Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Electromagnetic Waves

Multiple Choice Questions

1.

In case of linearly polarised light, the magnitude of the electric field vector

• does not change with time

• varies periodically with time

• increases and decreases linearly with time

• is parallel to the direction of propagation

B.

varies periodically with time

The orientation of a linearly polarized electromagnetic wave is defined by the direction of the electric field vector. For example, if the electric field vector is vertical ( alternately up and down as the wave travels ) the radiation is said to be vertically polarized.

The magnitude of electric field vector variesperiodically with time because it is the form ofelectromagnetic wave.

2.

The minimum wavelength of X-rays emitted by X-ray tube is 0.4125 Å. The accelerating voltage is :

• 30 kV

• 50 kV

• 80 kV

• 60 kV

A.

30 kV

${\mathrm{E}}{}{=}{}\frac{12400}{\mathrm{\lambda }}$

=$\frac{12400}{0.4125}$=30060≈ 30kV

3.

Light waves of wavelength$\mathrm{\lambda }$ is incident on a metal of work function$\phi$ . Maximum velocity of the electron is :

• ${\left[\frac{2\left(\mathrm{h\lambda }-\mathrm{\phi }\right)}{\mathrm{m}}\right]}^{1/2}$

• ${\left[\frac{2\left(\mathrm{hc}-\mathrm{\lambda \phi }\right)}{\mathrm{m\lambda }}\right]}^{1/2}$

• $\frac{2\left(\mathrm{hc}-\mathrm{\lambda \phi }\right)}{\mathrm{m}}$

• ${\left[\frac{\mathrm{hc}+\mathrm{\lambda \phi }}{\mathrm{m\lambda }}\right]}^{1/2}$

B.

${\left[\frac{2\left(\mathrm{hc}-\mathrm{\lambda \phi }\right)}{\mathrm{m\lambda }}\right]}^{1/2}$

For a photon of maximum velocity,

$\frac{1}{2}{\mathrm{mv}}^{2}+\mathrm{\phi }=\frac{\mathrm{hc}}{\mathrm{\lambda }}$

where$\mathrm{\phi }$= work function of metal

$\frac{1}{2}{\mathrm{mv}}^{2}=\frac{\mathrm{hc}-\mathrm{\lambda \phi }}{\mathrm{\lambda }}$

or v2 =$\frac{2\left(\mathrm{hc}-\mathrm{\lambda \phi }\right)}{\mathrm{m\lambda }}$

Hence, v =${\left[\frac{2\left(\mathrm{hc}-\mathrm{\lambda \phi }\right)}{\mathrm{m\lambda }}\right]}^{1/2}$

4.

X-ray beam are affected by

• electric field

• magnetic field

• both (a) and (b)

• none of these

D.

none of these

X rays aren't deflected by an electric and magnetic field because X rays do not carry any charge. They are electro-magnetic radiations and therefore cannot be deflected by an electric and magnetic field.

5.

X-rays are

• stream of electron

• stream of proton

• stream of uncharged particle

A.

X-rays consist of oscillating electric and magnetic field at right angles of each other and in the direction of propagation of the x-rays. Hence, X-rays are electromagnetic radiations.

6.

An electron beam is accelerated through a potential difference of V volt. The minimum wavelength of X-rays produced is :

• $\frac{\mathrm{he}}{\mathrm{cV}}$

• $\frac{\mathrm{cV}}{\mathrm{he}}$

• $\frac{\mathrm{eV}}{\mathrm{hc}}$

• $\frac{\mathrm{hc}}{\mathrm{eV}}$

D.

$\frac{\mathrm{hc}}{\mathrm{eV}}$

The energy of an electron accelerated through V volt is

$\frac{\mathrm{hc}}{\mathrm{\lambda }}=\mathrm{eV}$

$\therefore \mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{eV}}$

7.

In short wave communication, waves of whichof the following frequencies will be reflectedback by the ionospheric layer having electrondensity 1011 per m3 ?

