Physics

CBSE Class 12

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2.

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

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

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3.

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

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4.

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?

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6.

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?

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7.

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?

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8.

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?

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

Kirchhoff’s First Law or Junction Rule states -'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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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?

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