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CBSE

Subject

Physics

Class

CBSE Class 12
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CBSE Physics 2013 Exam Questions

Short Answer Type

1.

Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiations. Name the radiations and write the range of their frequency. 


The radiations are ultraviolet radiations.

The frequency range of UV rays is 1015 –1017Hz.


2.

A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor. 


Capacitance having a dielectric of thickness ‘t’ is given by, 

                                straight C space equals space fraction numerator straight epsilon subscript straight o space straight A over denominator straight d space minus space straight t space plus space begin display style straight t over straight K end style end fraction 

When the thickness of the plates is reduced to half, t = d/2 then,

 Capacitance becomes, 

straight C space equals space fraction numerator straight epsilon subscript straight o straight A over denominator straight d space minus space begin display style straight d over 2 end style plus begin display style fraction numerator straight d over denominator 2 straight K end fraction end style end fraction space

space space space equals space fraction numerator straight epsilon subscript straight o straight A over denominator begin display style straight d over 2 end style plus space begin display style fraction numerator straight d over denominator 2 straight K end fraction end style end fraction space

space space space space equals space fraction numerator straight epsilon subscript straight o straight A over denominator begin display style straight d over 2 end style open parentheses 1 plus begin display style 1 over straight K end style close parentheses end fraction space

space space space space equals space fraction numerator 2 straight epsilon subscript straight o AK over denominator straight d space left parenthesis straight K space plus space 1 right parenthesis thin space end fraction


3.

Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?


Internal resistance of the circuit is negligible.

 So, total resistance is R.

 Therefore, current across the circuit is given bystraight I space equals space E over R


4.

(a) For a given a.c.,i space equals space i subscript m sin space omega t comma show that the average power dissipated in a resistor R over a complete cycle is 1 half i squared subscript m R

(b) A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb. 


a) Average power consumed in resistor R is given by, 

straight P subscript av space equals space fraction numerator 1 over denominator integral subscript 0 superscript straight T dt end fraction space integral subscript 0 superscript straight T straight i squared space straight R space dt

space space space space space equals space fraction numerator straight i squared subscript straight m space straight R over denominator straight T end fraction space integral subscript 0 superscript straight T sin squared space ωt space dt space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis
space space space space space space equals fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction integral subscript 0 superscript straight T left parenthesis 1 space minus space cos space 2 ωt right parenthesis space dt space space space space space space space space space space space space space space

space space space space space space equals fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction space open square brackets integral subscript 0 superscript straight T dt space minus space integral subscript 0 superscript straight T cos space 2 ωt space dt close square brackets space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space

space space space space space space space equals fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction space left square bracket space straight T space minus space 0 right square bracket space

space space space space space space space equals space fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction 

b) In case of ac, we have

straight P subscript av space equals space straight v squared subscript rms over straight R space equals space straight V squared subscript eff over straight R

rightwards double arrow space straight R space equals space straight V squared subscript rms over straight P subscript av

space space space space space space space space space equals space fraction numerator 220 space straight x space 220 over denominator 100 end fraction

space space space space space space space space space equals space 484 space straight capital omega


5.

The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping? 


The changing magnetic flux through the plate produces a strong eddy current in the direction, which opposes the cause as the plate oscillates. 


6.

Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.


Potentiometer can be used to measure the internal resistance of the cell.

 

A cell of emf E is connected across the resistance box through key K1.

When key K1 is opened galvanometer shows deflection at the balancing length l1.

So, E = kError converting from MathML to accessible text. 
If both keys are closed, then balancing point is obtained at length l2 (l2 < l1).

So, V = kError converting from MathML to accessible text.
Now, using the relation, 

straight r space space equals straight R space open parentheses straight E over straight V space minus space 1 close parentheses

Therefore, we have

Error converting from MathML to accessible text.


7.

Two charges of magnitudes – 2Q and + Q are located at points (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin?        


Using gauss law, we have


phi space equals space q over epsilon subscript o

 

Electric flux is given by, 

straight ϕ space equals space fraction numerator negative 2 straight Q over denominator straight epsilon subscript straight o end fraction


8.

A capacitor, made of two parallel plates each of plate area A and separation d, is being charged by an external ac source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor.  


The charge on the plates is because of the conduction current flowing in the wires.

straight I subscript straight C space equals space dq over dtAccording to Maxwell’s equation, displacement current between the plated is given by, 


straight I subscript straight d space equals space straight epsilon subscript straight o dϕ subscript straight E over dt 
where, straight ϕ subscript straight E is the lectric flux. 

Now, using Gauss theorem, 

straight phi subscript straight E space equals space straight q over straight epsilon subscript straight o
So, 

straight I subscript straight d space equals space straight epsilon subscript straight o straight d over dt open parentheses straight q over straight epsilon subscript straight o close parentheses space

straight I subscript straight d space equals space dq over dt

Therefore, both the displacement current and conduction currents are equal.


9.

How does the mutual inductance of a pair of coils change when,

(i) distance between the coils is increased and

(ii) number of turns in the coils is increased?


Using the formula, 


straight ϕ space equals space MI space
i) Mutual inductance decreases when the distance between the coils is increased.

 

ii) Now using the formula,

                             straight M subscript 21 space equals space straight mu subscript straight o space straight n subscript 1 space straight n subscript 2 space straight A space straight l 

On increasing the number of turns in the coil, mutual inductance also increases.


10.

Explain the term ‘drift velocity’ of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of ‘drift velocity’.


The velocity gained by the accelerating electrons in uniform electric field inside the conductor is drift velocity. The average velocity, acquired by free electrons along the length of a metallic conductor, due to existing electric field is called drift velocity.



Let ‘n’ be the number density of free electrons in a conductor of length ‘l’ and area of cross-section ‘A’.

 

Total charge in the conductor, Q = Ne

                                          = (nAl)e

Time taken, t is given by, 

Error converting from MathML to accessible text.

Therefore, the current flowing across the conductor is given by, 

Error converting from MathML to accessible text.

That is, 

straight I space equals space neAv subscript straight d, which is the amount of current flowing through a conductor in terms of drift velocity.


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