Physics Test

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Physics Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Standing waves are formed on a string when interference occurs between two waves having :

A
The same amplitude travelling in the same direction with no phase difference between them

B
The same amplitude, travelling in the opposite direction with no phase difference between them

C
Different amplitudes travelling in the same direction

D
Different amplitudes travelling in the opposite direction

Solution

Standing waves are formed whenever two waves of identical frequency and same amplitude interfere with one another while travelling opposite directions along the same medium.

Hence, option B is correct answer.
Question 2
1 / -0

VA massless rod is suspended by two identical strings AB and CD of equal length. A block of mass mm is suspended from point O such that BO is equal to 'x'. Further, it is observed that the frequency of 1st harmonic (fundamental frequency) in AB is equal to 2nd harmonic frequency in CD. Then, length of BO is :

A
L/5

B
4L/5

C
3L/4

D
L/4

Solution

\(\frac{1}{2 l} \sqrt{\frac{T_{1}}{\mu}}=\frac{1}{l} \sqrt{\frac{T_{2}}{\mu}}\)
\(T_{2}=\frac{T_{1}}{4}\)
For rotational equilibrium \(T_{1} x=T_{2}(L-x) \Rightarrow x=\frac{L}{5}\)
Hence, the correct option is (A)
Question 3
1 / -0

Progressive waves are represented by the equation y₁ = a sin (ωt − x) and y₂= b cos (ωt − x) . The phase difference between waves is :

A
0°

B
45°

C
90°

D
180°

Solution

We know that \(\sin \left(\theta+90^{\circ}\right)=\cos \theta\)
So, \(y_{2}=b \cos (\omega t-x)=b \sin \left(\omega t-x+90^{\circ}\right)\)
Thus, the phase difference between the two waves is 90 degrees.
Hence, the correct option is (C)
Question 4
1 / -0

A standing wave in a pipe with a length \(L=1.2 \mathrm{m}\) is described by

\(y(x, t)=y_{0} \sin [(2 \pi / L) x] \sin [(2(\pi / L) x+\pi / 4)] .\) Based on above information, which one of

the following statement is incorrect.

(Speed of sound in air is \(300 \mathrm{ms}^{-1}\) ).

A
The pipe is closed at both ends

B
The wavelength of the wave could be 1.2m

C
There could be a node at x = 0 and antinode at x = L/2

D
The frequency of the fundamental mode of vibrations is 137.5 Hz

Solution

On comparing the equation with standard standing wave equation \(y=A_{o} \sin k x \cos \omega t\)
So \(k=\frac{2 \pi}{L}\)
as we know \(\omega \times k=2 \pi\) \(\operatorname{soc} x \frac{2 \pi}{L}=2 \pi\)
\(\omega=L=1.2 \mathrm{m}\)
On putting \(\mathrm{x}=0\) we get \(y=y_{o} \sin \left(\frac{2 \pi}{L} \times 0\right) \sin \left(\frac{2 \pi}{L} \times 0+\frac{\pi}{4}\right)\)
So \(y=0\) at \(\mathrm{x}=\mathrm{L}\)
\(y=y_{o} \sin \left(\frac{2 \pi}{L} \times L\right) \sin \left(\frac{2 \pi}{L} \times L+\frac{\pi}{4}\right)\)
\(y=0\)
so at \(\mathrm{x}=0,\) L we are getting nodes. We know velocity \(=\) wavelength \(\times\) frequency \(f=\frac{v}{\lambda}=\frac{300}{1.2}=250\)
Hence, the correct option is (D)
Question 5
1 / -0

Five sinusoidal waves have the same frequency \(500 \mathrm{Hz}\) but their amplitudes are in the ratio
\(2: \frac{1}{2}: \frac{1}{2}: 1: 1\) and their phase angles \(0, \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}\) and \(\pi\) respectively. The phase angle of
resultant wave obtained by the superposition of these five waves is:

A
30°

B
45°

C
60°

D
90°

Solution

The phasors of five waves can be represented as \(x_{1}=2 \hat{i}\)
\(x_{2}=\sqrt{3} / 4 \hat{i}+1 / 4 \hat{j}\)
\(x_{3}=1 / 4 \hat{i}+\sqrt{3} / 4 \hat{j}\)
\(x_{4}=\hat{j}\)
\(x_{5}=-\hat{i}\)
Resultant \(x=\sum_{i=1}^{5} x_{i}\)
\(x=\frac{5+\sqrt{3}}{4} \hat{i}+\frac{5+\sqrt{3}}{4} \hat{j}\)
Phase angle of resultant, \(\theta=\tan ^{-1} \frac{(5+\sqrt{3}) / 4}{(5+\sqrt{3}) / 4}\)
\(\Rightarrow \theta=45^{\circ}\)
Hence, the correct option is (B)
Question 6
1 / -0

A ripple is created in water. The amplitude at a distance of 5 cm from the point where the sound ripple was created is 4 cm. Ignoring damping, what will be the amplitude at the distance of 10 cm.

