Here are two multiple choice questions from the section, ‘work, energy and power’. The first question appeared in AIPMT 2010 question paper while the second question appeared in AIPMT 2008 question paper.

(i) An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 ms^{–1 }.The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?

(1) 800 W

(2) 400 W

(3) 200 W

(4) 100 W

The power of the engine is equal to the kinetic energy of water flowing out per second. The mass (M) of water flowing out per second is 2×100 = 200 kg/s

Kinetic energy of water flowing out per second = ½ Mv^{2} = ½ ×200×2^{2} = 400 W.

(ii) Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10 % of energy. How much power is generated by the turbine? (g = 10 ms^{–2})

(1) 7.0 kW

(2) 8.1 kW

(3) 10.2 kW

(4) 12.3 kW

The input power to the turbine is the potential energy due to the water falling per second and is equal to mgh = 15×10×60 =9000 watt.

The power generated by the turbine is 10 % less than the above value and is equal to (9000 – 900) watt = 8100 watt = 8.1 kW.

You will find similar useful questions in this section here as well as here.

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 R_{1} and R_{2} by the lens maker’s equation,

1/f = (n – 1) (1/R_{1} – 1/R_{2})

By Cartesian sign convention, R_{1} is positive and R_{2} is negative for a biconvex lens so that the above equation becomes

1/f = (n – 1) (1/R_{1} + 1/R_{2})

Substituting for f, R_{1} and R_{2 }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 R_{1} = R_{2} = 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 A_{1}. 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 A_{2}. What is the side of the square object?

(a) (√A_{1} +√A_{2})/2

(b) [(A_{1} + A_{2})/2]^{1/2}

(c) (A_{1}A_{2})^{1/2}

(d) (A_{1}A_{2})^{1/4}

(e) [√A_{1 }+√A_{2}]^{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 =√(a_{1}a_{2}) where a_{1 }and a_{2} 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 = a_{1}/a.

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

u/v = a_{2}/a.

From the above expressions, 1 = a_{1}a_{2}/a^{2} from which a =√(a_{1}a_{2})].

But a_{1 }= √A_{1 }and a_{2}= √A_{2}

Therefore, a =√(√A_{1}√A_{2}) = (A_{1}A_{2})^{1/4}

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

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 r_{1} and r_{2} respectively as shown in the figure. If the current flowing in the loop is i, the magnetic induction at the point O is

(1) (μ_{0}i/4)(1/r_{1 }– 1/r_{2})

(2) (μ_{0}i/4)(1/r_{1 }+ 1/r_{2})

(3) (μ_{0}i/2)(1/r_{1 }– 1/r_{2})

(4) (μ_{0}i/2)(1/r_{1 }+ 1/r_{2})

The straight portions PQ and RS do not produce any magnetic field at O. The semicircular portions of radii r_{1} and r_{2 }produce magnetic fields μ_{0}i/4r_{1} and μ_{0}i/4r_{2} respectively at the centre O.

[Remember that the magnetic flux density (magnetic induction) at the centre of a circular loop of radius ris μ_{0}i/2r and hence the magnetic flux density due to a semicircular loop is μ_{0}i/4r].

The field due to the portion of smaller radius r_{1} has greater magnitude μ_{0}i/4r_{1}and is directed perpendicular to the plane of the loop, towards the reader. The field due to the portion of larger radius r_{2} has smaller magnitude μ_{0}i/4r_{2}and is directed perpendicular to the plane of the loop, away from the reader.

The resultant flux density at O is therefore μ_{0}i/4r_{1}– μ_{0}i/4r_{2}, which is equal to (μ_{0}i/4)(1/r_{1 }– 1/r_{2}), 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 Am^{2}

(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 = πr^{2} = π/4