# Physics 9702 Doubts | Help Page 36

__Question 221: [Kinematics > Linear motion]__On a particular railway, train driver applies brake of the train at yellow signal, a distance of 1.0 km from a red signal, where the train stops.

Maximum deceleration of train is 0.20 m s

^{–2}.

Assuming uniform deceleration, what is maximum safe speed of train at the yellow signal?

For November 2009:

A 20 m s

^{–1}B 40 m s

^{–1}C 200 m s

^{–1}D 400 m s

^{–1}

For November 2013:

A 14 m s

^{–1}B 20 m s

^{–1}C 40 m s

^{–1}D 400 m s

^{–1}

**Reference:**

*Past Exam Paper – November 2009 Paper 11 Q5 & November 2013 Paper 13 Q8*

__Solution 221:__**For Nov 2009 Paper 11 Q5 – Answer: A.**

**For Nov 2013 Paper 13 Q8 – Answer: B.**

Consider the equation for uniformly accelerated motion: v

^{2}= u

^{2}+ 2as

Final velocity, v = 0

Acceleration, a = -0.2ms

^{-2}

Distance, s = 1000

Initial velocity, u = ???

So, 0 = u

^{2}+ 2(-0.2)(1000) giving u = 20ms

^{-1}

__Question 222: [Forces > Electric field]__Tiny oil droplet with mass 6.9 × 10

^{–13}kg is at rest in electric field of electric field strength 2.1 × 10

^{7}N C

^{–1}, as shown.

Weight of the droplet is exactly balanced by the electrical force on droplet.

What is the charge on droplet?

A 3.3 × 10

^{–20}C

B –3.3 × 10

^{–20}C

C 3.2 × 10

^{–19}C

D –3.2 × 10

^{–19}C

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q12*

__Solution 222:__**Answer: D.**

The weight of the droplet acts
downwards. For the droplet to be at rest, the electric force on the droplet
should act upwards. Electric field is from positive to negative. So, the
droplet is attracted towards the positive upper plate. The drop should be
negatively-charged so that it moves upwards, against the force of gravity.

The electric force F should
be equal in magnitude to the weight of the droplet (F = mg).

Electric field strength, E = F / Q

Charge, Q = F / E = mg / E = (6.9x10

^{-13}x 9.81) / 2.1x10^{7}= 3.2x10^{-19}C

__Question 223: [Forces > Moment]__
Uniform beam is pivoted at P as shown.
Weights of 10 N and 20 N are attached to its ends.

Length of beam is marked at 0.1 m
intervals. Weight of the beam is 100 N.

At which point should further weight
of 20 N be attached to achieve equilibrium?

**Reference:**

*Past Exam Paper – November 2005 Paper 1 Q12 & June 2014 Paper 13 Q15*

__Solution 223:__**Answer: D.**

Since the beam is uniform, its
weight acts at the centre, that is, 0.1m to the left of P (where B is).

For equilibrium, the sum
of clockwise moment should be equal to the sum of anticlockwise moment.

Clockwise moment = 20(0.4) Nm

Anticlockwise moment = 10(0.6) +
100(0.1) Nm

Since the anticlockwise moment is
greater, the further weight of 20N should contribute to the sum of clockwise
moments. That is, it should be to the
right of P. Let the position of the further weight of 20N be x to the right of
P.

Sum of clockwise moment = Sum of
anticlockwise moment

20(0.4) + 20(x) = 100(0.1) + 10(0.6)

Position x = 0.4m

__Question 224: [Waves > Phase difference]__Frequency of a certain wave is 500 Hz and its speed is 340 m s

^{–1}.

What is the phase difference between motions of two points on the wave 0.17 m apart?

A Ï€ / 4 rad B Ï€ / 2 rad C 3Ï€ / 4 rad D Ï€ rad

**Reference:**

*Past Exam Paper – June 2006 Paper 1 Q25*

__Solution 224:__**Answer: B.**

Speed, v = fÎ»

Wavelength, Î» = v / f = 340 / 500 =
0.68m

The wavelength Î»
corresponds to a phase difference of 2Ï€ rad (= 360 deg) – that is, 2 points
separated by a distance equal to Î» will have a phase difference of 2Ï€ rad –
they are in phase.

