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Physics

Class

CBSE Class 12

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CBSE Physics 2017 Exam Questions

Short Answer Type

21.
A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen.

Estimate the number of fringes obtained in Young's double slit experiment with fringe width 0.5 mm, which can be accommodated within the region of the total angular spread of the central maximum due to the single slit.

Angular width of central maximum

straight omega space equals space fraction numerator 2 straight lambda over denominator straight a end fraction
space equals space fraction numerator 2 space straight x 500 space straight x 10 to the power of negative 9 end exponent over denominator 0.2 space straight x 10 to the power of negative 3 end exponent end fraction
space equals space 5 space straight x 10 to the power of negative 3 end exponent space radian
straight beta space equals space λD over straight d
Linear width of central maxima in the diffraction pattern
straight omega apostrophe space equals space fraction numerator 2 λD over denominator straight a end fraction
Let ‘n’ be the number of interference fringes which can be accommodated in the central maxima
therefore space straight n space space straight x space straight beta space equals space straight omega apostrophe
straight n space equals space fraction numerator 2 λD over denominator straight a end fraction space straight x straight d over λD
straight n space equals space fraction numerator 2 straight d over denominator straight a end fraction

2890 Views

22.

A radioactive nucleus 'A' undergoes a series of decays as given below:
straight A space rightwards arrow with straight alpha on top space straight A subscript 1 space rightwards arrow with straight beta on top space straight A subscript 2 space rightwards arrow with straight alpha on top space straight A subscript 3 space rightwards arrow with straight gamma on top space straight A subscript 4
The mass number and atomic number of A2 are 176 and 71 respectively. Determine the mass and atomic number of A4 and A.


straight A subscript 74 superscript 180 space rightwards arrow with space straight alpha space on top space straight A subscript 72 superscript 176 space rightwards arrow with space straight beta on top space straight A subscript 71 superscript 176 space rightwards arrow with straight alpha on top space straight A subscript 69 superscript 172 space rightwards arrow with straight gamma on top straight A subscript 69 superscript 172
The mass number and atomic number of A4 is 172 and 69, respectively.
The mass number and atomic number of A is 180 and 74, respectively.

1621 Views

23.

Define the term 'amplitude modulation'. Explain any two factors which justify the need for modulating a low-frequency base band signal.


It is the process of superposition of information/message signal over a carrier wave in such a way that the amplitude of carrier wave is varied according to the information signal/message signal.

Direct transmission, of the low frequency base band information signal, is not possible due to the following reasons;
(i) Size of Antenna: For transmitting a signal, minimum height of the antenna should be λ/4; with the help of modulation wavelength of signal decreases, hence the height of antenna becomes manageable.
(ii) Effective power radiated by an antenna: Effective power radiated by an antenna varies inversely as λ2, hence effective power radiated into the space, by the antenna, increases.
(iii)To avoid mixing up of signals from different transmitters.

1926 Views

24.

The work function of the following metals is given : Na 2.75 ev, K = 2.3 eV, Mo = 4.17 eV and Ni = 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 Å from a laser source placed 1 m away from these metals? What happens if the laser source is brought nearer and placed 50 cm away?


straight E space equals space hc over straight lambda space equals space fraction numerator 6.63 space straight x 10 to the power of negative 34 end exponent space straight x space 3 space straight x space 10 to the power of 8 over denominator 3.3 space straight x space 10 to the power of negative 7 end exponent space straight x 1.6 space straight x 10 to the power of negative 19 end exponent end fraction eV
space equals space 3.77 space eV

The work function of Mo and Ni is more than the energy of the incident
photons; so photoelectric emission will not take place from these metals.
Kinetic energy of photo electrons will not change, only photoelectric current
will change.

1710 Views

25.

A narrow beam of unpolarized light of intensity I0 is incident on a Polaroid P1. The light transmitted by it is then incident on a second polaroid P2 with its pass axis making the angle of 60° relatives to the pass axis of P1. Find the intensity of the light transmitted by P2.


