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.

(i) Write a balanced equation showing nuclear fission of Uranium straight U presubscript 92 presuperscript 235

nucleus.

(ii) In a nuclear reactor, what is the function of:

(1) Cadmium rods? (2) Graphite rods?

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42. Starting with N = N0e-λt, obtain a relation between disintegration constant ‘λ’ of a radioactive element and its half life (T), various terms have their usual meaning.
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43.

On an energy level diagram of hydrogen, show by a downward or an upward arrow, a transition which results in:

(i) Emission line of Balmer series.

(ii) Emission line of Lyman series.

(iii) Absorption line of Lyman series.

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44. Calculate:

i) Mass defect of helium He presubscript 2 presuperscript 4 space nucleus  and 
ii) 

Its binding energy in MeV.

Mass of a proton = 1.007276 u

Mass of a neutron = 1.008665 u

Mass of He presubscript 2 presuperscript 4 nucleus = 4.001506 u

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45. Draw a labelled diagram of a common emitter amplifier. What is the phase angle between the input and output voltages? 
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46. Threshold wavelength of a certain metal is 792 nm. What is the maximum kinetic energy of photo-electrons emitted by this metal if it is exposed to ultraviolet light of wavelength 396 nm?


Maximum K.E of emitted electrons is given (K.E)max equals space hc space open square brackets 1 over straight lambda space minus space 1 over straight lambda subscript straight o close square brackets

where,

λ is wavelength of ultraviolet light λ0 is the threshold wavelength of certain metal c is the speed of light and h is Planck’s Constant.

λ0 = 792 nm, c = 3 x 108 m/sec

λ= 396 nm, h = 6.6x 10-34 Js.

On putting these value in the formula,

K.E = 6.6 x 10-34 x 3 x 108 x 10open square brackets 1 over 396 space minus space 1 over 792 close square brackets 

   
   = 6.6 x 10-34 x 3 x 1017 open square brackets fraction numerator 792 space minus space 396 over denominator 396 space minus space 792 end fraction close square brackets

      = 6.6 x 10-17 x 3 open square brackets fraction numerator 396 over denominator 396 space straight x space 792 end fraction close square brackets = 0.025 x 10-17 

      = 2.5 x 10-19 J

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47. The following combination of gates acts as a logic gate. With the help of a truth table, find out which logic gate the combination represents:


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