1 CDDCB
6 CCADC
11BDABC
15CBCAA
20CBADB
25DAABB
Check your answers and any queries will be answered next week during tutorial
Questions with difficulty:
4. Starting from rest hence x= 0.5 at2. After next interval means that the time is 2t. The value of x' = 4at2 which is 4x. During this second time interval the distance travelled is 4x-x= 3x.
9. The work done by the force F= elastic potential energy + heat energy. Hence it is not elastic potential energy unless no heat is produced.
12 . Power = Fv , P = kv3.
Writing ration of the two cases P2 can be found.
12. Work done by the external pressure = p ext(change in V). Work done by the gas = Net work done = (change in pressure)xchange in vol.
17. L = n1(24) = n2 (30) . Hence the n1=5 and n2 = 4. Length = 120 m
24. The broken lamp has 240 V reading and the unbroken one has 0 reading. use the rule that pd = IR. If I = 0 , pd = 0 eventhough the lamp is not broken.
Showing posts with label asp2010. Show all posts
Showing posts with label asp2010. Show all posts
Wednesday, September 15, 2010
Planning_2002_Q2
To investiagate how the magnetic flux density B at the centre of a flat coil is directly proportional to the number of turns in the coil.
Diag: Power supply , ammeter , coil , variable resistor in series.
Procedure:
1. Make a coil of N turns using a cylinder of radius r. Connect the circuit as shown.
2. Record the number of turns N. Adjust the variale resistor and note the current I.
3. Using a Hall probe connected to a data logger and computer , record the reading of magnetic flux density B .Ensure the hall probe is placed at the centre of coil.
4. Repeat steps 1 to 3 by varying N, keeping I constant to obtain 6 sets of readings of N and B.
5 Plot a graph of B vs N. A graph of B against N would be linear
Control of variables :
The radius of the coil is constant when N is varied
The current is kept constant by adjusting the variable resistor when N is varied.The ammeter reading should remain constant .
Reliabiltiy :
1. The Hall probe must be placed at the same location at the centre of the oil in the plane of the coil. It is also aligned perpendicular to the plane of the coil. Draw a diagram to shown this orientation.
2. The coil and probe should be kept away from other magnetic fields.
3. The Earth's magnetic field is negligible.
Diag: Power supply , ammeter , coil , variable resistor in series.
Procedure:
1. Make a coil of N turns using a cylinder of radius r. Connect the circuit as shown.
2. Record the number of turns N. Adjust the variale resistor and note the current I.
3. Using a Hall probe connected to a data logger and computer , record the reading of magnetic flux density B .Ensure the hall probe is placed at the centre of coil.
4. Repeat steps 1 to 3 by varying N, keeping I constant to obtain 6 sets of readings of N and B.
5 Plot a graph of B vs N. A graph of B against N would be linear
Control of variables :
The radius of the coil is constant when N is varied
The current is kept constant by adjusting the variable resistor when N is varied.The ammeter reading should remain constant .
Reliabiltiy :
1. The Hall probe must be placed at the same location at the centre of the oil in the plane of the coil. It is also aligned perpendicular to the plane of the coil. Draw a diagram to shown this orientation.
2. The coil and probe should be kept away from other magnetic fields.
3. The Earth's magnetic field is negligible.
Tuesday, September 14, 2010
Planning_2002_Q1
To investigate how the vol flow rate depends upon the separation d of threads in the material
Diagram 1: Stretch out a piece of material using clips and retort stand. Using a container the top of the material allow a constant flow of water onto the material. Place a beaker at the bottom of the cloth to collect the water.
Diagram 2: Use a retort stand to hold the laser . Use a holder to stretch out the material in front of the laser. The light that passes through should fall on a screen a distance D away from the material.
Procedure:
1. Set up expt as in Diag 1. Start the stopwatch and replace a clean beaker to collect the water.
2. After a time t, stop the stopwatch and remove the beaker. Record the time interval as shown.
3. Measure the vol of water V collected using a measuring cylinder.
4. Calculate the flow rate by using the formula v = V/t.
5. Repeat steps 1 to 4 by varying the material of different d value.
6. The distance between the threads of the material used , d can be measure by using set up shown in diag. 2
7. Turn off the light and turn on the power of the laser.
8. Stretch the material vertically in front of the laser .
9. Using a ruler , measure the fringe spacing on the screen, keeping the distance D from the laser to the screen constant.The angle can be calc using tan angle formula.