Therefore,

(Separation of Î» =) 0.68m
corresponds to phase difference of 2Ï€ rad

Separation of 0.17m will correspond to (0.17 / 0.68) x 2Ï€ = Ï€ / 2 rad

__Question 225: [Measurements > CRO]__Y-input terminals of cathode-ray oscilloscope (c.r.o.) are connected to supply of peak value 5.0 V and of frequency 50 Hz. Time-base is set at 10 ms per division and Y-gain at 5.0 V per division.

Which trace is obtained?

**Reference:**

*Past Exam Paper – November 2006 Paper 1 Q4*

__Solution 225:__**Answer: D.**

Y-gain is set at 5.0V per division.
Since the peak value of the supply is 5.0V, the amplitude should occupy only 1
division. [A and B are incorrect]

Time-base setting is 10ms per
division.

Frequency = 1 / Period

Period = 1 / 50 = 0.02s = 20ms

So, the period should occupy 2
divisions. [C is incorrect]

__Question 226: [Dynamics > Conservation of Linear momentum]__
A stationary thoron nucleus (A = 220,
Z = 90) emits an alpha particle with kinetic energy E

A E_{Î±}. Which is the kinetic energy of the recoiling nucleus?_{Î±}/ 108 B E

_{Î±}/ 110 C E

_{Î±}/ 54 D E

_{Î±}/ 55 E E

_{Î±}

{Note that the relative atomic mass of thoron, A is 220, not 200 as given in some books for this question}

**Reference:**

*Past Exam Paper – J84 / II / 37*

__Solution 226:__Answer: C.

A is the relative atomic mass and Z is the proton number.

Since the thoron nucleus is initially stationary (its speed is zero), it has zero momentum. From the conservation of momentum, the sum of momentum after the emission of the alpha particle should also be zero.

An alpha particle has A = 4 and Z = 2.

Mass of alpha particle = 4u where u is the unified atomic mass.

Let the velocity of the alpha particle = v

Momentum = mass x velocity

Kinetic energy = ½ mv

^{2}

Momentum of alpha particle = 4u(v) = 4uv

Kinetic energy of alpha particle = 0.5 (4u) v

^{2}= 2uv

^{2}= E

_{Î±}

The recoiling nucleus has A = 220 – 4 = 216 and Z = 90 – 2 = 88

Mass of recoiling nucleus = 216u

Let the velocity of the recoiling nucleus = y

Momentum of recoiling nucleus = 216u (y) = 216uy

From conservation of momentum,

Momentum of recoiling nucleus + Momentum of alpha particle = Momentum of thoron nucleus

216uy + 4uv = 0

Velocity, y = - 4v / 216 = - v/54

(The negative velocity means that the recoiling nucleus moves in a direction opposite to the particle particle)

Kinetic energy of recoiling nucleus = 0.5 (216u) (- v/54)

^{2}= uv

^{2}/ 27 = 2uv

^{2}/ 54

Since E

_{Î±}= 2uv

^{2},

Kinetic energy of recoiling nucleus = E

_{Î±}/ 54

__Question 227: [Measurements > Estimates]__Which estimate is realistic?

A Kinetic energy of a bus travelling on an expressway is 30 000 J.

B Power of a domestic light is 300 W.

C Temperature of a hot oven is 300 K.

D Volume of air in a car tyre is 0.03 m

^{3}.

**Reference:**

*Past Exam Paper – November 2013 Paper 13 Q1*

__Solution 227:__**Answer: D.**

The kinetic energy of the bus might
well be about 10

^{6}J (KE = ½ mv^{2}. Consider the mass of the bus and the square of its velocity on an expressway). The powers of domestic lights are usually around 60 W (often less) and an oven at 300 K is not hot (this temperature is 300 – 273 = 27 deg).
The correct answer, D, indicates
just how large a cubic metre is. By making a rough estimate of the size of a
tyre and the volume could be determined from it.

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