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ISC Class 12 Physics Solved Question Paper 2007

Short Answer Type

21. Derive an expression for the capacitance of a parallel plate capacitor in terms of the areas of the plates and the distance between them. Assume there is a vacuum in between the plates with the electric field between the plates E = σ/ ε₀. (where the symbols have their usual meaning.)

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22. Three equal charges of 5.0 μC each are placed at the three vertices of an equilateral triangle of side 5.0 cm each. Calculate the electrostatic potential energy of the system of charges.

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23. Define the emf of a cell in terms of the work done in moving a charge through the cell. Using Joule’s law obtain the equation, emf = 1(R + r). (The symbols have their usual meaning.)

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24. (i) State one important difference between the Joule effect and the Peltier effect.

ii) The principle of a deflection magnetometer is expressed by the formula B_m = B_H tan θ. What is represented by the symbol B_m and B_H?

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25.

Obtain the formula, I = k θ for a moving coil galvanometer, given, the deflecting torque where m is the magnetic dipole moment of the coil placed in the magnetic field B, I is the current in the galvanometer and θ is the deflection.

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26.

(i) Define the terms magnetization (M) and magnetic susceptibility (x_m).

(ii) What is the value of the magnetic susceptibility of aluminium if its relative permeability is 1.000022 ?

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27. Obtain the relation for the current I = I₀ sin (ωt - π/2) for a pure inductor across which an alternating emf ε = ε₀ sin ψt is applied.

Let,

E = E₀ sin ωt be the alternating voltage applied across an inductor (pure, ideal).

Due the applied e.m.f. a current flows through the coil.

This current produces a magnetic field.

Since the current is alternating the magnetic field will also be changing.

The changing magnetic field produces an e.m.f. in the coil.

According to Lenz’slaw this induced emf is opposes the cause producting it. We want a.c to flow through the coil so the applied emf must be equal and opposite to this induced back emf.

E = - L. dI/dt

Applied space AC space equals space minus space left parenthesis negative straight L space dI over dt right parenthesis space equals space straight L. space fraction numerator dI over denominator dt space end fraction straight E subscript straight o space sin space ωt space equals space straight L space. space dI over dt. space dI space equals space straight E subscript straight o over straight L space sin space ωt space. space dt Integrating space both space sides comma space integral dl space equals straight E subscript straight o over straight L integral space sin space straight omega space straight t space dt space straight I space equals straight E subscript straight o over straight L open parentheses negative fraction numerator cos space straight omega space straight t over denominator straight omega end fraction close parentheses space equals straight E subscript straight o over Lω space left parenthesis space minus space cos space ωt right parenthesis 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 equals space straight E subscript straight o over Lω space sin space left parenthesis ωt space minus space straight pi over 2 right parenthesis straight I space is space maximum space when space sin space left parenthesis space ωt space straight x space straight pi divided by 2 right parenthesis space is space maximum space which space is space one. So comma space straight I subscript straight o space equals space straight E subscript straight o over Lω space semicolon space straight I space equals space straight I subscript straight o space sin space open parentheses ωt space minus space straight pi over 2 close parentheses where comma space straight I subscript straight o space equals space straight E subscript straight o over Lω space

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28. A resistance of 150 Ω and capacitance of 15 μ F are connected in series with an AC source. The peak value of the current is 0.20 A. Calculate the average power consumed in the circuit. If the capacitor is removed but the current is kept the same, what is the average power consumed in the resistor alone ?

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29. In ac circuit resistance R, inductance I, and capacitance C are connected in series. The value of C is adjusted for resonance. State any three properties of this circuit which hold true for resonance only. (Give the properties in words or equations, relating X_L, Xc, Z, R, φ, power etc.)

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