Wednesday, September 30, 2015

Physics 9702 Doubts | Help Page 204

  • Physics 9702 Doubts | Help Page 204



Question 984: [Matter > Temperature]
A mercury-in-glass thermometer is to be used to measure the temperature of some oil.
The oil has mass 32.0 g and specific heat capacity 1.40 J g–1K–1. The actual temperature of the oil is 54.0 °C.
The bulb of the thermometer has mass 12.0 g and an average specific heat capacity of 0.180 J g–1K–1. Before immersing the bulb in the oil, the thermometer reads 19.0 °C.
The thermometer bulb is placed in the oil and the steady reading on the thermometer is taken.
(a) Determine
(i) the steady temperature recorded on the thermometer,
(ii) the ratio
change in temperature of oil / initial temperature of oil

(b) Suggest, with an explanation, a type of thermometer that would be likely to give a smaller value for the ratio calculated in (a)(ii).

(c) The mercury-in-glass thermometer is used to measure the boiling point of a liquid. Suggest why the measured value of the boiling point will not be affected by the thermal energy absorbed by the thermometer bulb.

Reference: Past Exam Paper – November 2006 Paper 4 Q2



Solution 984:
(a)
(i)
From the law of conservation of energy,
Heat lost (by oil) = Heat gained (by thermometer)
{The temperature of the oil decreases while that of the thermometer increases. Let the equilibrium temperature reached by both be T.
Amount of heat, H = mcΔθ}
32 × 1.4 × (54 – T) = 12 × 0.18 × (T – 19)
Temperature T = 52.4°C

(ii) EITHER ratio (= 1.6/54) = 0.030              OR (=1.6/327) = 0.0049
{Change in temperature of oil = 54 – 52.4 = 1.6°C = 1.6K
The ratio may be calculated in terms of °C or in K. The answers would be different because initial temperature of oil would be different. For a ‘change’ in temperature, the value would be the same in °C or K.
Initial temperature of oil = 54°C = 54+273 K = 327 K}

(b) A thermistor thermometer because it has a small mass / thermal capacity

(c) The boiling point temperature is constant. Heating of the bulb would affect only the rate of boiling
{Boiling point is constant, even if heat may is continuously being supplied. Thus, unlike (a)(i), there would be no change in temperature of the liquid even though the thermometer would be gaining heat from the liquid. Only the rate of boiling may be affected.}










Question 985: [Kinematics]
The diagram shows a velocity-time graph for a car.

What is the distance travelled during the first 4.0 s?
A 2.5 m                       B 3.0 m                       C 20 m                        D 28 m

Reference: Past Exam Paper – November 2008 Paper 1 Q6



Solution 985:
Answer: D.
The distance travelled during the first 4.0s is given by the area under the velocity-time graph for the first 4.0s.

Distance travelled = area of parallel of trapezium = ½ × sum of parallel sides × height
Distance travelled = ½ × (2 + 12) × 4 = 28m










Question 986: [Applications > Direct Sensing]
(a) The strain in a beam is to be monitored using a strain gauge.
The strain gauge is included in the potential divider circuit shown in Fig. 9.1.


The strain gauge has a resistance of 120.0 Ω when it is not strained. The resistance increases to 121.5 Ω when the strain is ε.
Calculate the potential difference between points A and B on Fig. 9.1 when the strain in the gauge is ε.  

(b) An inverting amplifier, incorporating an operational amplifier (op-amp), uses a high-resistance voltmeter to display the output. A partially completed circuit for the amplifier is shown in Fig. 9.2.

The voltmeter is to indicate a full-scale deflection of +6.0 V for an input potential VIN of 0.15 V.
(i) On Fig. 9.2,
1. complete the circuit for the inverting amplifier,
2. mark, with the letter P, the positive terminal of the voltmeter.

(ii) Suggest appropriate values for the resistors you have shown in Fig. 9.2.
Label the resistors in Fig. 9.2 with these values.

Reference: Past Exam Paper – June 2015 Paper 42 Q9



Solution 986:
(a)
{The p.d. across each of the parallel branch is 2000mV. Since the resistances are equal in the right branch, the potential at B would be exactly half of the p.d. across the branch.}
Potential at B, VB = 1000 mV
{The resistance of the strain gauge is 121.5 Ω when the strain is ε.
Note that the bottom junction of the strain gauge is connected to earth and is thus at a potential of 0V.
Potential difference across strain gauge = potential at A – potential at bottom junction
p.d across strain gauge = VA – 0 = VA
From the potential divider equation,}
when strained, potential at A, VA = 2000 × 121.5 / (121.5 +120.0) = 1006.2 mV
p.d. between points A and B {= 1006.2 – 1000} = 6.2 mV

(b)
(i)
1.
resistor between VIN and V and V+ is connected to earth
resistor between V and VOUT
{For an inverting amplifier, V+ is connected to earth (0V). The input voltage is applied to an input resistor between VIN and V. Negative feedback is applied by means of a resistor between V and VOUT.}

2. P / + sign shown on earth side of voltmeter
{The level connected to the ‘earth’ is at 0V. Providing a positive input potential to an inverting signal would give a negative output potential at the output of the amplifier. Thus, relative to the output of the amplifier, the earth side of the voltmeter is positive.}


(ii)
{Voltage gain = VOUT / VIN = - RF / RIN
As given in the question, when VIN = 0.15V, VOUT = +6.0 V
Voltage gain = 6.0 / 0.15 = 40}
Thus, neglecting the negative sign, ratio of RF / RIN = 40
For a realistic value, RIN should be between 100 Ω and 10 kΩ

Example: RIN = 200 Ω, RF = 40 × 200 = 8000 Ω = 8 kΩ












Question 987: [Waves > Diffraction]
A diffraction grating experiment is set up using yellow light of wavelength 600 nm. The grating has a slit separation of 2.00 μm.

What is the angular separation (θ2 – θ1) between the first and second order maxima of the yellow light?
A 17.5°                        B 19.4°                        C 36.9°                        D 54.3°

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q27



Solution 987:
Answer: B.
For a diffraction grating: d sinθ = nλ
Wavelength λ = 600nm = 600×10-9 m
Slit separation d = 2.00 μm = 2.0×10-6 m

Angle θn = sin-1 (nλ / d)
When n = 1, θ1 = sin-1 ({600×10-9} / {2.0×10-6}) = 17.458°
When n = 2, θ2 = sin-1 (2 × {600×10-9} / {2.0×10-6}) = 36.870°

Angular separation (θ2 – θ1) = 36.870 – 17.458 = 19.4°



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