You can access all posts involving magnetic fields and electromagnetic induction on this blog by trying a search using the search box at the top of this page or by clicking on the labels ‘magnetic field’ and ‘electromagnetic induction’ below this post. Today I give you a free response practice question on electromagnetism. This question is for AP Physics C aspirants:

A rectangular wire loop of length ℓ and breadth b havingnegligible resistance is arranged in the plane of an infinitely long straight vertical wire as shown, with the longer sides parallel to the wire. The wire loop contains a capacitor of capacitance C. A steady current ‘I’ flows upwards through the straight wire. Now answer the following questions:

(a) Calculate the magnetic flux through the wire loop.

(b) The loop is rotated through 180º about a central axis (fig) which is parallel to the straight wire. If the time taken for this rotation is ∆t, determine the magnitude of the emf induced in the loop.

(c) The loop is kept stationary and instead of the steady current I, a current ‘i’varying with time t as i = I_{m }sinωt (where I_{m} and ω are constants) is passed through the straight wire. Calculate the maximum value of the emf induced in the wire loop.

(d) Determine the maximum current induced in the wire loop under the conditions mentioned in part (c) above.

Try to answer this question. You can take 15 minutes for answering it and can score up to 15 points for the right answer. I’ll be back soon with a model answer for your benefit.

Equations to be remembered in the section ‘oscillations’ were discussed in the post dated 17^{th}April 2008.Some multiple choice practice questions in this section were discussed in the post dated 22^{nd} April 2008, followed by a free response practice question in the post dated 2^{nd}May 2008. You can access all these posts by clicking on the label ‘oscillation’ below this post. Today we will discuss a few multiple choice practice questions involving simple pendulum:

(1) A simple pendulum arranged inside a train has period of oscillation T when the train is at rest. When the train moves along straight horizontal rails with uniform acceleration of x ms^{–2}, the period of the pendulum is (assuming g = 10 ms^{–2})

(a) T ×g^{1/2}/(g+x)^{1/2}^{}

(b) T ×g^{1/4}/(g+x)^{1/4 }

(c) T ×g^{1/4}/(g^{2}+x^{2})^{1/4}

(d) T ×g^{1/2}/(g^{2}+x^{2})^{1/2}

(e)T ×g^{1/2}/(g^{2}+x^{2})^{1/4}

When the train is at rest the period of oscillation T of the pendulum is given by

T = 2π (ℓ/g)^{1/2} ………….(i)

where ℓ is the length of the pendulum and g is the acceleration due to gravity.

When the train moves forward with acceleration x ms^{–2}, an inertial backward force acts on the bob of the pendulum and supplies a backward acceleration of x ms^{–2} (fig.). The resultant acceleration of the bob is (g^{2}+x^{2})^{1/2}. The period T_{1}of the pendulum is now given by

T_{1} =2π[ℓ/(g^{2}+x^{2})^{1/2}]^{1/2}……..(ii)

Dividing eqn (ii) by eqn (i) we have

T_{1}/T = g^{1/2}/(g^{2}+x^{2})^{1/4} from which T_{1} = T ×g^{1/2}/(g^{2}+x^{2})^{1/4} The correct option is (e)

[Suppose the train is moving with uniform velocity of x ms^{–1}. What will be the period? No doubt, T itself since you cannot distinguish between state of rest and uniform motion].

(2) Two simple pendulums A and B have periods 2.1 s and 2 s respectively. They start oscillating at the same time in phase. They will be in phase instantly at the end of

(a) 42 s

(b) 40 s

(c) 22 s

(d) 21 s

(e) 20 s

Suppose the pendulums are in phase instantly at the end of n oscillations of pendulum A. Pendulum B should then execute n+1 oscillations. Therefore we have

n×2.1 = (n+1)×2

Therefore 0.1n = 2 so that n = 20

The time elapsed is therefore 20×2.1 = 42 s [Option (a)].

