Paper 1 Past Paper Questions

Question 1

The resistance of 1 resistor is 20.
The other is unknown.
But a student calculated the total resistance of the resistors.
The student answer was 26 ohms.

Explain why the student must be wrong.

Answer 1

The total resistance must be less than 20 ohms because the total resistance of the resistors its less than the resistance of the smallest resistor.

Question 2

The energy transferred by the National Grid in one second was 36 gigajoules (GJ)

Which of the following is the same as 36 GJ
- 36 x 10^3 J
- 36 x 10^6 J
- 36 x 10^9 J
- 36 x 10^12J

Answer 2

36 x 10^9J

Question 3

Which of the following is a renewable energy source?

• geothermal -Natural gas -Nuclear fuel

Answer 3

Geothermal

Question 4

V=IR
When the total resistance of the resistors was 7.5 ohms the current in the circuit was 480 mA.

Calculate the potential difference across the 2 resistors.

Answer 4

480/ 1000 = 0.48
V=IR
V= 0.48 x 7.5
V= 3.6

Question 5

When a student steps onto a scale the reading is 64.8kg.
The student steps off the scale and then immediately steps hack on.
The scales now read 64.1

Complete the sentence

The difference between the two values given by the scales is due to a _______________ error.

Answer 5

Random

Question 6

P=IV

This question is about the National Grid.
The potential difference across the primary coil in transformer Y is 400 000 V.
The potential difference across the secondary coil is 11 000V.
The current in the primary coil is 660 A.

Calculate the current in the secondary coil of transformer Y.

Answer 6

P=IV
P= ?
I1= 660 A
V1= 400 000 v2= 11 000
(Power will be equal for each coil)
I1 x v1 = I2 x v2
660 x 400 000 = I2 x 11 000

660 x 400 000/ 11 000 = I2
24 000 A = I2

Question 7

P=IV

This question is about the National Grid.
The potential difference across the primary coil in transformer Y is 400 000 V.
The potential difference across the secondary coil is 11 000V.
The current in the primary coil is 660 A.

Calculate the current in the secondary coil of transformer Y.

Answer 7

P=IV
P= ?
I1= 660 A
V1= 400 000 v2= 11 000
(Power will be equal for each coil)
I1 x v1 = I2 x v2
660 x 400 000 = I2 x 11 000

660 x 400 000/ 11 000 = I2
24 000 A = I2

Question 8

When a spring launches the into the air, the temperature of the air increases.

Explain why the spring on its own is not a closed system.

Answer 8

Energy from the spring is dissipated, but in a closed system the total energy remains constant.

Question 9

The temperature of the air is increased in a chamber.

Explain how the pressure in the temperature changes.

Answer 9

The temperature increasing causes the particles in the chamber to have more energy this causes them to move faster, causing more collisions against the wall of the chamber so the pressure increases