Quiz 14 of 14

2014 Physics WAEC Theory Past Questions

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Question 1

A particle is projected horizontally at 10ms-1 from the top of a tower 20m high. Calculate the horizontal distance travelled by the particle when it hits the level ground. [g = 10ms-2]

Solution

R = \( \scriptsize U \normalsize \sqrt{ \frac{2H}{g}} \)

R = \( \scriptsize 10 \normalsize \sqrt{ \frac{2 \: \times \: 20}{10}} \)

R = \( \scriptsize 10\: \times \: 2\)

= 20m

Question 2

A tennis ball projected at an angle θ attains a range R = 78m. If the velocity imparted to the ball by the racket is 30ms-1, calculate θ. [g = 10ms-2]

Solution 

Recall that:-

R = \( \frac{u^2sin2 \theta}{g} \)

Information given:- Range, R= 78m, acceleration due to gravity,g = 10ms-2,

θ = ?, U = 30ms-1

R = \( \frac{30^2 sin2 \theta}{10} \)

sin2θ =  \( \frac{780}{900} \)

sin2θ= 0.8667

2θ = sin-10.8667 = 60°

θ = \( \frac{60^o}{2} \)

θ = 30°

Question 3

An electron moves with a speed of 2.00 x 107ms-1in an orbit in a uniform magnetic field of 1.2 x 10-3 Calculate the radius of the orbit.
[Mass of an electron = 9.11 x 10-31kg, charge on an electron = 1.61 x 10-19C]

Solution

Remember that \(\frac{Mv^2}{r} \scriptsize = qvB \)

That is, centripetal force = magnetic field force

Making r the subject of the formula

r = \( \frac{mV^2}{Bvq} \)

∴ r = \( \frac{mV}{Bq} \)

r = \( \frac{9.11 \: \times \: 10^{-31} \: \times \: 2 \: \times \: 10^7}{1.2 \: \times \: 10^{-3} \: \times \: 1.61 \: \times 10^{-19}} \)

r = \( \scriptsize 9.43 \: \times \: 10^{-2}m\)

Question 4

A metallic bar 50cm long has a uniform cross-sectional area of 4.0cm2. If a tensile force of 35KN produces an extension of 0.25mm, calculate the value of young's modulus.

Answer

Remember that:

Young's modulus ( E ) =  \( \frac{\frac{F}{A}}{\frac{e}{l}} \)

F = 35KN, A = 4cm2 = 0.04m2, e = 0.25mm = 0.00025m

l = \( \frac{50}{100} \scriptsize = 0.5m \)

Area, A = 4.0cm2 = \( \frac{4}{100^2} \scriptsize = 0.0004m \)

Tensile Force, F = 35N

extension, e = 0.25mm = \( \frac{0.25}{1000m} \scriptsize = 0.00025m\)

Young modulus ( E ) =  \( \frac{\frac{F}{A}}{\frac{e}{l}}  \\ = \frac{F \: \times \: l}{A \: \times \:  e} \\ = \frac{35 \: \times \: 0.5}{0.0004 \: \times \:  0.00025} \\ \scriptsize = 1.75 \: \times \: 10^{11} Nm^{-2} \)

Question 5

(a) Explain how a gas can be made to conduct electricity.

Answer

The molecules /atom of gas must be ionized before the gas can conduct electricity. Ionisation of the gas requires that the gas pressure is very low within the closure and a very high voltage is applied to the enclosed gas.

 

(b)Name the electric charge carriers in gases.

Answer

  The electric charge carriers in gases are the

- electrons
- ions

Question 6

The diagram above represents the graph of electron energy against the frequency of the radiation incident on a metal surface. Interpret the:

(a) slope of the graph;

Answer

The slope of the graph is the Planck's constant, (h)

(b) intercept, OC;

Answer

Intercept, OC represents the work function (f) of the metal

(c) intercept, OK.

Answer

Intercept OK, depicts the threshold frequency, fo. of the incident's radiation

Question 7

(a) State two conditions under which photo-electrons can be emitted from the surface of a metal.

Answers

- The wavelength of the incidents radiation must be greater than the threshold wavelength for metal surfaces
- The frequency of the incidents radiation must be greater than the threshold frequency
- The energy of the incident radiation must be greater than the work function of the metal

 

(b) List two-particle characteristics of electromagnetic waves.

Answers

- Photoelectric effect
- Thermionic emission
- Radiations from the heated black bodies
- Emission and absorption of light [incandescence]
- Compton effects

Question 8

(a) (i) Give two examples of rotational motion

Answer

- The rotation of the earth about its axis.
- A cricket ball spinning about its axis.
- The wheel of a moving car.
- The rotation of the blades of an electric fan.