A
√16 cm

B
√8 cm

C
√4 cm

D
√2 cm

Solution

Spherical waves are created in
water. Energy of wave at distance \(r\) \(E_{r} \propto \frac{1}{r}\)
But \(E_{r} \propto A_{r}^{2}\)
where \(A_{r}\) is the amplitude of
wave \(\Longrightarrow A_{r} \propto \frac{1}{\sqrt{r}}\)
Given : \(A_{5}=4 \mathrm{cm}\)
\(r_{1}=5 \mathrm{cm} \quad r_{2}=10 \mathrm{cm}\)
\(\Rightarrow\)
\(A_{10}=\sqrt{\frac{r_{1}}{r_{2}}} A_{5}=\sqrt{\frac{5}{10}} \times 4=\sqrt{8}\)
\(\mathrm{cm}\)
Hence, the correct option is (B)
Question 7
1 / -0

The equation of sound wave is y = 0.0015 sin (62.4x + 316t) . Find the wavelength of this wave :

A
0.2 unit

B
0.1unit

C
0.3unit

D
none of these

Solution

The equation for sound wave is given as \(y=0.0015 \sin (62.4 x+316 t)\)
On comparing with \(y=A \sin (k x+w t),\) we get
\(k=62.4\)
Let the wavelength of the wave be \(\lambda\) \(\therefore \frac{2 \pi}{\lambda}=62.4\)
\(\Longrightarrow \lambda=0.1\) unit
Hence, the correct option is (B)
Question 8
1 / -0

It is possible to recognise a person by hearing his voice even if he is hidden behind a solid wall. This is due to the fact that his voice :

A
has a definite capacity

B
has definite quality

C
has definite pitch

D
can penetrate the wall

Solution

Each sound has a definite quality , due to this fact a person can be recognized without seeing him.
Hence, the correct option is (B)
Question 9
1 / -0

A sonometer consists of two wires of lengths 1.5m and 1m made up of different materials whose densities are 5g/cc and 8g/cc, and their respective radii are in the ratio 4:3. The ratio of tensions in these two wires, if their fundamental frequencies are equal is :

A
5 : 3

B
5 : 2

C
2 : 5

D
3 : 5

Solution
Question 10
1 / -0

Two tuning forks when sounded together produce 6 beats/s. The first fork has the frequency 3% higher than a standard one and the second has the frequency 2% less than the standard fork. The frequencies for the forks are :

A
126.3,120.3 Hz

B
162.7,156.7 Hz

C
136.2,130.2 Hz

D
123.6,117.6 Hz

Solution

\(\left(f+\frac{3 f}{100}\right)-\left(f-\frac{2 f}{100}\right)=6\)
\(\Rightarrow 5 f=600\) or \(f=120 \mathrm{Hz}\)
\(f+\frac{3 f}{100}=123.6 \mathrm{Hz}\)
\(f-\frac{2 f}{100}=120-\frac{2 \times 120}{100}=117.6 \mathrm{Hz}\)
Hence, the correct option is (D)

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Physics Test - 37

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Physics Test - 37

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SHARING IS CARING

Question 1

The same amplitude travelling in the same direction with no phase difference between them

The same amplitude, travelling in the opposite direction with no phase difference between them

Different amplitudes travelling in the same direction

Different amplitudes travelling in the opposite direction

Solution

Question 2

L/5

4L/5

3L/4

L/4

Solution

Question 3

0°

45°

90°

180°

Solution

Question 4

The pipe is closed at both ends

The wavelength of the wave could be 1.2m

There could be a node at x = 0 and antinode at x = L/2

The frequency of the fundamental mode of vibrations is 137.5 Hz

Solution

Question 5

30°

45°

60°

90°

Solution

Question 6

√16 cm

√8 cm

√4 cm

√2 cm

Solution

Question 7

0.2 unit

0.1unit

0.3unit

none of these

Solution

Question 8

has a definite capacity

has definite quality

has definite pitch

can penetrate the wall

Solution

Question 9

5 : 3

5 : 2

2 : 5

3 : 5

Solution

Question 10

126.3,120.3 Hz

162.7,156.7 Hz

136.2,130.2 Hz

123.6,117.6 Hz

Solution

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