Doppler Shift

8

a Which transition

Use of (

Δ)E = hc/λ OR (Δ)E = hf and f = c/λ 1

Use of 1.6 x 10

-19 1

Correct answer [1.9 eV]

1

C to B / -1.5 to – 3.4

1 4

[Accept reverse calculations to find wavelengths]

e.g.

(6.63 x 10

-34 J s)(3.00 x 108 m s-1)/

(656 x 10

-9 m)(1.6 x 10-19 J eV-1)

= 1.9 eV

b Explanation of absorption line

QOWC

1

Light of this wavelength is absorbed by hydrogen

1

In the outer part of the Sun (OR Sun’s atmosphere)

1

Absorbed radiation is reemitted in all directions

1

Transition from B to C (OR -3.4 to -1.5)

1 Max 4

c i Why galaxy receding

Wavelength increased (OR stretched) / red shift /

frequency decreased

1

c ii Calculation of speed

Use of

v/c = Δλ/λ 1

Correct answer [1.78 x 10

7 m s-1] 1 3

e.g.

(3.00 x 10

8 m s-1)(695 x 10-9 m – 656 x 10-9 m)/

(656 x 10

-9 m)

= 1.78 x 10

7 m s-1

[N.B. Use of wrong wavelength in denominator gives

1.68 x 10

7]

OR

Use of

v = fλ and v/c = Δf/f 1

Correct answer [1.67 x 10

7 m s-1] 1

[Look for

(4.573 x 10

14 Hz – 4.317 x 1014 Hz) / (4.537 x 1014 Hz)

N.B. Use of wrong frequency in denominator gives 1.78

x 10

7]

Total 11

Questions on resistivity

Resistivity

1. A coil of constantan wire of length 2 m has an area of cross-section of

7 x 10^-7 m². If its resistivity is 49 x 10^-8 Ω m, find its resistance.

2. A coil of wire 4 m long has a resistance of 20 Ω. If its resistivity is

108 x 10^-8 Ω m, find its area of cross-section.

3. A 1.75 m length of tin wire with a diameter of 0.70 mm is found to have a resistance of 0.5 Ω Find the resistivity of tin.

4. A resistor of 2 Ω is to be made from manganin wire of resistivity

42 x 10^-8 Ω m having a diameter of 0.42 mm. What length is required?

5. Find the resistance of a wire 2.8 m long with a resistivity of 22 x 10^-8 Ω m and a diameter of 0.56 mm.
6. A wire has a resistance of 2 Ω Find the resistance of a wire of the same material which is three times as long, but with half the diameter.

 

Mark Scheme February assessment

1.       The sum of the currents entering a point / junction is equal to the sum of the currents

leaving (the same point) Or ‘Algebraic sum of currents at a point = 0’                    B2

(–1 for the omission of ‘sum’ and -1 for omission of ‘point’/ ‘junction’)

(Do not allow I1 + I2 = I3 + I4 unless fully explained)

[2]

  2.     A:   gamma / γ (ray/radiation/wave)                                                                        B1
       λ = 10–16 to 10–10 (m)                                                                                       B1
B:    infrared / IR / i.r.                                                                                              B1
       λ = 7 × 10–7 to 10–3 (m)                                                                                    B1

[4]

  3.     (i)      v   = I / nAe = 0.0025 / (8.5 × 1028 × 1.1 × 10–7 × 1.6 × 10–19) (1)
     = 1.67 × 10–6 m s–1 (1)                                                                                2

  (ii)   Free electron concentration (or wtte) is much smaller in the thermistor than
in the wire.                                                                                                       1

[3]

  4.     (a)     R = R1 + R2 / R = 200 + 120 / R = 320                                                            C1
current =                                                                                                 C1
current = 2.5 ´ 10–2 (A)                                                                                 A0

  (b)   V = 25 ´ 10–3 ´ 120 / V =
V = 3.0 (V)                (Possible ecf)                                                                B1

  (c)   p.d. across the 360 (W) resistor = p.d. across the 120 (W) resistor /
There is no current between A and B / in the voltmeter                                 B1
(Allow ‘A & B have same voltage’ - BOD)

          The p.d. calculated across 360 W resistor is shown to be 3.0 V /
The ratio of the resistances of the resistors is shown to be the same.             B1

[5]

  5.     (i)      plane polarised light vibrates (travels) in one plane only                                B1
(look for reference to one plane of oscillation)

(ii)     only transverse waves can be polarised/AW                                                  B1
sound waves are longitudinal/not transverse/AW                                           B1

[3]