(ii) Give two examples of linear motion

Answer

- An aeroplane moving on the runway.
- A bee flying from one flower to another.
- A ball rolling.

(iii) Describe a laboratory experiment to determine the density of an irregularly shaped solid.

Answer

The mass m of the solid is determined using a chemical/beam balance, a graduated measuring cylinder is partially filled with water and the initial volume V1 is recorded the solid is completely immersed in the water and the final volume V2 is recorded.

Volume of the solid = V2 - V1

Density of the solid \( \scriptsize \rho = \normalsize \frac{m}{V_2 \: - \: V_1} \)

 

(c) State Newton's second law of motion

Answer

Newton's second law of motion states that the rate of change of momentum is directly proportional to the impressed force and takes place in the direction of that force.

 

(d) Explain the term inertia

Answer

Inertia is the term given to the tendency of an object in uniform motion to remain in this state of motion, or an object at rest to remain at rest.

- the more mass a body has ,the greater is its inertia

(e) (i)

The diagram above illustrates a body of mass 5.0kg being pulled by a horizontal force F. If the body accelerates at 2.0ms-2 and experiences a frictional force of 5N, calculate the

[g = 10ms-2]

(i)  net force on it

Answer

Information given; Mass = 5kg, Acceleration, a = 20ms-2, Force, F1 = 5N, F = ?

Net force = F - F1 = Ma

= 5 x 2 = 10N

 

(ii) magnitude of F

Answer

Magnitude of F = F1 + net force (ma)
= 5 + 10 = 15N

 

(iii) coefficient of kinetic friction.

Answer

Coefficient of kinetic friction
from the formula
F1 = μR
∴   μ =  \(\frac{F_1}{R} \)

μ = coefficient of kinetic friction

F1 = Frictional force

R = reaction = mg = 5 x 10 = 50N

μ  = \( \frac{5}{50} \)

∴  μ = 0.1

Question 9

(a) Define heat capacity and state its unit.

Answer

Heat capacity of a substance is the quantity of heat required to raise the temperature of unit mass of the substance by 1oC (or 1K). The unit is Joules per kilogram per Kelvin (J/kg.K)

 

(b) List two effects of heat on a substance.

Answer

- It leads to expansion of a substance /contraction

- It leads to change of state

- It leads to a change in colour

- It leads to a change in electrical resistance

 

(c) Explain how a tightly fitted glass stopper could be removed from a reagent bottle.

Answer

The neck of the glass bottle should be warmed gradually from outside using warm cloth this makes the glass to expand creating space for the cock to be removed.

 

(d) A quantity of pepper soup of mass 800g poured into a plastic container with a tight-fitting lid has a temperature of 30oC. The container is then placed in a microwave oven rated 1200W and operated for 3 minutes.

(i) Calculate the final temperature attained by the soup. (Assuming no heat losses)

Answer

Information given; mass = 800g = 0.8kg, θ1 = 30oc, θ2 = ?
θ = θ2 - θ1
Power = 1200W,  c = 4000JKg-1K-1
time = 3 mins = (3 x 60) = 180sec
It should be noted that; Heat gained by the pepper soup = mcθ

power = Heat/time = energy/time

∴ Heat = power x time

= 1200 x 180 = 216,000 = mcθ

∴ θ = \( \frac{216000}{mc} \)

θ = \( \frac{216000}{0.8 \: \times \: 4000} \scriptsize = 67.5^0C \)

Final temperature = 67.5 + 30 = 97.5oC

(ii) Explain why containers with tight-fitting lids are not suitable for use in microwave cooking.

Answer

The molecules of the the steam from the content are confined within the container as the temperature increases the rate of collision of the molecules with walls of the container increases thus the pressure inside the container increases If it is tightly fitted ,the pressure can can make the container burst thus releasing its hot content.

(iii) When the soup is brought out and allowed to cool, a dent is observed on the container. Explain

Answer

The container will be observed with a dent because the condensation of the steam in the container will lead to a decrease in pressure in the container making the atmospheric pressure greater than the pressure in the container

Question 10

(a) State the three characteristics of sound and the factor on which each of them depends.

Answer

- Pitch depends on the frequency of the soundwave
- Loudness depends on the mass/amplitude of the air which can be set into vibration
- Quality/timbre depends on the overtones/harmonics present in the note.

(b) Explain resonance as applied to sound

Answer

Resonance is a phenomenon which occurs whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses or signals, received from some other system or body which is vibrating with the same frequency.
When the imposed frequency is the same as the natural frequency, the vibration build-up to a large amplitude. Therefore, Resonance is an effect caused by a vibrating body setting another body vibrating both having the same natural frequency. Resonance is what happens when a book placed on a surface is blown by a rotating fan.

