Wednesday, February 25, 2015

Physics 9702 Doubts | Help Page 71

  • Physics 9702 Doubts | Help Page 71



Question 386: [Electricity > Electric field]
Two oppositely-charged parallel metal plates are situated in a vacuum, as shown in Fig.

Plates have length L.
Uniform electric field between plates has magnitude E. Electric field outside plates is zero.
Positively-charged particle has mass m and charge +q. Before particle reaches the region between the plates, it is travelling with speed v parallel to the plates.
Particle passes between plates and into the region beyond them.
(a)
(i) On Fig, draw path of the particle between the plates and beyond them.
(ii) For the particle in region between the plates, state expressions, in terms of E, m, q, v and L, as appropriate, for
1. Force F on the particle
2. time t for particle to cross the region between the plates

(b)
(i) State law of conservation of linear momentum.
(ii) Use answers in (a)(ii) to state an expression for change in momentum of the particle.
(iii) Suggest and explain whether law of conservation of linear momentum applies to the particle moving between the plates.

Reference: Past Exam Paper – November 2010 Paper 23 Q7



Solution 386:
(a)
(i) The path is a reasonable curve upwards between plates and straight and at a tangent to the curve beyond the plates


(ii)
1. Force F = Eq

2. {The force F only causes an acceleration vertically, it does not affect the horizontal motion.}
Time t = L / v

(b)
(i) The law of conservation of momentum states that the total momentum of a system remains constant OR the total momentum of a system before a collision equals the total momentum after collision
provided no external force acts on the system.

(ii) Change in momentum, Δp = {Ft =} EqL / v

(iii)
EITHER
The charged particle is not an isolated system, so the law does not apply

OR
The system is {consists of} the particle and the ‘plates’. There is an equal and opposite Δp {change in momentum} on the plates, so the law applies.










Question 387: [Current of Electricity > Power dissipated]
The resistors P, Q and R in the circuit have equal resistance.

The battery, of negligible internal resistance, supplies a total power of 12W.
What is power dissipated by heating in resistor R?
A 2 W                         B 3 W              C 4 W              D 6 W

Reference: Past Exam Paper – June 2010 Paper 11 Q33 & Paper 12 Q34 & Paper 13 Q35


 
Solution 387:
Answer: A.
EITHER
Power supplied by battery = VI = 12 W where I is the total current across the circuit and E is the e.m.f. of the battery.

We need to find the (relative) current flowing through each resistor and the potential difference across each of them.

Current flows from the positive terminal of the battery. So, the total current I would flow through resistor P. At the junction, the current splits equally into 2 since the resistances of Q and R are equal. So, the currents through Q and R are each I/2.

From Kirchhoff’s law, the sum of potential difference across any loop is equal to the sum e.m.f of the battery in the circuit. Also, the p.d. across the components in a parallel combination is the same.

The equivalent resistance of the parallel combination of resistors Q and R is half their original resistance. From the potential divider equation, the p.d. across resistor P is twice that across the parallel combination (and thus, across Q or R). Let the e.m.f of the battery be V. The p.d. across P is 2V/3 while the p.d. across Q or R is V/3 (p.d. across P is twice, giving the sum of p.d. in any loop = V).

Power supplied by battery = VI = 12 W
Resistor R: Current = I/2        p.d. = V/3
Power dissipated in resistor R = (I/2) (V/3) = (VI) / 6 = 12 / 6 = 2W

OR
This could alternatively be done by calculations.

Let the resistances of resistors P, Q and R be r each.
Equivalent resistance of parallel combination = [1/r + 1/r]-1 = 0.5r
Total resistance in circuit, rtotal = r + 0.5 = 1.5r = 3r / 2

Total power dissipated in circuit = I2rtotal where I is the total current in the circuit.
Total current I = (Power / rtotal)

As explained before, the total current I passes through P, but only half of the current passes through Q and R.
Current through P, Ip = (Power / rtotal) = √(12 / 1.5r) = √(8/r)

Half of this current passes through R.
Current IR through R = Ip/2 = 0.5 √(8/r) = √(2/r)

Power dissipated in resistor R = IR2r = (2/r) r = 2W









Question 388: [Gases > Kinetic theory of gases]
(a) Kinetic theory of gases is based on some simplifying assumptions.
Molecules of the gas are assumed to behave as hard elastic identical spheres.
State assumption about ideal gas molecules based on
(i) nature of their movement
(ii) their volume

(b) Cube of volume V contains N molecules of an ideal gas. Each molecule has a component cX of velocity normal to one side S of the cube, as shown.

Pressure p of the gas due to component cX of velocity is given by expression
pV = NmcX2
where m is mass of a molecule.
Explain how expression leads to the relation
pV = (1/3) Nm<c2>
where <c2> is mean square speed of the molecules.

(c) Molecules of an ideal gas have a root-mean-square (r.m.s.) speed of 520 m s–1 at a temperature of 27 °C.
Calculate r.m.s. speed of the molecules at a temperature of 100 °C.

Reference: Past Exam Paper – June 2012 Paper 41 Q2



Solution 388:
(a)
(i)
EITHER The gas molecules are in random motion
OR The gas molecules have constant velocity until they hit the wall / other molecules

(ii)
EITHER The (total) volume of the molecules is negligible compared to the volume of the containing vessel
OR The radius / diameter of a molecule is negligible compared to the average intermolecular distance.

(b)
{Consider a single molecule of the gas in motion. This molecule will have 3 components of velocity (since it is able to move in 3 dimensions).}
EITHER A molecule has components of velocity in 3 directions OR c2 = cX2 + cY2 + xZ2
{Now, consider all the gas molecules which are in motion. Since the motion of the gas molecules are random in nature and the number of molecules is large, the components of velocities of all the gas molecules can be averaged as <cX2> = <cY2> = <cZ2>}
Random motion and averaging, so <cX2> = <cY2> = <cZ2>
{Since <cX2> = <cY2> = <cZ2>, each of them can be written as the component of velocities in one specific direction. E.g. <cX2>. Now, replace this in the average of the equation c2 = cX2 + cY2 + xZ2 . <c2> = <cX2> + <cX2> + <cX2> = 3<cX2>}
<c2> = 3<cX2>
{But, since the pressure p of the gas due to component cX of velocity is given by expression pV = NmcX2. <cX2> = <c2> / 3}
So, pV = (1/3) Nm<c2>

(c)
{crms = <c2>}
<c2> α T          or crms α √T
(In kelvin,) the temperatures are (27+273 =) 300K and (100+273 =) 373K
{When T = 300K, crms = 520ms-1. When T = 373K, crms = (373/300) x 520 = 580 ms-1. }
crms = 580 ms-1






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