• Physics 9702 Doubts | Help Page 19

Question 100: [Current of Electricity > Potential difference]
Circuit is set up as shown, supplied by 3 V battery. All resistances are 1 kΩ.

What will be the reading on voltmeter?
A 0                  B 0.5 V                       C 1.0 V                       D 1.5 V

Reference: Past Exam Paper – June 2013 Paper 11 Q35



Solution 100:
Answer: B.
The circuit may be divided into 2 main loops. From Kirchhoff’s law, the potential difference across each loop is equal to the e.m.f. in the circuit.

For each loop, the potential at a point may be found by the potential divider equation. The potential difference across the voltmeter is equal to the difference in potential between points A and B indicated in the diagram.

In loop 1,
Total resistance = 1 + 1 = 2kΩ
Potential at point A = [1 / (1+1)] x 3 = 1.5V

In loop2,
Consider the 2 sections indicated.
Total resistance in section X = [1/1 + 1/1]^-1 + 1 = 1.5kΩ
Total resistance in section Y = [1/3 + 1/1]^-1 = 0.75kΩ

Total resistance in loop 2 = 1.5 + 0.75 = 2.25kΩ
Potential at point B = [1.5 / (1.5 + 0.75)] x 3 = 2.0V

Potential difference across voltmeter = 2.0 – 1.5 = 0.5V




Note 1:
{Note that for both loop 1 and 2, I have been considering the components on the left of A and B. If I had considered the components on the right of A and B, for the first loop, the potential would still be 0.5V since both resistances are the same. For loop 2, the potential at B would be [0.75 / (1.5 + 0.75)] x 3 = 1.0V. The potential difference across the voltmeter would still be (1.0 – 0.5 =) 0.5V. What’s important is that when you are considering the components on the left or the right in one loop, you need to do the same in the other loop. Potential difference is only a relative comparison. For example, the difference between 4 and 2 is 2 and the difference between 1000 and 1002 is still 2.}

Note 2:
{In fact, the explanation I provided is not totally good (but it’s still good.) [I’m not referring to note 1.] This is because current flows from the positive terminal towards the negative terminal. So, a component’s junction which is closer to the positive terminal would have a greater potential while the other end of the component would have a lower potential. The potential difference will nevertheless be the same across the component. Consider for example the component on the left of point A in loop 1. Its left end would be at a potential of 3.0V and its right end would be at a potential of 1.5V such that the potential difference across it is still (3.0 – 1.5 =) 1.5V (as calculated). If you consider the solution I gave, the potential at A would actually be 1.5V (here, it’s the same because the resistances are the same). Similarly, for loop 2, the actual potential at B would have been 3.0 – 2.0 = 1.0V. Well, since in both loops, I needed to subtract the answer I obtained from 3.0V, this would not have affected the potential difference between the voltmeter since as I said, potential difference is only relative. What I have explained in note 2 is very essential and important, but I think very few teachers make this clear to students.}









Question 101: [Kinematics > Conservation of energy]
Car travelling with speed 28 ms^-1 leaves motorway on an exit road. The end of the exit road is 22m higher than motorway.
If only the force of gravity is considered, what will be the speed of car at end of the exit road?
A. 7.3 ms^-1                   B. 19 ms^-1                    C 21 ms^-1                     D 24 ms^-1

Reference: Past Exam Paper – November 2012 Paper 13 Q18



Solution 101:
Answer: B.
From conservation of energy,
Gain in Potential energy = Loss in Kinetic energy

Gain in potential energy = mgh = m (9.81) (22) = 215.82(m)

Let speed of car at the end of the exit road = v
Loss in kinetic energy = ½m (28²) – ½mv2 = 0.5m(28² – v²)

From conservation of energy,
0.5m(28² – v²) = 215.82(m)
v² = 28² – (215.82 / 0.5)
Speed, v = 18.8 = 19ms^-1









Question 102: [Momentum]
Stationary nucleus has nucleon number A.
Nucleus decays by emitting proton with speed v to form new nucleus with speed u. New nucleus and proton move away from one another in opposite directions.
Which equation gives v in terms of A and u?
A v = (A/4 – 1)u                     B v = (A – 1)u             C v = Au                     D v = (A + 1)u

Reference: Past Exam Paper – June 2013 Paper 13 Q10



Solution 102:
Answer: B.
Nucleus is initially stationary, so its speed is zero. Its momentum (p = mv) is also zero initially.

