AIPMT Questions (MCQ) on Gravitation

Physics questions appearing in AIPMT question papers are generally simple. Here are two questions on gravitation which appeared in AIPMT 2010 question paper:

(1) A particle of mass M is situated at the centre of a spherical shell of the same mass and radius a. The gravitational potential at a point situated at a/2 distance from the centre will be

(1) – 4GM/a

(2) – 3GM/a

(3) – 2GM/a

(4) – GM/a

The gravitational potential V at a point situated at distance a/2 from the centre of the shell is equal to the sum of the gravitational potentials due to the particle of mass M and the shell of mass M.

Therefore, V = (– GM/a) + [– GM/(a/2)] = – 3GM/a

[Note that the gravitational potential is a negative quantity and that the potential due to the spherical shell is constant everywhere inside the shell and is equal to the surface value – GM/a].

(2) The radii of circular orbits of two satellites A and B of the earth are 4R and R, respectively. If the speed of satellite A is 3 V, then the speed of satellite B will be

(1) 3V/2

(2) 3V/4

(3) 6V

(4) 12V

The orbital speed v of a satellite is inversely proportional to the square root of the orbital radius r [since v =√(Gm/r)]. Therefore we have

v_A/v_B = √(r_B/r_A)

Here v_A = 3 V, r_A = 4 R and r_B = R (as given in the question).

Therefore 3 V/v_B = √(R/4R) = ½ so that v_B = 6 V

You will find some useful questions (with solution) on gravitation here as well as here.

EAMCET 2010 Multiple Choice Questions on Electronics

Here are the two questions on electronics which were included in EAMCET 2010 Agriculture-Medicine and Engineering question papers respectively:

(1) In the figures shown below

(1) In both Fig. (a) and Fig. (b) the diodes are forward biased

(2) In both Fig. (a) and Fig. (b) the diodes are reverse biased

(3) In Fig. (a) the diode is forward biased and in Fig. (b) the diode is reverse biased

(4) In Fig. (a) the diode is reverse biased and in Fig. (b) the diode is forward biased

In Fig. (a) the anode of the diode is at a positive potential of 10 V while the cathode is at zero volt and hence it is evidently forward biased. In Fig. (b) the anode of the diode is at a higher negative potential compared to the cathode and hence it is reverse biased [Option (3)].

(2) A transistor having a β equal to 80 has a change in base current of 250 μA. Then the change in collector current is

(1) 20,000 mA

(2) 200 mA

(3) 2000 mA

(4) 20 mA

The current gain β is given by

β = ∆I_c/∆I_b at constant collector voltage where ∆I_c is the change in collector current and ∆I_b is the change in base current.

Therefore, ∆I_c = β∆I_b = 80×250 μA = 20,000 μA = 20 mA.

Two Questions (EAMCET 2009 and EAMCET 2008) on Energy and Power

Today we will discuss two multiple choice questions involving energy and power. The first question appeared in the EAMCET 2009 (Engineering) question paper and the second question appeared in the EAMCET 2008 (Engineering) question paper:

(1) A motor of power P₀ is used to deliver water at a certain rate through a given horizontal pipe. To increase the rate of flow of water through the same pipe n times, the power of the motor is increased to P₁. The ratio of P₁ to P₀ is

(1) n : 1

(2) n² : 1

(3) n³ : 1

(4) n⁴ : 1

If the mass of water delivered per second is m when the power is P₀, we have

P₀ α ½ mv² where v is the velocity of water

[We do not equate P₀ to ½ mv² since the efficiency of the motor will not be 100%].

To increase the rate of flow of water through the same pipe n times, the velocity of water has to be made n times. The mass of water delivered per second then is nm. Therefore we have

P₁ α ½ nm (nv)²

From the above we obtain

P₁/ P₀ = n³ [Option (3)].

