Two AIPMT 2008 Questions (MCQ) from Current Electricity

Some of you may understand the principles in physics very well but your capacity for numerical manipulations may be poor. Practice can make you strong in solving questions involving numerical manipulations so that you will not waste your precious time on such questions. Here are two questions which appeared in AIPMT 2008 question paper:

(1) An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20º C? The temperature of boiling water is 100º C.

(1) 8.4 min

(2) 12.6 min

(3) 4.2 min

(4) 6.3 min

We have VIt =mcθ where V is the voltage, I is the current, t is the time of flow of the current, m c is the specific heat of water (which is approximately 4200 Jkg^–1 K^–1) and θ is the temperature rise. is the mass of water,

Therefore, 220×4×t = 1×4200×(100 – 20)

This will give t = 381 sec. (nearly) which is approximately 6.3 min.

(2) a galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

(1) 5550 Ω

(2) 6050 Ω

(3) 4450 Ω

(4) 5050 Ω

Since the current through an ordinary galvanometer is directly proportional to the deflection (remember that in a tangent galvanometer this is not the case) we have

3/(50+2950) = k×30 when the deflection is 30 divisions.

Here k is the proportionality constant (figure of merit of the galvanometer).

If the resistance in series for reducing the deflection to 20 divisions is X we have

3/(50+X) = k×20

On dividing the first equation by the second,

(50+X)/3000 = 3/2 from which X = 4450 Ω.

Two Questions from Electrodynamics

(1) A straight conductor carries a current of 10A. An electron traveling with a speed of 2.5×10^6m/s parallel to the conductor at a distance of 1cm from the conductor experiences a force of
(a) 8×10^-17N (b) 8×10^-18N (c) 8×10^-19N (d) 4×10^-20N (e) 4×10^-18N
Force on the electron, F=qvB, where B= μoI/2πr with usual notations.
Note that the direction of the magnetic field produced by the conductor is perpendicular to the direction of motion of the electron since the magnetic field lines are in the form of concentric circles.
Therefore, F= (1.6×10^-19×2.5×10^6×4π×10^-7×10) / 2π×0.01 = 8×10^-17N.
(2) A galvanometer having resistance 3Ω is converted into an ammeter by connecting a 3Ω shunt resistance across it. In order to double the range of this ammeter, the additional shunt resistance to be connected across it is
(a) 3Ω (b) 2.5Ω (c) 2Ω (d) 1.5Ω (e) 6Ω
With usual notations, Ig G = (I-Ig)S. Since G = S =3Ω, Ig = I/2. If S′ is the additional shunt required, the effective shunt for doubling the range is SS′/(S+S′) . Therefore, when the range is doubled, IgG = (2I-Ig) SS′/(S+S′) . Substituting G=S=3 and Ig= I/2 , 3S'/(3+S') =1. Therefore,S'=1.5Ω.
Instead of following the above mathematical steps, you can easily arrive at the answer as follows:
The ammeter with the original shunt has an effective resistance of 1.5Ω. If you want to pass twice the maximum allowed current into the ammeter, half of this doubled current has to be diverted through another path. So you have to provide a shunt path of resistance 1.5Ω.