Sunday, November 23, 2014

Physics 9702 Doubts | Help Page 22

  • Physics 9702 Doubts | Help Page 22

Question 116: [Moment > Equilibrium > Graph]
Horizontal bar is supported on pivot at its centre of gravity. Fixed load is attached to one end of the bar. To keep bar in equilibrium, a force F is applied at distance x from the pivot.

How does F vary with x?

Reference: Past Exam Paper – June 2013 Paper 11 Q14

Solution 116:
Answer: B.
The fixed load causes an anticlockwise moment on the bar.
Moment = Force x Perpendicular distance of force from pivot

Since the load is fixed (its distance from the pivot does not change), the magnitude of the anticlockwise moment is also constant (Let this magnitude be MA).

For equilibrium, the force F, acting at a distance x, causes a clockwise moment with a magnitude equal to the anticlockwise moment (MA).
So, for equilibrium, Fx = MA giving F = MA / x.

Therefore, by increasing distance s, a smaller value of force F is required. [C and D are incorrect]

Now, consider the following:
If distance x is doubled, only half the value of force F is needed and if this distance x is doubled, there cannot possibly be a distance where zero force is force. A force will always be needed to cause the moment, no matter how long the distance x is. Thus, the graph cannot be a straight line [A is incorrect] but should rather be a graph where force F never goes to zero.

Question 117: [Current of Electricity > Potential difference]
Constant 60 V d.c. supply is connected across two resistors of resistance 400 kΩ and 200 kΩ.

What is the reading on voltmeter, also of resistance 200 kΩ, when connected across 200 kΩ resistor as shown in diagram?
A 12 V                        B 15 V                        C 20 V                        D 30 V

Reference: Past Exam Paper – November 2011 Paper 12 Q38

Solution 117:
Answer: A.
Equivalent resistance of voltmeter and resistor in parallel = [1/200 + 1/200]-1 = 100

From Kirchhoff’s law, for components in series, the sum of potential differences across the components is equal to the e.m.f. in the circuit. Thus, the sum of the p.d. across the 400 kΩ and the p.d. across the equivalent resistance of 100 kΩ must be equal to the e.m.f of 60V in the circuit.

Additionally, there is the same p.d. across components in parallel.

From the potential divider equation, the p.d. across the components in parallel is
p.d. = [100 / (100+400)] x 60 = 12V

Question 118: [Waves > Interference > Double Slits]
Coherent monochromatic light illuminates two narrow parallel slits and interference pattern that results is observed on screen some distance beyond the slits.
Which change increases separation between the dark lines of interference pattern?
A using monochromatic light of higher frequency
B using monochromatic light of longer wavelength
C decreasing the distance between screen and the slits
D increasing distance between the slits

Reference: Past Exam Paper – November 2002 P1 Q28

Solution 118:
Answer: B.
For a double slit: x = λD / a
x is the separation between fringes
λ is the wavelength of the light
D is the distance between the screen and the slits
a is the distance between the slits

A light of higher frequency would have a smaller wavelength (λ = c / f). This would cause the separation between the dark fringes of the interference pattern to decrease.

Using monochromatic light of a longer wavelength increases the separation of fringes.

Decreasing the distance between the screen and the slits or increasing the distance between the slits would each cause the separation of the fringes to decrease

Question 119: [Dynamics > Momentum]
Object of mass 4.0 kg moving with speed of 3.0 m s–1 strikes a stationary object in an inelastic collision.
Which statement is correct?
A After collision, total kinetic energy is 18 J.
B After collision, total kinetic energy is less than 18 J.
C Before collision, total kinetic energy is 12 J.
D Before collision, total kinetic energy is less than 12 J.

Reference: Past Exam Paper – June 2014 Paper 11 Q9

Solution 119:
Answer: B.
In an inelastic collision, kinetic energy is not conserved.

Before collision,
Kinetic energy, KE = ½ mv2 + 0 = ½ (4.0)(3)2 = 18J             [C and D are incorrect]

Since collision is inelastic, kinetic energy is less than 18J after collision. 

