# Physics 9702 Doubts | Help Page 27

__Question 156: [Current of Electricity > Kirchhoff’s laws]__In circuit shown, all the resistors are identical and all ammeters have negligible resistance.

Reading on ammeter A

_{1}is 0.6 A.

What are the readings on other ammeters?

**Reference:**

*Past Exam Paper – June 2014 Paper 11 Q34*

__Solution 156:__**Answer: D.**

Current flows from the positive
terminal of the battery. So, current in A

_{2}should have the highest value.
Let the resistance of 1 resistor be
R.

The left-hand side of the circuit {(A

_{1}– A_{2}) loop} consists of 2 resistors in series, with a total resistance of (R+R =) 2R. Current in (A_{1}– A_{2}) loop = 0.6A, as read by ammeter A_{1}.
So, from Ohm’s law [V = IR], the
potential difference in (A

_{1}– A_{2}) loop = 0.6 (2R) = 1.2R.
From Kirchhoff’s law, the
potential different is the same as the e.m.f. in the circuit for any loop. So, the e.m.f. in the circuit is 1.2R and this is also the
potential difference across ANY loop.

Consider the (A

_{2}– A_{3}) loop,
Current in A

_{3}= V / R = 1.2R / R = 1.2A.
Similarly consider the (A

_{2}– A_{4}) loop,
Current in A

_{4}= 1.2R / 3R = 0.4A.
Current in A

_{2}= total current = 0.6 + 1.2 + 0.4 = 2.2A.

__Question 157:__What is the current in the 40 Ω resistor of the circuit shown?

A zero B 0.13 A C 0.25 A D 0.50 A

**Reference:**

*Past Exam Paper – June 2014 Paper 13 Q37*

__Solution 157:__**Answer: A.**

The answer to this question can be
reasoned by the symmetry of the circuit.

Current flows from the positive
terminal of the battery.

At the left junction, the current
divides to flow through the 2 resistors and since the resistances of the
resistors are equal, the same current flows through each of them. So, the
potential difference across each resistor is the same.

Therefore, the potential at the top
junction and the potential at the bottom junction of the 40Ω
resistor are the same – that is, the

**potential**across the 40Ω resistor is__difference__**.**__zero__
Thus, no current flows through the
40Ω
resistor.

__Question 158: [Work Done > Efficiency]__Constant force F, acting on car of mass m, moves the car up slope through a distance s at constant velocity v. Angle of the slope to horizontal is α.

Which expression gives efficiency of the process?

A mgs sinα / Fv B mv / Fs C mv

^{2}/ 2Fs D mg sinα / F

**Reference:**

*Past Exam Paper – June 2010 Paper 12 Q15*

__Solution 158:__**Answer: D.**

The force applied (F) does a work (total energy) = Fs.

**Efficiency = useful energy / total energy**

Component of weight against F = mg sinα

For constant velocity, the resultant force is zero. So, the force to move the car up slope should be equal to the component of weight along slope.

Work done to move car a distance s = (mgsinα)s

Efficiency = (mgsinα)s / Fs = mgsinα / F

__Question 159: [Current of Electricity > Resistance]__What is total resistance between points P and Q in this network of resistors?

A 8 Ω B 16 Ω C 24 Ω D 32 Ω

**Reference:**

*Past Exam Paper – June 2014 Paper 11 Q36*

__Solution 159:__**Answer: A.**

The circuit consists of a
parallel combination of 3 parts.

First, consider the point
P, then go up and then to the right. It
can be seen that the 3 resistors of 16Ω,
8Ω and 8Ω are in series with a combined resistance equal to 16 + 8 + 8 = 32Ω.
This is the first part.

Now, if from P, we go to
the right and then up, the 3
resistors of 8Ω,
8Ω and 16Ω are in series with a combined resistance equal to 8 + 8 + 16 = 32Ω.
This is the second part.

So, the 2
‘corners’ (part 1 and 2) of the circuit each
have resistance 32Ω.

The third part is the 16Ω diagonally between P and Q.

Therefore, the circuit consists of 3
parts which are in parallel combination with each other.

Total resistance = [(1/32) + (1/16)
+ (1/32)]

^{-1}= 8Ω

__Question 160: [Dynamics > Momentum > Pressure]__Beam of α-particles collides with lead sheet. Each α-particle in beam has a mass of 6.6 × 10

^{–27}kg and a speed of 1.5 × 10

^{7}m s

^{–1}.

5.0 × 10

^{4}α-particles per second collide with area of 1.0 cm

^{2}of lead. Almost all of the α-particles are absorbed by lead so that they have zero speed after collision.

What is an estimate of average pressure exerted on the lead by the α-particles?

A 5.0 × 10

^{–15}Pa

B 5.0 × 10

^{–13}Pa

C 5.0 × 10

^{–11}Pa

D 5.0 × 10

^{–9}Pa

**Reference:**

*Past Exam Paper – November 2013 Paper 11 & 12 Q11*

__Solution 160:__**Answer: C.**

**Pressure = Total Force / Area**

Let number of particles per second =
N/t = 5.0 x 10

^{4}
Area = 1.0cm

^{2}= 1 x 10^{-4}m^{2}
Mass of one particle, m = 6.6 x 10

^{-27}kg
Velocity, v = 1.5 x 10

^{7}ms^{-1}
Force is defined as the
rate of change of momentum (momentum p = mv)

Force due to 1 particle = mv / t

(Total) Force due to N particles = N
(mv / t) = (N/t) (mv)

Pressure = Total Force/Area =
(N/t)(mv) / Area = 4.95 x 10

^{-11}Pa

__Question 161: [Waves > Stationary waves]__Sound waves, emitted by small loudspeaker, are reflected by wall.

Frequency f of waves is adjusted until stationary wave is formed with the antinode nearest the wall at a distance x from wall.

