# Physics 9702 Doubts | Help Page 131

__Question 655: [Pressure]__
Graph shows how pressure exerted by
a liquid varies with depth below the surface.

What is density of the liquid?

A 600 kgm

^{–3}B 760 kgm^{–3}C 5900 kgm^{–3}D 7500 kgm^{–3}**Reference:**

*Past Exam Paper – November 2003 Paper 1 Q19*

__Solution 655:__**Answer: B.**

Pressure P = hρg

Density ρ = P / hg = (P/h) / g

From the graph (consider points
(0.4, 3000) and (0, 0)),

Graph = (3000 – 0) / (0.4 – 0) =
7500

Density = 7500 / 9.81 = 760 kgm

^{-3}

__Question 656: [Current of Electricity]__**(a)**Define the

*ohm*.

**(b)**Determine SI base units of

**resistivity**.

**(c)**Cell of e.m.f. 2.0 V and negligible internal resistance is connected to a variable resistor R and a metal wire, as shown in Fig.1.

Wire is 900 mm long and has area of
cross-section of 1.3 × 10

^{–7}m^{2}. Resistance of the wire is 3.4 Ω.
(i) Calculate resistivity of the
metal wire.

(ii) Resistance of R may be varied
between 0 and 1500 Ω.

Calculate maximum potential
difference (p.d.) and minimum p.d. possible across the wire.

(iii) Calculate power transformed in
the wire when the potential difference across the wire is 2.0 V.

**(d)**Resistance R in (c) is now replaced with a different variable resistor Q. State power transformed in Q, for Q having

(i) zero resistance,

(ii) infinite resistance.

**Reference:**

*Past Exam Paper – November 2011 Paper 21 Q5*

__Solution 656:__**(a)**Ohm = volt / ampere

**(b)**

Resistivity ρ = RA / l or unit is Ω m

{Voltage = Energy /
Charge. Energy (= work done) = Force × distance.

Units: V = NmC

^{-1}}
Units of ρ: VA

^{–1}m^{2}m^{–1}= (NmC^{–1}) A^{–1}m^{2}m^{–1}
{Force = ma. Units: N = kg
ms

^{-2}. Nm = kg m^{2}s^{-2}
Charge Q = It. Units: C =
As. C

^{-1}= A^{-1}s^{-1}.}
Units of ρ: = (kg
m

^{2}s^{–2}) (A^{–1}s^{–1}) A^{–1}m^{2}m^{–1}= kg m^{3}s^{–3}A^{–2}**(c)**

(i) Resistivity ρ = [3.4 × 1.3 × 10

^{–7}] / 0.9 = 4.9 × 10^{–7}(Ω m)
(ii)

{Resistance R = 0 Ω} Maximum p.d. = 2.(0) V

{Resistance R = 1500 Ω} Minimum p.d. = 2 × (3.4 /1503.4) = 4.5 × 10

^{–3}V
(iii)

Power P = V

^{2}/ R or P = VI__and__V = IR
Power P = (2)

^{2}/ 3.4 = 1.18 (allow 1.2) W**(d)**

(i) Power in Q is zero when R = 0

{P = I

^{2}R. When R = 0, P = 0}
(ii) Power in Q = 0 / tends to zero
as R = infinity

{As the total resistance
in the circuit increases, the current decreases. I = V / R. So, if R =
infinite, current I = 0}

__Question 657: [Matter > Density]____June 2004 Paper 1 Q21 & November 2007 Paper 1 Q17:__

Two solid substances P and Q have
atoms of mass M

_{P}and M_{Q}respectively. They have N_{P}and N_{Q}atoms per unit volume.
It is found by experiment that
density of P is greater than that of Q.

Which of the following deductions
from this experiment must be correct?

A M

_{P}> M_{Q}
B N

_{P}> N_{Q}
C M

_{P}N_{P}> M_{Q}N_{Q}
D M

_{P }/ N_{P}> M_{Q }/ N_{Q}__June 2011 Paper 12 Q21:__

Two solid substances P and Q have
atoms of mass M

_{P}and M_{Q}respectively. They have n_{P}and n_{Q}atoms per unit volume.
The density of P is greater than the
density of Q.