• 2 MHz

• 10 MHz

• 12 MHz

• 18 MHz

A.

2 MHz

The layer of the earth's atmosphere which contains high concentration of ions and free electrons and is able to reflect radio waves. It lies above the mesosphere.

We know that the artificial frequency of asky wave for reflection from an ionospheric layer ofthe atmosphere is given by

νc= 9 N1/2

where n is the number density of electrons /m3

It is given that

N = 1011 m3

νc= 9× ( 1011 )1/2

= 2.8× 106 Hz

νc= 2.8 MHz

Hence the wave of frequency 2 MHz will be reflected back.

8.

Assertion: There are very small sporadic changesin the period of rotation of the earth.

Reason: Shi fling of large air masses in the earth'satmosphere produce a change in the moment ofinertia of the earth causing its period of rotationto change.

• If both assertion and reason are true and reason is the correct explanation of assertion

• If both assertion and reason are true but reason is not the correct explanation of assertion

• If assertion is true but reason is false

• If both assertion and reason are false

A.

If both assertion and reason are true and reason is the correct explanation of assertion

Normally when a body is moving in a straightline, whatever be the internal changes, the centre ofmass continuous in its trajectory with the same velocityand a system that is rotating continues to rotate withthe same angular velocity and changes occur in sucha way that the angular momentum is conserved.Classical physics, taking the whole of the earth as asystem does not allow sporadic changes in the speedof the earth.

However, according to the information on the earth'srotation by U.S. Naval Observatory, sporadic ( occurring at irregular interval ), changesof the speed of rotation of the earth have been observedand this is attributed to the influence of the moon on the tides. This makes it necessary to take the earth and the moon ( at least ) as one system.

In the reason given, though normally when takingthe earth as a single system, internal changes do notaffect the motion, now that one has to take the earth-moon system, the reason that the earth's angularvelocity can also change due to the events on earthalone cannot be ruled out. Then the total angularmomentum of the earth-moon system will be conserved.The reason is also right, though according to therules of the game for the 12th class examinations, onecan take only the earth as a whole as a system. In thatcase, both the assertion and reason are wrong.

9.

The maximum distance upto which TV transmission from a TV of height h can be received is proportional to

• h1/2

• h

• h3

• h2

A.

h1/2

If the height of theantenna is Ii, then the maximumdistance upto which the TV
transmission for a TV tower canbe received is proportional h1/2.

Suppose that the height of theTV antena be h and the radiusof the earth be R and R > > h. Let A be a receiving station. Inthe limit h < <R, we can assume that BA is a tangentto the surface of the earth. Then L.BAO = 90° and so

BO2 = AB2 + AO2

⇒ ( R + h )2 = AB2 + R2

⇒ AB2 = R2 + h2 + 2Rh$-$R2

⇒ AB2 = 2 Rh + h2

as h<< R we neglect h2 and so

AB2 = 2Rh

⇒ AB =$\sqrt{\mathrm{Rh}}$

∴ AB∝ h1/2

10.

Assertion: The true geographic north direction is found by using a compass needle.

Reason:The impurities meridian of the earth is along the axis of rotation of the earth.

• If both assertion and reason are true and reason is the correct explanation of assertion

• If both assertion and reason are true but reason is not the correct explanation of assertion

• If assertion is true but reason is false

• If both assertion and reason are false

D.

If both assertion and reason are false

From thecompass we are ableto know the directionof the magneticpoles. The north ofcompasstowardspointsthemagnetic south pole.If we know themagnetic declinationat that particularplace (which is angle between geographic meridianand magnetic meridian) we can easily find out thetrue geographic north-south direction.Imaginary lines drawn along the earth's surface inthe direction of the horizontal component of themagnetic field of the earth at all points passing throughthe north and south magnetic poles. This is similarto the longitudes of the earth, which pass throughthe geographic north and south poles.