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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 varies periodically with time because it is the form of electromagnetic wave.

2.

**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 the compass we are able to know the direction of the magnetic poles. The north of compass towards points the magnetic south pole. If we know the magnetic declination at that particular place (which is angle between geographic meridian and magnetic meridian) we can easily find out the true geographic north-south direction. Imaginary lines drawn along the earth's surface in the direction of the horizontal component of the magnetic field of the earth at all points passing through the north and south magnetic poles. This is similar to the longitudes of the earth, which pass through the geographic north and south poles.

3.

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

4.

**Assertion:** There are very small sporadic changes in the period of rotation of the earth.

**Reason:** Shi fling of large air masses in the earth's atmosphere produce a change in the moment of inertia of the earth causing its period of rotation to 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 straight line, whatever be the internal changes, the centre of mass continuous in its trajectory with the same velocity and a system that is rotating continues to rotate with the same angular velocity and changes occur in such a way that the angular momentum is conserved. Classical physics, taking the whole of the earth as a system does not allow sporadic changes in the speed of the earth.

However, according to the information on the earth's rotation by U.S. Naval Observatory, sporadic ( occurring at irregular interval ), changes of the speed of rotation of the earth have been observed and 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 taking the earth as a single system, internal changes do not affect the motion, now that one has to take the earth-moon system, the reason that the earth's angular velocity can also change due to the events on earth alone cannot be ruled out. Then the total angular momentum of the earth-moon system will be conserved. The reason is also right, though according to the rules of the game for the 12^{t}^{h} class examinations, one can take only the earth as a whole as a system. In that case, both the assertion and reason are wrong.

5.

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

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

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

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

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

B.

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

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 v^{2} = $\frac{2(\mathrm{hc}-\mathrm{\lambda \phi})}{\mathrm{m\lambda}}$

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

6.

X-rays are

electromagnetic radiation

stream of electron

stream of proton

stream of uncharged particle

A.

electromagnetic radiation

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.

7.

In short wave communication, waves of which of the following frequencies will be reflected back by the ionospheric layer having electron density 10^{11} per m^{3} ?

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 a sky wave for reflection from an ionospheric layer of the atmosphere is given by

ν_{c} = 9 N^{1/2}

where n is the number density of electrons /m^{3}

It is given that

N = 10^{11 }m^{3}

ν_{c} = 9 × ( 10^{11} )^{1/2}

= 2.8 × 10^{6} Hz

ν_{c} = 2.8 MHz

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

8.

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

h

^{1/2}h

h

^{3}h

^{2}

A.

h^{1/2}

If the height of the antenna is Ii, then the maximum distance upto which the TV

transmission for a TV tower can be received is proportional h^{1/2}.

Suppose that the height of the TV antena be h and the radius of the earth be R and R > > h. Let A be a receiving station. In the limit h < <R, we can assume that BA is a tangent to the surface of the earth. Then L.BAO = 90° and so

BO^{2} = AB^{2} + AO^{2}

⇒ ( R + h )^{2} = AB^{2} + R^{2}

⇒ AB^{2} = R^{2} + h^{2} + 2Rh $-$ R^{2}

⇒ AB^{2} = 2 Rh + h^{2}

as h << R we neglect h^{2} and so

AB^{2} = 2Rh

⇒ AB = $\sqrt{\mathrm{Rh}}$

∴ AB ∝ h^{1/2}

9.

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.

10.

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

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

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