Tuesday, June 9, 2015

Physics 9702 Doubts | Help Page 168

  • Physics 9702 Doubts | Help Page 168

Question 841: [Work, Energy and Power]
Electrical generator is started at time zero. Total electrical energy generated during the first 5 seconds is shown in the graph.

What is maximum electrical power generated at any instant during these first 5 seconds?
A 10 W                       B 13 W                        C 30 W                        D 50 W

Reference: Past Exam Paper – June 2005 Paper 1 Q16 & November 2013 Paper 13 Q19

Solution 841:
Answer: C.
Power = Energy / time

From the energy-time graph, the power generated is given by the gradient. Maximum electrical power generated at any instant is given by the gradient at that instant.

The steeper the graph, the greater the value of gradient and thus the greater the power generated. The graph is steepest between times t = 2s and t = 3s.

Consider the points: (2, 10) and (3, 40)
Maximum power = gradient = (40 – 10) / (3 – 2) = 30 W

Question 842: [Matter > Kinetic theory of Gases]
A student is studying Brownian motion.
Using a microscope, she observes particles of smoke in a glass container, illuminated by strong light. Particles of smoke have a zig-zag path, constantly changing speed and direction.

What happens to the smoke particles if air in the container is heated?
A The smoke particles become easier to see.
B The smoke particles change direction more frequently.
C The smoke particles increase in volume.
D The smoke particles move further apart.

Reference: Past Exam Paper – November 2014 Paper 13 Q21

Solution 842:
Answer: B.
The zig-zag path of the smoke particles is due to their collisions with the air molecules present in the container (through these air molecules cannot be seen even with a microscope).

The air molecules are themselves in random motion. By heating the air, the molecules gain heat energy which is converted into kinetic energy of the molecules as they move faster and randomly and thus collide with the smoke particles more frequently. Hence, the smoke particles change direction more frequently.

Question 843: [Work, Energy and Power]
The forward motion of motor-boat is opposed by forces F which vary with the boat’s speed v in accordance with the relation F = kv2, where k is a constant.
Effective power of the propellers required to maintain the speed v is P.
Which expression relates k, P and v?
A k = P / v                   B k = P / v2                  C k = P / v3                  D k = P / v4

Reference: Past Exam Paper – June 2009 Paper 1 Q14

Solution 843:
Answer: C.
Power = Force × speed = Fv
Force F = kv2

Power P = kv2 × v = kv3
k = P / v3

Question 844: [Measurement > Graph]
Uncalibrated analogue voltmeter P is connected in parallel with another voltmeter Q which is known to be accurately calibrated. For a range of values of potential difference (p.d.), readings are taken from the two meters.
Diagram shows the calibration graph obtained.

The graph shows that meter P has a zero error. This meter is now adjusted to remove this zero error. When meter is recalibrated, the gradient of the calibration graph is found to be unchanged.
What is the new scale reading on meter P when it is used to measure a p.d. of 5.0 V?
A 6.6                           B 6.7                           C 7.2                           D 7.4

Reference: Past Exam Paper – November 2013 Paper 13 Q5

Solution 844:
Answer: D.
A zero error can caused the value measured to be either greater or less than the correct value.

The calibrated meter Q gives the correct value. The uncalibrated meter P contains a zero error.

Consider the points (3, 4) and (0.3, 0) on the calibration graph given.
Gradient = Δy / Δx = (4 – 0) / (3 – 0.3) = 1.48

Note the scale on the axes are not that same. That is, 1cm (5 squares) on the x-axis does not represent to same amount of volts as 1cm (5 squares) on the y-axis. The value of the gradient may be interpreted as such: 1V on the x-axis corresponds to 1.48V on the y-axis.

The zero error is indicated by the fact that the line does not pass through the origin. In graph, we would say that the y-intercept is not zero when the zero error is present.

With the zero error removed,
When value of x-axis is 5.0V (as read by meter Q), the y-intercept would be zero.
We are told that the gradient is the same as when the zero error was still present.

Equation of a straight line: y = mx + c
When the zero error is removed, c = 0. Gradient = m = 1.48

When the calibrated meter Q reads 5.0V (x = 5.0),
New scale reading on meter P (= y) = mx + c = 1.48 (5.0) + 0 = 7.4V [D is correct]

From the graph shown (which still contains zero error), when meter Q = 5.0V, the uncalibrated meter P = 7.0V. So, the zero error is 7.4 – 7.0 = 0.4V less than the actual value.
This can be verified by considering a point on the graph shown and replace in y = mx + c. The zero error is given by c.
Consider point (0.3, 0)
0 = 1.48 (0.3) + c
Zero error = - 0.44 ≈ - 0.4

Question 845: [Dynamics > Collisions]
A small glider moves along friction-free horizontal air track as shown below.

At each end of the air track there is a perfectly elastic buffer.
Which graph represents variation with time t of the velocity v of the glider as it moves between the two buffers?

Reference: Past Exam Paper – June 2011 Paper 12 Q9

Solution 845:
Answer: D.
A ‘friction-free horizontal air track’ means that no work needs to be done against friction. Thus, the kinetic energy of the glider is not reduced due to friction.

A ‘perfectly elastic buffer’ means that the collision of the glider is perfectly elastic. Again, both momentum and energy is conserved.

Thus, in this experiment, energy is not lost in any form. The kinetic energy of the glider remains constant during its motion and so does the magnitude of its velocity.

The magnitude of the velocity is given by the vertical distance from the time-axis. This remains constant. [A and B are incorrect]
The horizontal line indicates the time that the glider takes to move from one buffer to the other. Since the speed is constant, this time should the same during its motion. [C is incorrect]

Question 846: [Matter > Solids]
What is correct name for a material containing long-chain molecules that are tangled and coiled?
A amorphous metal
B amorphous polymer
C crystalline metal
D crystalline polymer

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q17

Solution 846:
Answer: B.
Crystalline materials have well defined shapes (e.g cubic shape, …). The stated material is said to be ‘tangled and coiled’. [C and D are incorrect]

Metals consists of atoms / molecules bonded to each one to another. They do not contain long-chain molecules. Polymers are materials that contain long-chain molecules (e.g rubber). [A is incorrect]

No comments:

Post a Comment

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 168