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This is second of our many quizzes on Types of waves . Click on this to access more such competitive quizzes on SHM Waves and Optics.

Types of waves Quiz 2

physics

Which type of wave maintains average temperature on the earth surface by means of greenhouse effect? 

The process of absorption of waves which are reflected back by the earth surface due to the greenhouse gases present in the atmosphere causes greenhouse effect and hence the temperature increases. It is found that much of the infrared radiations at longer wavelengths are absorbed by these gases. 

  Greenhouse effect is a natural process of absorption of infrared waves that are reflected back by the earth surface due to the greenhouse gases present in the atmosphere. If earth is not having an atmosphere, its temperature will be lower than that of it is now. Therefore, infrared rays of the sun keeps the earth’s surface warm.

Radio waves are used for communication purposes. But which band of frequency in it is used for mobile phone communication? 

Radio waves are electromagnetic waves which are having a frequency range from high 300 GHz to low 3 kHz. So, they are further divided into some sub categories. Mobile phones usually work under a frequency range of 840 MHz to 935 MHz. 

  Radio waves are usually used for communication purposes. But they are further classified into specific groups based on its frequencies. In which UHF, Ultra high frequency band which ranges from 300 MHz to 3 GHz. Mobile phones use portions of radio frequency spectrum allocated to the mobile service to receive and transmit the signals. They are included precisely in the range of 840 MHz to 935 MHz.

A rope of mass 2 kg is tied between two trees 8 m apart for placing wet washed clothes. Now, a person pulls the rope downward and releases. What type of wave will be formed inside the rope and find the velocity of that wave in that rope if it is having a tension of 16 N.

The velocity of a wave formed on a stretched string is directly proportional to the tension developed inside the string and inversely proportional to the mass per unit length of that string. Velocity is given by the formula $v=\sqrt{\dfrac{T}{\left( \dfrac{m}{L} \right)}}$. By using this formula, we can easily find the velocity of the wave. 

 $\text{The type of wave formed on a stretched string}\\ \text{ if we create a disturbance on it will be a transverse wave.}\\ \text{The velocity of this wave can be found using this formula.}\\ v=\sqrt{\dfrac{T}{\left( \dfrac{m}{L} \right)}}\\ \text{Then, velocity of the wave will be,}\\ v=\sqrt{\dfrac{T}{\left( \dfrac{m}{L} \right)}} \\ \Rightarrow v =\sqrt{\dfrac{16}{\left( \dfrac{2}{8} \right)}}\\ \Rightarrow v =\sqrt{\dfrac{16\times 4}{1}}\\ \Rightarrow v =4\times 2\\ \therefore v =8m/s \\ \text{Therefore, the speed of the transverse wave formed on the rope will be 8m/s.}$ 

What is the angle formed between wave particles and energy particles in a pressure wave?

First of all we will need to find what type of a wave is a pressure wave. It is a longitudinal wave which travels by creating pressure variation in the medium as compressions and rarefaction. So, the angle between the eave and the particle motion will be the same as that of a longitudinal wave. 

 Pressure waves are longitudinal waves. Sound wave is a pressure wave. They travel in a medium by the particles vibrating to and fro in a medium. As the wave and the particles in the medium are vibrating in the same direction, the angle between the wave particles and energy particles will be zero. Therefore, the angle formed between wave particles and energy particles in a pressure wave is equal to zero.

An excited electron has an energy of 2 eV. Then, what type of wave does it emit while coming back to ground state?  

We can find the nature of waves emitted during the emission by the electron using the equation which connects energy of the electron and frequency of the wave produced by it. By finding the frequency, we can compare it with the known frequencies and find out the type of wave. The formula is given as $E=h\upsilon $.

 $\text{We must know that energy of a wave is directly proportional to its frequency.}\\ \text{Energy is given by,}\\ E=h\upsilon \\ \text{Then, frequency is given by,}\\ \upsilon =\dfrac{E}{h} \\ \upsilon =\dfrac{2\times 1.6\times {{10}^{-19}}}{6.626\times {{10}^{-34}}}=0.483\times {{10}^{15}}Hz \\ \therefore \upsilon =4.83\times {{10}^{14}}Hz \\ \text{This frequency comes at a range under the visible red light region.}\\ \text{The frequency of visible red light ranges from}\, \sim 4\times {{10}^{14}}Hz\text{ to }4.84\times {{10}^{14}}Hz.\\ \text{Therefore, the wave emits visible red light}\\ \text{while coming back to ground state from the excited state.}$ 