6.       (a)     (i)      amplitude = 3.75 cm {allow 3.7 to 3.8)                                                B1

(ii)     when t = 1.8 ms displacement = ANY negative value                           B1

(–)3.35 cm (ALLOW 3.3 TO 3.4)                                                         B1

(iii)    period = 2.64 ms (allow 2.64 to 2.68)                                                   B1

(iv)    frequency = 1/period                                                                            C1

= 1/(2.64 × 10 – 3) = 379 Hz (379 to 373 or 380) {ecf for T}              A1

  (b)   recall of v = fl                                                                                               C1

l = v/f = 300/379 = 0.79 m (or 0.8 m) {allow ecf from (iv)}                         A1

[8]

  7.     Any three properties from:                  (–1 for each error or contradiction)        B1 ´ 3

1.      Travel at the speed of light / c / 3 ´ 108 m s–1 (NOT ‘same speed’)

2.      Travel through vacuum / ‘free space’

3.      Have oscillating electric and magnetic fields

4.      They are (all) transverse waves / can be polarised

5.      Allow: ‘They show diffraction / reflection / refraction / interference’

6.      Allow: ‘Consist of photons’

Any three regions from the list below:                                                               B1 ´ 3

Gamma (rays / radiation) / g (rays) ; X-rays ; u.v ; ir ; microwaves ; radio waves

(NOT ‘radio’)

One suitable application for the opted region.                                                         B1

(E.g.: Gamma rays for radiotherapy / sterilisation;

X-rays for taking pictures of skeleton / bones; u.v for tanning; ir for TV remote
control; microwaves for cooking / mobile phones; radio waves for communication)

(Note: Reference to alpha, beta and gamma can only score the last marking point)

[7]

  8.     (i)      evidence of knowledge of:
full/max transmission when the (transmission axis of) polarising sheet is
parallel to the light’s plane of polarisation/vibrations                                     B1
no transmission when the (transmission axis of) polarising sheet is at right
angles to light’s plane of polarisation/vibrations                                            B1

(ii)     reflected light from surface is partially plane polarised                                  B1
polarising sheet is placed at right angles to reflected light’s polarisation        B1
plane/AW

[4]

  9.       (b)                                                    (Allow other subject, with or without D)     C1
(charge =) 7800 ´ 0.23                                                                                            C1
1.794 ´ 103 » 1.8 ´ 103 (C)                                                  (Ignore minus sign)     A1
(1.8 ´ 106 (C) scores 2/3)

(c)     (number =)           (Possible ecf)                                                    C1
(number =) 1.12 ´ 1022 » 1.1 ´ 1022                                                              A1

[6]

Answers to questions on probing the nucelus

Y13 Answers on capacitance

markscheme (projectiles & moments)

Charge and Discharge of Capacitors

2 a) 470 s
b) 564 uF
c) 3.8 uC
d) 0.081V

3 a) 600 uC
b) 1.8 mJ
c) 2 s
d) 10 s

4 a) 90 uC
b) 7.5 s
c) 6.0 V
d) 12 uA
e) 7.5 s

5 1.39 s

6 a) 0.90 uC
b) 0.50 uC
c) 5.0 V
d) 4.0 V
e) 1.8 uA

7 120 ohms

y12 mock

Boyle's Law

Note. Any convenient units can be used for pressure and volume, provided that the units for P₁ are the same as those for P₂, and that the units for V₁ are the same as those for V₂.

(Use g = 9.8 N kg^-1)     1 atm =  Atmospheric pressure = 100 000 Pa = 100 000 Nm^-2

1. A fixed mass of gas at constant temperature has a volume of 500 cm² at a pressure of 1 atm. What will its volume be at a pressure of 4 atm?  125 cm³

2. A gas has a volume of 600cm³ at a pressure of 3.0 x 10⁵ Pa.  What is its volume at a pressure of 6.0 x 10⁵ Pa?  300 cm³

3. A gas has a volume of 1000 cm³ at a pressure of 1 atm. At what pressure will it occupy 100 cm³?   10 atm

4. The pressure exerted by a gas is due to the molecules of the gas bombarding the walls of the container. Why does the pressure increase when the volume is decreased at constant temperature? Less volume, molecules closer together, more collisions etc

5. A cylinder of compressed air has a volume of 51 cm³ at a pressure of 8 atm. When the cylinder is opened, what volume of air will escape?                                           

P₁V₁ = P₂V₂

rearranging V₂ = P₁V₁/ P₂           

V₂ = 8 x 51/ 1 = 408 cm³ minus the 51cm³ left in the cylinder = 357 cm³

 

