Physics

CBSE Class 12

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1.

Two spherical bobs, one metallic and the other of glass, of the same size are allowed to fall freely from the same height above the ground. Which of the two would reach earlier and why?

The glass bob would reach the ground earlier. Glass bob which is non-conducting in nature will only experience Earth’s gravitational pull unlike the metallic bob which is conducting.

Since the metallic bob is conducting in nature, eddy current is induced as it falls through the magnetic field of the Earth. As per Lenz’s law, current is induced in a direction opposite to the motion of the metallic bob. Hence, there is a delay.

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2.

Write the expression, in a vector form, for the Lorentz magnetic force due to a charge moving with velocity in a magnetic field B. What is the direction of the magnetic force?

Lorentz magnetic force is given by:

; q is the magnitude of the moving charge.

The direction of the magnetic force is perpendicular to the plane containing the velocity vector and the magnetic field vector.

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3.

Given a uniform electric field E =5 ×10^{3 }N/C, find the flux of this field through a square of 10 cm on a side whose plane is parallel to the y-z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis?

Given, Electric flux,

We have, Electric flux, ϕ = E.A =

= 50 Weber

When the plane makes a 30° angle with the *x*-axis, the area vector makes 60° with the *x*-axis

ϕ = E. A

⇒ ϕ=EA cos θ

⇒ ϕ= (5×10^{3})(10^{−2}) cos 60°

⇒ ϕ=

⇒ ϕ = 25 Weber

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4.

"For any charge configuration, equipotential surface through a point is normal to the electric field." Justify.

Work done (*W*) in moving a test charge along an equipotential surface is zero.

Work done is given by,

F is the electric force and s is the magnitude of displacement.

For non-zero displacement, this is possible only when cos is equal to 0.

= 90°

Thus, the force acting on the point charge is perpendicular to the equipotential surface.

Electric field lines give us the direction of electric force on a charge.

Thus, for any charge configuration, equipotential surface through a point is normal to the electric field.

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5.

Out of the two magnetic materials, 'A' has relative permeability slightly greater than unity while 'B' has less than unity. Identify the nature of the materials 'A' and 'B'. Will their susceptibilities be positive or negative?

A is a paramagnetic material because its permeability is greater than unity and its susceptibility is positive. The relative permeability of a paramagnetic material is between and susceptibility is

For a diamagnetic material, the relative permeability lies between 0 ≤ μ_{r} < 1 and its susceptibility lies between −1< χ< 0.

Therefore, 'B' is a diamagnetic material and its susceptibility is negative.

This is because its relative permeability is less than unity.

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6.

The carrier wave is given by C (*t*) = 2sin (8π*t*) Volt.

The modulating signal is a square wave as shown. Find modulation index.

The generalized equation of a carrier wave is given by,

The generalized equation of a modulating signal is given by,

On comparing the given equation of carrier wave with the generalized equation, we get,

Amplitude of modulating signal, A_{m} = 1 V

Amplitude of carrier wave, A_{c} = 2 V

Modulation index is the ratio of the amplitude of modulating signal to the amplitude of carrier wave .

It is denoted by,

So,

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Define the term 'Mobility' of charge carriers in a conductor. Write its S.I. unit

Drift velocity per unit applied electric field is known as the mobility of charge carriers in a conductor.

Mathematically,

Mobility,

S.I. unit of mobility: or

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8.

State Kirchhoff's rules. Explain briefly how these rules are justified.

Kirchhoff’s First Law or Junction Rule states that “The sum of the currents flowing towards a junction is equal to the sum of currents leaving the junction.”

This is in accordance with the conservation of charge which is the basis of Kirchhoff’s current rule.

Here*, I*_{1}, *I*_{2} *I*_{3}, and *I*_{4} are the currents flowing through the respective wires.

Convention: The current flowing towards the junction is taken as positive and the current flowing away from the junction is taken as negative.

*I*_{3} + (− *I*_{1}) + (− *I*_{2}) + (− *I*_{4}) = 0

Kirchhoff’s Second Law or Loop Rule states that In a closed loop, the algebraic sum of the *emf*s is equal to the algebraic sum of the products of the resistances and the currents flowing through them.

OR

“The algebraic sum of all the potential drops and emfs along any closed path in a network is zero.”

For the closed loop BACB:

*E*_{1} − *E*_{2} = *I*_{1}*R*_{1} + *I*_{2}*R*_{2} − *I*_{3}*R*_{3}

For the closed loop CADC:

*E*_{2} = *I*_{3}*R*_{3} + *I*_{4}*R*_{4} + *I*_{5}*R*_{5}

This law is based on the law of conservation of energy.

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9.

Show variation of resistivity of copper as a function of temperature in a graph.

The graph below shows variation of resistivity of copper with temperature. The graph is parabolic in nature.

'

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10.

A capacitor 'C', a variable resistor 'R' and a bulb 'B' are connected in series to the ac mains in circuit as shown. The bulb glows with some brightness. How will the glow of the bulb change if (i) a dielectric slab is introduced between the plates of the capacitor, keeping resistance R to be the same; (ii) the resistance R is increased keeping the same capacitance?

(i) When a dielectric slab is introduced between the plates of the capacitor, its capacitance will increase. Hence, the potential drop across the capacitor will decrease (V= Q/C ) thereby decreasing the potential drop across the bulb because, both the bulb and capacitor is connected in series. So, brightness of the bulb will increase.

(ii) When resistance (R) is increased keeping the capacitance same, the potential drop across the resistor will increases. Therefore, the potential drop across the bulb will decrease because both are connected in series. So, brightness of the bulb will decrease.

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