# Physics 9702 Doubts | Help Page 155

__Question 767: [Current of Electricity > Resistance]__
Which graph best represents the way
in which current I through a thermistor depends upon the potential difference V
across it?

**Reference:**

*Past Exam Paper – June 2011 Paper 12 Q35*

__Solution 767:__**Answer: A.**

As the potential difference V across
the thermistor increases, the temperature of the thermistor also increases,
causing its resistance R to decrease.

Ohm’s law: V = IR

As the p.d. V increases, the current
increases. This is represented by a positive gradient in the graph. [D is incorrect]

Resistance R = V / I

I / V = 1 / R

As the resistance R decreases, the
value of (1/R) increases. So, the ratio V / I increases.

So, current I increases with the p.d
V, with an increasing gradient since the resistance R is also decreasing as the
p.d. V increases.

It is important to note though that
the gradient only tell us the WAY the resistance is changing. We cannot use the
value of the gradient to obtain the VALUE of the resistance. We need to
consider the ratio of V / I of a single point.

__Question 768: [Waves > Diffraction Grating]__
Light of wavelength Î» passes through
diffraction grating with slit spacing d. A series of lines is observed on a
screen.

What is angle Î± between the two
first order lines?

A sin

^{-1 }(Î» / 2d) B sin^{-1 }(Î» / d) C 2sin^{-1 }(Î» / 2d) D 2sin^{-1 }(Î» / d)**Reference:**

*Past Exam Paper – November 2013 Paper 11 & 12 Q28*

__Solution 768:__**Answer: D.**

Diffraction
grating: d sinÎ¸ = nÎ»

The angle Î¸
used in the diffraction grating formula is taken from the 0

^{th}order (along the line of incidence of the light) to the other orders. So, we need to divide the angle Î± shown in the diagram by 2. This can be done by drawing a line bisecting Î±.
d sin(½ Î±) = nÎ»

For first
order, n = 1

½ Î± = sin

^{-1}(Î»/d)
Î± = 2sin

^{-1 }(Î» / d)

__Question 769: [Dynamics > Equilibrium]__
Diagram shows a solid cube with
weight W and sides of length L. It is supported by a frictionless spindle that
passes through the centres of two opposite vertical faces. One of these faces
is shaded.

Spindle is now removed and replaced
at a distance L / 4 to the right of its original position.

When viewing the shaded face, what
is torque of the couple that will now be needed to stop the cube from toppling?

A WL / 2 anticlockwise

B WL / 2 clockwise

C WL / 4 anticlockwise

D WL / 4 clockwise

**Reference:**

*Past Exam Paper – June 2012 Paper 12 Q15*

__Solution 769:__**Answer: D.**

We need to find the torque needed to
maintain equilibrium. This is not the torque produced by the centre of mass of
cube. This force (weight – the weight is at the centre) causes an anticlockwise
moment at a distance of (L / 4) from the spindle.

Anticlockwise moment due to weight =
W × (L/4) = WL / 4

For equilibrium, the (sum of)
clockwise moment should be equal to the (sum of anticlockwise moment.

Now, the question asks for the
torque of the COUPLE. For a couple, the distance of the other force (causing
the clockwise moment) from the pivot (the spindle here) should be the same
distance from which the weight acts (to produce the anticlockwise moment).

A distance of (L/4) at the other
side of the spindle is actually located at the vertical edge of the shaded
face.

So, a force equal to W, in magnitude
should act along the vertical edge of the shaded face, which is a distance L/4
to the right of the new pivot position. It should be clockwise so as to
maintain equilibrium.

Anticlockwise moment due to weight =
W × (L/4) = WL / 4

So, the force to be applied at the
vertical edge should be equal to W. Additionally, the weight is providing an
anticlockwise moment, so the torque at the vertical edge should produce a
clockwise moment.

Torque of couple = WL / 4 clockwise

__Question 770: [Current of Electricity]__
When four identical resistors are connected
as shown in diagram 1, ammeter reads 1.0 A and voltmeter reads zero.

Resistors and meters are reconnected
to the supply as shown in diagram 2.

What are meter readings in diagram
2?

voltmeter
reading / V ammeter reading /
A

A 0
1.0

B 3.0
0.5

C 3.0
1.0

D 6.0
0

**Reference:**

*Past Exam Paper – November 2007 Paper 1 Q35*

__Solution 770:__**Answer: A.**

In diagram 1, the voltmeter reads
zero because the potential difference across it is zero – that is, both of its
junctions are at the same potential. Current splits at a junction. Since the
resistors are identical, the current splits equally into 2. The ammeter reads a
current of 1.0A, so the total current in the circuit is 1.0 + 1.0 = 2.0A.

