• Physics 9702 Doubts | Help Page 169

Question 847: [Current of Electricity]

In circuit shown, the ammeters have negligible resistance and the voltmeters have infinite resistance.

The readings on the meters are I₁, I₂, V₁ and V₂, as labelled on diagram.

Which statement is correct?

A I₁ > I₂ and V₁ > V₂

B I₁ > I₂ and V₁ < V₂

C I₁ < I₂ and V₁ > V₂

D I₁ < I₂ and V₁ < V₂

Reference: Past Exam Paper – November 2014 Paper 13 Q37

Solution 847:

Answer: A.

This question required careful working.

Combined resistance of the left-hand section = [1/3 + 1/6]^-1 = 2 Ω

Combined resistance of the right-hand section = [1/2 + 1/2]^-1 = 1 Ω

From Kirchhoff’s law, the sum of V₁ and V₂ should be equal to the e.m.f. in the circuit.

So, potential difference V₁ > potential differenceV₂. [B and D are incorrect]

In the right-hand section, each parallel branch has half of the total current since the resistances are both 2 Ω, while in the left-hand section there is more current in the 3 Ω resistor than in the 6 Ω resistor (from Ohm’s law, I = V / R – the greater the resistance, the smaller the current).

This gives I₁ > I₂, so the correct answer is A.

In terms of calculations, the total current leaving the right-hand section is I₂ + I₂ = 2I₂. As for the left-hand section, the current I₁ is twice the current flowing through the 6 Ω resistor. Thus, the total current entering the left-hand section is I₁ + ½ I₁ = 1.5I₁.

But the total current leaving the supply should be equal to the total current entering the supply.

1.5I₁ = 2I₂

I1 = 2I₂ / 1.5 = 4I₂ / 3 = 1.33I₂

So, I₁ > I₂

Question 848: [Measurement > Uncertainty]

The speed of a car is calculated from measurements of distance travelled and the time taken.

Distance is measured as 200 m, with an uncertainty of ± 2 m.

Time is measured as 10.0 s, with an uncertainty of ± 0.2 s.

What is the percentage uncertainty in the calculated speed?

A ± 0.5 %                    B ± 1 %                       C ± 2 %                       D ± 3 %

Reference: Past Exam Paper – November 2012 Paper 12 Q7

Solution 848:

Answer: D.

Speed v = Distance d / Time t

Percentage uncertainty in the calculated speed = (Δv / v) × 100%

Δv / v = (Δd/d) + (Δt/t)

Δv / v = (2/200) + (0.2/10) = 0.03 = ±3%

Question 849: [Matter > Hooke’s law]

A beam, the weight of which may be neglected, is supported by three identical springs. When weight W is hung from the middle of the beam, the extension of each spring is x.

The middle spring and the weight are removed.

What is the extension when weight of 2W is hung from the middle of the beam?

A 3x / 2                       B 4x / 3                       C 2x                            D 3x

Reference: Past Exam Paper – June 2003 Paper 1 Q22

Solution 849:

Answer: D.

This is quite a tricky question which needs careful working, done on paper and not mentally.

Hooke’s law: F = kx

Since the system consist of more than one springs, we need to consider the effective spring constant of the springs, k_eff.

Each of the 3 springs exerts an upward vertical force (tension – from Newton’s 3^rd law) on the beam such that these forces balances the downward weight W after extension.

When weight = W, extension = x

W = k_eff x

Effective spring constant of the 3 springs, k_eff = W / x

Let the spring constant of 1 spring be k. For 3 springs in parallel,

k_eff = k + k + k

Spring constant k of 1 spring = k_eff / 3 = (W/x) / 3 = W / 3x

Now, the middle spring is removed and a weight of 2W is used.

Effective spring constant of 2 springs, k_eff = (W/3x) + (W/3x) = 2W / 3x

Let the new extension be e.

From Hooke’s law,

2W = (2W/3x) e

New extension, e = 3x

Question 850: [Dynamics > Collisions]

Two equal masses travel towards each other on frictionless air track at speeds of 60 cm s^–1 and 40 cm s^–1. 

They stick together on impact.

What is the speed of the masses after impact?

A 10 cm s^–1                 B 20 cm s^–1                  C 40 cm s^–1                  D 50 cm s^–1

Reference: Past Exam Paper – June 2005 Paper 1 Q11 & June 2010 Paper 11 Q10 & Paper 12 Q11 & Paper 13 Q12

Solution 850:

Answer: A.

Let the mass of 1 trolley = M

From the conservation of momentum, the sum of momentum before collision should be equal to the sum of momentum after collision.

Sum of momentum before collision = M(60) + M(-40) = 20M

After collision, momentum = 2M(v)

2Mv = 20M

Final speed, v = 10cms^-1