Subject

Physics

Class

CBSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

11.

Calculate the de-Broglie wavelength of the electron orbitting in the n=2 state of hydrogen atom.


Given,

n =2

K.E for the second state, Ek,

 fraction numerator 13.6 over denominator straight n squared end fraction space equals space fraction numerator 13.6 over denominator 4 end fraction
space space space space space space space space space space space equals space 3.4 space x space 1.6 space x space 10 to the power of negative 19 end exponent space J

De-Broglie wavelength, straight lambda space equals space fraction numerator straight h over denominator square root of 2 mE subscript straight k end root end fraction

   equals space fraction numerator 6.63 space straight x space 10 to the power of negative 34 end exponent over denominator square root of 2 space straight x space 9.1 space straight x space 10 to the power of negative 31 end exponent straight x 3.4 straight x 1.6 straight x 10 to the power of negative 19 end exponent end root end fraction

equals space 0.067 space straight m

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12.

Name the essential components of a communication system.


Essential components of communication system:

1. Transmitter
2. Channel
3. Receiver

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13.

(i) State law of Malus.

(ii) Draw a graph showing the variation of intensity (I) of polarised light transmitted by an analyser with angle (straight theta) between polarizer and analyser.
(iii) What is the value of refractive index of a medium of polarising angle 60o?


i) Law of Malus states that when a completely plane polarised light beam is incident on an analyser, the intensity of the emergent light varies as the square of the cosine of the angle between the plane of transmission of the analyser and the polariser.

                               I = Io cos2straight theta
ii) The variation of intensity (I) of polarised light transmitted by an analyser with angle straight theta.


iii) According to the Brewster's law, we have

straight mu space equals space tan space straight i subscript straight p

straight mu space equals space tan space 60 to the power of straight o

straight mu space equals space square root of 3

The refractive index of the material is 1.73
 
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14.

Why does sun appear red at sunrise and sunset?


Sun appears red at sunrise and sunset because of least scattering of red light as it has the lowest wavelength.
As per Rayleigh scattering, the amount of light scattered proportional to space 1 over straight lambda to the power of 4.

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15.

a) Write the basic nuclear process involved in the emission of beta plus in a symbolic form, by a radioactive nucleus.

b) In the reactions given below:

i right parenthesis space straight C presubscript 6 presuperscript 11 space rightwards arrow space straight B presubscript straight y presuperscript straight z plus space straight x plus straight nu
ii right parenthesis space straight C presubscript 6 presuperscript 12 space plus straight C presubscript 6 presuperscript 12 space rightwards arrow space Ne presubscript straight a presuperscript 20 space plus space He presubscript straight b presuperscript straight c
Find the values of x, y and z and a, b and c.


a)
The basic nuclear process involved in the emission of beta plus in a symbolic form, by a radioactive nucleus,

straight p space rightwards arrow space straight n space plus space straight beta to the power of plus plus straight nu
In a beta-plus decay, a proton transforms into a neutron within the nucleus.

b)

i) straight C presubscript 6 presuperscript 11 space rightwards arrow space straight B presubscript 5 presuperscript 11 space plus space straight beta presubscript 1 presuperscript 0 space plus space straight nu
The corresponding y and z are 5 and 11 respectively.
Here, x is the positron.

ii) straight C presubscript 6 presuperscript 12 space plus straight C presubscript 6 presuperscript 12 space rightwards arrow space Ne presubscript 10 presuperscript 20 space plus space He presubscript 2 presuperscript 4
The corresponding values of a, b and c are 10, 2 and 4 respectively.

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16.

Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies straight nuA > straight nuB.

(i) In which case is the stopping potential more and why?

(ii) Does the slope of the graph depend on the nature of the material used? Explain.


Graph showing the variation of stopping potential with frequency of incident radiations:



From the above graph, we can observe that,

i) The stopping potential is inversely proportional to the threshold frequency. Hence, the stopping potential is higher for metal B.

ii)The slope of the graph does not depend on the nature of the material used:

We know that,

straight K subscript max space equals space hν space minus straight ϕ subscript straight o space equals space eV subscript straight o
Dividing the equation throughout by e, we get

hν over straight e minus ϕ subscript o over e space equals space V subscript o
On comparing the above equation with the straight line equation, we get

Slope of the graph = straight h over straight e
That is, the slope of the graph does not depend on the nature of the material used.

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17.

Calculate the shortest wavelength of the spectral lines emitted in Balmer series. [ Given Rydberg constant, R = 107 m-1]


The formula is,

1 over straight lambda space equals space R space open parentheses 1 over 2 squared minus 1 over infinity squared close parentheses
For shortest wavelength,

Therefore,

space space space space space 1 over straight lambda equals R over 4
rightwards double arrow space straight lambda space equals space 4 over straight R space equals space 4 space straight x space 10 to the power of negative 7 end exponent space straight m

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18.

A ray PQ incident normally on the refracting face BA is refracted in the prism BAC made of material of refractive index 1.5. Complete the path of ray through the prism. From which face will the ray emerge ? Justify your answer.


Here,

sin ic2 over 3 space equals space 0.66

rightwards double arrow space sin space 30 to the power of straight o space equals space 0.5

space space space space space space space space straight i subscript straight c space greater than space 30 to the power of straight o

The light will emerge out from face AC.

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19.

Define modulation index. Why is it kept low ? What is the role of a bandpass filter?


Modulation index is the ratio of the amplitude of modulating signal to that of carrier wave.

straight mu space equals space straight A subscript straight m over straight A subscript straight C
Modulation index is kept low in order to avoid distortion. The low frequency modulating signal is mixed with high-frequency carrier wave, the distortion is restricted due the high-frequency carrier wave for modulation index lying between 0 and 1.

Bandpass filter rejects low and high frequencies and allows a band of frequencies to pass through.

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20.

Define ionisation energy.

How would the ionisation energy change when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge ?


The minimum energy, required to free the electron from the ground state of the hydrogen atom, is known as ionization energy.

Ionisation energy is given by,

space space space space space space space straight E subscript straight o space equals space fraction numerator me to the power of 4 over denominator 8 straight epsilon subscript straight o squared straight h squared end fraction
straight i. straight e. comma space straight E subscript straight o space proportional to space straight m

Since, ionisation energy is directly proportional to mass, the Ionization energy will become 200 times.

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