[The difference between the periods of the two pendulums is 0.1 s. So you require 20 oscillations of pendulum A to obtain a time difference equal to one period (2 s) of pendulum B so that the two pendulums will be instantly in phase].

(3) The bob of a simple pendulum is a hollow metal sphere filled with water. There is a small hole at the bottom of this hollow sphere and water drains out through the hole as the pendulum oscillates. After completely filling the bob with water, if this pendulum is made to oscillate for a long time, its period of oscillation will

(a) increase first and will reach a final constant value

(b) decrease first, reach a minimum value, then increase and will finally settle at the initial value

(c) increase first, reach a maximum value, then decrease and will finally settle at the initial value

(d) decrease first and will reach a final constant value

(e) remain unchanged

The correct option is (c). Initially the centre of gravity of the spherical bob is at its centre since it is completely filled with water. When the water flows out, the centre of gravity of the bob moves gradually downwards, reaches a minimum level and then moves up. When the water is fully drained out, the centre of gravity of the bob once again reaches the centre of the bob and remains there.

The length of the pendulum is the distance between the centre of gravity of the bob and the point of suspension. Therefore, the length of the pendulum gets increased initially, becomes a maximum, then gets decreased and finally settles at the initial value. Therefore, the period of the pendulum will increase first, reach a maximum value, then decrease and will finally settle at the initial value.

(4) A simple pendulum is taken to a location where the acceleration due to gravity is decreased by 0.1 %. If the period of oscillation is to be unaltered

(a) the mass of the bob is to be increased by 0.1 %

(b) the mass of the bob is to be decreased by 0.1 %

(c) the length of the pendulum is to be decreased by 0.316 %

(d) the length of the pendulum is to be decreased by 0.2 %

(e) the length of the pendulum is to be decreased by 0.1 %

Since the period of oscillation T is given by T = 2π√(ℓ/g), the ratio ℓ/g should be unaltered. Therefore, the length ℓ should be decreased by 0.1 %

The following questions are for AP Physics C aspirants only:

(5) The period of a simple pendulum is decreased by 0.02 s when the length of the pendulum is decreased by 1 cm. The original length of the pendulum is nearly (g = 10 ms^{–2})

-->

(a) 0.25 m

(b) 0.5 m

(c) 1 m

(d) 1.01 m

(e) 2 m

The period of oscillation T is given by T = 2π√(ℓ/g).

Therefore, ∆T/T = ½ ∆ℓ/ℓ – ½ ∆g/g

[Here ∆T,∆ℓ, and∆g represent the increments in the period, length and acceleration due to gravity respectively. We have written the above equation by taking the logarithm of the expression for period and then differentiating it]

The increment in T is – 0.02 s and the increment in ℓ is – 0.01 m (negative signs are because T and ℓ are decreased). There is no change in g.

Therefore we have – 0.02/T = – ½ × 0.01/ℓ

This gives T = 4ℓ.

But T = 2π√(ℓ/g) so that 2π√(ℓ/10) = 4ℓ

Or, π^{2}/10 = 4ℓ

Since π^{2} is nearly equal to 10, we obtain ℓ = 0.25 m, nearly.

Equations to be remembered for answering questions on direct current circuits and RC transients were posted on this site on 31^{st} March 2008. You can access them by clicking on the label ‘direct current circuit’ below this post or by trying a search using the search box at the top of this page. Today we will discuss some more multiple choice practice questions in this section:

(1) A uniform copper wire of radius r and length L has resistance R at temperature T.If this wire is carefully stretched so that its radius becomes kr where k is a constant, its resistance at the temperature T will be

(a) kR

(b) R/k^{2}

(c) k^{2}R

(d) k^{4}R

(e)R/k^{4}^{}

If A is the area of cross section of the wire before the stretching, we have

R = ρL/A where ρ is the resistivity (specific resistance) of the material of the wire.

