Minimum time of travel of an electric car- M.C.Q. from one dimensional motion

An experimental electric car can produce a maximum acceleration of 2m/s². Its brakes can produce a maximum retardation of 8m/s². What is the minimum time required by this electric car to start from rest and come to a stop, covering a distance of 500m?

(a) 10s (b) 15s (c) 20s (d) 25s (e) 30s

This question may appear to be a bit difficult at the first sight, but it is a simple one which can be solved using the basic equations of motion. The maximum velocity ‘v’ attained during the motion is given by

v² = 0+2a₁s₁ during the accelerated motion and

0 = v²-2a₂s₂ during the decelerated motion. From these equations, s₁ = v²/2a₁ and s₂ = v²/2a₂

Here s₁+ s_{2 =} 500m, a₁= 2m/s² and a₂ = 8m/s².

Therefore, 500 = v²/2a₁ + v²/2a_2. On substituting for a₁ and a_2, v = 40m/s.

But we have, v=0+a₁t₁ and 0=v-a₂t₂ for the accelerated and decelerated parts of the motion respectively so that t₁=v/a₁=40/2 = 20 s and t₂=v/a₂ = 40/8 = 5 s.

The total minimum time therefore is 25 s, given in option (d).

Two Questions from Newton’s Laws of Motion

(1)The breaking strength of a rope is one and a half times the weight of a stone. The minimum time in which this stone can be raised through 10m using this rope is nearly
(a) 0.5s (b)1s (c) 2s (d) 2.5s (e) 3s
The minimum time will be obtained when the acceleration with which the stone is pulled up is maximum. But the maximum possible acceleration ‘a’ is that which will make the apparent weight of the stone(as in the case of a body in a lift) one and a half times its real weight. Therefore we have, m(g+a) = 1.5mg from which a=0.5g.
Substituting this value of ‘a’ in the equation, s = ut + ½ at^2, we have
10 = ½ ×0.5g t^2 since u=0. Taking ‘g’ to be nearly 10m/s^2 we obtain t = 2s.
(2) A rocket with a lift off mass 3.5×10^4 kg is blasted upwards with an initial acceleration of 10m/s^2. Then the initial thrust of the blast is
(a) 1.75×10^5N (b) 3.5×10^5N (c) 7×10^5N (d) 14×10^5N
The initial thrust of the blast is equal in magnitude to the apparent weight of the rocket while moving up with the given acceleration(as in the case of a body in a lift).Therefore, thrust = m(g+a) = 3.5×10^4(10+10) = 7×10^5N [Option (c)]This question appeared in the A.I.E.E.Exmination question paper of 2003.

Intrinsic and Extrinsic Semiconductor

Intrinsic semiconductor is pure (un-doped) semiconductor whereas extrinsic semiconductor is impure (doped) semiconductor. The intrinsic carrier number density Ni and extrinsic carrier number densities densities Ne and Nh are related by the law of mass action as
Ni^2 = NeNh
Note that Ni is the number density of mobile electrons as well as holes in the pure semiconductor where as Ne and Nh are their number densities in the doped semiconductor. Simple problems based on this simple law often find place in Medical and Engineering entrance test papers. Consider the following M.C.Q. which appeared in Kerala Engineering entrance test paper of 2006:
The number densities of electrons and holes in pure germanium at room temperature are equal and its value is 3×10^16 per m^3. On doping with aluminium the hole density increases to 4.5×10^22 per m^3. Then the electron density in doped germanium is

(a) 2×10^10/m^3 (b) 5×10^9/m^3 (c) 4.5×10^9/m^3 (d) 3×10^9/m^3 (e) 4×10^10/m^3

From the law of mass action we have, Ne = Ni^2 /Nh = (9×10^32)/4.5×10^22 = 2×10^10. The correct option therefore is (a).

Two Multiple Choice Questions from Electrostatics
Let us consider the following questions:
(1) What fraction of the energy drawn from the charging battery is stored ina capacitor?

(a) 100% (b) 50% (c)75% (d) 70.7% (e) 41.4%

With usual notations, the energy of a charged capacitor is ½CV^2 = ½QV. Since the charge Q is flowing under the action of the e.m.f. V of the charging battery, the energy supplied by the battery is QV. Therefore, the fraction of the energy stored is ½QV/QV = ½ = 50%.

(2) A charge ‘q’ is placed at the centre of the line joining two equal point charges, each equal to +Q. This system of three charges will be inequilibrium if q is equal to

(a) +Q (b) +Q/2 (c) –Q/2 (d) +Q/4 (e) –Q/4

+Q____q_____+Q Since the force on the charge ‘q’ placed at the centre will be zero irrespective of the sign of ‘q’, it is enough to set the condition for zero net force on the charge +Q. Therefore, 1/4πε0[Q^2/d^2 + Qq/(d/2)^2] = 0 where ‘d’ is the separation between the charges Q&Q. From this we get q = -Q/4, making (e) the correct option.

Magnetic Monopole

In the post dated 22-6-06 it was stated that we don’t get free magnetic poles. However, there are scientists who believe that the possibility of the existence of free magnetic poles (monopoles) can not be ruled out completely. Now consider the following M.C.Q.:
If magnetic monopoles are discovered, which one of the following will have to be modified?
(a) Lenz’s law
(b) Faraday’s law of electromagnetic induction
(c) Gauss theorem in electrostatics
(d) Gauss theorem in magnetism
(e) Ampere’s circuital law.
The correct option is (d). Gauss theorem in magnetism states that the total magnetic flux over a closed surface is zero. This zero value is obtained since we find magnetic dipoles only in nature and the outward flux due to the positive pole and the inward flux due to the negative pole get cancelled, to give a net zero flux. If we have a monopole of strength ‘p’ the flux will be μ0p and not zero.