Thursday, May 28, 2015

Physics 9702 Doubts | Help Page 155

  • 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 λ passes through diffraction grating with slit spacing d. A series of lines is observed on a screen.

What is angle α between the two first order lines?
A sin-1 (λ / 2d)             B sin-1 (λ / d)               C 2sin-1 (λ / 2d)                       D 2sin-1 (λ / d)

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q28



Solution 768:
Answer: D.
Diffraction grating: d sinθ = nλ

The angle θ used in the diffraction grating formula is taken from the 0th order (along the line of incidence of the light) to the other orders. So, we need to divide the angle α shown in the diagram by 2. This can be done by drawing a line bisecting α.

d sin(½ α) = nλ 

For first order, n = 1
½ α = sin-1(λ/d)
α = 2sin-1 (λ / 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.




18 comments:

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

    ReplyDelete
    Replies
    1. Check solution 777 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-157.html

      Delete
  2. in solution 767 why is option b and c incorrect?

    ReplyDelete
    Replies
    1. The 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]

      Now, 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]

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

    ReplyDelete
    Replies
    1. A linear graph means that the resistance is constant, which is not the case here.

      Delete
  4. in 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..

    ReplyDelete
    Replies
    1. g is always downwards (towards the ground) and has a constant value (9.81).

      If 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.

      Delete
  5. 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?
    I 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?

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

      As 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.

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

    ReplyDelete
    Replies
    1. The weight is downwards and is on the left of the pivot. This produces an anticlockwise moment.

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

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

      Delete
  7. Replies
    1. Check at
      http://physics-ref.blogspot.com/2014/08/9702-november-2011-paper-12-worked.html

      Delete
  8. 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.

    ReplyDelete
    Replies
    1. the answer is B and it's negative

      Delete

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