• Physics 9702 Doubts | Help Page 163

Question 812: [Electric field]

Two parallel metal plates are at potentials of +800 V and +1300 V.

Which diagram best shows electric field between the metal plates?

Reference: Past Exam Paper – June 2006 Paper 1 Q29

Solution 812:

Answer: B.

Electric field strength E = V / d

As long as the p.d. V between the plates and the separation d of the plates are constant, the electric field strength is constant between them ideally.

The direction of the electric field is drawn from a positive (or relatively more positive) potential to a negative (or relatively less positive) potential. So, the direction of the electric field should be from right to left. [C and D are incorrect]

When the electric field is constant, the field lines are equally spaced from each other such that in any areas of similar size, the same amount of field lines is seen and they are equally spaced.

However, due to imperfection in reality, the electric field is less and varies at the edges of the plates. When a field is weaker in a region, the field lines spacing would be greater in that region. In choice A, the field lines are closer at the edges. [A is incorrect]

Question 813: [Dynamics > Resultant force]

A sphere is released from rest in viscous fluid.

Which graph represents variation with time t of the acceleration a of the sphere?

Reference: Past Exam Paper – June 2012 Paper 12 Q16

Solution 813:

Answer: D.

The weight of the sphere causes the sphere to sink in the fluid. But since the fluid is viscous, there will be a viscous drag in the sphere that opposes its downward motion. This causes the resultant force on the sphere to decrease.

Resultant force, F = ma

Acceleration decreases until a terminal velocity is reached due to viscous force. Terminal velocity is reached when the resultant force F on the sphere is zero, causing the resultant acceleration to be zero.

Thus, the acceleration is highest initially and then decreases to zero.

Question 814: [Nuclear Physics > Atoms]

When α-particles are fired at a thin metal foil, most of the particles pass straight through but a few are deflected by large angle.

Which change would increase the proportion of α-particles deflected by a large angle?

A using α-particles with greater kinetic energy

B using a foil made of a metal with fewer protons in its nuclei

C using a double thickness foil

D using an alpha source with a higher activity

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q39

Solution 814:

Answer: C.

The fact that the α-particles are deflected by a large angle indicated that they seem to have struck something massive – which we know is the nucleus of the atom.

We want to know which of these changes would increase the proportion of α-particles deflected by a large angle – that is, compared with the total number of α-particles incident, more α-particles are deflected by a large angle. For this to occur, we need to increase the number of nuclei present. This can be done by using a double thickness foil. [C is correct]

With the other choices (A, B or D), the number of nuclei remains constant.

Question 815: [Electric field]

In each electric field diagram, a positively charged particle is moved from X to Y.

In which diagram would the particle experience an increasing repulsive force?

Reference: Past Exam Paper – June 2011 Paper 12 Q31

Solution 815:

Answer: B.

Electric field is drawn from positive to negative. That is, it shows the direction of the electric force on a positive charge which is along the direction shown by the arrow.

For a positive charge to experience a repulsive force, it should be moved towards the positive – that is, against the direction of the field since the field is drawn FROM the positive to negative. Like charges repel. [A and C are incorrect]

Now, as it is moved from X to Y, the repulsive force should increase. The strength of an electric field is given by the spacing between the field lines. The closer the field lines, the stronger is the field, and thus the force.

For choice D, the spacing between the field lines is uniform, so the electric force is constant. As for choice B, if we draw circles (of constant radii) closer and closer to the centre, it can be seen that the spacing between the field lines ON the circle becomes closer.

Question 816: [Current of Electricity]

B₁, B₂ and B₃ are three identical lamps. They are connected to battery with zero internal resistance, as shown.

Initially the switch is closed. Switch is then opened and lamp B₃ goes out.

What happens to the brightness of lamps B₁ and B₂ when switch is opened?

brightness of lamp B₁              brightness of lamp B₂

A         decreases                                 decreases

B         decreases                                 increases

C         increases                                  decreases

D         increases                                  increases

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q30

Solution 816:

Answer: B.

This question must be answered with reference to p.d.

When the switched is still closed, B₂ and B₃ are in parallel with each other. The combined resistance of a parallel combination is less than the individual resistance. So, the whole circuit consists of B₁ in series with the parallel combination of B₂ and B₃.

Ohm’s law: V = IR

Initially the p.d. across B₂ is low and therefore the p.d. across B₁ is high since from Kirchhoff’s law, the sum of p.d. in any loop should be equal to the e.m.f.

On opening the switch, B₃ goes out. Now, the whole circuit consists of B₁ in series with B₂. The p.d. across these become equal. As for the individual lamp, the p.d. across B₁ has decreased while that across B₂ has increased.

Question 817: [Units]

The product of pressure and volume has same SI base units as

A energy

B force

C force / area

D force / length

Reference: Past Exam Paper – June 2010 Paper 11 Q2 & Paper 12 Q7 & Paper 13 Q1

Solution 817:

Answer: A.

Pressure = Force / Area

Unit of Pressure: kg ms^-2 m^-2 = kg m^-1 s^-2  

Unit of pressure × volume: kgm^-1s^-2 m³ = kg m² s^-2

Energy (Work done) = Force × distance

Unit of energy: kg ms^-2 m = kg m² s^-2

[A is correct]

Alternatively consider,

Work done on gas = pΔV

Pressure × volume = (force / area) × volume = force × distance = energy