Year 10 Paper 6: Applications
Covers cell biology, organisation, atomic structure, energy and electricity.
Year 10 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 10, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.
Questions
Question 1 [4 marks]
Atomic Structure and the Periodic Table
An ion of oxygen, O2-, has a mass number of 16 and an atomic number of 8.
State the number of protons, neutrons and electrons in this ion of oxygen.
Question 2 [4 marks]
Energy
A crane lifts a load of mass 250 kg through a height of 12 m.
Calculate the gravitational potential energy gained by the load. Use GPE = m g h, with g = 9.8 N/kg.
Question 3 [4 marks]
Organisation
Villi increase the surface area of the small intestine lining for the absorption of digested food.
The surface area of the small intestine's lining without villi would be about 0.4 m^2. Villi increase this surface area by a factor of 30.
Calculate the surface area of the small intestine's lining including villi, and explain why a greater surface area is useful for digestion.
Question 4 [5 marks]
Cell Biology
A student observes an onion cell under a microscope at a magnification of x600.
The length of the cell in the image is 3.0 mm.
Calculate the actual (real) length of the cell, in micrometres. Use real size = image size / magnification.
Question 5 [5 marks]
Electricity
Two resistors, of resistance 6 ohms and 3 ohms, are connected in parallel.
Calculate the total (combined) resistance of the parallel combination. Use 1 / Rtotal = 1 / R1 + 1 / R2.
State how the combined resistance compares with the smallest individual resistance.
Question 6 [4 marks]
Organisation
Coronary heart disease is a non-communicable disease linked to diet.
Explain how a diet high in saturated fat can increase the risk of coronary heart disease.
Question 7 [5 marks]
Energy
An electric motor is supplied with 800 J of energy each second and usefully transfers 600 J of energy each second to kinetic energy.
Calculate the efficiency of the motor.
Calculate the power wasted as heat by the motor.
Question 8 [5 marks]
Electricity
A 2.5 kW electric heater is used for 3 hours.
Electricity costs 28p per kWh.
Calculate the total cost of using the heater for this time.
Question 9 [6 marks]
Cell Biology
In the investigation described above, the potato chip lost mass in the salt solution.
Explain, in terms of water potential and the cell membrane, why the potato chip lost mass in the salt solution.
Question 10 [6 marks]
Energy
A 1.5 kg block of an unknown metal is heated from 20 degrees C to 95 degrees C using 87750 J of energy.
Calculate the specific heat capacity of the metal, in J/kg degrees C. Use change in thermal energy = m c (change in temperature).
Question 11 [6 marks]
Cell Biology
A potato chip is placed in distilled water.
Water moves into the chip by osmosis, increasing its mass by 8% over 24 hours.
The mass of the chip after 24 hours is 5.94 g.
Calculate the original mass of the potato chip before it was placed in the water, and explain why the chip gained mass.
Question 12 [6 marks]
Energy
An electric heater with a power of 1.5 kW is used to heat 2.0 kg of water. The specific heat capacity of water is 4200 J/kg degrees C.
Calculate the energy transferred by the heater in 4 minutes, and calculate the resulting temperature rise of the water, assuming all the energy is usefully transferred to the water. Use energy = power x time and change in thermal energy = m c (change in temperature).
Model solutions
| Question 1[4 marks] | |
|---|---|
| Answer or working | Marks |
| protons = atomic number = 8 | M1 |
| neutrons = mass number - atomic number, 16 - 8 | M1 |
| 8 neutrons | A1 |
| electrons = 10, since a 2- ion has 2 more electrons than protons | B1 |
| Final answer: 8 protons, 8 neutrons, 10 electrons | |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| using GPE = m g h | M1 |
| substituting 250 x 9.8 x 12 | M1 |
| correct evaluation | M1 |
| 29400 J (29.4 kJ) | A1 |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| using surface area = base area x factor | M1 |
| substituting 0.4 x 30 | M1 |
| 12 m^2 | A1 |
| explaining a greater surface area allows more digested food to be absorbed into the blood in a given time | B1 |
| Final answer: 12 m^2; a greater surface area allows faster (more efficient) absorption of digested food | |
| Question 4[5 marks] | |
|---|---|
| Answer or working | Marks |
| using real size = image size / magnification | M1 |
| substituting 3.0 / 600 | M1 |
| 0.005 mm | A1 |
| converting mm to micrometres by multiplying by 1000 | M1 |
| 5 micrometres | A1 |
| Final answer: 5 micrometres (0.005 mm) | |
| Question 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| using 1 / Rtotal = 1 / R1 + 1 / R2 | M1 |
| 1/6 + 1/3 | M1 |
| combining the fractions to 1/2 | M1 |
| Rtotal = 2 ohms | A1 |
| stating the combined resistance in parallel is less than the smallest individual resistance | B1 |
| Final answer: 2 ohms (less than the smallest individual resistance, 3 ohms) | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| a diet high in saturated fat raising blood cholesterol levels | B1 |
| cholesterol building up as fatty deposits (plaques) in artery walls | B1 |
| this narrowing the lumen of the coronary arteries | B1 |
| reduced blood flow / oxygen supply to the heart muscle, increasing the risk of a heart attack | B1 |
| Final answer: Saturated fat raises cholesterol, which narrows the coronary arteries and reduces blood flow to the heart | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| using efficiency = useful output energy / total input energy | M1 |
| substituting 600 / 800 | M1 |
| 0.75 (75%) | A1 |
| wasted energy per second = 800 - 600 | M1 |
| 200 W | A1 |
| Final answer: 75% efficient; 200 W wasted | |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| energy in kWh = power in kW x time in hours | M1 |
| substituting 2.5 x 3 | M1 |
| 7.5 kWh | A1 |
| cost = 7.5 x 28 | M1 |
| 210p (2.10 pounds) | A1 |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| reference to osmosis | B1 |
| movement of water across a partially permeable membrane | B1 |
| stating the salt solution has a lower water concentration than the potato cells | B1 |
| water moving down a concentration gradient | B1 |
| water moving out of the potato cells, into the salt solution | B1 |
| a clear concluding statement linking water loss to the fall in mass | B1 |
| Final answer: Water moves out of the potato cells by osmosis, down a water potential gradient, into the more concentrated salt solution | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| finding the temperature change, 95 - 20 = 75 degrees C | M1 |
| rearranging to c = E / (m x change in temperature) | M1 |
| substituting 87750 / (1.5 x 75) | M1 |
| evaluating 1.5 x 75 = 112.5 | M1 |
| 780 J/kg degrees C | A1 |
| stating this value could be used to identify the metal, by comparing it to known specific heat capacities | B1 |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the final mass is 108% of the original mass | M1 |
| original mass = final mass / 1.08 | M1 |
| substituting 5.94 / 1.08 | M1 |
| 5.5 g | A1 |
| stating distilled water has a higher water potential than the contents of the potato cells | B1 |
| stating water moved into the cells by osmosis, down the water potential gradient, increasing the mass | B1 |
| Final answer: Original mass 5.5 g; water moved in by osmosis because distilled water has a higher water potential than the cell contents | |
| Question 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| converting 4 minutes to seconds, 4 x 60 | M1 |
| energy = power x time, 1500 x 240 | M1 |
| 360000 J | A1 |
| rearranging E = m c (change in temperature) to change in temperature = E / (m c) | M1 |
| substituting 360000 / (2.0 x 4200) | M1 |
| a temperature rise of 42.9 degrees C (3 sf) | A1 |
| Final answer: 360000 J transferred; a temperature rise of about 42.9 degrees C | |