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Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - NEET MCQ


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10 Questions MCQ Test - Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28)

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Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 1

Which of the following equations represents a travelling wave?

Detailed Solution for Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 1

A traveling wave is a wave that moves through space or a medium and propagates in a particular direction. It can be represented by an equation of the form y = f(x - vt) or y = f(x + vt), where x is the spatial coordinate, t is time, v is the wave velocity, and f is a function of x and t.

Option A does not represent a traveling wave because it has an exponential term with x, which is not a characteristic of a traveling wave.

Option B also does not represent a traveling wave because it has an exponential term with x squared, which is not a characteristic of a traveling wave.

Option C represents a traveling wave because it has the form y = Asin(15x - 2t), where the wave is propagating in the positive x-direction with a speed of 2/15 m/s.

Therefore, the correct answer is Option C.

Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 2

Linear density of a string is 1.3 × 10?? kg/m and wave equation is y = 0.021 sin (x + 30t). Find the tension in the string when x is in meter and t is in second.

Detailed Solution for Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 2

Explanation:

- The wave equation given is y = 0.021 sin (x + 30t), which is in the form of y = A sin(kx - ωt) where A is the amplitude, k is the wave number, and ω is the angular frequency.
- From the given wave equation, we can identify that ω = 30 rad/sec.
- The linear density (μ) of the string is given as 1.3 x 10-3 kg/m.
- The tension (T) in the string can be found using the formula ω2 = T/μ.
- Substituting the given values into the formula, we get T = ω2 * μ.

Calculation:

- Substituting the given values, we get T = (302) * (1.3 x 10-3) kg m/s2.
- Simplifying this, we get T = 0.12 N.

So, the tension in the string is 0.12 Newtons. Thus, the correct answer is A: 0.12 N.

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Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 3

What will be the wavelength of a wave that is produced in air and travels at a speed of 300 m/s at the frequency of 60,000 Hz?

Detailed Solution for Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 3

Explanation:

To find the wavelength of a wave, we use the formula:

Wavelength = Speed / Frequency
 

  • First, we need to plug in the given values into the formula. The speed (v) is given as 300 m/s and the frequency (f) is 60,000 Hz (or 60,000 s-1).

  • So, the formula becomes Wavelength = 300 / 60,000.

  • Upon calculating this, we get the wavelength as 0.005 m.

  • However, the options are given in centimeters. To convert meters to centimeters, we multiply by 100 (since 1 meter = 100 centimeters).

  • So, 0.005 m * 100 = 0.5 cm.


Conclusion:

Therefore, the wavelength of the wave that is produced in air and travels at a speed of 300 m/s at the frequency of 60,000 Hz is 0.5 cm or Option A.

For more detailed explanations and practice problems, you can refer to the wave physics modules on EduRev.

Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 4

Equation of a wave is y = 15 x 10-² sin (300t – 100x) where x is in meter and t is in seconds. The wave velocity is:

Detailed Solution for Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 4

**Understanding the Wave Equation**

The wave equation you provided, y = 15 x 10-² sin (300t – 100x), is a standard equation for a wave propagating in the x direction.

It can be written in the form: y = A sin(ωt - kx), where:

- A is the amplitude of the wave,
- ω (omega) is the angular frequency,
- t is time,
- k is the wave number,
- x is the position.

**Finding the Wave Velocity**

The velocity (v) of a wave is related to its frequency (f) and wavelength (λ) by the equation v = fλ.

However, in our case, we don't have the frequency and the wavelength directly. We have the angular frequency (ω) and wave number (k) instead. But they are related to the frequency and wavelength by the equations ω = 2πf and k = 2π/λ.

So we can write the wave velocity as v = ω/k.

From the given wave equation, we can see that ω = 300 and k = 100.

So, the wave velocity v = ω/k = 300/100 = 3 m/s.

So, the correct answer is:
B: 3 m/s.

Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 5
Sound waves are not transmitted to long distance because
Detailed Solution for Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 5
Sound waves are not transmitted over long distances primarily because they are absorbed by the atmosphere. As sound waves travel through the air, they encounter air molecules that absorb some of the energy of the waves, causing the sound to become weaker and eventually dissipate. This absorption is a significant factor in limiting the range of audible sound.

