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Question 739: [Current of Electricity > Resistance]

A pencil is used to draw line of length 30 cm and width 1.2 mm. Resistivity of the material in the pencil is 2.0 × 10^–5 Ω m and the resistance of the line is 40 kΩ.

What is the thickness of the line?

A 1.25 × 10^–10 m

B 1.25 × 10^–8 m

C 1.25 × 10^–7 m

D 1.25 × 10^–5 m

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q32

Solution 739:

Answer: C.

Resistance R = ρL / A

where ρ is the resistivity of the material, L is the length and A is the cross-sectional A

The line drawn can be considered to be a cuboid (in 3-dimension).

Let the thickness of the line be y.

Cross-sectional A = y × (1.2×10^-3)

R = ρL / A

A = ρL / R

y × (1.2×10^-3) = ρL / R

Thickness y = (2.0 × 10^–5) (0.3) / (40000 × 1.2×10^-3) = 1.25×10^-7 m

Question 740: [Kinematics]

Diagram shows a velocity-time graph for a mass moving up and down on the end of a spring.

Which point represents velocity of the mass when at the lowest point of its motion?

Reference: Past Exam Paper – November 2014 Paper 13 Q7

Solution 740:

Answer: D.

At the lowest and highest point in the motion, the mass is momentarily at rest. So, at these points, the velocity should be zero. [A and C are incorrect]

Additionally at the extremes of the motion, the velocity changes direction.

If the mass was moving upwards towards the highest point, at the highest point, the velocity would changes from being previously positive to negative. That is, the mass will then be moving downwards afterwards. So, at the highest point, the velocity changes from positive to negative. This is point B in the graph.

If the mass was moving downwards towards the lowest point, at the lowest point, the velocity would changes from being previously negative to positive. That is, the mass is about to move upwards afterwards. So, at the lowest point, the velocity changes from negative to positive. This is point D in the graph.

Question 741: [Measurement > C.R.O]

Diagram shows a trace of a wave on a cathode-ray oscilloscope.

The vertical and horizontal gridlines have spacing of 1.0 cm. Voltage scaling is 4 V cm^–1 and time scaling is 5 ms cm^–1.

What are the amplitude and period of the wave?

amplitude / V              period / ms

A         1.5                               4

B         5.0                               10

C         6.0                               20

D         12.0                             20

Reference: Past Exam Paper – June 2011 Paper 11 Q4

Solution 741:

Answer: C.

The amplitude is given by the vertical spacing from the equilibrium position and the period is given by the horizontal spacing.

Note that the equilibrium position is found at the middle of the 3^rd square along the vertical.

Amplitude = 1.5cm = 1.5 (4) = 6V

Period = 4cm = 4 (5) = 20ms

Question 742: [Pressure]

A child drinks liquid of density ρ through a vertical straw.

Atmospheric pressure is p₀ and the child is capable of lowering the pressure at top of the straw by 10%. The acceleration of free fall is g.

What is maximum length of straw that would enable the child to drink the liquid?

A p₀ / 10ρg                  B 9p₀ / 10ρg                C p₀ / ρg                      D 10p₀ / ρg

Reference: Past Exam Paper – June 2003 Paper 1 Q20

Solution 742:

Answer: A.

Imagine a beaker containing the liquid and a vertical straw (not an inclined straw).

Atmospheric pressure p₀ acts on the surface of the liquid.

Now, the child is sucking at top of the straw with a pressure of 0.9p₀ (since the child is capable of lowering the pressure at the top of the straw by 10%). As the child sucks up the liquid rises up the straw of length h.

The difference in pressure at the surface of the liquid in the beaker (= p₀) and the pressure at the top of the straw (= 0.9p₀) is represented by the pressure due to the column of liquid in the straw (= hρg).

p₀ – 0.9p₀ = hρg

hρg = 0.1p₀ = p₀ / 10

Length of straw, h = p₀ / 10ρg

Question 743: [Waves > Intensity]

Intensity I of a sound at a point P is inversely proportional to the square of the distance x of P from the source of the sound. That is

Air molecules at P, a distance r from S, oscillate with amplitude 8.0 μm.

Point Q is situated a distance 2r from S.

What is the amplitude of oscillation of air molecules at Q?

A 1.4 μm                     B 2.0 μm                     C 2.8 μm                     D 4.0 μm

Reference: Past Exam Paper – June 2008 Paper 1 Q26

Solution 743:

Answer: D.

Intensity I is proportional to the square of the amplitude (A²).

Now, the intensity I is also inversely proportional to the square of the distance x form the source (1/x²).

So, doubling the distance x causes the intensity to be (I / 4).

Recall: Intensity I is proportional to the square of the amplitude (A²).

Intensity I is proportional to 8.0²

Intensity I/4 is proportional to (8.0² / 4)

So, at distance = 2r,

(Amplitude)² = (8.0² / 4)

Amplitude = √(8.0² / 4) = 8.0 / 2 = 4.0μm

OR

Since I α A²     and I α 1 / x²,

It can be concluded that A² α 1 / x²

So, amplitude A α 1 / x

Thus, if distance x is doubled, amplitude A is halved.

OR

At distance = r, Intensity I_P α 8²

At distance = 2r, Intensity I_Q = I_P / 4 (as calculated above)

At distance = 2r, Intensity (I_P / 4) α A_Q²

OR

Since we are dealing with the same quantities in both cases, we can take the ratio of intensities and amplitudes at P and at Q.

I_P / I_Q = A_P² / A_Q²

I_P / (I_P/4) = 8² / A_Q²

4 = 8² / A_Q²

Amplitude A_Q = √(8.0² / 4) = 8.0 / 2 = 4.0μm

Question 744: [Electric field]

Two vertical conducting plates X and Y are positioned so that they are separated by distance of 6.0 mm in air. A 60 V d.c. supply is connected as shown.

What is electric field strength at E, a point midway between the plates?

A 1.0 × 104 V m^–1 towards X

B 1.0 × 104 V m^–1 towards Y

C 2.0 × 104 V m^–1 towards X

D 2.0 × 104 V m^–1 towards Y

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q31

Solution 744:

Answer: B.

The direction of the electric field is from positive to negative. Since plate X is connected to the positive terminal of the supply, it is also positive. So, the direction of the electric field is from plate X to plate Y. [A and C are incorrect]

Electric field strength E = V / d = 60 / 0.006 = 1 × 10⁴ Vm^-1

The electric field strength between 2 parallel plates is constant anywhere between them.