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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 m². 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 m² m^–1 = (NmC^–1) A^–1 m² m^–1

{Force = ma. Units: N = kg ms^-2. Nm = kg m²s^-2

Charge Q = It. Units: C = As. C^-1 = A^-1s^-1.}

Units of ρ: = (kg m²s^–2) (A^–1s^–1) A^–1 m² m^–1 = kg m³ 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 = V² / 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 = I²R. 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_PN_P > M_QN_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_PN_P

Density of Q = M_QN_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 × 10² pm

{1 pico = 10^-12}

(b)

Time in seconds = 24 × 3600 s

Time in Ms = 0.086 (0.0864) Ms

(c) Time = distance / speed = (1.5×10¹¹) / (3×10⁸) = 500 (s) = 8.3 min

{1 tera = 1T = 10¹²}

(d) distance     energy             momentum                  weight                         work

(e)

(i) An arrow should be drawn to the right of plane direction (about 4° to 24°)

(ii)

EITHER a scale diagram is drawn

OR use of cosine formula v² = 250² + 36² – (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,

v² = 250² + 36² – (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.