# Physics 2007 Set 2 Sol

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```					2/3/2011                                               Subjective Test Paper - Physics - Meritn…

Physics 2007 Set 2                                                                               Close

Subjective Test

These are additional set 2 questions.

Question 1 ( 1.0 marks)
What is the purpose of modulating a signal in transmission? (1)

Solution:
The purpose of modulating a signal in transmission is

(i) to increase the effective power of the signal

(ii) to keep the height of antenna or aerial small

(iii) not to allow mixing up of signals from different transmitters

Question 9 ( 2.0 marks)
A cylindrical metallic wire is stretched to increase its length by      . Calculate the percentage
change in its resistance. (2)

Solution:
Initial length of cylindrical metallic wire is L.

Change in length, ∆l=          of L = 0.1L

New length, L′ = L + ∆l = L + 0.1L = 1.1L

Let A be the initial area of cross-section of the coil and A′ be the new area of

cross-section of the coil.

The volume of cylindrical metallic wire remains constant.

A L = A′ L′

Let R be the initial resistance of the coil and R′ be the new resistance of the coil.

Resistivity ρ remains the same in both the cases.

Percentage increase in resistance is

Therefore, the percentage change in resistance is

Question 10 ( 2.0 marks)
Distinguish between the terms ‘effective value’ and ‘peak value’ of an alternating current. An
alternating current from a source is represented by (2)

I = 10sin (314t)

Write the corresponding values of
meritnation.com/…/28S6\$cMa81Ju87b6…                                                                        1/5
2/3/2011                                            Subjective Test Paper - Physics - Meritn…
Write the corresponding values of

(i) Its ‘effective value’

(ii) Frequency of the source

Solution:
Effective value of alternating current is the value of direct current, which produces the same heating
effect in a given resistor as is produced by the given ac when passed for the same time.

Peak current is the maximum value attained by an alternating current in either of its half cycles.

Given, I = 10 sin (314 t)

Standard equation for current is, I = I0 sin (2πft)

Where,

I0 = Peak current = 10 A

f = Frequency of source

t = Time period

(i)

(ii)

Question 12 ( 2.0 marks)
A convex lens of refractive index 1.5 has a focal length of 20 cm in air.

Calculate the change in its focal length when it is immersed in water of refractive index   . (2)

Solution:
Reflective index of convex lens in air, µ= 1.5

Focal length of convex lens, fa = 20 cm

Reflective index of water, µ′=

When the lens is in air:

Where R1 and R2 are the radius of curvature of the convex lens

When the lens is in air:

Where, fw is the focal length of the lens, when immersed in water

meritnation.com/…/28S6\$cMa81Ju87b6…                                                                         2/5
2/3/2011                                        Subjective Test Paper - Physics - Meritn…

Question 13 ( 3.0 marks)
If the nucleons bound in a nucleus are separated apart from each other, the sum of their masses is
greater than the mass of the nucleus. W here does this mass difference come from? Explain briefly.
(2)

Solution:
Protons and neutrons are bound together inside a nucleus by means of strong attractive nuclear
forces. Therefore, a definite amount of work is required to be done to break up the nucleus into its
constituent particles and to separate them at infinite distance from one another. This work done
gets converted into mass in accordance with Einstein’s mass energy relation, E = mc2 . Thus, when
the nucleons bound in a nucleus are separated from each other, extra mass is generated from the
energy required to separate the nucleons.

Because of this extra mass, the sum of the masses of daughter nuclei is greater than the mass of
the mother nucleus.

Question 14 ( 3.0 marks)
The output of 2- input AND gate is fed to a NOT gate. Draw the logic circuit of this combination of
gates and write its truth table. (2)

Solution:

The above gate functions as a 2-input NAND gate.

Truth Table

A   B

0   0    0      1

0   1    0      1

1   0    0      1

1   1    1      0

Question 24 ( 3.0 marks)
Radiations of frequency 1015 Hz are incident on tw o photosensitive surfaces A and B. Following
observations are recorded:

Surface A: No photo-emission takes place.

Surface B: Photo-emission takes place but photo-electrons have zero energy.

Explain the above observations on the basis of Einstein’s photo-electric equation.

How will the observation with surface B change when the wavelength of incident radiations is
decreased? (3)

Solution:
(a) The threshold frequency of surface A is greater than 1015 Hz. Therefore, no photoemission takes
place.

(b) The threshold frequency of surface is equal to 10 15 Hz. Therefore, photoemission takes place,
but photoelectrons have zero kinetic energy.

When the wavelength of the incident radiation is decreased, the kinetic energy of photoelectrons
emitted from surface B increases.

Question 25 ( 5.0 marks)
Draw a labelled ray diagram of an astronomical telescope. W rite mathematical expression for its
magnifying power. How does the magnifying power get affected on increasing the aperture of the
objective lens and why? (3)

Solution:
meritnation.com/…/28S6\$cMa81Ju87b6…                                                                       3/5
2/3/2011                                         Subjective Test Paper - Physics - Meritn…
Least distance of distinct vision:

Magnifying power, m

The magnifying power is not affected on increasing the aperture of the objective.

Question 27 ( 5.0 marks)
State Gauss’s theorem in electrostatics. Using it, deduce an expression for electric field intensity at a
point near a thin infinite plane sheet of electric charge. (3)

Solution:
Gauss’s theorem states that the total electric flux through a closed surface is 1/ε0 times the charge
enclosed by the surface.

Electric field intensity   on either side of the sheet must be perpendicular to the plane of sheet
having same magnitude at all points equidistant from the sheet.

Let P be any point at a distance r from the sheet. Let the small area element

ds = πr2 , where r is the radius.

and     are perpendicular on the surface of the cylinder. Therefore, the electric flux is zero.

and     are parallel on the two cylindrical edges P and Q, which contributes to electric flux.

Electric flux over the edges P and Q of the cylinder is:

Charge density,σ =

∴q =πr2 σ … (1)

From equation (1),

meritnation.com/…/28S6\$cMa81Ju87b6…                                                                            4/5
2/3/2011                              Subjective Test Paper - Physics - Meritn…

meritnation.com/…/28S6\$cMa81Ju87b6…                                               5/5

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