Subject

Physics

Class

ICSE Class 12

Pre Boards

Practice to excel and get familiar with the paper pattern and the type of questions. Check you answers with answer keys provided.

Sample Papers

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 Multiple Choice QuestionsShort Answer Type

41.

A sinusoidal voltage e = e0 Sin (wt) is fed to a common emitter amplifier. Draw neatly labelled diagrams to show:

(i) Signal voltage

(ii) Output voltage of the amplifier.

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42. Make a truth table showing inputs at A and B and outputs at X, Y and Z for the combination of gates shown in figure below:


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 Multiple Choice QuestionsLong Answer Type

43. With the help of a labelled diagram, obtain an expression for the electric field intensity ‘E’ at a point P in broad side position (i.e., equatorial plane) of an electric dipole. 
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44. What is meant by ‘Chromatic aberration’? A thin convex lens of focal length 30 cm and made of flint glass (dispersive power = 0.03) is kept in contact with a thin concave lens of focal length 20 cm and made of crown glass. Calculate the dispersive power of crown glass if the above said combination acts as an achromatic doublet. 


Chromatic Aberration: When white light is passed through a prism it splits up into seven colours, with violet deviated towards the base of the prism, with maximum deviation and red suffering minimum deviation.

A lens can be assumed to be made up of a number of prisms.

The refracting angle of the prism goes on decreasing at a uniform rate from its centre outwards to its periphery. When white light is passed through a lens it is dispersed into its constituent colours.

The deviation is different for different colours and hence they come to focus at different points on the principal axis. Due to this we get a coloured and blurred image. This defect is called chromatic aberration.

Given,

f1 = 30 cm, w1 = 0.03

f2 = 20 cm, w2 = ?

Focal length of the combination,

straight f space equals space fraction numerator straight f subscript 1 straight f subscript 2 over denominator straight f subscript 2 space minus straight f subscript 1 space end fraction space equals space fraction numerator 30 space straight x space 20 over denominator 20 space minus space 30 end fraction space equals space fraction numerator 600 over denominator negative 10 end fraction

straight f space equals space minus 60

We know, 

straight f subscript 2 space equals space open parentheses fraction numerator straight w subscript 1 space minus space straight w subscript 2 over denominator straight w subscript 1 end fraction close parentheses space straight f

20 space equals space open parentheses fraction numerator 0.03 space minus space straight w subscript 2 over denominator 0.03 end fraction close parentheses left parenthesis negative 60 right parenthesis

space space space space space equals space fraction numerator negative 20 over denominator 60 end fraction space equals space 1 space minus space fraction numerator straight w subscript 2 over denominator 0.03 end fraction

rightwards double arrow negative space 8 over 6 space equals space fraction numerator negative straight w over denominator 0.03 end fraction

rightwards double arrow space straight w subscript 2 space equals space 8 over 6 straight x space 0.03 space equals space 0.04 

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45.

Electrons, initially at rest, are passed through a potential difference of 2 kV. Calculate their:  

(i) Final velocity and

(ii) de Broglie wavelength.

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