What

**must**be correct?
A M

_{P}> M_{Q}
B n

_{P}> n_{Q}
C M

_{P }n_{P}> M_{Q }n_{Q}
D M

_{P }/ n_{P}> M_{Q }/ n_{Q}**Reference:**

*Past Exam Paper – June 2004 Paper 1 Q21 & November 2007 Paper 1 Q17 & June 2011 Paper 12 Q21*

__Solution 657:__**Answer: C.**

Mass of 1 atom of P and Q are M

_{P}and M_{Q}respectively.
Number of atoms per unit volume of P
and Q are N

_{P}and N_{Q}respectively.
Total mass per unit volume = MN

Density = total mass / volume

Density of P = M

_{P}N_{P}
Density of Q = M

_{Q}N_{Q}
The density of P is greater than
that of Q.

__Question 658: [Measurement]__**(a)**Spacing between two atoms in a crystal is 3.8 × 10

^{–10}m. State this distance in pm.

**(b)**Calculate time of one day in Ms.

**(c)**The distance from Earth to the Sun is 0.15 Tm. Calculate time in minutes for light to travel from the Sun to the Earth.

**(d)**Underline all the vector quantities in list below.

distance energy momentum
weight work

**(e)**Velocity vector diagram for an aircraft heading due north is shown to scale in Fig.1. There is a wind blowing from the north-west.

Speed of the wind is 36 m s

^{–1}and speed of the aircraft is 250 m s^{–1}.
(i) Draw an arrow on Fig.1 to show
direction of the resultant velocity of the aircraft.

(ii) Determine magnitude of the
resultant velocity of the aircraft.

**Reference:**

*Past Exam Paper – November 2012 Paper 23 Q1*

__Solution 658:__**(a)**Spacing in pm= 380 or 3.8 × 10

^{2}pm

{1 pico = 10

^{-12}}**(b)**

Time in seconds = 24 × 3600 s

Time in Ms = 0.086 (0.0864) Ms

**(c)**Time = distance / speed = (1.5×10

^{11}) / (3×10

^{8}) = 500 (s) = 8.3 min

{1 tera = 1T = 10

^{12}}**(d)**distance energy

__momentum__

__weight__work

**(e)**

(ii)

EITHER a scale diagram is drawn

OR use of cosine formula v

^{2}= 250^{2}+ 36^{2}– (2×250×36×cos 45°)
OR resolving v = [(36 cos 45°)

^{2}+ (250 – 36 sin 45°)^{2}]^{1/2}
{I recommend using
calculation to solve this question as the answer obtained will be more accurate
(+ it has not been imposed to use vector diagram here).

From the laws of angles,
the angle between the wind direction and the aircraft motion is 45°. The
direction if the wind is from north-west, so the wind will always be at 45° to both
the vertical and the horizontal.

From the cosine formula,

v

^{2}= 250^{2}+ 36^{2}– (2×250×36×cos 45°)
Resultant velocity v = 226
ms

^{-1}}
{Alternatively, this can
be solved by resolving the perpendicular components of the vectors involved and
then finding the resultant vector.

The motion of the aircraft
is shown to be vertically up, so it has no horizontal component.

Horizontal component of
wind (to the right) = 36 cos 45°

Vertical component of win
(down) = 36 sin45°

Resultant vertical component
of velocity = 250 – 36sin45°

Resultant horizontal
component of velocity = 36cos45°

Resultant velocity vector v
= [(36 cos 45°)

^{2}+ (250 – 36 sin 45°)^{2}]^{1/2}}
Resultant velocity = 226 m s

^{–1}

__Question 659: [Matter > Phases of matter]__
Student writes some statements about
solids, liquids and gases.

1 Solids are rigid because the
molecules in a solid vibrate.

2 Liquids flow because the molecules
in a liquid are closer than in a gas.

3 Gases are less dense than liquids
because the molecules in a gas move randomly.

Which statements are correct?

A 1 only

B 1 and 3 only

C 2 and 3 only

D none of the above

**Reference:**

*Past Exam Paper – November 2010 Paper 11 Q20 & Paper 13 Q22*

__Solution 659:__**Answer: D.**

None of the above statements are
correct.

Solids are rigid because the atoms /
molecules in the solids are closely packed together and are held together by
strong forces of attraction.

Both liquids and gases (fluids) can
flow. The reason for the flow is not because the molecules in a liquid are closer
than in a gas.

Gases are less dense than liquids
because the molecules in a liquid are closer to each other. So, in an equal
volume of a gas and a liquid, the amount of molecules (matter) in a liquid is
greater.

We can't use this formula /V^2=250^2+26^2.../ cuz velocity's vector isn't hypotenuse

ReplyDeleteit's the cosine formula

Delete