6. A diver is working in sea-water of density 1025 kg m^-3 on a day when atmospheric pressure is 100 450 Pa. A bubble from his mask is three times the volume when it reaches the surface. At what depth is he working? 20m

7. A lake is 92.7 m deep. A bubble at the bottom of the lake has a volume of 0.5 cm³. What will its volume be at the surface if a water barometer would read 10.3 m (i.e. atmospheric pressure is the same as the pressure at this depth)? 5cm³

8. A uniform cylinder 0.60 m long is sealed at one end. It is lowered open end downwards into a lake so that the mouth just touches the bottom. When retrieved the inside is wet up to 0.50 m from the mouth. Find the depth of the lake if a water barometer would read 10.4 m of water. 52.5m

9. A cylindrical diving bell 8 m high is lowered into water so that the top of the bell is 0.75 m below the surface. If the atmospheric pressure is the same as that exerted by a column of water 10.25 m high, find how far the water rises inside the bell. 6m

A vessel of volume 500 cm³ containing air at a pressure of 8.0 x 10⁴ Pa is connected by a very narrow tube fitted with a tap, to a vessel of volume 700 cm³ containing air at a pressure of 2.0 x 10⁵ Pa. What is the resulting pressure in the vessels when the tap is opened?                     1.5 x10⁵ Pa

A compression pump is connected to a vessel of volume 500 cm³. For each stroke of the pump the piston sweeps 50 cm³ of air at atmospheric pressure into the vessel. If the initial pressure in the vessel is 1 atm, find the pressure in the vessel after 2 strokes.

1.2 atm

An exhaust pump is connected to a flask of volume 100 cm³. The volume swept out by the piston is 25 cm³ for each stroke. The pressure in the vessel is originally 1 atm. By considering one stroke at a time, find the pressure in the vessel after two strokes. 0.64atm

Conservation of energy answers

Conservation of Energy

(Use g = 9.8 N kg^-1)

1. A waterfall is 120 m high. If 75 % of the potential energy available is converted into heat energy, find the difference in temperature between the top and bottom of the fall. The specific heat capacity of water is 4200 J kg^-1 K^-1.                                       .21K

2. A lead bullet moving at 210 m s^-1 is brought to rest on hitting a target. If 60 % of its energy is converted into heat energy, find the rise in temperature of the bullet. The specific heat capacity of lead is

140 J kg^-1 K^-1.                                   94.5K

3. A lump of lead falls to the ground and its temperature rises by 1.4 °C. If all the potential energy is converted into heat energy, find the height fallen, if the specific heat capacity of lead is 140 J kg^-1 K^-1.                20m

4. If 60 % of the potential energy available in a waterfall is converted into heat energy, find the height of the waterfall when the temperature difference between the top and the bottom of the fall is 0.21 °C. (The specific heat capacity of water = 4200 J kg^-1 K^-1.)                  150m

5. A paddle wheel is made to rotate inside a calorimeter containing water by a mass of 12 kg falling through a height of 2 m. The total heat capacity of the calorimeter and its contents is 800 J K^-1. Find the rise in temperature of the water.                        .294K

6. A car of mass 800 kg is moving at 108 km h^-1. What heat energy is produced in the brake drums when it is brought to rest by the brakes?             360 kJ

7. On entering the earth's atmosphere the velocity of a meteorite of mass 6 kg is reduced rapidly from 300 m s^-1 to 100 m s^-1. How much heat energy is produced?                  240 kJ

8. A lead bullet moving at 70 m s^-1 is brought to rest on hitting a target. If 80% of its energy is converted into heat energy, find the rise in temperature. (The specific heat capacity of lead = 140 J kg^-1 K^-1.)                  14K

9. A lead bullet at 40.5 °C before impact just melts when it hits a target without rebounding. If all the energy of the bullet is converted into heat energy, find the velocity of the bullet. (The specific heat capacity of lead = 140 J kg^-1 K^-1. The melting point of lead = 328 °C. The specific latent heat of fusion of lead = 21 000 J kg^-1.)                  350ms^-1

