Multiple Choice Questions on Lenses

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Here are a few questions from optics involving the lens maker’s equation and the concept of magnification in the case of lenses:

(1) A biconvex lens has the same radius of curvature R for its faces. If the focal length of the lens in air is R/2, the refractive index of the material of the lens is

(a) 1.2

(b) 1.33

(c) √2

(d) √3

(e) 2

The refractive index n is related to the focal length f and the radii of curvature R₁ and R₂ by the lens maker’s equation,

1/f = (n – 1) (1/R₁ – 1/R₂)

By Cartesian sign convention, R₁ is positive and R₂ is negative for a biconvex lens so that the above equation becomes

1/f = (n – 1) (1/R₁ + 1/R₂)

Substituting for f, R₁ and R₂ we obtain

1/(R/2) = (n – 1) (2/R)

Or, 2/R = (n – 1) (2/R) from which n = 2

(2) A biconvex lens has the same radius of curvature R for its faces. If the focal length of the lens in air is R, what will be its focal length in a liquid of refractive index 1.5?

(a) R/2

(b) R

(c) 2R

(d) 4R

(e) Infinite

Many of you might have noted that an equiconvex lens of refractive index 1.5 has focal length equal in magnitude to the radius of its faces.

Some of you might have noted the converse of the above fact: If the refractive index is 1.5 in the case of an equiconvex (or equiconcave) lens, the magnitude of its focal length will be the same as that of the radius of curvature of its faces

[You can check this by substituting n = 1.5 and R₁ = R₂ = R in the lens maker’s equation].

Since the refractive index of the material of the lens is 1.5, its focal length in a liquid of refractive index 1.5 will be infinite.

(3) A converging lens produces a real, magnified and well defined image of a small illuminated square on a screen. The area of the image is A₁. When the lens is moved towards the screen without disturbing the object and the screen, the area of the well defined image obtained on the screen is A₂. What is the side of the square object?

(a) (√A₁ + √A₂)/2

(b) [(A₁ + A₂)/2]^1/2

(c) (A₁A₂)^1/2

(d) (A₁A₂)^1/4

(e) [√A₁ +√A₂]^1/4

The two positions (of the lens) for which well defined images of the square are obtained, are the conjugate positions and hence we have

a =√(a₁a₂) where a₁ and a₂ are the sides of the images in the two cases and a is the side of the square object.

[The linear magnification in the first case is v/u = a₁/a.

Since the object distance u and the image distance v are interchanged in conjugate positions, we have, in the second case,

u/v = a₂/a.

From the above expressions, 1 = a₁a₂/a² from which a =√(a₁a₂)].

But a₁ = √A₁ and a₂ = √A₂

Therefore, a =√(√A₁√A₂) = (A₁A₂)^1/4

You will find similar useful multiple choice questions on optics here as well as here.

EAMCET (Medical) 2010 Questions on Magnetic Fields Due to Current Carrying Conductors

Today we will discuss two questions on magnetic fields due to current carrying wires. These questions appeared in EAMCET (Agriculture-Medicine) 2010 question paper:

(1) A wire loop PQRS is constructed by joining two semicircular coils of radii r₁ and r₂ respectively as shown in the figure. If the current flowing in the loop is i, the magnetic induction at the point O is

(1) (μ₀i/4)(1/r₁ – 1/r₂)

(2) (μ₀i/4)(1/r₁ + 1/r₂)

(3) (μ₀i/2)(1/r₁ – 1/r₂)

(4) (μ₀i/2)(1/r₁ + 1/r₂)

The straight portions PQ and RS do not produce any magnetic field at O. The semicircular portions of radii r₁ and r₂ produce magnetic fields μ₀i/4r₁ and μ₀i/4r₂ respectively at the centre O.

[Remember that the magnetic flux density (magnetic induction) at the centre of a circular loop of radius r is μ₀i/2r and hence the magnetic flux density due to a semicircular loop is μ₀i/4r].

The field due to the portion of smaller radius r₁ has greater magnitude μ₀i/4r₁ and is directed perpendicular to the plane of the loop, towards the reader. The field due to the portion of larger radius r₂ has smaller magnitude μ₀i/4r₂ and is directed perpendicular to the plane of the loop, away from the reader.

The resultant flux density at O is therefore μ₀i/4r₁– μ₀i/4r₂, which is equal to (μ₀i/4)(1/r₁ – 1/r₂), as given in option (1).

[The resultant field at O is directed normal to the plane of the loop, towards the reader. This point too can be incorporated in the question to make it a little more difficult].

(2) A wire of length 6.28 m is bent into a circular coil of 2 turns. If a current of 0.5 A exists in the coil, the magnetic moment of the coil is, in Am²

(1) π/4

(2) ¼

(3) π

(4) 4π

The magnetic moment m of a plane circular coil of area A with n turns carrying current i is given by

m = niA

The radius r of the coil is given by

2×2πr = 6.28

Therefore, r = 6.28/4π = ½ and area A = πr² = π/4

Substituting, m = 2×0.5 × π/4 = π/4.

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.