The nucleus contains an amount of A nucleons.
Nucleus decays by emitting proton with speed v to form new nucleus with speed u.
Number of nucleons in new nucleus = A – 1
Let the mass of 1 nucleon = m

After decay,
Momentum of proton = mv    {the proton is a nucleon}
Momentum of new nucleus = (A – 1)m(–u) 
{negative is taken for the speed because the new nucleus and the proton travel in opposite direction. The direction of motion of the proton is taken as positive [v is taken as positive], so u should be taken as negative.}

Total momentum after decay = mv – (A – 1)mu

From the conservation of momentum,
Sum of Momentum before decay = Sum of Momentum after decay
0 = mv – (A – 1)mu
v = (A – 1)u








Question 103: [Vectors > Equilibrium]
Diagrams represent systems of coplanar forces acting at a point. Lengths of the force vectors represent magnitudes of the forces.
Which system of forces is in equilibrium?

Reference: Past Exam Paper – November 2002 Paper 1 Q15



Solution 103:
Answer: A.
For the forces to be in equilibrium, the resultant horizontal and vertical components of the force vectors should be zero.

The lengths of the vectors represent the magnitudes. Note that the diagonal vectors in the diagrams will have vertical and horizontal components less than the magnitude of the diagonal vector itself {this is obvious since the diagonal vector and its components form a right-angled triangle with the diagonal vector as the hypotenuse. The hypotenuse has the longest length in a right-angled triangle}.

Therefore, in these diagrams, the horizontal and vertical components of the diagonal vectors should be equal in length (and opposite in direction) as the horizontal and vertical vectors already given.

Diagram A:
The horizontal and vertical components of the diagonal vector are equal in magnitude and opposite in direction to the respective vectors already given. So, in diagram A, the system of forces is in equilibrium.

Diagram B:
The sum of the upward vertical components of the 2 diagonal vectors is greater than the downward vertical vector already present. So, the system of forces is not in equilibrium.

Diagram C:
The sum of the horizontal components (toward the left) of the 2 diagonal vectors is smaller than the horizontal vector (toward the right) already present. So, the system of forces is not in equilibrium.

Diagram D:
The system is not balanced vertically since the downward vertical component of one of the diagonal vectors is greater than the upward vertical component of the other diagonal vector.









Question 104: [Ohm’s law > Resistance > Resistivity]
Wire PQ is made of three different materials, with resistivities ρ, 2ρ and 3ρ. There is current I in this composite wire, as shown.

Which graph best shows how potential V along the wire varies with distance x from P?

Reference: Past Exam Paper – November 2010 Paper 12 Q34



Solution 104:

Answer: B.

Ohm’s law: Potential difference, V = IR

The resistance R of a wire proportional to the resistivity ρ of the material from which the wire is made [R = ρL / A].

When the resistivity of the material is greater, its resistance is also greater and thus, the potential difference across it is greater.  So, the potential difference is greater across the 3ρ material.

Note that the graph gives the potential V along the wire on the y-axis while the V from V = IR represents the potential difference – that is, the change in potential.

Therefore, when the resistivity of the material is greater, the potential difference across it is greater. This means that the reduction is potential V is greatest for the part where the resistivity is 3ρ.

The change in potential {potential difference, ΔV} as the distance x changes (Δx) is represented by the gradient. Note that for a specific value of resistivity {along a specific length of wire, L}, the potential difference (gradient) is constant {and non-zero unlike the graph of D where the gradient is zero}. [D is incorrect]

{Resistance of the wire also depends on the length if wire being considered. So, as x increases, the length of wire being considered increases and so, as x increases, the difference in potential V at that particular value of x is greater. That is, in general as x increases, V decreases, even if the resistivity was constant}

As the resistivity of the wire increases, the potential difference across it also increases. This is represented by an increase in gradient of the straight lines as distance x reaches the materials of higher resistivity. [A is incorrect since the gradient is constant everywhere] [C is incorrect as the gradient of the straight lines decreases as x increases]