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CBSE

Subject

Physics

Class

CBSE Class 12

CBSE Physics 2013 Exam Questions

Short Answer Type

1.

A capacitor has been charged by a dc source. What are the magnitudes of conduction and displacement currents, when it is fully charged?


When capacitor is charged by a dc source, 

Conduction current is equal to the displacement current.

i.e.,                           straight I subscript straight C space equals space straight I subscript straight D space equals space fraction numerator straight epsilon subscript straight o dφ subscript straight E over denominator dt end fraction
Because, straight phi subscript straight E is maximum when, capacitor is fully charged.

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2.

The emf of a cell is always greater than its terminal voltage. Why? Give reason.  


V= E – I r;

The emf of a cell is greater than terminal voltage because there is some potential drop across the cell due to its small internal resistance.

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3.

State Lenz’s Law. A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer.


The polarity of the induced emf at the open ends of a closed loop is such that it tends to produce a current which opposes the change in magnetic flux that produced it.



Emf will be induced at the ends. Magnetic flux changes in a metallic rod falling freely under gravity and hence, emf is induced.

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4.

What ate permanent magnets? Give one example.      


Substances that can retain their magnetism for a long duration of time at room temperature are called permanent magnets. They have high retentivity and coercivity.

Eg: Steel, cobalt 

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5.

An ammeter of resistance 0.80 can measure current upto 1.0A. 

(i) What must be the value of shunt resistance to enable the ammeter to measure current upto 5.0A?

(ii) What is the combined resistance of the ammeter and the shunt?   


Given, Resistance, R = 0.80

Maximum current, Imax = 1 A

So, Voltage across ammeter, V = IR = 1.0 0.80 = 0.8 V

i) If a shunt is connected in parallel,

Current I flowing through S is (I – I1)

For parallel combination of resistors, 

I1. RA = ((I – I1) S

Error converting from MathML to accessible text. 

This is the required value of shunt resistance. 

ii) Combined resistance of the ammeter and the shunt, 

1 over straight R space equals space 1 over R subscript A plus 1 over S

Error converting from MathML to accessible text.

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6.

What is the geometrical shape of equipotential surfaces due to a single isolated charge?  


The equipotential surface for an isolated charge is concentric circles. The distance between the shell decreases with increase in electric field.





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7.

A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38 straight capital omega as shown in the figure. Find the value of the current in circuit. 


 


Since the cells are in opposition,

E net = E1 – E2 = (200-10) = 190 V

So, current in the circuit is given by, I = straight E subscript net over straight R subscript eq space equals space 190 over 38 space equals space 5 space straight A


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8.

(a) An em wave is traveling in a medium with a velocity . Draw a sketch showing the propagation of the em wave, indicating the direction of the oscillating electric and magnetic fields.

(b) How are the magnitudes of the electric and magnetic fields related to the velocity of the em wave?

E with rightwards harpoon with barb upwards on top cross times B with rightwards harpoon with barb upwards on top represents the direction of propagation of wave. The figure below shows us the propagation of the em wave.

a) The direction of oscillating field is straight i with hat on top space equals space j with hat on top space cross times k with hat on top
b) Speed of EM wave is given by |c| = straight E subscript straight o over straight B subscript straight o ; Eo is the magnitude of electric field and is the magnitude of magnetic field.
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9.

A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature?         


Given, a convex lens placed coaxially in contact with a concave lens. 

Power of convex lens, P1fraction numerator 1 over denominator plus space straight f subscript 1 end fraction space equals space fraction numerator 100 over denominator f subscript 1 left parenthesis space i n space c m right parenthesis end fraction space equals space fraction numerator 100 over denominator plus 25 end fraction space equals space 4 D
Power of concave lens, P2fraction numerator 1 over denominator plus space straight f subscript 2 end fraction space equals space fraction numerator 100 over denominator f subscript 2 left parenthesis space i n space c m right parenthesis end fraction space equals space fraction numerator 100 over denominator negative 20 end fraction space equals space minus 5 space D

So, Power of combination, P = P1 + P

                                                       =  4 D space plus space left parenthesis negative 5 D right parenthesis
                                                      equals space minus 1 D
Since, power has negative magnitude; system of lenses is diverging in nature.

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10.

A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 C . When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 C .

Calculate:

(i) The potential V and the unknown capacitance C.

(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120V?

OR

A hollow cylindrical box of length 1m and area of cross-section 25cm2 is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by, where E is in NC-1 and x is in metres. Find:

(i) Net flux through the cylinder.

(ii) Charge enclosed by the cylinder.


The charge on an unknown capacitor is given by,

Q = CV

CV = 360 C                                  ... (1) 

On reducing the potential by 120 V, charge on the capacitor is reduced which is given by,

Q’ = C(V-120)

C(V-120) = 120 C                          … (2)

On solving equation (1) and (2), we have

space space space space fraction numerator 360 space μC over denominator straight V end fraction equals space fraction numerator 120 mu C space over denominator V space minus space 120 end fraction

rightwards double arrow space V space equals space 180 space V 

Unknown capacitance from equation (1),

Q = CV

space space space space space 360 space μC space equals space straight C cross times 180 space straight V

rightwards double arrow space straight C space equals space fraction numerator 360 space μC over denominator 180 space straight V end fraction equals space 2

rightwards double arrow space straight C space equals space 2 space μF 

ii) Charge on the capacitor, if voltage is increased by 120V

Q = C (V+120)

   = 2 (180+120)

   = 600 C

                                                          OR




i) Electric flux through a surface, straight phi space equals space straight E with rightwards harpoon with barb upwards on top. straight S with rightwards harpoon with barb upwards on top
Flux through the left surface, phi subscript L space equals space minus vertical line E vertical line vertical line S italic space
italic space italic space italic space italic space italic equals negative 50 x. space vertical line S vertical line

Since x = 1m,

Error converting from MathML to accessible text.   

ii) Using Gauss Theorem, we can calculate the charge inside the cylinder.

ϕ subscript space n e t end subscript space space space equals q over epsilon subscript o

Error converting from MathML to accessible text.

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