# 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

^{–1}K^{–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

^{–1}K^{–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 V

_{IN}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, V

_{B}= 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 =
V

_{A}– 0 = V_{A}
From the potential divider
equation,}

when strained, potential at A, V

_{A}= 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 V

_{IN}and V^{–}and V^{+}is connected to earth
resistor between V

^{–}and V_{OUT}
{For an inverting
amplifier, V

_{+}is connected to earth (0V). The input voltage is applied to an input resistor between V_{IN}and V^{–}. Negative feedback is applied by means of a resistor between V^{–}and V_{OUT}.}
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 = V

_{OUT}/ V_{IN}= - R_{F}/ R_{IN}
As given in the question,
when V

_{IN}= 0.15V, V_{OUT}= +6.0 V
Voltage gain = 6.0 / 0.15
= 40}

Thus, neglecting the negative
sign, ratio of R

_{F}/ R_{IN}= 40
For a realistic value, R

_{IN}should be between 100 Ω and 10 kΩ
Example: R

_{IN}= 200 Ω, R_{F}= 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 (θ

_{2}– θ_{1}) 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 θ

_{n}= sin^{-1}(nλ / d)
When n = 1, θ

_{1}= sin^{-1}({600×10^{-9}} / {2.0×10^{-6}}) = 17.458°
When n = 2, θ

_{2}= sin^{-1}(2 × {600×10^{-9}} / {2.0×10^{-6}}) = 36.870°
Angular separation (θ

_{2}– θ_{1}) = 36.870 – 17.458 = 19.4°
Please consider answering ALL of the following questions:

ReplyDelete4/O/N/02 Q.5(b),Q.6(c)(i)

6/O/N/02 Q.11(a)(b)

6/O/N/03 Q.9

04/M/J/04 Q.8(a),(b)(i),(ii)1.

06/M/J/04 Q.9(b)(iii),Q.11(b)

06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)

04/M/J/05 Q.7(a)

06/O/N/05 Q.8(b),Q.10(a)

04/M/J/06 Q.6(a),(c),Q.7(b)

06/M/J/06 Q.14(b)

04/O/N/06 Q.3(c)

06/O/N/06 Q.3(b)

05/M/J/07 Q.2(d)

04/O/N/07 Q.7(b)(i),(c),Q.10(c)

04/M/J/08 Q.5(b),Q.9(b),Q.11(a)(ii)

04/O/N/08 Q.7(c)

41/O/N/09 Q.6(a),(b)(i),Q.10

41/M/J/10 Q.6(a),Q.7(a)

51/M/J/10 Q.2(d)

For 04/O/N/08 Q.7(c), see solution 988 at

Deletehttp://physics-ref.blogspot.com/2014/09/9702-november-2008-paper-4-worked.html

Can u plzz give sloution of all the questions of october november 2008 paper 1 ??? And by the way..this blog is really veryyy helpful..thank you!

ReplyDeleteGo to

Deletehttp://physics-ref.blogspot.com/2015/06/9702-november-2008-paper-1-worked.html

No problem.

Hey I didn't quite get what happened in q986 part A, what formula was used and how did you go about it. Thanks

ReplyDeleteit's the potential divider formula

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