According to Malus’ Law:

straight I space equals space straight I subscript straight o space cos squared space straight theta
therefore space straight I space equals space open parentheses straight I subscript 0 over 2 close parentheses Cos squared straight theta
Where space straight I subscript 0 space is space the space intensity space of space unpolarized space light
straight theta space equals space 60 to the power of 0 space left parenthesis given right parenthesis
straight I space equals space straight I subscript 0 over 2 cos squared 60 space equals space straight I subscript straight o over 2 space straight x open parentheses 1 half close parentheses squared space equals space straight I subscript straight o over 8

1726 Views

26.

For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Given the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ. 


Given, Rc = 2000Ω
RB = 1000 Ω,
 β = 100,
voltage gain can given as,

straight V subscript 0 over straight V subscript straight i space equals space straight beta straight R subscript straight c over straight R subscript straight i
2 over straight V subscript straight i space equals space 100 space straight x 2000 over 1000
rightwards double arrow space straight V subscript straight i space equals space 0.01 space straight V
Now comma space collector space current space is space given space as
straight I subscript straight C space equals space straight V subscript 0 over straight R subscript straight C space equals space 2 over 2000 space equals space 0.001 space straight A
straight beta space equals space straight I subscript straight C over straight I subscript straight B
100 space equals space fraction numerator begin display style 0.001 end style over denominator straight I subscript straight B end fraction
rightwards double arrow straight I subscript straight B space equals space 10 straight mu space straight A
1431 Views

27.

Write the basic nuclear process underlying β+ and β- decays.


Basic process underlying  β+ and  β- decay are
During a weak interaction an atomic nucleus converts into a nucleus with one higher atomic number while emitting one electron and an electron antineutrino this is called beta minus decay.

straight X subscript straight Z superscript straight A space rightwards arrow with space on top space straight Y subscript straight Z plus 1 end subscript superscript straight A space plus straight e to the power of minus space plus top enclose straight v subscript straight e end enclose

During a weak interaction an atomic nucleus converts into a nucleus with one Lower atomic number while emitting one electron and an electron neutrino this is called beta minus decay.
straight X subscript straight Z superscript straight A space rightwards arrow with space on top space straight Y subscript straight Z plus 1 end subscript superscript straight A space plus straight e to the power of plus space plus straight v subscript straight e

1545 Views

28.

Mrs Rashmi Singh broke her reading glasses. When she went to the shopkeeper to order new specs, he suggested that she should get spectacles with plastic lenses instead of glass lenses. On getting the new spectacles, she found that the new ones were thicker than the earlier ones. She asked this question to the shopkeeper but he could not offer a satisfactory explanation for this. At home, Mrs Singh raised the same question to her daughter Anuja who explained why plastic lenses were thicker.

(a) Write two qualities displayed each by Anuja and her mother.

(b) How do you explain this fact using lens maker's formula?


(a) Two qualities displayed each by Anuja and her mother:
Anuja has good knowledge of lens and she is very co-operative.
Mrs Rashmi is curious to know about lenses and science behind it.

(b)The lens maker formula is given as  1 over straight f space space equals open parentheses straight mu minus 1 close parentheses space open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses

where f is the focal length, μ is the refractive index and R1 and R2 is the radius of curvature of the lens.
Since μgp
where g stands for glass and p stands for plastic
Therefore, we get (μ-1)> (μp-1)
Now, using the lens maker formula, we see that focal length is inversely proportioned to (μ-1)
Hence, fp>fg
Thus, in the case of the plastic lens, the thickness of the lens should be increased to keep the same focal length as that of the glass lens to give the same power.

2137 Views

29.

How does one explain the emission of electrons from a photosensitive surface with the help of Einstein's photoelectric equation?

 


Einstein’s Photoelectric equation is
hv space equals space straight phi subscript 0 space plus straight K subscript max
When a photon of energy is incident on the metal, some part of this
energy is utilized as work function to eject the electron and remaining
energy appears as the kinetic energy of the emitted electron.

6871 Views

30.

Explain two features to distinguish between the interference pattern in Young's double slit experiment with the diffraction pattern obtained due to a single slit.


Interference pattern Diffraction pattern 
1) All fringes are of equal width.  1) The width of central maxima is twice the width of higher order band.
2) Intensity of all bright bands is equal. 2) The intensity goes on decreasing for a higher order of diffraction bands.
1467 Views

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