## Monday, September 20, 2010

### Karnataka CET 2010 Questions on Alternating Currents

Science without religion is lame; religion without science is blind.

– Albert Einstein

Questions on alternating currents are generally interesting and often not tough. By clicking on the label ‘alternating current’ below this post, you can access all posts related to this section on this site.

Here are two questions on alternating currents which appeared in Karnataka CET 2010 question paper:

(1) In a series R-L-C circuit, the voltage across R is 100 V and the value of R = 1000 Ω The capacitance of the capacitor is 2×10–6 F; angular frequency of A.C. is 200 rad s–1. Then the P.D. across the inductance coil is

(a) 100 V

(b) 40 V

(c) 250 V

(d) 400 V

Since the voltage drop across the 1000 Ω resistor is 100 V, the current (I) in the R-L-C circuit is 100 V/1000 Ω = 0.1 A.

At resonance the magnitude of the voltage across the inductance (VL) is the same as that across the capacitor (VC).

Or, VL = VC = I/Cω where ω is the angular frequency of A.C.

Therefore, VC = 0.1/(2×10–6×200) = 250 V.

(2) A capacitor and an inductance coil are connected in separate AC circuits with a bulb glowing in both the circuits. The bulb glows more brightly when

(a) an iron rod is introduced into the inductance coil

(b) the number of turns in the inductance coil is increased

(c) separation between the plates of the capacitor is increased

(d) a dielectric is introduced into the gap between the plates of the capacitor

The reactance of the inductor is where L is the inductance and ω is the angular frequency of AC. In options (a) and (b) the value of L is increased and hence the inductive reactance is increased. The current through the bulb is therefore decreased, thereby decreasing the brightness.

The capactive reactance is 1/ where C is the capacitance. When the separation between the plates is increased as given in option (c), the capacitance is decreased. The capacitive reactance is therefore increased, thereby decreasing the current and the brightness.

On introducing a dielectric slab between the plates, the capacitance is increased, thereby decreasing the capacitive reactance. The brightness of the bulb is then increased because of the increased current. So the correct option is (d).

## Saturday, September 11, 2010

### IIT-JEE 2010 Questions on Thermodynamics (Multiple Correct Answer Type and Integer Type)

The following questions on thermodynamics appeared in the IIT-JEE 2010 question paper:

[Question No.1 is Multiple Correct Answer Type. Question No.2 is Integer Type, the answer to which is a single digit integer ranging from 0 to 9].

(1) One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, As shown in figure. Its pressure at A is P0. Choose the correct option(s) from the followin g:

(A) Internal energies at A and B are the same

(B) Work done by the gas in process AB is P0V0 ln 4

(C) Pressure at C is P0/4

(D) Temperature at C is T0/4

AB is an isothermal process (A and B are at the same temperature). Therfore the internal energies at A and B are the same.

The work (W) done by the gas in the isothermal process AB is given by

W = nRT ln(V2/V1) where R is universal gas constant, ‘n’ is the number of moles in the sample of the gas, T is the temperature at which the process occurs, V1 is the initial volume and V2 is the final volume.

Therefore W = 1×RT0 ln (V2/V1) = RT0 ln(4V0/V0)

Since RT0 = P0V0 we obtain

W = P0V0 ln 4

Thus options A and B are correct.

We have PV/T = constant

Assuming that the line BC passes through the origin, the temperature at C must be T0/4. Considering the states at A and C we have

P0V0 /T0 = PCV0/(T0/4)

The pressure at C is therefore given by

PC = P0/4

So all options are correct.

(2) A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. If the initial temperature of the gas is Ti (in Kelvin) and the final temperature is aTi, the value of a is:

In the case of an adiabatic process we have

TVγ–1 = constant where γ is the ratio of specific heats of the gas.

Therefore, TiVγ–1 = aTi(V/32)γ–1

For an ideal diatomic gas γ = 7/5 and hence we have

TiV2/5 = aTi(V/32)2/5

This gives a = 322/5 = 2/5 = 22 = 4.