Now, consider diagram 2. Current
flows from the positive terminal of the supply and is split equally at a
junction. So, the ammeter still reads the current that has been split equally
into 2. This current is 2.0 / 2 = 1.0A. [B and D
are incorrect]

From that junction connected to the
positive terminal of the supply (the junction at the bottom), one of the
current flows through 1 resistor to reach the other junction (on the right).
The other current flows through the other resistor to reach the other junction
(on the left).

Since the same amount of current (=
1.0A) flows through identical resistors (as described above), the potential of
the left and right junctions would be the same. These are connected to the
voltmeter. Thus, the potential difference across the voltmeter (difference in
potential at its junctions) is zero. The voltmeter reading is zero.

__Question 771: [Kinematics]__
A boy throws a ball vertically
upwards. It rises to maximum height, where it is momentarily at rest, and then
falls back to his hands.

Which row gives acceleration of the
ball at various stages in its motion? (Take vertically upwards as positive.
Ignore air resistance.)

rising at maximum height falling

A –9.81
m s

^{–2}0 +9.81 m s^{–2}
B –9.81
m s

^{–2}–9.81 m s^{–2}–9.81 m s^{–2}
C +9.81
m s

^{–2}+9.81 m s^{–2}+9.81 m s^{–2}
D +9.81
m s

^{–2}0 –9.81 m s^{–2}**Reference:**

*Past Exam Paper – November 2011 Paper 11 Q8 & Paper 13 Q9*

__Solution 771:__**Answer: B.**

The acceleration of free, g is a
constant and is directed towards the surface (downwards). Since the positive
direction is taken as being vertically upwards, the value of g should be
negative.

So, g is negative and constant at
any position.

Salam! can I get help with s12/11 question 15?

ReplyDeleteCheck solution 777 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-157.html

in solution 767 why is option b and c incorrect?

ReplyDeleteThe resistance of the thermistor decreases as the p.d V or current I through it increases. So, the graph is not linear. A linear graph means a constant resistance. [B incorrect]

DeleteNow, since the resistance decreases, the current I should increase with an increasing amount as V increases. This is represented by an increasing gradient, not a decreasing gradient. [C is incorrect]

in option 767 i didn't understand why is the graph not linear?

ReplyDeleteA linear graph means that the resistance is constant, which is not the case here.

Deletein q 771 the value of g is considered positive or negative when the ball is thrown upwards? And that at max height should the value of g be -ve or +ve? i understand the value of should be negative when it falls down..

ReplyDeleteg is always downwards (towards the ground) and has a constant value (9.81).

DeleteIf the upwards direction is taken as +ve, then g is -ve. Always, at any position.

If the upwards direction is taken as -ve, then g is +ve. Always, at any position.

The sign depends on how the directions are defined. g is a vector.

in q77, then why is the sign negative for rising, at maximum height and at falling? since option b is correct? Isn't rising suppose to be positive?

ReplyDeleteI understand that there is no option as such of +ve g at rising, and -ve g at falling..but what would be the sign of g when it is at max hieght?

g is always downwards and has a constant value of 9.81. With the sign, its -9.81.

DeleteAs the ball is rising, it moves upwards with a certain speed which decreases with time since the acceleration of free fall is always downwards. Motion is in one direction while the acceleration on the ball is in the opposite direction. So, the ball decelerates.

At the maximum height, the speed of the ball is momentarily zero. Then, as the ball falls, it accelerates downwards with an acceleration equal to g.

Hello! In solution 769, how do you know that weight causes an anticlockwise moment?

ReplyDeleteThe weight is downwards and is on the left of the pivot. This produces an anticlockwise moment.

DeleteSo what if it is at the left of the pivot,it might even produce a clockwise moment.

Deletethe weight acts downwards and when at the left of the pivot, it would produce an anticlockwise moment.

DeleteWhere is Q35 Nov11 p12 ya?

ReplyDeleteCheck at

Deletehttp://physics-ref.blogspot.com/2014/08/9702-november-2011-paper-12-worked.html

I wanted to ask that if we calculate max height of a body in throwing it vertically upward , g is taken negative but how come the answer is positive.

ReplyDeletethe answer is B and it's negative

Deletein question 77, the upward acceleration is taken as negative beacuse the direction of motion is in the opposite direction of acceleration, but when it is falling down the direction of motion is in the same direction as acceleration doesn't the acceleration have to be positive in this case?

ReplyDeleteNo, as stated in the question, the vertically upwards direction is taken as positive. Since the acceleration due to gravity is always downwards it is always negative.

Delete