The distance d can be measured using the formula dsin(angle) = n(wavelength) .
10. Repeat steps 7 to 9 by varing the material of different d . Matching the material and the flow rate it yields.
11. Plot a graph of lg v vs lg d . If the graph is linear , the equation v=kdn is valid ......
Control of variables:
1. The height from which the water was dropped should be kept constant .
2. The water falling onto material at a constant rate.
3. The tension on the material should be constant.
Reliability:
Repeat the measurement of fringe separation and averaging the value to calc angle .
Ensure the cloth is dry when used in diffraction expt
Wet the material before starting the volume rate measurement.
Diagram 1: Stretch out a piece of material using clips and retort stand. Using a container the top of the material allow a constant flow of water onto the material. Place a beaker at the bottom of the cloth to collect the water.
Diagram 2: Use a retort stand to hold the laser . Use a holder to stretch out the material in front of the laser. The light that passes through should fall on a screen a distance D away from the material.
Procedure:
1. Set up expt as in Diag 1. Start the stopwatch and replace a clean beaker to collect the water.
2. After a time t, stop the stopwatch and remove the beaker. Record the time interval as shown.
3. Measure the vol of water V collected using a measuring cylinder.
4. Calculate the flow rate by using the formula v = V/t.
5. Repeat steps 1 to 4 by varying the material of different d value.
6. The distance between the threads of the material used , d can be measure by using set up shown in diag. 2
7. Turn off the light and turn on the power of the laser.
8. Stretch the material vertically in front of the laser .
9. Using a ruler , measure the fringe spacing on the screen, keeping the distance D from the laser to the screen constant.The angle can be calc using tan angle formula.
The distance d can be measured using the formula dsin(angle) = n(wavelength) .
10. Repeat steps 7 to 9 by varing the material of different d . Matching the material and the flow rate it yields.
11. Plot a graph of lg v vs lg d . If the graph is linear , the equation v=kdn is valid ......
Control of variables:
1. The height from which the water was dropped should be kept constant .
2. The water falling onto material at a constant rate.
3. The tension on the material should be constant.
Reliability:
Repeat the measurement of fringe separation and averaging the value to calc angle .
Ensure the cloth is dry when used in diffraction expt
Wet the material before starting the volume rate measurement.
Sunday, September 12, 2010
H1_N08_P2
1a No external force acting on the system
b The force is opposite and hence the gradient is positive and negative
ii) The initial momentum is equal to final momentum = 36KNs
iii) The force is change in momentum divide by time taken F= 2.7 kN
iv) The KE is not conserved as intial is 11.4 kJ and final is 10.8kJ. Hence change of 6kJ. The collision is inelastic.
3a Force multiply by perpendicular distance
b) The three forces are Wbricks , W beam and Rx and Ry at the hinge.
ii) The tension increases as the W bricks is moved towards the end P as the moment increase. To remain balance T must remain the same
iii) the Tension = 2880N
4 Energy level is the energy state an electron can occupy
b) A: Emission line spectra
B: line absorption spectrum.
A :cooler gas excited by the white light will emit photon of specific wavelength
B: White light is absorped by cooler gas has some dark lines missing over bright backgd .
ii) 1 The freq f= 6.2 x 1014Hz 2 E2-E1 = 4.10 x 10-19 - hf = 3.0 x 10-19. f= 4.52 x 1014 and wavelength = 6.65 x10-7m
5. change MeV to Joule
b) Photon E = hf.
c) the loss in ke = 550 (photon energy) divide by energy of proton
d) Use photoelectric equation max KE = hf - work function
e) No of electron = 3 power of 10 = 59049
f) I = Q/t = 550e /time = 4.1 x 10-7
7. ii) R = 3 ohm using P = V2/R
2. length of wire is 9.5 cm
3. I = E/R+r
where R = parallel of 1.5 and 6 ohm .
3 I = 9.8 A
4 When a large current is drawn from the battery , the pd across the battery is E - Ir which is much smaller . Hence the lamps will be less bright as pd across the lamps is reduced.
