# Physics 9702 Doubts | Help Page 119

__Question 602: [Dynamics]__
A 1.2 kg mass is supported by
person’s hand and two newton-meters as shown.

When person’s hand is removed, what
is the initial vertical acceleration of the mass?

A 0.6 m s

^{–2}B 2 m s^{–2}C 4 m s^{–2}D 6 m s^{–2}**Reference:**

*Past Exam Paper – November 2012 Paper 12 Q14*

__Solution 602:__**Answer: D.**

The newton-meters gives the
magnitude of the forces in the direction shown.

We are asked for the initial vertical
acceleration of the mass, so we do no need to consider the horizontal
components of forces involved.

Resultant upward force = 4cos(37) +
3cos(53) = 4.999.. = 5.0N

Resultant vertical force = ma = 12 –
5 = 7N (downwards)

Acceleration a = 7 / 1.2 = 5.8 = 6ms

^{-2}

__Question 603: [Operational Amplifier]__**(a)**State three properties of ideal operational amplifier (op-amp).

**(b)**Circuit incorporating an ideal op-amp is to be used to indicate whether a door is open or closed.

Resistors, each of resistance R, are
connected to the inputs of the op-amp, as shown in Fig.1.

Switch S is attached to the door so
that, when the door is open, the switch is open. The switch closes when door is
closed.

(i) Explain why polarity of the
output of the op-amp changes when the switch closes.

(ii) A red light-emitting diode
(LED) is to be used to indicate when door is open. A green LED is to indicate
when the door is closed.

On Fig.1,

1. draw symbols for the LEDs to show
how they are connected to output of the op-amp,

2. identify the green LED with
letter G.

**Reference:**

*Past Exam Paper – November 2010 Paper 43 Q10*

__Solution 603:__**(a)**Examples: (choose any 3)

Infinite input impedance /
resistance

Zero output impedance / resistance

Infinite gain

Infinite bandwidth

Infinite slew rate

**(b)**

(i)

With the switch S opened, V

^{–}is less (positive) than V^{+}, so the output is positive.
{When the switch S is
opened, current flows through the resistor R at the top. Note that switch S is
connected only to the vertical connection on the left, not to that on the right
too. From the potential divider equation, for V

^{–}, the p.d. is the same across each resistor R, and so, V^{–}is at +1.5V.
As for V

^{+}, there are 3 resistors R and the p.d. across each of them is the same (= 1.0V each). V^{+}is connected after 2 resistors R, so the potential at V^{+}is 2(1.0) = 2.0V.
V

_{OUT}= A_{0}(V^{+}– V^{–})
So, V

_{OUT}is positive.}
With the switch S closed, V

^{–}is more (positive) than V^{+}, so the output is negative.
{When the switch S is
closed, current flows through the switch S instead of the top resistor R on the
left vertical connection. So, V

^{–}is at a potential of 3.0V. The potential at V^{+}is still 2.0V. From the equation above, V_{OUT}is now negative.}
(ii)

1. The diode should be connected
correctly between the output and earth.

{When the door is opened,
the switch is opened. When the switch is opened (as explained above), the output
is positive. So, the red LED should be directed from the output towards the earth.
Similarly, the green LED is in the opposite direction}

2. The green LED should be identified
correctly

__Question 604: [Kinematics]__
The velocity-time graph below is for
a stone thrown vertically up into air. Air resistance is negligible.

The stone is thrown up at time zero.

Area X represents a distance of 5 m.
Area Y represents a distance of 3 m.

What is the displacement of the
stone from its initial position at time t?

A 2 m B 3 m C
5 m D 8 m

**Reference:**

*Past Exam Paper – November 2010 Paper 11 Q8 & Paper 13 Q6*

__Solution 604:__**Answer: A.**

At time zero, the stone is thrown
up. When the graph intercepts the time axis (x-axis), the velocity of the stone
is zero. This means that the stone has reached its maximum height which is
given by the area X = 5m.

Then, the stone falls down due to
the force of gravity. In the graph, this is represented by the velocity being
negative. So, area Y represents the distance that the stone has fallen after
reaching its maximum. This distance is 3m from the maximum height, not 3m from
the initial position.

Displacement from initial position =
5 – 3 = 2m

__Question 605: [Current of Electricity]__
Graph shows how current I varies
with voltage V for a filament lamp.

Since the graph is not a straight
line, resistance of the lamp varies with V.

Which row gives the correct
resistance at stated value of V?

V /
V R / Î©

A 2.0
1.5

B 4.0
3.2

C 6.0
1.9

D 8.0
0.9

**Reference:**

*Past Exam Paper – June 2013 Paper 13 Q33*

__Solution 605:__**Answer: C.**

The I-V graph of a filament lamp is
not a straight line.

A filament lamp does not follows Ohm’s
law in that the variation of I with V is not linear (a straight line is not
obtained), but the resistance is still obtained as follows.

Resistance R = V / I

{The values of current I
read from the graph may not be exact for all the points, but they are close to
the correct value. The resistance would be approximately the same though.}

Choice A:

At V = 2.0 V, current I = 1.8 A

Resistance R = 2.0 / 1.8 = 1.11 Î©

Choice B:

At V = 4.0 V, current I = 2.5 A

Resistance R = 4.0 / 2.5 = 1.6 Î©

Choice C: [Correct
choice]

At V = 6.0 V, current I = 3.2 A

Resistance R = 6.0 / 3.2 = 1.875 =
1.9 Î©

Choice D:

At V = 8.0 V, current I = 4.0 A

Resistance R = 8.0 / 4.0 = 2.0 Î©

__Question 606: [Work, energy and Power > Gravitational Potential Energy]__
Mass m is situated in a uniform
gravitational field.

When the mass moves through
displacement x, from P to Q, it loses an amount of potential energy E.

Which row correctly specifies
magnitude and direction of the acceleration due to gravity in this field?

magnitude
direction

A E
/ mx →

B E
/ mx ←

C E
/ x →

D E
/ x ←

**Reference:**

*Past Exam Paper – November 2014 Paper 11 & 12 Q14*

__Solution 606:__**Answer: A.**

Gravitational Potential Energy E =
mgh

To easily answer this question,
compare this situation (where the gravitational field lines are horizontal) with
our usual case on Earth (where the gravitational field lines are vertical).
Both are gravitational fields, so the same principles apply.

At a specific location, the force of
gravity acts downwards (towards the centre if we look at the Earth as a whole).
The acceleration due to gravity is in the same direction as the force of
gravity. The gravitational field lines point downwards towards the surface of
the Earth (towards its centre). The direction of the field lines show the
direction that an object would move when placed in that specific field.

We know that as the height of an
object on Earth increases, its gravitational potential energy increases. In
other words, as the object moves against the gravitational field, it gains
potential energy. Conversely, if the object moves in the direction of the
field, its potential energy decreases.

The question states that when the
mass moves through displacement x, from P to Q, it loses an amount of potential
energy E. So, the direction of the field is from P to Q – that is to the right.
[B and D are incorrect]

Gravitational Potential Energy E =
mgh

[h = x here]

E = mgx

Acceleration due to gravity, g = E /
mx

november 2006 paper 6 question 15 b

ReplyDeletehow to attempt it?

thanks in advance

See question 618 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-122.html

Can you explain why is the acceleration 9.81 in November 2011 Paper 43 Q3 (c) (ii)?

ReplyDeleteIt's explained as solution 616 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-121.html

nov 2012 variant 11 q 10 14 23 29

ReplyDeleteFor Q10, see solution 1061 at

Deletehttp://physics-ref.blogspot.com/2015/11/physics-9702-doubts-help-page-223.html