When the radius becomes kr, the area of cross section of the wire becomes k^{2}A.Since the volume of the wire is unchanged, the length of the wire becomes L/k^{2}. The resistance of the wire after stretching therefore becomes R’ given by

R’ = ρ(L/k^{2})/k^{2}A = ρL/ k^{4}A = R/k^{4}

[Note that the resistance of the wire after stretching is greater than R since k < 1].

(2) In the above question suppose the wire was stretched so that its length becomes 1.01 L. The resistance of the wire after stretching then becomes nearly

(a) 1.01R

(b) R/1.02^{}

(c) 1.02R

(d) 1.04R

(e)R/1.04^{}

Note that if a wire of resistance R is stretched so that its length becomes n times its original length, its area of cross section becomes (1/n) times its original area of cross section. The resistance of the wire therefore becomes n^{2}R.

Since n = 1.01 the resistance after stretching is (1.01)^{2}R =1.02R, nearly [Option (c)].

Here is a variation of question no.1:

(3) If the radius of a uniform metal wire of resistance R is changed by 1% by stretching it carefully under isothermal conditions, its resistance becomes

(a) 1.04R

(b) R/1.04^{}

(c) 1.02R

(d) R/1.02

(e)1.01R

We have R’ = R/k^{4} as shown in question no.1. Since the change (decrease) in the radius is 1%, k = 0.99.

(4) The resistance of an ammeter of range 2 A is R. The shunt resistance required to make its range 6 A is

(a) R

(b) R/5

(c) R/6

(d) R/2

(e)R/3

Let S be the shunt resistance required. When a current of 6 A is passed through the ammeter- shunt combination, 2 A should pass through the ammeter and the remaining 4 A should pass through the shunt (fig.). Therefore we have

2×R = 4×S, on equating the potential difference across the ammeter resistance R to the potential difference across the shunt resistance S.

This gives S = R/2

The following questions are for AP Physics C aspirants (even though AP Physics B aspirants also can try them):

(5) A power supply having internal resistance r drives a current through a load of resistance P. The heat developed in the load resistance is H. When the same power supply drives a current through another load of resistance Q for the same time t, the heat developed in the load resistance is H itself. The internal resistance of the power supply is for a time

(a) √(PQ)

(b) √[(P^{2 }+ Q^{2})/2]

(c) (P + Q)/2

(d) (P – Q)/2

(e) √[(P^{2 }– Q^{2})/2]

Since H = I^{2}Rt where I is the current through the load resistance R, we have

H = [V/(P+r)]^{2}Pt = [V/(Q+r)]^{2}Qt where V is the emf of the power supply.

[V/(P+r) is the current in the load resistance P and V/(Q+r) is the current in the load resistance Q].

From the above equation we have

P/(P+r)^{2} = Q/(Q+r)^{2}

Or, P/(P^{2 }+r^{2 }+2Pr) = Q/(Q^{2}+r^{2 }+ 2Qr)

Or, PQ^{2} + Pr^{2} + 2PQr = QP^{2} + Qr^{2} + 2PQr from which

r^{2}(P – Q) = QP^{2} – PQ^{2}

Or, r^{2}(P – Q) = PQ(P – Q) so that r = √(PQ)

(6) A 12 V battery having internal resistance 0.2 Ω is being charged at a current of 2 A using the circuit shown in the figure. The generator which charges the battery has a terminal voltage of 16 V at this charging current. What is the value of the resistance R connected in the circuit?

(a) 0.2 Ω

(b) 0.8 Ω

(c) 1.8 Ω

(d) 2.6 Ω

(e) 3.4 Ω

The voltage between the terminals of the battery (terminal potential difference) while the battery is being charged (which is equal to the sum of the emf of the battery and the potential drop across its internal resistance) = 12 + 0.4 = 12.4 V.

[0.4 V is the voltage drop (2×0.2 volt) across the internal resistance of the battery].

Therefore, the voltage drop across the series resistance R is (16–12.4) V = 3.6 V.

Since the charging current is 2 A, we have 2 ×R = 3.6 from which R = 1.8 Ω.

You will find some useful multiple choice questions (with solution) in this section here.