5. 2 % of 48 W divide by photon energy =
rate of photon emitted .N = 2.7 x 1018 s-1
8 Force is perpendicular to velocity hence the force in the dir of vel is zero . The direction of motion changes but the speed remain constant,
9 qV = 0.5 mv2
v2 = 2eV/m where V = 2.8 kV
v = 3.1 x 107 ms-1
ii) 1 a = QE/m = 3.5 x 1015 ms-2
2. L = v/t, t = 3.87 x10-9 s
3. W = Fs where s = 0.5 at2
W = Fs= 3.2 x 10-15 x 0.5 (3.51 x 1015( x( 3.9 10 -19 ) =0.263 J
4. v = 1.07 x 104 ms-1
5 angle = 23.7 deg.
iii Proton deflect below the horizonal and the velocity and angle of deflection will be smaller than that of electron as the acc is small
b The force is opposite and hence the gradient is positive and negative
ii) The initial momentum is equal to final momentum = 36KNs
iii) The force is change in momentum divide by time taken F= 2.7 kN
iv) The KE is not conserved as intial is 11.4 kJ and final is 10.8kJ. Hence change of 6kJ. The collision is inelastic.
3a Force multiply by perpendicular distance
b) The three forces are Wbricks , W beam and Rx and Ry at the hinge.
ii) The tension increases as the W bricks is moved towards the end P as the moment increase. To remain balance T must remain the same
iii) the Tension = 2880N
4 Energy level is the energy state an electron can occupy
b) A: Emission line spectra
B: line absorption spectrum.
A :cooler gas excited by the white light will emit photon of specific wavelength
B: White light is absorped by cooler gas has some dark lines missing over bright backgd .
ii) 1 The freq f= 6.2 x 1014Hz 2 E2-E1 = 4.10 x 10-19 - hf = 3.0 x 10-19. f= 4.52 x 1014 and wavelength = 6.65 x10-7m
5. change MeV to Joule
b) Photon E = hf.
c) the loss in ke = 550 (photon energy) divide by energy of proton
d) Use photoelectric equation max KE = hf - work function
e) No of electron = 3 power of 10 = 59049
f) I = Q/t = 550e /time = 4.1 x 10-7
7. ii) R = 3 ohm using P = V2/R
2. length of wire is 9.5 cm
3. I = E/R+r
where R = parallel of 1.5 and 6 ohm .
3 I = 9.8 A
4 When a large current is drawn from the battery , the pd across the battery is E - Ir which is much smaller . Hence the lamps will be less bright as pd across the lamps is reduced.
5. 2 % of 48 W divide by photon energy =
rate of photon emitted .N = 2.7 x 1018 s-1
8 Force is perpendicular to velocity hence the force in the dir of vel is zero . The direction of motion changes but the speed remain constant,
9 qV = 0.5 mv2
v2 = 2eV/m where V = 2.8 kV
v = 3.1 x 107 ms-1
ii) 1 a = QE/m = 3.5 x 1015 ms-2
2. L = v/t, t = 3.87 x10-9 s
3. W = Fs where s = 0.5 at2
W = Fs= 3.2 x 10-15 x 0.5 (3.51 x 1015( x( 3.9 10 -19 ) =0.263 J
4. v = 1.07 x 104 ms-1
5 angle = 23.7 deg.
iii Proton deflect below the horizonal and the velocity and angle of deflection will be smaller than that of electron as the acc is small
SPA Q4_N99
Aim: To investigate how the vol flow rate of a concentrated salt soln depends upon the magnetic field strength, keeping the concentration constant .
Diagram: Do not copy diagram again. Draw a circuit diagram with coil , battery, rheostat and ammeter.
Procedure:
1. Set up the circuit as above.
2. Set the variable resistor to a maximum and measure the magnetic field strength B between the coils using a Hall probe connected to a data logger and computer.
3. Open the tap at the end of the pipe and collect the soln at the start of the stopwatch.
4. Measure the vol of soln v, using a measuring cylinder and time taken, t from the stopwatch.
5. Calculate the vol flow rate V = v /t
6. Repeat steps 2 to 5 by varying the value of B to obtain 6 sets of v and t.
7. Plot a graph of lg B vs lgV assuming the equation is B = kVn . If the graph is linear then equation is valid, the grad = n and intercept = lg k.
8. To investigate the how the flow rate depends on concentration of soln , the B is kept constant.The concentration can be varied by adding more mass of salt into the same amount of water. C = m/vol.
Reliability :
The Hall probe is placed perpendicular to the coil.
The current through the sol is kept oonstant.
The expt is repeated and average value is calc .
Safety :
The current through the coils is large , hence do not touch with bare hands.
