emit ultrasonic waves of large frequencies. When these waves are reflected from the obstacles in their path, they give them the idea about the distance, direction, size and nature of the obstacle.

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Question 522 Marks

A transverse harmonic wave on a string is described by
$\text{y}(\text{x, t})=3.0\sin\big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\big)$
where x and y are in cm and t in s. The positive direction of x is from left to right.
What is the least distance between two successive crests in the wave?

Answer

3.49m
Explanation:
Given,
$\text{y}(\text{x, t})=3\sin\big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\big)$
the smallest distance between two adjacent crests in the wave, $\lambda=\frac{2\pi}{\text{k}}=\frac{2\pi}{0.018}=349\text{cm}$

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Question 532 Marks

An observer at a sea-coast observes waves reaching the coast. What type of waves does he observe? Why?

Answer

Elliptical waves while the waves on the surface of water are transverse, the waves just below the surface of water are longitudinal. So the resultant waves are elliptical.

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Question 542 Marks

Why bells are made of metal and not of wood?

Answer

Metals have less damping than wood. So, bells are made of metals.

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Question 552 Marks

Explain graphically that number of beats formed per second is $n=v_1-v_2$ where $v_1$ and $v_2$ be the frequency of two superimposing waves.

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Question 562 Marks

The apparent frequency of the whistle of an engine changes in the ratio 3 : 2 as the engine passes a stationary observer. If the velocity of sound is 330m/s, calculate the velocity of the engine.

Answer

$\frac{\text{v}_1}{\text{v}_2}=\frac{\text{c}+\text{u}_0}{\text{c}-\text{u}_0}=\frac{3}{2}$
Solving we get, $\text{u}_0=\frac{\text{c}}{5}=\frac{330}{5}=66\text{ms}^{-1}.$

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Question 572 Marks

Explain your answers.
What is the amplitude of a point 0.375m away from one end?

Answer

Amplitude at a point x = 0.375m will be obtained by putting $\cos(120\pi\text{ t})$ as + 1 in the wave equation
$\therefore\ \text{A(x)}=0.06\sin\Big(\frac{2\pi}{3}\times0.375\Big)\times1=00.06\sin\frac{\pi}{4}=0.042\text{m}$

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Question 582 Marks

Given below are some functions of x and t to represent the displacement of an elastic wave.
$\text{y}=5\cos(4\times)\sin(20\text{t})$

Answer

A travelling wave along (-x) direction must have $+\text{kx}\ \text{i}.\text{e}.,$ in $(\text{iv})\text{y}=100\cos(100\pi\text{t}+0.5)$ so (a) (iv).

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Question 592 Marks

Two sound sources produce 20 beats in 5s. By how much do their frequencies differ?

Answer

Number of beats per second $=\frac{20}{5}=4$
$\therefore \text{V}_1-\text{V}_2=4.$

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Question 602 Marks

What is the ratio of frequencies of fundamental tone and various overtones formed in a vibrating string?

Answer

The frequencies of various harmonics (or fundamental tone and overtones) in vibrating string are in the ratio:
$V_1: V_2: V_3 \ldots \ldots=1: 2: 3: 4 \ldots \ldots$

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Question 612 Marks

If a balloon is filled will $\text{CO}_2$ gas, then how can it behave as a lens for sound waves? If it was filled with hydrogen gas, then what will happen?

Answer

Velocity of sound in $\text{CO}_2$ is less than that in air. Therefore, balloon will behave as a convex lens for sound waves. In hydrogen, velocity of sound is greater than that in air. Therefore, balloon filled with hydrogen will behave as a concave lens.

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Question 622 Marks

A steel wire has a length of 12.0m and a mass of 2.10kg. What should be the tension in the wire so that the speed of transverse wave on the wire equals the speed of sound in dry air at 20°C? Take the speed of sound at 20°C = 343m/s.

Answer

$\text{l}=12\text{m}$ and $\text{M}=2.10\text{kg}$
$\therefore\mu=\frac{2.10}{12}=0.175\text{kg/m}$
and $\text{v}=343\text{m/s}$
Using, $\text{v}=\sqrt{\frac{\text{T}}{\mu}}$
$\text{T}=\text{v}^2\mu=(343)^2\times0.175$
$=20.5\times10^4\text{N}$

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Question 632 Marks

In an experiment, it was found that a tuning fork and a sonometer wire gave 5 beats per second, both when the length of the wire was 1m, and 1.05m. Calculate the frequency of the fork.

Answer

We know $\text{v}=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\mu}}$
I case: $\text{v}_1=\frac{1}{2\times1}\sqrt{\frac{\text{T}}{\mu}}\ ... (\text{i})$
II case: $\text{v}_2=\frac{1}{2\times1.05}\sqrt{\frac{\text{T}}{\mu}}\ ...\text{(ii)}$
Let v be the frequecy of the tuning fork, therefore,
$\text{v}_1-\text{v}=5$ and $\text{v}-\text{v}_2=5$
$\text{v}_1=\text{v}+5$
and $\text{v}_2=\text{v}-5$
$\frac{\text{v}_1}{\text{v}_2}=\frac{5+\text{v}}{\text{v}-5}$
From (i) and (ii), we have
$\frac{\text{v}+5}{\text{v}-5}=\frac{1}{2}\times2\times1.05=1.05$
$\frac{\text{v}+5}{\text{v}-5}=\frac{105}{100}=\frac{21}{20}$
$\Rightarrow\text{v}=205\text{Hz}$

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Question 642 Marks

The transverse displacement of a string (clamped at its both ends) is given by
$\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\text{x}\Big)\cos(120\pi\text{ t})$
where x and y are in m and t in s. The length of the string is 1.5m and its mass is $3.0 \times 10^{–2}kg$.
Answer the following:
Does the function represent a travelling wave or a stationary wave?

