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CBSE Class 12

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CBSE Class 12 Physics Solved Question Paper 2015 Short Answer Type

1. 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  2873 Views

2. 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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3.

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.
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4.

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, where,

V is the potential difference across the terminals of the voltmeter. is current through galvanometer and

G is the resistance of the galvanometer.

When resistance R1 is connected in series with the galvanometer, When resistance R2  is connected in series with the galvanometer, ... (2)

From equations (1) and (2), we get Resistance R3 required to convert galvanometer into voltmeter of range 0 to 2V is given by, Therefore, R1 - 2Ris the galvanometer resistance in terms of R1 and R2

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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.

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6.

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, Inductance is given by, As number of turns decreases, L decreases.

Inductance is given by, Therefore, 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, ; 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.

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7.

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, Energy stored in capacitors, E  When the capacitors are connected in series,  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, C1 Charge on capacitor 2, C2 4147 Views

8. 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 Applying Kirchoff’s law to loop BCDB, we get When the bridge is balanced, Then, the equations can be written as, ... (1) ... (2)

On dividing equation (1) by (2), we get , which is the balanced condition of a Wheatstone bridge.

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9. 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, where, is the angular frequency of the source, and
C, is the capacitance of the capacitor.

SI Unit is ohm.

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10.

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 , the current is 1 A and when R is increased to 9 , 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, 5271 Views