# 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 = kv

^{2}, 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 / v

^{2}C k = P / v^{3}D k = P / v^{4}**Reference:**

*Past Exam Paper – June 2009 Paper 1 Q14*

__Solution 843:__**Answer: C.**

Power = Force × speed = Fv

Force F = kv

^{2}
Power P = kv

^{2}× v = kv^{3}
k = P / v

^{3}

__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 = Î”y
/ Î”x = (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]

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