# Physics 9702 Doubts | Help Page 97

__Question 500: [Thermodynamics]__**(a)**Resistance of thermistor at 0

^{o}C is 3840Ω. At 100

^{o}C resistance is 190Ω. When thermistor is placed in water at a particular constant temperature, resistance is 2300Ω.

(i) Assuming that resistance of thermistor
varies linearly with temperature, calculate the temperature of water

(ii) Temperature of water, as
measured on thermodynamic scale of temperature, is 286K. By reference to what
is meant by thermodynamic scale of temperature, comment on answer in (i)

**(b)**Polystyrene cup contains mass of 95g of water at 28

^{o}C.

Cube of ice of mass 12g is put into
water. Initially, ice is at 0

^{o}C. The water, of specific heat capacity 4.2x10^{3}Jkg^{-1}K^{-1}, stirred until all ice melts.
Assuming that cup has negligible
mass and that there is no heat exchange with atmosphere, calculate final
temperature of water

(Specific latent heat of fusion of
ice is 3.3x10

^{5}Jkg^{-1})**Reference:**

*Past Exam Paper – June 2010 Paper 42 & 43 Q3*

__Solution 500:__**(a)**

(i)

A change of 1

^{o}C corresponds to (3840 – 190) / 100 Ω (= 36.5Ω)
So, for a resistance of 2300Ω,

Temperature = 100 x [(2300 - 3840) /
(190 - 3840)] = 42

^{o}C.
(ii)

EITHER 286K = 13

^{o}C OR 42^{o}C = 315K
The thermodynamic scale of
temperature does not depend on the property of a substance. So, the change in
resistance (of the thermistor) with temperature is non-linear.

**(b)**

{Heat gained by the ice in
melting = mass x specific latent heat of fusion of ice}

Heat gained by the ice in melting =
0.012 x 3.3x10

^{5}= 3960J
Heat lost by the water (= mcΔθ) = 0.095 (4.2x10

^{3}) (28 – θ)
{Heat energy of MELTED ice
to reach temperature θ = 3960 + [0.012 (4.2x10

^{3}) (θ – 0)])
The heat lost by the water
is used by the ice to reach temperature θ}

3960 + [0.012 (4.2x10

^{3}) (θ – 0)] = 0.095 (4.2x10^{3}) (28 – θ)
Temperature θ = 16

^{o}C

__Question 501: [Thermistor]__
Variation with temperature of
resistance R

_{T}of thermistor shown in Fig.
Thermistor connected in circuit of
Fig.

Battery has e.m.f. 9.0V and
negligible internal resistance. Voltmeter has infinite resistance.

**(a)**For the thermistor at 22.5

^{o}C, calculate

(i) total resistance between points
A and B on Fig

(ii) reading on the voltmeter

**(b)**Temperature of the thermistor is changed. Voltmeter now reads 4.0V.

Determine

(i) total resistance between A and B
on Fig

(ii) Temperature of the thermistor

**(c)**Student suggests that the voltmeter, reading up to 10V, could be calibrated to measure temperature.

Suggest 2 disadvantages of using the
circuit of Fig with this voltmeter for measurement of temperature in range 0

^{o}C to 25^{o}C**Reference:**

*Past Exam Paper – November 2010 Paper 22 Q6*

__Solution 501:__**(a)**

(i)

At temperature = 22.5

^{o}C, R_{T}= 1600Ω or 1.6kΩ
Total resistance = (1/1600 + 1/1600)

^{-1}= 800Ω
(ii)

Either use potential divider formula or Current = 9 / 2000 (= 4.5mA)

V = (0.8/2.0) x 9 = 3.6V V = IR =
(9/2000) x 800 = 3.6V

(Total resistance = 0.8 +
1.2 = 2.0kΩ)

**(b)**

(i)

{Voltmeter reading = 4.0V.
So, p.d. across the 1.2kΩ is
9 – 4 = 5.0V.

Ohm’s law: V = IR. (that
is, p.d. V is proportional to resistance R}

A p.d. of 5.0V corresponds to 1.2k
Ω. So, 4.0V would correspond to (4/5) x 1200)

Total resistance between A and B =
(4/5) x 1200 = 960Ω

(ii)

For parallel combination, 1/960 =
(1/1600 + 1/R

_{T})
R

_{T}= 2400Ω / 2.4kΩ
Temperature = 11

^{o}C**(c)**

Any 2 sensible suggestions:

e.g.: Only a small part of the scale
is used / small sensitivity

Non-linear

{A linear graph is a
straight line graph. If they do no depend linearly on each other, a similar
change in voltage would correspond to different ranges of temperature at the
different regions of temperature. The separation of the scale would small in
one region and large at another region.}

__Question 502: [Current of Electricity]__Diagram shows an arrangement of resistors.

What is total electrical resistance between X and Y?

A less than 1 Ω

B between 1 Ω and 10 Ω

C between 10 Ω and 30 Ω

D 40 Ω

**Reference:**

*Past Exam Paper – November 2010 Paper 11 Q36*

__Solution 502:__**Answer: B.**

The arrangement represents a parallel combination of a 10 Ω with a set of three 10 Ω resistors that are connected in series.

Total resistance of the set of three resistors = 10 + 10 + 10 = 30 Ω

Let the equivalent resistance across X and Y be R.

1 / R = 1/10 + 1/30

Equivalent resistance R = [1/10 + 1/30]

^{-1}= 7.5 Ω

__Question 503: [Waves > Diffraction grating]__Parallel beam of red light of wavelength 700 nm is incident normally on diffraction grating that has 400 lines per millimetre.

What is the total number of transmitted maxima?

A 3 B 4 C 6 D 7

**Reference:**

*Past Exam Paper – June 2013 Paper 13 Q27*

__Solution 503:__**Answer: D.**

Diffraction grating: d sinθ = nλ

Order of maxima, n = d sinθ / λ

Slit separation, d = 1x10

^{-3}/ 400 = 2.5x10

^{-6}m

Wavelength, λ = 700x10

^{-9}m

The greatest angle (if possible) for a maxima to form is at 90

^{o}.

Order of maxima, n = (2.5x10

^{-6}) (sin90

^{o}) / (700x10

^{-9}) = 3.57 = 3

Since the order n can only be an integer, the greatest number of maxima on one side is 3.

But diffraction occurs on both sides of the incident light, so there are 3 maxima on each side of the incident light. Additionally, one maxima is (undeflected) obtained at the centre, where the light is incident.

Total number of maxima = 3 + 3 + 1 = 7

Q501) c) do you mind explaining this part a little bit?

ReplyDeleteAn explanation on 'linearity' is already provided above.

DeleteAs for the sensitivity part,

We can see, from the graph, that the resistance varies from 0 to 4. However, the curve only occupies about half of the graph (from 1.5 to 3.5). The other regions are useless.

Similarly, a voltmeter reading up to 10V will be useless as this range of temperature corresponds to a smaller range of voltages.

From the previous parts of the question, when the p.d. is 4V, this corresponds to 11 deg C and when the p.d. is 3.6V, it corresponds to 22.5 deg C.

We can see that a range of temperature 11 - 22.5 only corresponds to a change in voltage of 4 - 3.6 = 0.4 V. That is, for the range for temperature 11 - 22.5, only 0.4V of the scale of the voltmeter is used.

So, even if the voltmeter can read up to 10V, only a very small section of the scale would be useful (which corresponds to the temperature range asked in the question). Most of the scale won't be necessary. Thus, it would have a low sensitivity. That is, a change in temperature would only correspond to a very small change in voltage.