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.

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Two AIPMT 2008 Questions from Thermal Physics

The following questions appeared in the All India Pre-Medical / Pre-Dental Entrance 2008 Examination:

(1) At 10º C the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110º C this ratio is

(1) 383x/283

(2) 10x/110

(3) 283x/383

(4) x

We have PV = rT where r is the gas constant for the given mass.

Since the volume V = M/ρ where M is the mass and ρ is the density, we have

P/ρ = rT/M

Since r and M are constants for a given mass of gas, we have

(ρ₁/P₁) /(ρ₂/P₂) = T₂/T₁ where the suffix 1 is for the quantities at temperature 10º C and suffix 2 is for the quantities at temperature 110º.

It is given that (ρ₁/P₁) = x.

Therefore x/(ρ₂/P₂) = 383/283 since the temperatures are in degree kelvin.

From this (ρ₂/P₂) = 283x/383

(2) On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39º W and 239º W respectively. What will be the temperature on the new scale, corresponding to a temperature of 239º C on the Celsius scale?

(1) 117º W

(2) 200º W

(3) 139º W

(4) 78º W

We have 0º C = 39º W and 100º C = 239º W

A temperature difference of 100º C is therefore equivalent to a temperature difference of 200º W. A temperature difference of 1º C is thus equivalent to a temperature difference of 2º W.

Therefore, a temperature 39º C is equal to [39 + (39×2)]º W = 117º W

All India Pre-Medical / Pre-Dental Entrance Examination (AIPMT)-2009

CBSE (Central Board of Secondary Education), Delhi has invited applications for All India Pre-Medical / Pre-Dental Entrance Examination -2009 for admission to 15% of the total seats for Medical/Dental Courses in all Medical/Dental colleges run by the Union of India, State Governments, Municipal or other local authorities in India except in the States of Andhra Pradesh and Jammu & Kashmir, as per the following schedule :-

(i). Preliminary Examination - 5th April, 2009 (Sunday)

(ii). Final Examination - 10th May, 2009 (Sunday)

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MCQ on Newton’s Laws of Motion- KEAM (Engineering) 2008 Questions

The following multiple choice questions appeared in Kerala Engineering Entrance 2008 question paper:

(1) A particle of mass 2 kg is initially at rest. A force acts on it whose magnitude changes with time. The force time graph is shown below

The velocity of the particle after 10 s is

(a) 20 ms^–1

(b) 10 ms^–1

(c) 75 ms^–1

(d) 26 ms^–1

(e) 50 ms^–1

The area under the force-time graph gives the impulse received which is equal to the change in momentum of the particle. Since the initial momentum of the particle is zero, the area under the graph gives the momentum of the particle after 10 s.

The area under the curve is made of rectangles and triangles and is equal to100 newton second. [Note that Ns is the same as kg ms^–1].

The velocity of the particle after 10 s = (100 kg ms^–1)/2 kg = 50 ms^–1

(2) An object of mass 5 kg is attached to the hook of a spring balance and the balance is suspended vertically from the roof of a lift. The reading on the spring balance when the lift is going up with with an acceleration of 0.25 ms^–2 is (g = 10 ms^–2)

(a) 51.25 N

(b) 48.75 N

(c) 52.75 N

(d) 47.25 N

(e) 55 N

The weight of a body of mass ‘m’ in a lift can be remembered as m(g-a) in all situations if you apply the proper sign to the acceleration ‘a’ of the lift. The acceleration due to gravity ‘g’ always acts vertically downwards and its sign is to be taken as positive. Since the lift is moving upwards the sign of its acceleration is negative so that the weight of the object as indicated by the spring balance is m[g-(-a)] = m(g+ a) = 5×10.25 = 51.25 N.

(3) A bullet of mass 0.05 kg moving with a speed of 80 ms^–1 enters a wooden block and is stopped after a distance of 0.4 m. The average resistance force exerted by the block on the bullet is

(a) 300 N

(b) 20 N

(c) 400 N

(d) 40 N

(e) 200 N

The bullet is retarded within the wooden block. The retardation is given by the usual equation of uniformly accelerated motion, v² = u² + 2as. Here v = 0, u = 80 ms^–1 and s = 0.4 m.

Therefore, 0 = 80² + 2a×0.4 from which a = – 8000 ms^–2

The retardation is 8000 ms^–2 and the resistance force exerted by the wooden block is 0.05×8000 = 400 N.