Place bricks on retort stand to prevent the apparatus from toppling.,
Diagram: Do not copy diagram again. Draw a circuit diagram with coil , battery, rheostat and ammeter.
Procedure:
1. Set up the circuit as above.
2. Set the variable resistor to a maximum and measure the magnetic field strength B between the coils using a Hall probe connected to a data logger and computer.
3. Open the tap at the end of the pipe and collect the soln at the start of the stopwatch.
4. Measure the vol of soln v, using a measuring cylinder and time taken, t from the stopwatch.
5. Calculate the vol flow rate V = v /t
6. Repeat steps 2 to 5 by varying the value of B to obtain 6 sets of v and t.
7. Plot a graph of lg B vs lgV assuming the equation is B = kVn . If the graph is linear then equation is valid, the grad = n and intercept = lg k.
8. To investigate the how the flow rate depends on concentration of soln , the B is kept constant.The concentration can be varied by adding more mass of salt into the same amount of water. C = m/vol.
Reliability :
The Hall probe is placed perpendicular to the coil.
The current through the sol is kept oonstant.
The expt is repeated and average value is calc .
Safety :
The current through the coils is large , hence do not touch with bare hands.
Place bricks on retort stand to prevent the apparatus from toppling.,
Sunday, July 18, 2010
Design question
A : To investigate how P varies with Q keeping Z constant
B: Diagram
C: Procedure:
1. Set up
2. Measure P using M
3. Measure Q using N
4. Repeat steps 2 to 3 by varying P to obtain 6 sets of P and Q keeping Z constant.
5. Calculate R using the formula ...
6. Plot a graph of ln P vs ln Q assuming .....the gradient = n and intercept = lnk
D: Safety precautions
E: Reliability:
1.How you measure P and Q accurately
2.How you keep Z constant .
3.Any other variables kept constant.
4.Check question for any unanswered parts from a) to e) .
B: Diagram
C: Procedure:
1. Set up
2. Measure P using M
3. Measure Q using N
4. Repeat steps 2 to 3 by varying P to obtain 6 sets of P and Q keeping Z constant.
5. Calculate R using the formula ...
6. Plot a graph of ln P vs ln Q assuming .....the gradient = n and intercept = lnk
D: Safety precautions
E: Reliability:
1.How you measure P and Q accurately
2.How you keep Z constant .
3.Any other variables kept constant.
4.Check question for any unanswered parts from a) to e) .
ASP1- 16 july 2010
1. Path looks like parabolic and not circular. It goes with the field lines so had to be positive.
2. direct application of fromula for de Broglie wavelength where it is h divide by momentum.
3.The experiment does not tell me how the nucleus is arranged. Hence it it not structure of nucleus but size of nucleus only.
4. Current can tell me number of ions per s if I divide current by charge . If the number of alpha is known then I can find number of ions produced by one alpha. Number per s = No of ions per s divide by number of alpha per s .
5. Magnetic force = centripetal force.
b) the ratio can be obtained by just considering mass and charge. The charge of alpha is 2X the charge of beta. The mass of alpha is 4u and the electron is 9.11x 10-31 kg. Answer is 3640.
c) The radius of alpha is 25.9 m. Subs. the correct mass of alpha (4u) and the charge is 2e.
The radius of electron is 7.12 x 10-3 m. The mass of electron is 9.11 x 10-31 and charge is e.
d ) Draw the alpha path of large radius (nearly straight) out of the vertical face of box. Label A.
Draw eelctron path of 0.71 cm radius , hence exit the bottom horizontal face of box. Label B.
For both path draw straight line once it exit the box.
6. a) Define ...
b) Gravitational force is attractive in nature. Hence a system does work by its own field to move closer to the source (mass) . By def. of potential which is external work done must therefore assume negative values.
c) The change in gravitationa potential is the difference of the values of potential at the two location , surface and at an altitude.
ii) the speed of the projection to reach that alitude must therefore have KE to overcome the increase in PE. Hence m(change in potential) = 1/2 mv2. Cancel m , v can be found. Many mistaken the speed to be escape speed which is projection into outer space.
d) the acceleration is not a constant around the Earth and hence the formula given cannot be used .
