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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

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:
Go to
A uniform beam of mass 1.4 kg is pivoted at P as shown. The beam has a length of 0.60 m and P is 0.20 m from one end.








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






10 comments:

  1. In question 453: why is the distance covered not 2Ï€r?

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

      Delete
    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 ??

      Delete
  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?

    ReplyDelete
    Replies
    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,

      Delete
  3. In Q451, do we really need to multiply mass by g? Because we can simplify it too.
    6*9.81*0.15 = (1.4*9.81*0.1) + (3*9.81*0.35)
    9.81(0.9) = 9.81(0.14+1.05)
    This 9.81 if cancelled out, gives incorrect ans. BUT they can be cancelled out. Puzzled!

    ReplyDelete
    Replies
    1. if you cancel it out during the working, the final value obtained would have the same unit of mass times distance. You will need to multiply by g to obtain the torque.

      Delete
  4. JazakAllahuKhairan.

    ReplyDelete

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