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ISC Class 12 Physics Solved Question Paper 2008

Short Answer Type

41. Alpha particles having kinetic energy of 1.8 MeV each are incident on a thin gold foil, form a large distance. Applying the principle of conservation of energy, find the closest distance of approach of the alpha particle from the gold nucleus.

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42. Calculate the minimum wavelength of X rays emitted by an X ray tube operating at a tube potential of 40 kV. How will this wavelength change if target is made of another metal ?

391 Views

43. Draw a rough sketch showing the variation of binding energy per nucleon with mass number of various nuclei. Label the graph and state the region where the nuclei are most stable.

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44. Draw a labelled circuit diagram of a N-P-N transistor in a common-emitter configuration to study its characteristics.

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45. Why is a NAND gate called a universal gate ? Show how an AND gate can be obtained using one or more NAND gates.

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Long Answer Type

46. In the circuit given below, E₁ and E₂ are two cells of emfs 4 V and 6 V respectively, having negligible internal resistances. Applying KirchofFs laws of electrical networks, find the values of I₁ and I₂.

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47. A 30 mF capacitor, 0.2 H inductor and a 50 Ω resistor are connected in series to an A.C.

source whose emf is given by E = 310 Sin (314 t) where E is in volt and t is in second. Calculate:

(i) the impedance of the circuit and

(ii) peak value of current in the circuit.

It is given that

E = 310 Sin (314 t).

Comparing this equation with standard equation

E = E₀ Sin wt

We have E₀ = 310 V and w = 314

(i) Impedance of the circuit is,

$straight Z space equals space square root of left parenthesis straight X subscript straight L space minus space space straight X subscript straight C right parenthesis squared plus straight R squared end root space space equals space square root of open parentheses ωL space minus space 1 over ωL close parentheses squared plus space straight R squared end root space space space equals square root of open parentheses 314 space straight x space 0.2 space minus fraction numerator 1 over denominator 314 space straight x space 30 space straight x space 10 to the power of negative 6 end exponent end fraction close parentheses squared plus space left parenthesis 50 right parenthesis squared end root space space space equals square root of 4379.96 end root space space space space equals space 66.18 space straight capital omega ii right parenthesis space Peak space value space of space current space is comma space straight i subscript straight o space equals space straight E subscript straight o over straight Z space equals space fraction numerator 310 over denominator 66.18 end fraction space equals space 4.68 space straight A with straight o on top$