Question 120: [Current of Electricity > Potentiometer]
In circuit below, P is potentiometer of total resistance 10 Ω and Q is fixed resistor of resistance 10 Ω.

Battery has an e.m.f. of 4.0 V and negligible internal resistance. Voltmeter has very high resistance.

Slider on potentiometer is moved from X to Y and graph of voltmeter reading V is plotted against slider position.
Which graph would be obtained?

Reference: Past Exam Paper – November 2010 Paper 11 Q37 & Paper 13 Q35 & November 2013 Paper 11 & 12 Q36

Solution 120:
Answer: A.
When the slider is at X {at X, resistance of the potentiometer is taken to be zero}, the voltmeter must read 4 V.
{The flow of current is as follows: from the positive terminal – through the moving jack until the point it is connected on resistor P – from P to end X – through the voltmeter and back to the negative terminal of the supply (no current goes through Q).}

When the slider is moved to Y, the potentiometer P and the voltmeter are in series {no current flows through Q – resistor Q can be said to be short circuited in both cases} with almost all the resistance in the voltmeter {as stated in the question, the voltmeter has a very high resistance – Ohm’s law: V = IR – since the total resistance of the potentiometer is negligible compared to that of the voltmeter}, and hence the p.d. across the voltmeter is still 4 V.
{This can be verified using the potential divider equation: V1 = [R1 / (R1+R2)] V. Let R1 be the resistance of the voltmeter. Since the voltmeter has a very high resistance, R1 >> R2, where R2 is the resistance of the potentiometer. Hence, (R1 + R2) R1 and V1 = V}

Therefore, the voltmeter reading remains 4V as the slider is moves from X to Y.

In a proper circuit, a voltmeter is usually connected in parallel to a component if we want to know the p.d. across that component. Connecting a voltmeter in series results in the above situation where the voltmeter always reads 4V.

Question 121: [Current of Electricity > Potentiometer]
12 V battery is in series with ammeter, 2 Ω fixed resistor and a 0 – 10 Ω variable resistor. High-resistance voltmeters P and Q are connected across variable resistor and fixed resistor respectively, as shown.

Resistance of variable resistor is changed from its maximum value to zero.
Which graph shows variation with current of voltmeter readings?

Reference: Past Exam Paper – June 2013 Paper 12 Q36

Solution 121:
Answer: D.
Since the variable resistor and the fixed resistor are in series, the total e.m.f. in the circuit is divided between them. The high-resistance voltmeters P and Q are connected across (in parallel) them.

Components with high resistance tend to prevent current from flowing. So, at a junction for components in parallel, most current will tend to flow in the component with smaller resistance (considering P and the variable resistor, most current would flow in the variable resistor).

Additionally, from Kirchhoff’s law, the potential difference across P and the variable is the same. This is also true for the potential difference across the 2Ω resistor and Q.

The potential difference across each of these resistors may be obtained from the potential divider equation: V1 = [R1 / (R1+R2)] V.
Let R1 be the resistance of the variable resistor and R2 be the resistance of the fixed resistor. As R1 is decreased, the p.d. across it decreases. When R1 = 0, V1 = 0. Thus, the voltmeter reading of P should give zero. [Graph D is correct since it shows that P is decreasing and going to zero]
Decreasing R1 causes the p.d. across the fixed resistor to increases. This same p.d. is across Q.


  1. You are amazing. You have solved so many of my doubts. Thank you so very much. Really.

  2. Replies
    1. see solution 960 at

  3. Hey Thank you so much for the explanation. I have doubt on Q120. Why would the current choose the path through the voltmeter with very high resistance instead of resistors P & Q? At first glance I thought there isn't any reading on the voltmeter

    1. P is a potentiometer and this work in a different way than does a fixed resistor. In this case, current would always flow form the jack to end X. Then, for the current to flow in a complete circuit, the only possible path is through the voltmeter.

      For any position of the jack, the voltmeter reading would remain the same due to its very high resistance compared to that of the potentiometer.


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 22