# Physics 9702 Doubts | Help Page 21

__Question 109: [Energy > Kinetic energy]__Car driver adjusts the pressure on car’s brakes so that car travels at constant speed down a hill from P to Q.

Magnitude of change in the car’s kinetic energy is ΔE

_{k}. Magnitude of change in its gravitational potential energy is ΔE

_{p}.

Which statement is correct?

A ΔE

_{k}> ΔE

_{p}B ΔE

_{k}= ΔE

_{p}C ΔE

_{p}> ΔE

_{k}> 0 D ΔE

_{k}= 0

**Reference:**

*Past Exam Paper – November 2002 Paper 1 Q17*

__Solution 109:__**Answer: D.**

The question clearly states that the pressure on the car’s brakes is adjusted so that the car travels at

**.**

__constant velocity__Kinetic energy = ½mv

^{2}

If velocity is constant (not changing), the kinetic energy will also be constant. Therefore, the

**change**in kinetic energy ΔE

_{k}is

**zero**.

Additionally, application of the car’s brakes uses the frictional force between the hill and the wheels (work is being done). So, the change in kinetic energy is not equal to the change in potential energy.

__Question 110: [Kinematics > Linear motion]__Graph shows how the acceleration of an object moving in straight line varies with time.

Object starts from rest.

Which graph shows variation with time of velocity of the object over the same time interval?

**Reference:**

*Past Exam Paper – June 2014 Paper 11 Q8*

__Solution 110:__**Answer: A.**

Area under the (acceleration – time) graph represents the velocity. The gradient of the (acceleration – time) graph represents the rate of change [can be either an increase or a decrease] of acceleration.

As the choices deal with velocity – time graphs, we may need to consider the area under the acceleration – time graph, not its gradient {we actually won’t be considering any of these as the question is straight-forward}.

Both acceleration and velocity are vectors. Acceleration is the rate of change of velocity. For a positive acceleration (in a straight line motion as in this case), the velocity increases in the same direction.

Since the acceleration is always positive, it means that the velocity keeps on increasing. [B and C are wrong]

Also, at the end of the time interval being considered, the acceleration is zero. So, the gradient of the velocity – time graph {which gives the acceleration} should become zero at the end of the time interval being considered. [D is incorrect since the gradient is not zero at the end]

__Question 111: [Waves > Stationary Wave]__Horizontal glass tube, closed at one end, has layer of dust laid inside it on its lower side. Sound is emitted from loudspeaker that is placed near the open end of tube.

Frequency of sound is varied and, at one frequency, a stationary wave is formed inside tube so that the dust forms small heaps.

Distance between four heaps of dust is 30 cm.

What is the frequency of sound emitted by loudspeaker?

A 1650 Hz B 2200 Hz C 3300 Hz D 6600 Hz

**Reference:**

*Past Exam Paper – November 2012 Paper 12 Q29*

__Solution 111:__**Answer: A.**

A heap formed from the
dust corresponds to a node (a point with zero displacement).

A wavelength of the sound wave would
form 3 heaps (a wavelength of the sound wave contains 3 nodes – so, the

__distance formed__by 3 successive heaps would be equal to 1 wavelength).
Note that the heaps
are only POINTS and so, the distance between 2 successive heaps actually gives
a distance equal to

**half**the wavelength. Thus, the distance formed between 3 successive heaps is equal to 1 wavelength.
The 30cm distance shown
contains 4 heaps. Label these heaps from 1 to 4.

Distance between 1

^{st}and 3^{rd}heap = 1 wavelength (1λ)
Distance between 3

^{rd}and 4^{th}heap = 0.5 wavelength (0.5λ)
Therefore, 30cm (= 0.3m) corresponds
to (1 + 0.5 =) 1.5λ.

Wavelength of sound wave, λ = 0.3 /
1.5 = 0.2m

Speed of wave, v = fλ

Frequency, f = v / λ = 330 / 0.2 =
1650Hz

__Question 112: [Kinematics > Linear motion]__In order that train can stop safely, it will always pass signal showing a yellow light before it reaches a signal showing red light. Drivers apply brake at the yellow light and this results in uniform deceleration to stop exactly at the red light.

Distance between red and yellow lights is x.

What must be the minimum distance between lights if train speed is increased by 20 %, without changing the deceleration of the trains?

A 1.20 x B 1.25 x C 1.44 x D 1.56 x

**Reference:**

*Past Exam Paper – November 2010 Paper 11 Q9*

__Solution 112:__**Answer: C.**

**For 1**

^{st}case:
Let initial speed = u. Final speed =
0. Acceleration (deceleration) = a. Distance between the lights = x.

