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CBSE

Subject

Physics

Class

CBSE Class 12
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CBSE Physics 2015 Exam Questions

Short Answer Type

1.

Name the parts of the electromagnetic spectrum which is

(a) suitable for radar systems used in aircraft navigation.

(b) used to treat muscular strain.

(c) used as a diagnostic tool in medicine.

 Write in brief, how these waves can be produced.     

a)  Microwaves are suitable for Radar system used in air navigation. They are produced by special vacuum tubes namely klystron valve or magnetron valve.

b) Infrared waves are used to treat muscular strain. These ways are produced by the vibration of atoms and molecules.

c) X-rays are used as diagnostic tool in medicine. X-rays are produced when high energy electrons are suddenly stopped on metal of high atomic number and by the radio-active decay of nucleus. 

2. What is the electric flux through a cube of side 1 cm which encloses an electric dipole? 

Given, cube encloses an electric dipole.

So, the total charge enclosed by the cube is zero.

That is, Q= 0.

Now, using the formula for electric flux as per the Gauss’s law of electrostatics, we have

ϕ subscript E space equals space contour integral E. d s space equals space Q over epsilon subscript o

ϕ subscript E space equals 0


3.

State the principle of working of a galvanometer.

A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance R1 in series with the coil. If a resistance R2  is connected in series with it, then it can measure upto V/2 volts. Find the resistance, in terms of R1 and R2, required to be connected to convert it into a voltmeter that can read upto 2 V. Also find the resistance G of the galvanometer in terms of R1 and R2

Principle:

When a current carrying loop or coil is placed in the uniform magnetic field, moving coil galvanometer experiences a torque.

A high resistance is connected in series with the galvanometer to convert it into a voltmeter.

The value of resistance is given by, 

straight R straight space equals straight space straight V over straight I subscript straight g minus straight space straight G

where, 

V is the potential difference across the terminals of the voltmeter. 

bold italic I subscript bold italic g bold italic spaceis current through galvanometer and

G is the resistance of the galvanometer. 

When resistance R1 is connected in series with the galvanometer, 

straight R subscript 1 straight space end subscript equals straight space straight V over straight I subscript straight g minus straight G space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space

When resistance R2  is connected in series with the galvanometer, 

straight R subscript 2 straight space equals straight space fraction numerator straight V over denominator 2 straight I subscript straight g end fraction minus straight G straight space              ... (2) 

From equations (1) and (2), we get 

fraction numerator straight V over denominator 2 straight I subscript straight g end fraction equals straight space straight R subscript 1 straight space end subscript minus straight R subscript 2 straight space comma straight space and

straight G space equals space straight R subscript 1 space minus space 2 straight R subscript 2 

Resistance R3 required to convert galvanometer into voltmeter of range 0 to 2V is given by, 

straight R subscript 3 straight space equals straight space fraction numerator 2 straight V over denominator straight I subscript straight g end fraction minus straight space straight G space

rightwards double arrow    straight R subscript 3 straight space end subscript equals straight space 4 left parenthesis straight R subscript 1 straight space end subscript minus straight R subscript 2 right parenthesis minus left parenthesis 2 straight R subscript 1 straight space end subscript minus straight R subscript 2 right parenthesis

space space space space space space space space space space space space space equals straight space 3 straight R subscript 1 straight space end subscript minus 2 straight R subscript 2

Therefore, R1 - 2Ris the galvanometer resistance in terms of R1 and R2


4.

An inductor L of inductance Xis connected in series with a bulb B and an ac source.

How would brightness of the bulb change when (i) number of turn in the inductor is reduced, (ii) an iron rod is inserted in the inductor and (iii) a capacitor of reactance XC = XL  is inserted in series in the circuit. Justify your answer in each case.   

i) Net resistance in the circuit is given by, 

straight z straight space equals straight space square root of straight capital chi subscript straight L squared plus straight R squared end root
Inductance is given by, 

straight L thin space equals straight space straight mu subscript straight o straight N squared over straight l straight A
As number of turns decreases, L decreases.

Inductance is given by, straight X subscript straight L space equals space straight omega space straight L

Therefore, straight capital chi subscript straight L will also decrease thereby, reducing the net resistance in the circuit. Thus, current increases and brightness of the bulb is increased.

ii) When a soft iron rod is inserted in the circuit, L increases. Therefore, inductive reactance also increases. Net resistance increases and flow of current in the circuit decreases. Thus, the brightness of the bulb will decrease.

 

iii) When a capacitor of reactance XL = XC is connected in series with the circuit net resistance becomes,

straight Z straight space equals straight space square root of left parenthesis straight capital chi subscript straight L minus straight capital chi subscript straight C right parenthesis squared plus straight R squared end root ; where R is the resistance of the bulb.

Here, we will have Z= R which is the condition of resonance.

At resonance, maximum current will flow through the circuit. Therefore, the brightness of the bulb will increase.


5.

Graph showing the variation of current versus voltage for a material GaAs is shown in the figure. Identify the region of

i) negative resistance

ii) where Ohm’s law is obeyed.    