Latent Heat Answers

Latent Heat

Latent Heat of fusion of ice = 3.36 x 10⁵ J kg^-1

Latent Heat of vaporization of water = 2.25 x 10⁶ J kg^-1

Specific heat capacity of water = 4200 J kg^-1 K^-1

1. How much heat energy is needed to change 0.010 kg of water at 20^oC to steam at 100^oC?                    25860J
2. How much heat energy is needed to change 0.020 kg of ice at 0°C into steam at 100°C?           60 120J  
3. How much heat energy is needed to melt 0.010 kg of ice at 0^o C and then heat the water formed to 20^o C? 4200j
4. What mass of ice at 0 °C is needed to cool 0.086 kg of water from      54 °C to 6 °C?                     .048kg
5. When a kettle boils a space can be seen between the spout and the so called `cloud of steam'. If your finger were placed in the space a much more serious scald would result than if it were placed in the `cloud of steam'. Explain why, and criticize the expression `cloud of steam'.
6. A heating coil takes 14 min to heat 0.20 kg of water from 20 °C to boiling point. How much longer will it take to boil away 0.02 kg of the boiling water?     562.5s
7. A 1 kW immersion heater takes 225 s to boil away 0.1 kg of water at 100 °C. Find the specific latent heat of vaporization of water.                     2.25x 10⁶ J kg^-1
8. How long will it take an immersion heater of power 500 W to boil away a mixture of 0.05 kg of water and 0.02 kg of ice initially at 0 °C? 387 s
9. A mixture of ice and water at 0 °C contains 0.060 kg of ice. A 750 W immersion heater takes 26.8 s to melt all the ice. Deduce a value for the specific latent heat of fusion of ice.        336000J kg^-1
10. A mixture of ice and water at 0 °C has a total mass of 1 kg, and con­tains 0.5 kg of ice. An immersion heater melts all the ice in 168 s, and then needs a further 84 s to heat the mixture to 20 ^°C. Find the specific latent heat of fusion of ice.            336000 Jkg^-1
11. A vessel of heat capacity 115 J K^-1 contains 0.30 kg of water and 0.20 kg of ice at 0 °C. An immersion heater takes 134.4 s to melt the ice and a further 88.6 s to heat the vessel and its contents to 20 °C. Find the specific latent heat of fusion of ice. 336000 Jkg^-1
12. A heating coil takes 115 s to heat 0.05 kg of a liquid from 20 °C to its boiling-point of 80 °C. It needs a further 340 s to boil away 0.02 kg of the liquid. If the specific heat capacity of the liquid is 2300 J kg^-1 K^-1, find its specific latent heat of vaporization. 1.02 x 10⁶ Jkg^-1

 

Specific Heat Capacity Answers

Specific Heat Capacity

Data

Use the following for Specific Heat capacities

Substance	Specific Heat Capacity c	Units
Water	4200	J kg-1 K-1
Copper	380	J kg-1 K-1
Methylated Spirit	2500	J kg-1 K-1
Turpentine	1800	J kg-1 K-1

Notes

• The heat capacity of an object is the heat required to raise the all of the object’s temperature by 1K
• When heating a liquid you have to heat the vessel that it is in as well.
• When an object at high temperature is dropped into a liquid at low temperature there is a net heat flow from the high temperature object to the low temperature object until they are both at the same temperature.
• Unless you state “Ignoring heat losses” in each question you will lose a mark!

1 Find the quantity of heat needed to heat 0.50 kg of copper from 20^oC to 50 °C.              5700J

2 When 0.25 kg of ice are heated from -10 °C to -2 °C the heat supplied is 4200 J. Find the specific heat capacity of ice.                      2100 Jkg^-1K^-1

3 A vessel of heat capacity 30 J K^-1 contains 0.30 kg of water at 10 °C. When heated by a Bunsen burner for 5 min the temperature rises to 90 °C. At what rate does the Bunsen supply heat energy?     344W

4 A piece of iron of mass 0.20 kg is heated to 100 °C and dropped into 0.15 kg of water at 20 °C. If the temperature of the mixture is 30 °C, find a value for the specific heat capacity of iron.     450 Jkg^-1K^-1

5 A piece of iron of mass 0.05 kg is heated in a flame and then quickly transferred to a calorimeter of mass 0.05 kg and specific heat capacity 360 J kg^-1 K^-1 containing 0.18 kg of water at 18 °C. If the temperature of the water rises to 43 °C, find the temperature of the flame. The specific heat capacity of the iron is 450 J kg^-1 K^-1.                 903 ^oC

6 Which would result in the more serious burn if dropped onto your outstretched hand a white-hot spark from a firework, or a lump of red-hot iron? Explain your answer.

The white hot spark has a higher temperature than the lump of iron

The lump of iron has a greater mass than the white hot spark

They both have the same specific heat capacity.

Lump of iron has greater store of heat energy

The rate of heat transfer from the white hot spark will be greater for the spark than the lump of iron.

A greater quantity of heat will be transferred from the iron

Causing more serious skin damage.