Therefore, the correct answer is Option A.
Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 6

A radar is able to detect the reflected waves from an enemy's aeroplane after a time interval of 0.02 × 10-3 sec. If the velocity of the waves is 3 x 108 m/s. The distance of the aeroplane from the Radar will be:

Detailed Solution for Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 6

Understanding the Problem

  • We are given the time it takes for the radar to detect the reflected waves from the plane.
  • We are also given the speed of the waves.
  • We need to find the distance of the plane from the radar.

Calculations

  • The total time taken for the wave to travel to the airplane and back is given as 0.02 x 10-3 seconds.
  • Since the wave has to travel to the plane and back, the distance we want is half the total distance covered by the wave.
  • The total distance covered by the wave is the speed of the wave multiplied by the total time taken, which is 3 x 108 m/s x 0.02 x 10-3 sec = 6000 meters.
  • Therefore, the distance of the airplane from the radar is half of this, or 6000 meters / 2 = 3000 meters = 3 kilometers.

Answer
Therefore, the answer is C: 3.0 km. The plane is 3 kilometers away from the radar.

Please note: This solution is based on the assumption that the speed of the wave is constant and that it travels in a straight line to the plane and back. Actual conditions may vary.

This answer is provided by EduRev, a comprehensive educational platform that offers a wide range of learning resources for students of all levels.

Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 7
In which of the following medium sound would have maximum speed?
Detailed Solution for Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 7
The speed of sound in a medium depends on the properties of the medium, particularly its density and elasticity (stiffness). In general, sound travels fastest in a medium with high elasticity and low density.

Among the options:

Hydrogen gas (Option A) has low density but relatively low elasticity.
Air (Option B) has moderate density and elasticity.
Tap water (Option D) has higher density but lower elasticity compared to sea water.
Sea water (Option C) typically has higher density and higher elasticity compared to the other options, which results in a higher speed of sound. Therefore, sound would have the maximum speed in sea water.
Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 8
A pulse travels through a uniform medium, the speed of the pulse will be
Detailed Solution for Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 8
When a pulse travels through a uniform medium, such as a uniform string or a uniform medium in which the wave speed remains constant, the speed of the pulse remains constant. This is because the properties of the medium do not change along the path of the pulse, leading to a constant propagation speed for the pulse.

Therefore, the correct answer is Option C.
Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 9
Sound travels fastest in
Detailed Solution for Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 9
Sound travels fastest in solids among the given options. The speed of sound in a medium depends on its density and elasticity. Solids typically have higher elasticity and higher density compared to liquids and gases. This combination of properties results in a higher speed of sound in solids.

In a vacuum (Option D), sound cannot propagate at all because there are no particles (atoms or molecules) to transmit the sound waves.

Therefore, the correct answer is Option A.
Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 10

Which of the following will not travel with the speed of light?

Detailed Solution for Test: The Speed of a Travelling Wave & Principle of Superposition of Waves (September 28) - Question 10

Explanation

The answer is A: de-Broglie waves. Here's why:

de-Broglie waves

  • de-Broglie waves, or matter waves, are associated with moving particles. The speed of de-Broglie waves is dependent on the momentum of the associated particle, and not on the speed of light. In fact, the de-Broglie wavelength is inversely proportional to the momentum of a particle. Thus, they do not necessarily travel at the speed of light.

X-rays

  • X-rays are a type of electromagnetic radiation, like light waves. They have a much shorter wavelength than visible light, which enables them to penetrate materials that light cannot. But, like all electromagnetic waves, X-rays travel at the speed of light.

γ-rays (Gamma rays)

  • Gamma rays are also a type of electromagnetic radiation, with even shorter wavelengths than X-rays. Due to their high energy, gamma rays can cause significant damage to living tissues and DNA. However, like all electromagnetic waves, gamma rays travel at the speed of light.


So, only de-Broglie waves among the options given do not travel with the speed of light.

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