                         

i) The region of negative resistance is DE because, the slope is negative for this part of curve.

ii) BC is the part of the curve where Ohm’s law is obeyed because here, current is varying linearly with with the voltage. This gives us direct proportionality between current and voltage.


6.

Two capacitors of unknown capacitances C1  and C2  are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C1  and C2. Also, calculate the charge on each capacitor in parallel combination.


When the capacitors are connected in parallel, 

Equivalent resistance, straight C subscript straight P space equals space straight C subscript 1 space plus space straight C subscript 2 

Energy stored in capacitors, E1 half straight C subscript straight P straight V squared 
therefore straight space straight E subscript straight P straight space equals straight space 1 half left parenthesis straight C subscript 1 plus straight C subscript 2 right parenthesis left parenthesis 100 right parenthesis squared straight space equals straight space 0.25 straight space straight J

rightwards double arrow straight space straight C subscript 1 straight space plus straight space straight C subscript 2 straight space equals straight space 0.5 straight space cross times straight space 10 to the power of – 4 end exponent straight F space

straight C subscript 1 straight space plus straight space straight C subscript 2 straight space equals 50 straight space μF space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space

When the capacitors are connected in series,

Error converting from MathML to accessible text.
Error converting from MathML to accessible text.

Putting this in equation (2), we get 

C2 = 50 – 11.2 = 38.2 μF

The charge on capacitor is given by, Q = CV

Charge on capacitor 1, C1straight Q subscript 1 straight space equals straight C subscript 1 straight V straight space equals straight space 35 straight space cross times straight space 10 to the power of negative 6 end exponent cross times 100 straight space equals straight space 35 straight space cross times straight space 10 to the power of negative 4 end exponent straight space straight C

Charge on capacitor 2, C2 straight Q subscript 2 straight space equals straight space straight C subscript 2 straight V straight space equals straight space 15 straight space cross times 10 to the power of negative 6 end exponent cross times 100 straight space equals straight space 15 cross times straight space 10 to the power of negative 4 end exponent straight space straight C


7. Define capacitor reactance. Write its S.I. units. 

The resistance offered by the capacitor when connected to an electrical circuit is known as capacitive reactance.

It is given by,

chi subscript C space equals space 1 over omega subscript c

where, 

omega italic comma italic spaceis the angular frequency of the source, and 
C, is the capacitance of the capacitor.

SI Unit is ohm.


8.

A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current I.

It is found that when R = 4 straight capital omega, the current is 1 A and when R is increased to 9 straight capital omega , the current reduces to 0.5 A. Find the values of the emf E and internal resistance r.

Plot for V vs. R  



Plot for V vs. I


Current in the circuit is given by, 


straight I straight space equals straight space fraction numerator straight epsilon over denominator straight R plus straight space straight r end fraction

Terminal straight space voltage comma straight space straight V straight space equals straight space IR straight space equals straight space straight V subscript straight A minus straight space straight V subscript straight B

straight V straight space equals straight space fraction numerator εI over denominator straight R plus straight space straight r end fraction straight space equals straight epsilon straight space – straight space Ir

straight I straight space equals straight space fraction numerator straight epsilon over denominator 4 plus straight space straight r end fraction

rightwards double arrow straight space straight r straight space plus 4 straight space equals straight space straight epsilon straight space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis space

0.5 straight space equals straight space fraction numerator straight epsilon over denominator 9 plus straight space straight r end fraction

rightwards double arrow straight space 0.5 straight r straight space plus straight space 4.5 straight space equals straight space straight epsilon space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space

On space solving space equations space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space have space space

Internal space resistance comma space straight r space equals space 1 space space comma space and space space

EMF comma space straight epsilon space space equals space 5 space Volt


9. Use Kirchhoff’s rules to obtain conditions for the balance condition in a Wheatstone bridge.

A Wheatstone bridge arrangement is shown as below: 

 

Using Kirchoff’s second  law to the loop ABDA, we get 

straight I subscript 1 straight P space minus space straight I subscript straight g straight G space minus space straight I subscript 2 straight R space equals space 0 space semicolon space straight G space is space the space galvanometer space resistance. space

Applying Kirchoff’s law to loop BCDB, we get

left parenthesis straight I subscript 1 space – space straight I subscript straight g right parenthesis straight Q space – space left parenthesis straight I subscript 2 space plus space straight I subscript straight g right parenthesis straight S space – space GI subscript straight g space equals space 0 space

When the bridge is balanced, straight I subscript straight g space equals space space 0
Then, the equations can be written as, 

straight I subscript 1 straight P space – space straight I subscript 2 straight R space equals space 0 space orI subscript 1 straight P space equals space straight I subscript 2 straight R                     ... (1) 

straight I subscript 1 straight Q space – space straight I subscript 2 straight S space equals space 0 space orI subscript 1 straight Q space equals space straight I subscript 2 straight S space space                   ... (2) 

On dividing equation (1) by (2), we get

fraction numerator straight space straight P over denominator straight Q end fraction equals space straight R over straight S, which is the balanced condition of a Wheatstone bridge.


10. A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens ? 

Here, a concave lens is placed in a medium of refractive index greater than the refractive index of the material of the lens. 

So, the given lens will behave as a diverging lens. 


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