• Physics 9702 Doubts | Help Page 27

Question 156: [Current of Electricity > Kirchhoff’s laws]
In circuit shown, all the resistors are identical and all ammeters have negligible resistance.

Reading on ammeter A₁ is 0.6 A.
What are the readings on other ammeters?

Reference: Past Exam Paper – June 2014 Paper 11 Q34



Solution 156:

Answer: D.

Current flows from the positive terminal of the battery. So, current in A₂ should have the highest value.

Let the resistance of 1 resistor be R.

The left-hand side of the circuit {(A₁ – A₂) loop} consists of 2 resistors in series, with a total resistance of (R+R =) 2R. Current in (A₁ – A₂) loop = 0.6A, as read by ammeter A₁.

So, from Ohm’s law [V = IR], the potential difference in (A₁ – A₂) loop = 0.6 (2R) = 1.2R.

From Kirchhoff’s law, the potential different is the same as the e.m.f. in the circuit for any loop. So, the e.m.f. in the circuit is 1.2R and this is also the potential difference across ANY loop.

Consider the (A₂ – A₃) loop,

Current in A₃ = V / R = 1.2R / R = 1.2A.

Similarly consider the (A₂ – A₄) loop,

Current in A₄ = 1.2R / 3R = 0.4A.

Current in A₂ = total current = 0.6 + 1.2 + 0.4 = 2.2A.

Question 157:
What is the current in the 40 Ω resistor of the circuit shown?

A zero                         B 0.13 A                     C 0.25 A                     D 0.50 A

Reference: Past Exam Paper – June 2014 Paper 13 Q37



Solution 157:

Answer: A.

The answer to this question can be reasoned by the symmetry of the circuit.

Current flows from the positive terminal of the battery.

At the left junction, the current divides to flow through the 2 resistors and since the resistances of the resistors are equal, the same current flows through each of them. So, the potential difference across each resistor is the same.

Therefore, the potential at the top junction and the potential at the bottom junction of the 40Ω resistor are the same – that is, the potential difference across the 40Ω resistor is zero.

Thus, no current flows through the 40Ω resistor.

Question 158: [Work Done > Efficiency]
Constant force F, acting on car of mass m, moves the car up slope through a distance s at constant velocity v. Angle of the slope to horizontal is α.

Which expression gives efficiency of the process?
A mgs sinα / Fv                       B mv / Fs                     C mv² / 2Fs                 D mg sinα / F

Reference: Past Exam Paper – June 2010 Paper 12 Q15



Solution 158:
Answer: D.
The force applied (F) does a work (total energy) = Fs.
Efficiency = useful energy / total energy

Component of weight against F = mg sinα

For constant velocity, the resultant force is zero. So, the force to move the car up slope should be equal to the component of weight along slope.
Work done to move car a distance s = (mgsinα)s

Efficiency = (mgsinα)s / Fs = mgsinα / F











Question 159: [Current of Electricity > Resistance]
What is total resistance between points P and Q in this network of resistors?

A 8 Ω              B 16 Ω                        C 24 Ω                        D 32 Ω

Reference: Past Exam Paper – June 2014 Paper 11 Q36



Solution 159:

Answer: A.

The circuit consists of a parallel combination of 3 parts.

First, consider the point P, then go up and then to the right. It can be seen that the 3 resistors of 16Ω, 8Ω and 8Ω are in series with a combined resistance equal to 16 + 8 + 8 = 32Ω. This is the first part.

Now, if from P, we go to the right and then up, the 3 resistors of 8Ω, 8Ω and 16Ω are in series with a combined resistance equal to 8 + 8 + 16 = 32Ω. This is the second part.

So, the 2 ‘corners’ (part 1 and 2) of the circuit each have resistance 32Ω.

The third part is the 16Ω diagonally between P and Q.

Therefore, the circuit consists of 3 parts which are in parallel combination with each other.

Total resistance = [(1/32) + (1/16) + (1/32)]^-1 = 8Ω

Question 160: [Dynamics > Momentum  > Pressure]
Beam of α-particles collides with lead sheet. Each α-particle in beam has a mass of 6.6 × 10^–27kg and a speed of 1.5 × 10⁷m s^–1.
5.0 × 10⁴ α-particles per second collide with area of 1.0 cm² of lead. Almost all of the α-particles are absorbed by lead so that they have zero speed after collision.
What is an estimate of average pressure exerted on the lead by the α-particles?
A 5.0 × 10^–15Pa
B 5.0 × 10^–13Pa
C 5.0 × 10^–11Pa
D 5.0 × 10^–9Pa

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q11



Solution 160:

Answer: C.

Pressure = Total Force / Area

Let number of particles per second = N/t = 5.0 x 10⁴

Area = 1.0cm² = 1 x 10^-4 m²

Mass of one particle, m = 6.6 x 10^-27kg

Velocity, v = 1.5 x 10⁷ms^-1

Force is defined as the rate of change of momentum (momentum p = mv)

Force due to 1 particle = mv / t

(Total) Force due to N particles = N (mv / t) = (N/t) (mv)

Pressure = Total Force/Area = (N/t)(mv) / Area = 4.95 x 10^-11 Pa

Question 161: [Waves > Stationary waves]
Sound waves, emitted by small loudspeaker, are reflected by wall.
Frequency f of waves is adjusted until stationary wave is formed with the antinode nearest the wall at a distance x from wall.
Which expression gives f in terms of x and speed of sound c?
A f = 4c / x                  B f = 2c / x                  C f = c / 2x                  D f = c / 4x

Reference: Past Exam Paper – June 2008 Paper 1 Q27



Solution 161:
Answer: D.
Speed of wave, c = fλ
Frequency, f = c / λ

For a stationary wave to form, a node must be present at the wall. The distance between  a node and the nearest antinode is equal to λ / 4.
{In these types of questions, a quick diagram may help.}
So, λ / 4 = x giving the wavelength λ = 4x

Frequency = c / λ = c / 4x









Question 162: [Dynamics > Elastic collision]
Two spheres travel along same line with velocities u₁ and u₂. They collide and after collision their velocities are v₁ and v₂.

Which collision is not elastic?

Reference: Past Exam Paper – June 2013 Paper 13 Q11



Solution162:

Answer: A

For an elastic collision, the relative speed of approach is equal to the relative speed of separation. (Note that approach means coming towards each other, and separation means going away from each other.)

In this question, we need to identify which one is NOT elastic.

Let the sphere on the back be sphere 1 and the front sphere be sphere 2.
{Before proceeding with any calculations, you need to think how the spheres are moving before and after collision from the signs (+ve or -ve) of the velocities.}

Consider A:
Before collision sphere 1 is moving forward (u₁ = 2) while sphere 2 is moving backward (u₂ = -5). So, they are approaching.
Relative speed of approach = 2 + 5 = 7ms^-1

After collision, both spheres 1 and 2 are moving backward, but sphere 1 is moving faster.
Relative speed of separation = 5 – 2 = 3ms^-1

So, choice A is correct here

Consider B:
Before collision, both spheres are approaching.
Relative speed of approach = 3 + 3 = 6ms^-1

After collision, sphere 1 is at rest and sphere 2 is moving forward.
Relative speed of separation = 6ms^-1

Consider C:
Relative speed of approach = 3 – 2 = 5ms^-1
Relative speed of separation = 6 – 1 = 5ms^-1

Consider D:
Relative speed of approach = 5 – 2 = 3ms^-1
Relative speed of separation = 6 – 3 = 3ms^-1