7. Draw the resultant graph having the same freq as the first graph but having wavy edges due to wave 2. Wave 2 have small amplitude and hence does not change the freq. This is in fact what we see in resonance tube where the fundamental freq is dominant and overtones are produced to enhance the sound of the musical instrument.
b) Draw equal spacing. For large gap the straight lines passes through and only curve at its edges. The small gap produces circular waves. Draw two lines to show how it spread out. The big gap spread out with small angle but the small gap spread out nearly 180 degrees.
c) the length is calculated in terms of number of wavelengths. Hence the pd difference is in terms of number of wavelengths. 33.3 - 30.8 = 2.5
Hence for iii) the intensity is zero (minima) when the two waves meet with p.d 2.5 wavelength which is out of phase.
As the dectector moves from P to O , the detector receives 3 maxima including the maxima at O. You can deduce the maxima by counting from O , n=1 (M) n=2(M) again then n=2.5 is not a maxima . Not wrong to count minima . P is minima , as the detector moves towards P it will encouter n=1.5 (m) and n=0.5 (m). Hence 2 minima only from P to O.
8. a)The equation shows energy released hence Y must have higher binding energy than Sr.
b) def of decay constant
c) i) the decay constant must be in s-1 as the activity is in Bq.
ii) the mass can be calculated either using the Avogadro constant and molar mass formula or mass = N ( 90u) as each atom is approx 90 nucleons.
iii) the ratio is 0.882 . Can leave the decay constant in yr-1 as time is 5 yrs.
2. direct application of fromula for de Broglie wavelength where it is h divide by momentum.
3.The experiment does not tell me how the nucleus is arranged. Hence it it not structure of nucleus but size of nucleus only.
4. Current can tell me number of ions per s if I divide current by charge . If the number of alpha is known then I can find number of ions produced by one alpha. Number per s = No of ions per s divide by number of alpha per s .
5. Magnetic force = centripetal force.
b) the ratio can be obtained by just considering mass and charge. The charge of alpha is 2X the charge of beta. The mass of alpha is 4u and the electron is 9.11x 10-31 kg. Answer is 3640.
c) The radius of alpha is 25.9 m. Subs. the correct mass of alpha (4u) and the charge is 2e.
The radius of electron is 7.12 x 10-3 m. The mass of electron is 9.11 x 10-31 and charge is e.
d ) Draw the alpha path of large radius (nearly straight) out of the vertical face of box. Label A.
Draw eelctron path of 0.71 cm radius , hence exit the bottom horizontal face of box. Label B.
For both path draw straight line once it exit the box.
6. a) Define ...
b) Gravitational force is attractive in nature. Hence a system does work by its own field to move closer to the source (mass) . By def. of potential which is external work done must therefore assume negative values.
c) The change in gravitationa potential is the difference of the values of potential at the two location , surface and at an altitude.
ii) the speed of the projection to reach that alitude must therefore have KE to overcome the increase in PE. Hence m(change in potential) = 1/2 mv2. Cancel m , v can be found. Many mistaken the speed to be escape speed which is projection into outer space.
d) the acceleration is not a constant around the Earth and hence the formula given cannot be used .
7. Draw the resultant graph having the same freq as the first graph but having wavy edges due to wave 2. Wave 2 have small amplitude and hence does not change the freq. This is in fact what we see in resonance tube where the fundamental freq is dominant and overtones are produced to enhance the sound of the musical instrument.
b) Draw equal spacing. For large gap the straight lines passes through and only curve at its edges. The small gap produces circular waves. Draw two lines to show how it spread out. The big gap spread out with small angle but the small gap spread out nearly 180 degrees.
c) the length is calculated in terms of number of wavelengths. Hence the pd difference is in terms of number of wavelengths. 33.3 - 30.8 = 2.5
Hence for iii) the intensity is zero (minima) when the two waves meet with p.d 2.5 wavelength which is out of phase.
As the dectector moves from P to O , the detector receives 3 maxima including the maxima at O. You can deduce the maxima by counting from O , n=1 (M) n=2(M) again then n=2.5 is not a maxima . Not wrong to count minima . P is minima , as the detector moves towards P it will encouter n=1.5 (m) and n=0.5 (m). Hence 2 minima only from P to O.
8. a)The equation shows energy released hence Y must have higher binding energy than Sr.
b) def of decay constant
c) i) the decay constant must be in s-1 as the activity is in Bq.
ii) the mass can be calculated either using the Avogadro constant and molar mass formula or mass = N ( 90u) as each atom is approx 90 nucleons.
iii) the ratio is 0.882 . Can leave the decay constant in yr-1 as time is 5 yrs.
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