Which expression gives f in terms of x and speed of sound c?

A f = 4c / x B f = 2c / x C f = c / 2x D f = c / 4x

**Reference:**

*Past Exam Paper – June 2008 Paper 1 Q27*

__Solution 161:__**Answer: D.**

**Speed of wave, c = fλ**

Frequency, f = c / λ

For a stationary wave to form, a node must be present at the wall. The distance between a node and the nearest antinode is equal to λ / 4.

{In these types of questions, a quick diagram may help.}

So, λ / 4 = x giving the wavelength λ = 4x

Frequency = c / λ = c / 4x

__Question 162: [Dynamics > Elastic collision]__Two spheres travel along same line with velocities u

_{1}and u

_{2}. They collide and after collision their velocities are v

_{1}and v

_{2}.

Which collision is not elastic?

**Reference:**

*Past Exam Paper – June 2013 Paper 13 Q11*

__Solution162:__**Answer: A**

For an elastic collision, the
relative speed of approach is equal to the relative speed of separation. (Note that approach means coming towards each other, and
separation means going away from each other.)

In this question, we need to identify which one is NOT elastic.Let the sphere on the back be sphere 1 and the front sphere be sphere 2.

{Before proceeding with any calculations, you need to think how the spheres are moving before and after collision from the signs (+ve or -ve) of the velocities.}

Consider A:

Before collision sphere 1 is moving forward (u

_{1}= 2) while sphere 2 is moving backward (u

_{2}= -5). So, they are approaching.

Relative speed of approach = 2 + 5 = 7ms

^{-1}

After collision, both spheres 1 and 2 are moving backward, but sphere 1 is moving faster.

Relative speed of separation = 5 – 2 = 3ms

^{-1}

So, choice A is correct here

Consider B:

Before collision, both spheres are approaching.

Relative speed of approach = 3 + 3 = 6ms

^{-1}

After collision, sphere 1 is at rest and sphere 2 is moving forward.

Relative speed of separation = 6ms

^{-1}

Consider C:

Relative speed of approach = 3 – 2 = 5ms

^{-1}

Relative speed of separation = 6 – 1 = 5ms

^{-1}

Consider D:

Relative speed of approach = 5 – 2 = 3ms

^{-1}

Relative speed of separation = 6 – 3 = 3ms

^{-1}

Salam!

ReplyDeleteIn question 157, if the resistances of the two 20 ohm resistors had been different, would that have any effect?It shoudn't because the voltage is the same for any two resistors connected in parallel. So this should mean, that whatever the value of the four resistances other than the 40 ohm resistor, the current flowing through the 40 ohm resistor should always be 0.

Yes, but as long as the 2 resistors on ONE side (on the left side to be more precise) of the 40 ohm resistance have the SAME resistance.

Deletebut why would resistors on one side of the 40 ohm resistor with different resistances have any effect on the current flowing through 40 ohm resistor?no matter what the case, the potential difference across the 40 ohm resistor will always be zero, so even if the resistances are different, the current will be zero as p.d. is zero.

Deleteit's explained below

Delete^ no but what if the resistance on the left side of 40 ohm resistor is 20 ohm and 50 ohm. still, each of these get the same potential and so the current across the resistor should be O right?

ReplyDeletethis is not a simple parallel combination of 2 components. It's more complex. On each side of the parallel combination, there is more than 1 component.

DeleteSo, you need to consider the whole thing. one of the 20 ohm and 50 ohm terminals are connected to each other, but the other ones are connected to different points - one to the upper terminal of the 40ohm resistor and the other to the other end of the 40 ohm resistor.

as soon as there is a difference in potential at these 2 points of the 40 ohm resistor, a current would flow

in solution 162, for the speed of separation of each, I dont understand why speed at which it moves decides the sign, isnt the direction all that matters so why would it be 5-2 and 6-1 or 6-3

ReplyDeleteWhat is also (and more) important here is the relative speed of APPROACH (before collision) and SEPARATION (after collision).

Delete'Approach' means coming towards each other. If car A is moving towards car B with speed u1 and car B is moving towards car A with speed u2, the relative speed of approach is u1 + u2, even if the directions are different.

Similarly, 'separation' means going away from each other. We calculate the speed of separation accordingly.

I hope this helped clear your doubts. Try to apply it in the question above by yourself and check the answers again. If you are still having doubts, let me know then, I try to see if I can help.

I still don't understand :/ if you're saying the speead of approach would be u1+u2 even if direction is different then why on the above question we're not doing 2+(-5) and 3+(-3) and 3+(-2). I know it may sound silly but I really don't get it :S

DeleteI said this was an example when they are approaching each other.

DeleteDifferent directions may be in 2 ways.

One way: car A is moving towards car B which is to the right and car B is moving towards car A which is in the left. Here the directions are different but they are approaching each other.

Second way: Car A is on the left and car B is on the right. Car A is moving to the left and car B is moving to the right. Again, their directions of motion are different and now, they are separating.