Answer

The general equation representing a stationary wave is given by the displacement function:
$\text{y}(\text{x, t})=2\text{a}\sin\text{kx}\cos\omega\text{t}$
This equation is similar to the given equation:
$\text{y}(\text{x, t})=0.06\sin\Big(\frac{2}{3}\text{x}\Big)\cos(120\pi\text{ t})$
Hence, the given function represents a stationary wave.

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Question 652 Marks

If the frequency of a tuning fork is 400Hz and the velocity of sound in air is 320m/s, find how far does the sound travel when the fork executes 30 vibrations.

Answer

$\text{v}=320\text{m/s};$
$\text{v}=400\text{Hz};$
$\lambda=\frac{\text{v}}{\text{v}}=\frac{320}{400}=0.8\text{m}$
$\therefore$ The required distance travelled during the time the fork makes 30 vibrations = 30 × 0.8 = 24m.

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Question 662 Marks

For the harmonic travelling wave $\text{y}=2\cos2\pi(10\text{t}-0.0080\text{x}+3.5)$ where x and y are in cm and t is second. What is the phase difference between the oscillatory motion at two points separated by a distance of:
4m

Answer

$\text{y}=2\cos2\pi(10\text{t}-0.0080\text{x+3.5})$
$\text{y}=2\cos(20\pi\text{t}-0.0016\pi\text{x}+7.0\pi)$
Wave is propagated in $+\text{x}$ direction because $\omega\text{t}$ and kx are in with opposite sign standard equation $\text{y}=\text{a}\cos(\omega\text{t}-\text{kx}+\phi)$
a = 2, $\omega=20\pi,\ \text{k}=0.016\pi$ and $\phi=7\pi$
Path difference p =4 m (given) = 400cm
Phase difference $\Delta\phi=\frac{2\pi}{\lambda}\times\text{p}=\frac{2\pi}{\lambda}\times400$
$\Delta\phi=\text{k}\times400=0.016\pi\times400$
Phase difference $\Delta\phi=6.4\pi$ rad.

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Question 672 Marks

A child blows air at one end of a straw and slowly cuts pieces of the straw from the other end. What will be the outcome that will be observed?

Answer

As the pipe gets cut, the length of the resonating column varies, and so at a particular length, there will be an audible frequency that will be heard.

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Question 682 Marks

Show graphically, the intensity variation of beats formation?

Answer

When beats are formed by two sources having same amplitude but slightly different frequencies, the amplitude is given by $2\text{A}\cos2\pi\text{v}_\text{m}\text{t}$ where $\text{v}_\text{m}=\frac{\text{v}_1-\text{v}_2}{2}.$ Intensity $=4\text{A}^2\cos^2(2\pi\text{v}_\text{m}\text{t})$ and is shown graphically here.

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Question 692 Marks

Define the terms 'node' and 'antinode'?

Answer

Node: It is a point on stationary wave at which amplitude of vibration of the particle is zero.
Antinode: It is a point on stationary wave at which amplitude of vibration of the particle is maximum.

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Question 702 Marks

An open pipe resonates with a frequency v. When half of it is immersed in a dense liquid, what is the fundamental frequency?

Answer

$\text{V}=\frac{\nu}{2\text{l}}$ with fundamental frequency. When half immersed in a denser liquid, it will act as a closed pipe of length $\frac{\text{l}}{2}.$
$\therefore \text{V}'=\frac{\nu}{4\text{l}'}=\frac{\nu}{\frac{4\text{l}}{2}}=\frac{\nu}{2\text{l}}$
So, V = V'

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Question 712 Marks

If oil of density higher than that of water is used in place of water in a resonance tube, how does the frequency change?

Answer

The frequency is governed by the air column and does not depend upon the nature of the liquid. So frequency would not change.

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Question 722 Marks

What is the condition to be obeyed for receiving an echo?

Answer

The separation between the reflecting surface and source of sound ‘d’ should be greater than $\Big(\frac{1}{20}\Big)^{\text{th}}$ of the velocity of sound, since persistence of sound in human ears is $\frac{1}{10}\text{sec}.$

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Question 732 Marks

The frequencies of the two tuning forks A and B are 250Hz and 255Hz respectively. Both are sounded together. How many beats will be heared in 5s?

Answer

Given: $\mathrm{v}_{\mathrm{A}}=250 \mathrm{~Hz}, \mathrm{v}_{\mathrm{B}}=255 \mathrm{~Hz}, \mathrm{t}=5 \mathrm{~s}$
$\mathrm{v}_{\mathrm{B}}-\mathrm{v}_{\mathrm{A}}=255 \mathrm{~Hz}-250 \mathrm{~Hz}=5 \mathrm{~Hz}=$ Beat frequency
$\therefore$ No. of beats heard in $5 s=5 \times 5=25$

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Question 742 Marks

Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is $29.0 \times 10^{-3} kg$.

Answer

We know that 1 mole of any gas occupies 22.4 litres at STP. Therefore, density of air at STP is:
$
\begin{aligned}
\rho_o= & \text { (mass of one mole of air)/ (volume of one } \\
& \text { mole of air at STP) } \\
& =\frac{29.0 \times 10^{-3} kg }{22.4 \times 10^{-3} m ^3} \\
& =1.29 kg m ^{-3}
\end{aligned}
$
According to Newton's formula for the speed of sound in a medium, we get for the speed of sound in air at STP,
$
v=\left[\frac{1.01 \times 10^5 N m ^{-2}}{1.29 kg m ^{-3}}\right]^{1 / 2}=280 m s ^{-1}
$

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Physics 2 Marks Questions