• Physics 9702 Doubts | Help Page 152

Question 751: [Current of Electricity > Resistance]

A conductor consists of three wires connected in series. The wires are all made of same metal but have different cross-sectional areas. There is current I in the conductor.

Point Y on the conductor is at zero potential.

Which graph best shows variation of potential V with distance along the conductor?

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q33

Solution 751:

Answer: A.

Resistance R = ρL / A

Ohm’s law: V = IR

The resistance R of the wire is inversely proportional to the cross-sectional area. So, the thickest conductor has the least resistance.

Now from Ohm’s law, the potential difference V is proportional to the resistance R. The thickest conductor has the least resistance, so the smallest voltage drop occurs across the thickest conductor.

Consider a part of the wire with constant cross-sectional area – say, the thickest conductor.

Resistance R = ρL / A

So, along the thickest conductor, as the distance from X increases, the length L of the conductor also increases. Since the resistance is directly proportional to the length L, as we go along the same conductor (of constant cross-sectional area), the resistance R increases and so, the potential difference also changes. That is, the potential V is not constant (not a straight line) for any of the conductors of constant cross-sectional area. The stepped changes in C and D are unphysical. [C and D are incorrect]

As explained above, the thickest conductor has the least resistance, so the smallest voltage drop occurs across the thickest conductor.

The potential drop with distance is represented by the gradient of the graph. So, the value of the gradient should be smallest for the thickest conductor. [B is incorrect] Since there is a potential DROP, the gradient is negative.

Question 752: [Measurement > Estimates]

What is approximate kinetic energy of an Olympic athlete when running at maximum speed during a 100 m race?

A 400 J                        B 4000 J                      C 40 000 J                   D 400 000 J

Reference: Past Exam Paper – November 2012 Paper 12 Q3

Solution 752:

Answer: B.

In this question, we need to estimate the maximum speed and mass of an Olympic athlete.

A 100m race in an Olympic competition is usually completed in about 10s by the fastest persons.

Let the maximum speed of the Olympic athlete be about 10ms^-1.

Let the mass of the Olympic athlete be about 80kg.

Kinetic energy = ½ mv² = ½ (80)(10)² = 4000J

Question 753: [Waves > Diffraction]

A diffraction grating with 500 lines per mm is used to observe diffraction of monochromatic light of wavelength 600 nm.

Light is passed through a narrow slit and grating is placed so that its lines are parallel to the slit. Light passes through the slit and then the grating.

Observer views the slit through the grating at different angles, moving his head from X parallel to the grating, through Y, opposite the slit, to Z parallel to the grating on the opposite side.

How many images of the slit does he see?

A 3                              B 4                              C 6                              D 7

Reference: Past Exam Paper – June 2011 Paper 11 Q27

Solution 753:

Answer: D.

For diffraction grating: d sinθ = nλ

The largest value of θ is 90° on both sides of Y. We need to identify the largest order n for which an image can be seen (constructive interference).

There are 500 lines in 1×10^-3m.

Slit separation, d = (1×10^-3) / 500 m

Value of n = d sinθ / λ = (1×10^-3) (sin90°) / (500) (600×10^-9) = 3.33

Since n can only be an integer, the largest order is n = 3. So, there are 3 images on both sides of Y. An image will also be seen at Y, which is directly along the source.

Number of images = 3 + 3 + 1 = 7

Question 754: [Waves > Period]

Diagram shows a square-wave trace on the screen of a cathode-ray oscilloscope. A grid of 1 cm squares covers the screen. Time-base setting is 10 ms cm^–1.

What is approximate frequency of the square wave?

A 70 Hz                      B 140 Hz                     C 280 Hz                     D 1400 Hz

Reference: Past Exam Paper – June 2010 Paper 11 Q4 & Paper 12 Q5 & Paper 13 Q3

Solution 754:

Answer: B.

In the diagram shown, there are approximately 8 periods shown. This takes 6 squares (= 6cm).

Time-base setting is 10 ms cm^–1.

Time corresponding to 6 cm = 6 x 10 = 60ms

1 period T = 60 /8 = 7.5ms

Frequency = 1 / T = 1/ (7.5x10^-3) = 133.3Hz.

So, answer: B.

Question 755: [Current of Electricity]

An extension lead is used to connect a 240 V electrical supply to heater as shown.

A voltmeter measures potential difference (p.d.) across the heater as 216 V and an ammeter measures the current through the heater as 7.7 A.

What is the total resistance of the extension lead?

A 3.1 Ω                       B 6.2 Ω                       C 28 Ω                        D 31 Ω

Reference: Past Exam Paper – November 2014 Paper 13 Q36

Solution 755:

Answer: A.

The extension lead is connected in series with the heater and the supply.

The same current of 7.7 A flows through both the heater and the extension lead.

From Kirchhoff’s law, the sum of p.d. in a circuit is equal the e.m.f. in the circuit here.

Let the p.d across the extension lead be V.

V + 216 = 240

p.d. V = 240 – 216 = 24 V

Ohm’s law: V = IR

Total resistance R = V / I = 24 / 7.7 = 3.1 Ω

Question 756: [Matter > Deformation]

A simple crane consists of rigid vertical pillar supporting a horizontal beam.

Weight W is lifted by a rope at the end of the beam.

What are the forces at points X, Y and Z due to the weight W?

force at X                    force at Y                    force at Z

A         tension                         compression                tension

B         tension                         tension                         compression

C         compression                tension                         compression

D         compression                compression                compression

Reference: Past Exam Paper – June 2007 Paper 1 Q20

Solution 756:

Answer: B.

With the weight W placed there, the horizontal beam would very slightly be curved up (like the upper part of a semi-circle). This is only slightly since it is a rigid pillar. It may not even be observed by our eyes.

Thus, the part around point Z would be compressed. Force at Z: compression

As the horizontal beam is now slightly curved, the part around Y would be extended. Force at Y: tension.

The same is true for the area around X, which is on the outside of the pillar. Force at X: tension.