• Physics 9702 Doubts | Help Page 88

Question 454: [Current of Electricity]
Two cells X and Y are connected in series with resistor of resistance 9.0 Ω, as shown.

Cell X has electromotive force (e.m.f.) of 1.0 V and internal resistance of 1.0 Ω. Cell Y has e.m.f. of 2.0 V and internal resistance of 2.0 Ω.
What is current in the circuit?
A 0.25 A                     B 0.17 A                     C 0.10 A                     D 0.083 A

Reference: Past Exam Paper – June 2013 Paper 13 Q31



Solution 454:
Answer: D.
Ohm’s law: V = IR
Current I = V / R

Total resistance in circuit = 9.0 + 1.0 + 2.0 = 12.0Ω

Current flows from the positive terminals of the cells. In the circuit, cell X and Y are connected in such a way that the currents flowing due to each of them are opposing each other.
Resultant e.m.f. in circuit = 2.0 – 1.0 = 1.0V

Current I = V / R = 1.0 / 12.0 = 0.083A











Question 455: [Current of Electricity]
Four resistors of resistance R, 2R, 3R and 4R are connected to form network.
Battery of negligible internal resistance and voltmeter are connected to the resistor network as shown.

Voltmeter reading is 2 V.
What is electromotive force (e.m.f.) of the battery?
A 2 V                         B 4 V                         C 6 V                         D 10 V

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q38



Solution 455:

Answer: C.

2 loops can be drawn in the network and from Kirchhoff’s laws, both of them will have the same total voltage (= emf of battery).

1^st loop: from a terminal of the battery, through the 4R resistor and back to the other terminal of the battery.

2^nd loop: from a terminal of the battery, goes through the 3R, 2R and 1R resistors and back to the other terminal of the battery.

Ohm’s law: V = IR

At the 2R resistor, the voltmeter reading is 2V. So, current I through it = 1 A and current I is constant through any specific loop (since the resistors are connected in series).

So, through the 2^nd loop, total resistance = 3 + 2 + 1 = 6R

e.m.f. = total p.d. = IR = 1 x 6 = 6V

Question 456: [Dynamics > Weight]
Gravitational field strength on surface of planet P is one tenth of that on surface of planet Q.
On surface of P, a body has mass of 1.0 kg and a weight of 1.0 N.
What are the mass and weight of same body on surface of planet Q?

mass on Q / kg            weight on Q / N

A         1.0                                           0.1
B         1.0                                           10
C         10                                            10
D         10                                            100

Reference: Past Exam Paper – November 2010 Paper 11 Q10



Solution 456:

Answer: B.

The mass is a property of the body and is constant anywhere. [C and D are incorrect]

Weight of a body = mg

The gravitational field strength g is 10 times greater on planet Q than on planet P. So, the weight is 10 times greater.

Question 457: [Dynamics > Torque]
A cupboard is attached to wall by a screw.
Which force diagram shows the cupboard in equilibrium, with weight W of the cupboard, the force S that the screw exerts on the cupboard and force R that the wall exerts on the cupboard?

Reference: Past Exam Paper – November 2013 Paper 13 Q14



Solution 457:

Answer: B.

For equilibrium, both the resultant force and the resultant torque should be zero.

For zero torque the line of action of all three forces must go through a single point so that the distance of the force from that point is zero. Thus, torque = Fd = F (0) = 0.

So, we must extend each force and see if all of them pass through a single point.

Only choice B satisfies this condition. The forces pass through a single point which is a bit above the centre of gravity of the cupboard.