(2) A river of salty water is flowing with a velocity 2 ms^–1. If the density of water is 1.2 g cm^–3, then the kinetic energy of each cubic metre of water is

(1) 2.4 J

(2) 24 J

(3) 2.4 kJ

(4) 4.8 kJ

The density of the water in the river is 1.2 g cm^–3, which is equal to 1200 kg m^–3. Mass of one cubic metre of the water is 1200 kg and its kinetic energy E is given by

E = ½ mv² = ½ ×1200×2² = 2400 J = 2.4 kJ.

Multiple Choice Practice Questions involving Power in Direct Current Circuits for AP Physics B Exam.

If you grasp fundamental principles thoroughly, you will be able to answer complicated questions without much difficulty. The following questions meant for AP Physics aspirants are relatively simple but they will definitely help you to test your understanding of fundamental principles.

(1) For transferring 50 coulomb of charge through a 10 Ω resistor, the work required is 150 J. The potential difference across the resistor is

(a) 500 V

(b) 50 V

(c) 10 V

(d) 5 V

(e) 3 V

We have W = VQ where W is the work done in transferring Q coulombs of charge through a potential difference of V volt.

Therefore, V = W/Q = 150/50 volt = 3 V.

(2) Four resistors R₁, R₂, R₃ and R₄ having resistances 12 Ω, 16 Ω, 26 Ω and 36 Ω respectively are connected in parallel with a battery of negligible internal resistance. The current in the 12 Ω resistor is 0.5 A. What is the electric power dissipated in the 36 Ω resistor?

(a) 0.5 W

(b) 1 W

(c) 6 W

(d) 18 W

(e) 36 W

Since the resistors are in parallel, the potential difference across them will be the same. (This is true even if the battery has internal resistance).

The potential difference across the 12 Ω resistor is 12×0.5 volt = 6 V.

The potential difference across the 36 Ω resistor also is 6 V.

Therefore, the power dissipated in the 36 Ω resistor = V²/R = 6²/36 = 1 W.

(3) Two 24 V, 16 W electric lamps are connected in series and this series combination is connected across a 12 V battery of negligible internal resistance. The power consumed by each lamp is

(a) 1 W

(b) 2 W

(c) 4 W

(d) 8 W

(e) 16 W

If the resistance of each lamp is R, we have (from power, P = V²/R)

16 = 24²/R ………..(i)

When the series combination of the lamps is connected across a 12 V battery, each lamp has 6 volt across it. The power (X) consumed by each lamp is therefore given by

X = 6²/R …………(ii)

From equations (i) and (ii) X = 1 W.

(4) A 12 V battery of internal resistance 3 Ω is connected in series with a 1 Ω resistor and a variable resistor. The power dissipated in the variable resistor will be maximum when the current through it is

(a) 0.5 A

(b) 1 A

(c) 2 A

(d) 3 A

(e) 4 A

According to maximum power transfer theorem maximum power transfer occurs when the external resistance is equal to the internal resistance of the battery. Therefore, the variable resistor and the fixed 1 Ω resistor together must be 3 Ω for maximum power transfer. The current in the circuit then is 12 V/6 Ω = 2 A [Option (c)].

You will find similar useful questions at AP Physics Resources and at physicsplus.

AIPMT 2010 Multiple Choice Questions on Vectors

The following questions (MCQ) on vectors were included in the AIPMT 2010 question paper:

(1) Six vectors, a through f have the magnitudes and directions indicated in the figure. Which of the following statements is true?

(1) b + e = f

(2) b + c = f

(3) d + c = f

(4) d + e = f

The correct option is (4) as is clear from the adjoining figure

(2) A particle has initial velocity (3î + 4ĵ) and has acceleration (0.4î + 0.3ĵ). Its speed after 10 s is

(1) 10 units

(2) 7 units

(3) 7√2 units

(4) 8.5 units

We have v = u + at with usual notations.

Therefore the velocity after 10 s is given by

v = (3î + 4ĵ) + (0.4î + 0.3ĵ)10 = 7î + 7ĵ

The speed after 10 s is the magnitude of the above velocity vector and is equal to √(7² + 7²) = 7√2 units [Option (3)].

Now see these questions (MCQ) on vectors.