# Physics 9702 Doubts | Help Page 96

__Question 495: [Pressure]__Horizontal plate of area 0.036 m

^{2}is beneath the surface of liquid of density 930 kg m

^{–3}. Force on the plate due to the pressure of liquid is 290 N.

What is the depth of plate beneath the surface of the liquid?

A 0.88 m B 1.13 m C 8.7 m D 9.1 m

**Reference:**

*Past Exam Paper – November 2011 Paper 12 Q20*

__Solution 495:__**Answer: A.**

Pressure P = hρg

where h is the depth, ρ is the
density of the liquid and g is the acceleration of free fall.

But,

Pressure P = Force / Area = F / A

Therefore,

Depth h = P / ρg = F / (Aρg) = 290 /
[0.036 (930) (9.81)] = 0.88m

__Question 496: [Waves]__Diagram shows a sketch of wave pattern, over a short period of time.

Which description of this wave is correct?

A The wave is longitudinal, has a wavelength of 20 cm and is stationary.

B The wave is transverse, has a wavelength of 20 cm and is stationary.

C The wave is transverse, has a wavelength of 40 cm and is progressive.

D The wave is transverse, has a wavelength of 40 cm and is stationary.

**Reference:**

*Past Exam Paper – November 2013 Paper 11 & 12 Q25*

__Solution 496:__**Answer: D.**

As stated, the sketch is that of a
wave pattern over a period of time. So, the sketch contains different waves at
different times in a single diagram.

The wave is transverse since the displacement
is vertical while the direction of travel is horizontal. Longitudinal waves are
waves in which the displacement of the medium is the same direction as, or the
opposite direction to, the direction of travel of the wave. [A is incorrect]

The 1.0m distance indicated
consisted of 2.5 waves. (2 ‘loops’ represent 1 complete wave)

2.5λ = 100cm

Wavelength λ =
40cm [B
is incorrect]

Over the period of time, it can be
seen that the nodes (points of zero displacement) remain still. Therefore, the
wave is not travelling forward or backward. The wave is stationary. [C is incorrect]

__Question 497: [Kinematics > Linear motion]__A goods train passes through station at steady speed of 10 m s

^{–1}. An express train is at rest at station. The express train leaves station with a uniform acceleration of 0.5 m s

^{–2}just as the goods train goes past. Both trains move in same direction on straight, parallel tracks.

How much time passes before the express train overtakes the goods train?

A 6 s B 10 s C 20 s D 40 s

**Reference:**

*Past Exam Paper – June 2013 Paper 11 Q8*

__Solution 497:__**Answer: D.**

Let the initial position of the
express train (when it is a rest) be the origin, from which the distance
travelled will be considered. Since the express train leaves the station just
as the goods train passes it, this moment is taken as the origin for time.

For the express train to overtake
the goods train, the distance travelled by each of the trains must be equal.
Let this distance = s.

Consider the goods train.

Let the time taken for the train to
travel distance s be t.

Speed = Distance / Time

Distance, s = 10t

Consider the express train.

The express train will overtake the
goods train at time t, since the express train leaves the station as soon as
the goods train passes it.

Acceleration, a = 0.5 ms

^{-2}
Initial speed, u = 0 ms

^{-1}
Distance s, = ut + ½ at

^{2}= 0 + 0.5 (0.5) t^{2}
Since the distance s is the same for
both trains,

10t = 0.25t

^{2}
Time t = 10 / 0.25 = 40s

__Question 498: [Work, Energy and Power]__Box of weight 200 N is pushed so that it moves at a steady speed along ramp, through a height of 1.5 m. Ramp makes an angle of 30° with the ground. Frictional force on the box is 150 N while the box is moving.

What is the work done by the person?

A 150 J B 300 J C 450 J D 750 J

**Reference:**

*Past Exam Paper – November 2010 Paper 12 Q14*

__Solution 498:__**Answer: D.**

Work done = Force x distance moved
in direction of force

Distance travelled along the ramp =
1.5 / sin30 = 3m

Component of weight along ramp =
200sin30 = 100N

This component of the weight, along
with the frictional force opposes the motion. But since the box moves at a
steady speed, the resultant force on the box is zero (there is no acceleration).
This means that that the person is providing a force equal to the sum of these
2 forces.

Force provided by person = 100 + 150
= 250N

Work done = 250 x 3 = 750J

__Question 499: [Work, Energy and Power]__Van driver adjusts the force on van’s brakes so that the van travels at constant speed down a hill from P to Q.

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

_{k}. Magnitude of the 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 2014 Paper 13 Q19*

__Solution 499:__**Answer: D.**

As stated in the question, ΔE

_{k}is change in the van’s kinetic energy and ΔE_{p}is the change in the van’s gravitational potential energy.
Kinetic energy = ½ mv

^{2}
Since the van travels at constant speed,
its kinetic energy is constant, that is it does not change. So, ΔE

_{k}= 0.Only the potential energy changes as the height of the van above the ground changes.

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