Subject

Physics

Class

CBSE Class 12

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 Multiple Choice QuestionsShort Answer Type

1.

The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°.

What is the value of vertical component of earth’s magnetic field at equator?     

Vertical component of magnetic field at the equator is zero. At the equator, the direction of magnetic field is horizontal.

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2.

Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 Å in vacuum. 


Both microwaves and UV radiations are EM waves. Therefore, their speed ( c = 3 space cross times 10 to the power of 8 space m divided by s) will remain same in vacuum. 

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3.

A cell of emf E and internal resistance r is connected to two external resistances R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations:

(i) without any external resistance in the circuit.

(ii) with resistance R1 only

(iii) with R1 and R2 in series combination

(iv) with R1 and R2 in parallel combination.

The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in that order. Identify the currents corresponding to the four cases mentioned above.      

The current in the circuit to corresponding situations is given by:

 

i) When there is no external resistance in the circuit, 

               straight I subscript 1 straight space equals straight space straight E over straight r

The current in this case will be maximum because effective resistance is minimum. So, I1 = 4.2 A 

ii) In the presence of resistance R1 only, we have

             I subscript 2 space equals space fraction numerator E over denominator r plus R subscript 1 end fraction

Here, effective resistance is more than (i) and (iv) but less than (iii).

So, I2 = 1.05 A

iii) When R1 and R2 are in series combination, we have

            straight I subscript 3 straight space equals straight space fraction numerator straight E over denominator straight r plus left parenthesis straight R subscript 1 plus straight R subscript 2 right parenthesis end fraction

In this case, effective resistance is maximum, so current is minimum.

Thus, I3 = 0.42 A

iv) When R1 and R2 are in parallel combination, we have

Error converting from MathML to accessible text.

Here, effective resistance is more than (i) but less than (ii) and (iii).

So, I4 = 1.4 A

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4.

Why should electrostatic field be zero inside a conductor? 


In a conductor, charge resides on the surface. So, charge inside the conductor is zero.

Then, according to the Gauss Theorem, electrostatic field is zero. 


ϕ space equals space contour integral E with rightwards harpoon with barb upwards on top. space stack d s with rightwards harpoon with barb upwards on top space equals space q over epsilon subscript o
rightwards double arrow space E space equals space fraction numerator q over denominator italic 4 pi epsilon subscript o r end fraction equals 0

950 Views

5.

Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily. 




The direction of induced current is predicted using Lenz’s law. So, current is induced in a direction so as to oppose the increasing magnetic flux.

Therefore, direction of current is clockwise in ring 1 and anticlockwise in ring 2.

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6.

The susceptibility of a magnetic material is –2.6 × 10–5. Identify the type of magnetic material and state its two properties.     


Negative susceptibility is the property of diamagnetic material.

Two properties of diamagnetic material are:

i) This material expels the magnetic field lines and do not obeys Curies Law.

ii) They have the tendency to move from stronger to weaker part of the external magnetic field.

2342 Views

7.

When electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction?


No. When electric field is applied, the net drift of the electrons is from lower to higher potential. But, locally electrons collide with ions and may change its direction during the course of their motion.

5451 Views

8.

When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used, the current flows continuously. How does one explain this, based on the concept of displacement current? 


When an ideal capacitor is charged by dc battery, charge flows momentarily till the capacitor gets fully charged. Ideal capacitor means infinite resistance for dc.

When an ac source is connected to the capacitor, conduction current, i subscript c space equals space fraction numerator d q over denominator d t end fractioncontinuously flows in the connecting wire. Due to changing current, charge deposited on the plates of the capacitor changes with time. This causes change in electric field between the plates of the capacitor which causes the electric flux to change and gives rise to a displacement current in the region between the plates of the capacitor. 

Displacement current is given by, 

straight i subscript straight d straight space equals straight space straight epsilon subscript straight o dϕ subscript straight E over dt space and space straight i subscript straight d space equals space straight i subscript straight c space comma space at space all space instant

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9.

Show on a graph, the variation of resistivity with temperature for a typical semiconductor. 


For a semiconductor, resistivity decreases rapidly with increase in temperature.


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10.

Two identical circular wires P and Q each of radius R and carrying current ‘I’ are kept in perpendicular planes such that they have a common center as shown in the figure. Find the magnitude and direction of the net magnetic field at the common center of the two coils. 


Coils P and Q each of radius R have a common centre:



BP is directed vertically upwards and BQ is directed horizontally.

Therefore, resultant magnetic field is given by,

straight B subscript straight R straight space equals straight space square root of straight B subscript straight P squared plus straight B subscript straight Q squared end root
We have,

straight B subscript straight P straight space equals straight space straight B subscript straight Q equals fraction numerator straight mu subscript straight o straight I over denominator 2 straight R end fraction

rightwards double arrow                         straight B straight space equals straight space square root of 2 straight B subscript straight P equals square root of 2 fraction numerator straight mu subscript straight o straight I over denominator 2 straight R end fraction
rightwards double arrow                         straight space straight B straight space equals straight space fraction numerator straight mu subscript straight o straight I over denominator square root of 2 straight R end fraction

This is the required of the resultant magnetic field. 

Direction of magnetic field is given by, 

tan straight space straight theta straight space equals straight space straight B subscript straight P over straight B subscript straight Q equals 1
theta equals pi over 4
The resultant magnetic field is directed at angle of 45° with either of the fields.

3743 Views

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