Tuesday, March 17, 2015

Physics 9702 Doubts | Help Page 87

  • Physics 9702 Doubts | Help Page 87

Question 450: [Measurements]
Angular deflection of the needle of ammeter varies with the current passing through the ammeter as shown in the graph.

Which diagram could represent the appearance of scale on this meter?

Reference: Past Exam Paper – November 2010 Paper 11 Q4

Solution 450:
Answer: A.
The gradient of the graph decreases as current increases. This means that an equivalent change in current at large values of current results in a smaller angular deflection than it would for smaller values of current.

Therefore, the scales of the ammeter should be further to each other at small values because a change in current ΔI in current gives a large deflection.

At larger values on the scale, a similar change in current ΔI gives only a small deflection. So, the scales should be closer there.

Question 451: [Forces > Moments]
Uniform beam of mass 1.4 kg is pivoted at P as shown. Beam has a length of 0.60 m and P is 0.20 m from one end. Loads of 3.0 kg and 6.0 kg are suspended 0.35 m and 0.15 m from pivot as shown.

What torque must be applied to beam in order to maintain it in equilibrium?
A 0.010 N m               B 0.10 N m                 C 0.29 N m                 D 2.8 N m

Reference: Past Exam Paper – June 2013 Paper 12 Q14

Solution 451:
Answer: D.
First of all, the weights masses, which are given in kg, must be calculated (that is, multiply by g). The weight due to the centre of mass of the beam should also be included.

For equilibrium, the sum of clockwise moments must be equal to the sum of anti-clockwise moments.

Clockwise moment = 0.15 x (6.0x9.81) = 8.829Nm

Since the uniform beam has a length of 0.60mm, its centre of mass will act at the 0.30m mark, a distance of (0.30 – 0.20 =) 0.10m from the pivot. This causes an anticlockwise moment.
Anticlockwise moments = [0.10 x (1.4x9.81)] + [0.35 x (3.0x9.81) = 11.6739Nm

The torque that must be applied should act in such a way that the 2 moments calculated become equal. Therefore, a (clockwise) torque that must be applied = 11.6739 – 8.829 = 2.8449 2.8Nm.

Question 452: [Current of Electricity > Potential difference]
A 2 Ω resistor and 4 Ω resistor are connected to a cell.

Which graph shows how potential V varies with distance between X and Y?

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q37

Solution 452:
Answer: D.
The cell has some e.m.f., say E. The negative terminal of the cell is usually taken to be at a potential of 0V while the positive terminal is taken to be at a potential of E. So, the positive terminal is at a higher potential than the negative terminal.

No component is connected between the positive terminal and point X, so they are at the same potential. Thus, point X is at a potential equal to the e.m.f. of the cell. Similarly, point Y is at the same potential as the negative terminal of the cell. So, the potential at Y is zero. [A and B are incorrect]

Ohm’s law: potential difference V = IR
Therefore, the p.d. across the 2Ω resistor is less than the p.d. across the 4Ω resistor. That is, the change in potential across the 4Ω resistor is more significant than that across the 2Ω resistor.

In the graph, this is represented by the gradient. The magnitude of the gradient across the 4Ω resistor (which is just before point Y) is greater than that across the 2Ω resistor (which is just after point X). [C is incorrect]

Question 453: [Work, Energy, Power]
Diagram shows arrangement used to find the output power of an electric motor.
The wheel attached to motor’s axle has a circumference of 0.5 m and the belt which passes over it is stationary when the weights have the values shown.

If wheel is making 20 revolutions per second, what is the output power of the motor?
A 300 W                     B 500 W                      C 600 W                      D 700 W

Reference: Past Exam Paper – J91/I/5 & J96/I/6 & June 2013 Paper 13 Q16

Solution 453:
Answer: A.
Assume that the motor is switched off.
There is a resultant anticlockwise force on the belt which is equal to 50 – 20 = 30N

Now, switch on the motor. It is said that the belt is stationary. Thus, it can be concluded that the motor exerts a force equal to 30N on the belt, so that the resultant force on the belt is now zero.

Power = Work done / time = Force x velocity

The wheel makes 20 revolutions per seconds.
Total distance covered in 1 second = 20 (0.5) = 10 m

Speed v = distance / time = 10 / 1 = 10 ms-1

Power output of motor = Fv = 30 (10) = 300W


  1. In question 453: why is the distance covered not 2πr?

    1. the circumference is 0.5m but it makes 20 revolutions per second. So, we should account for this.

    2. in qtn 453 why didnt we use the eqn of motion s=0.5(v+u)t ? since we know u is zero , s covered in one sec is 10 and t=1 ??

  2. in question 453 , we know u=0 , time for a revolution=1 , s= 10m. why cant we use s=0.5 (v+u)*t to find the velocity?

    1. equations of motion should be used when the acceleration is constant (in both magnitude and direction). we do not have such info.

      in fact, the acceleration could be towards the centre,


If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 87