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CBSE

Subject

Physics

Class

CBSE Class 12

CBSE Physics 2014 Exam Questions

Short Answer Type

1.

Why is the use of A.C. voltage preferred over D.C. voltage? Give two reasons. 


The reason why AC voltage is preferred over DC is because of the following reasons:

i) AC can be transmitted with much lower energy losses as compared to DC.

ii) Alternating current can be generated easily.

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2.

Two equal balls having equal positive charge ‘q’ coulombs are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two? 


On inserting the plastic sheets between the two balls the force will decrease.

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3.

Using the concept of force between two infinitely long parallel current carrying conductors, define one ampere of current. 


One ampere is the value of steady current which when maintained in each of the two very long, straight, parallel conductors of negligible cross section and placed one metre apart in vacuum, would produce on each of these conductors a force equal of 2 x 10-7 N/m of its length.

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4.

Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere’s circuital law to include the term due to displacement current.


Ampere’s circuital law states that,

contour integral B with rightwards harpoon with barb upwards on top space. space stack d l with rightwards harpoon with barb upwards on top space equals space mu subscript o space I

Electric flux across a parallel capacitor is given by, 

straight ϕ subscript straight E space space equals space EA space equals space 1 over straight epsilon subscript straight o straight Q over straight A xA space equals space straight Q over straight epsilon subscript straight o 

Current in the plates of the capacitor is given by, 

i space equals space fraction numerator d Q over denominator d t end fraction
italic therefore italic space fraction numerator d ϕ subscript E over denominator d t end fraction italic space italic equals italic space fraction numerator d over denominator d t end fraction open parentheses Q over epsilon subscript o close parentheses italic space italic equals italic space italic 1 over epsilon subscript o fraction numerator d Q over denominator d t end fraction

italic rightwards double arrow italic space fraction numerator d ϕ subscript E over denominator d t end fraction italic space italic equals italic space fraction numerator d Q over denominator d t end fraction italic equals i

This is the missing term in the ampere's law. 

Therefore, total current in the conductor is the sum of displacement current and conduction current.

straight i space equals space straight i subscript straight c space plus space straight i subscript straight d space equals space space straight i subscript straight c space plus space straight epsilon subscript straight o space dϕ over dt

Putting space the space value space of space straight I space in space Amper apostrophe straight s space Circuital space law comma space we space have

contour integral straight B with rightwards harpoon with barb upwards on top. dl with rightwards harpoon with barb upwards on top space equals space straight mu subscript straight o space straight I subscript straight c space plus space straight mu subscript straight o element of subscript straight o space dϕ subscript straight E over dt

This space is space the space required space generalized space form space of space Ampere ’ straight s space law.

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5.

The electric current flowing in a wire in the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept above the wire as shown. 



The direction of induced current is so as to oppose the change in magnetic flux through the wire.

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6.

To which part of the electromagnetic spectrum does a wave of frequency 5  x 1019 Hz belong? 


The part of the electromagnetic spectrum is X-rays.

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7.

A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.              


Energy stored in a capacitor is given by, E space equals space 1 half Q squared over C

When capacitor is connected in parallel combination, equivalent capacitance = 2C

Here, total charge Q remains the same.

Initial energy is given by, straight E subscript straight i space equals space 1 half straight Q squared over straight C

Final energy is given by, straight E subscript straight f space equals space 1 half fraction numerator straight Q squared over denominator 2 straight C end fraction

So, the ratio of energy is given by,  

fraction numerator F i n a l space e n e r g y over denominator I n i t i a l space e n e r g y end fraction equals space 1 half

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8.

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 x 10–7 m2 carrying a current of 1.5 A. Assume the density of conduction electrons to be 9 x 1028 m–3


Area of cross-section, A = 1.0  x 10–7 m2

Current = 1.5 A

Density of conduction electrons = 9 x 1028 m–3

Drift velocity of electrons is given by, 

straight v subscript straight d space equals space 1 over neA space
space
space space space equals fraction numerator 1 over denominator 9 space straight x space 10 to the power of 28 space straight x space 1.6 space straight x space 10 to the power of negative 19 end exponent space straight x space 1.0 space straight x space 10 to the power of negative 7 end exponent end fraction space straight m divided by straight s

space space equals space 1.048 space straight x space 10 to the power of negative 3 end exponent space straight m divided by straight s

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9.

Show diagrammatically the behaviour of magnetic field lines in the presence of (i) paramagnetic and (ii) diamagnetic substances. How does one explain this distinguishing feature?        

                                                   


A paramagnetic material tends to move from weaker to stronger regions of the magnetic field and hence increases the number of lines of magnetic field passing through it. Relative permeability is greater than 1. So, magnetic lines of force pass through the substance. 

A diamagnetic material tends to move from stronger to weaker regions of the magnetic field and hence, decreases the number of lines of magnetic field passing through it. Relative permeability is 1 and magnetic lines of force do not pass through the substance.

Distinguishing feature is because of the difference in their relative permeability. 

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10.

A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable resistor ‘R’. Plot a graph showing variation of terminal voltage ‘V’ of the cell versus the current ‘I’. Using the plot, show how the emf of the cell and its internal resistance can be determined.


The relation between V and I is given by,

V = E – Ir

Thus, the graph between V and I is as shown below:

 

Emf is given by the intercept on the vertical axis i.e., the V axis.

Internal resistance is given by the slope of the line i.e.. slope of V vs. I graph.

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