Consider the equation for uniformly
accelerated motion:

This becomes: 0 = u**v**^{2}= u^{2}+ 2as^{2}+ 2ax

Since the deceleration is constant in both cases, make it the subject of formula.

Deceleration, a = - (u

^{2}/ 2x)

**For 2**

^{nd}case:Initial speed = 1.2u {increased by 20%}. Final speed = 0. Acceleration (deceleration) = a = - (u

^{2}/ x). Let minimum distance = s.

Considering the same equation for
uniformly accelerated motion:

0 = (1.2u)**v**^{2}= u^{2}+ 2as^{2}+ 2as

Minimum distance, s = - [1.44u

^{2}/ 2a] = [1.44u

^{2}(2x) / 2u

^{2}] = 1.44x

__Question 113: [Current of Electricity > Variable resistor]__Diagram shows potential divider circuit which, by adjustment of contact X, can be used to provide a variable potential difference between the terminals P and Q.

What are the limits of this potential difference?

A 0 and 5 V B 0 and 20 V C 0 and 25 V D 5 V and 25 V

**Reference:**

*Past Exam Paper – November 2011 Paper 12 Q37*

__Solution 113:__**Answer: B.**

Let the resistance of the variable
resistor be R

_{1}and let the resistance of the fixed resistor be R_{2}. R_{1}varies from 0V – 4kΩ and R_{2}= 1kΩ. The e.m.f. in the circuit is V = 25V.
Let the potential difference between
the terminals P and Q be V

_{2}.
From the

**,**__potential divider equation__**V**

_{2}= [R_{1}/ (R_{1}+ R_{2})] x V
The p.d. across a
component is proportional to its resistance. The higher the resistance, the
greater is the p.d.

For V

_{2}to be maximum, the resistance of the variable resistor, R_{1}should be 4kΩ.
Maximum V

_{2}= [4 / (1+4)] x 25 = 20V [A, C and D are incorrect]
For V

_{2}to be minimum, the smallest resistance is chosen. So, here R_{1}is zero {the connections of a variable resistor can be used such as no resistance is taken}.
Minimum V

_{2}= [0 / (1+0)] x 25 = 0V [B is confirmed]

__Question 114: [Matter > Density]__
A mass of liquid of density ρ is
thoroughly mixed with an equal mass of another liquid of density 2ρ. No change
of total volume occurs.

What is the density of the liquid
mixture?

A 4/3ρ B 3/2ρ C
5/3ρ D 3ρ

**Reference:**

*Past Exam Paper – June 2002 Paper 1 Q21*

__Solution 114:__**Answer: A.**

**Density = mass / volume**

Volume = mass / density

**For 1**Let the mass of the 1

^{st}liquid:^{st}liquid = m.

Volume of 1

^{st}liquid = m / ρ**For 2**mass = m and density = 2ρ

^{nd}liquid:
Volume of 2

^{nd}liquid = m / 2ρ**For mixture:**

Total volume of mixture = (m / ρ) +
(m / 2ρ) = 3m / 2ρ

Total mass of mixture = m + m = 2m

Density of mixture = mass / volume =
2m / [3m / 2ρ] = 4ρ / 3

__Question 115: [Energy > Potential energy]__Bow of mass 400 g shoots arrow of mass 120 g vertically upwards. Potential energy stored in the bow just before release is 80 J. System has an efficiency of 28%.

What is the height reached by arrow when air resistance is neglected?

A 4 m B 19 m C 187 m D 243 m

**Reference:**

*Past Exam Paper – June 2013 Paper 12 Q16*

__Solution 115:__**Answer: B.**

Initial (elastic) potential energy stored = 80J

Efficiency = 28% = 0.28

Energy available for arrow = 0.28x80 = 22.4J.

Only the arrow will be moving upwards, so we consider the mass of arrow only. Since air resistance is negligible, as the arrow moves upwards, its energy is converted to gravitational potential energy.

At maximum height h,

Gravitational potential energy = mgh = 22.4

Height h = 22.4 / (0.12x9.81) = 19.0m

Salame admin,

ReplyDeleteIn as 133 how do I know and for other exam qs...how do I know that R1 is a variable resistor..? Can you explain how moving the contact can make R1 have 0 resistance leaving total resistance to 1 ohms.

Thank you

Wslm.

DeleteThe resistor with an arrow (the moving jack) at the middle is itself the symbol for a variable resistor.

When the jack is at one end, all the resistance is being considered and at the other end, no resistance is being considered. It's basically how a variable resistor works.

I remember explaining it in greater details but have forgotten for which solution it was.

Thank you so much for all the help. :)

ReplyDelete