(c) What role does echo play in the construction of a concert hall?

Answer

Echo sensitizes the construction engineers on the need to line the interior wall with appropriate sound-absorbing materials.

(d) The surface of an ear drum (assumed circular) has a radius 2.1 mm. It resonates with an amplitude of 0.8 x 10-7m as a result of impulses received from an external body vibrating at 2400 Hz. If the resulting pressure change on the ear drum is 3.6 x 10-5NM-2,calculate the:

(i) period of oscillation;

Answer

Period of oscillation T = \( \frac{1}{F} \)

T = \( \frac{1}{2400} \)

= 0.00042s

(ii) velocity;

Answer

Velocity = V = fλ = 2400 x 0.8 x 10-7
= 2.4 x 10³ x 0.8 x 10 -7
= 1.2 x 104-7
= 1.2 x 10 -3m/s

(iii) acceleration;

Answer

Acceleration
a =w2 r = (2πf)2r , here r = Amplitude
= (2 x 3.14 x  2400)2 x 0.8 x 10-7

= 18.2 ms -2

(iv) force

Answer

To calculate force

Force = Pressure x Area

radius = 2.1mm = 0.0021m

Area of circle = 2πr = 3.14 x (0.0021)2

Area = 1.38 x 10-5m 2

Pressure = 3.6 x 10-5NM-2

Force = 3.6 x 10-5 x 1.38 x 10-5

= 4.968 x 10(-5 - 5)

= 4.968 x 10-10N

Question 11

(a) Define electromotive force.

Answer

Electromotive force of a cell is the energy needed to take a unit positive charge around a circuit in which the cell is connected

- the p.d across the terminal of the cell when the cell/battery/generator is on open circuit

-the energy per unit charge transformed in the cell.

 

(b) (i) State the principle of operation of a potentiometer.

Answer

- The p.d of a length of wire is directly proportional to the length of the wire

-when a steady current flows through a uniform wire, equal lengths of the wire will have equal potentials

(ii) State two advantages that a potentiometer has over a voltmeter in measuring the potential difference.

Answer

- It is very accurate/has no errors
- It is more sensitive
- It eliminates errors due to the internal internal resistance of the cell

 

(c) (i) Sketch and label a diagram of a gold-leaf electroscope

Answer

Gold leaf electroscope

(ii) Give one use of a gold-leaf electroscope.

Answer

-Detects the [electrostatic] charge
-Detects the nature of [electrostatic] charge
-Determines the magnitude of the induced charges

 

(d) (i) Explain the action of a magnetic relay.

Answer

The mode of action of magnetic relay relies on the principle of electromagnetic induction, a small current passing through one circuit controls another circuit containing devices which require larger current.

(ii) List two factors which determine the magnitude of an induced emf in a coil.

Answer

- Speed of rotation of the coil
- Cross section area of the coil
- Number of turns

(iii) A current of 5 A passes through a straight wire in a uniform magnetic field of flux density 2.0 x 10-3 T. Calculate the force per unit length exerted on the wire when it is inclined at 30° to the field.

Answer

Information provided

B = \( \scriptsize 2.0 \: \times \: 10^{-3} \)

I = 5A

θ = 30º

L = ?

F = ?

From the formula

F = BIL sinθ

To get force per unit length, divide through by L

= \( \frac{F}{L} = \frac{BILsin \theta}{L}  \)

= \( \scriptsize BIsin \theta \)

Force per unit length =  \( \frac{F}{L}  \)

= \( \scriptsize 2.0 \: \times \: 10^{-3} \: \times \: 5sin30^o \)

= \( \scriptsize 10 \: \times \: 10^{-3} \: \times \: 0.5\)

= \( \scriptsize 5 \: \times \: 10^{-3} Nm^{-1}\)

Question 12

(a) Write Einstein's photoelectric equation and identify each component of the equation.

Answer

Einsten photoelectric equation is given as: E = W0  +  KEmax
E = Energy of radiation incidence
W0 = Work function
K.Emax = maximum kinetic energy of photoelectron

(b) For a photocell, state, one factor each that is responsible for the:

(i) emission

Answer

Emission - frequency/wavelength of incident radiation

(ii) rate of emission;

Rate of emission - intensity of incident radiation /number of protons

Answer

(iii) energy of photoelectrons.

Answer

Energy of photoelectrons - frequency/wavelength of incident radiation

 

(c)(i) Two nuclear equations are given below:

\( \scriptsize ^{222}_{p}RN \to \:\: ^{218}_{84}Po + \:\: ^q_2He\)  ...................A

\(\scriptsize  ^{214}_{83}RN \to\:\: ^{214}_{84}Po + \:\: ^m_nX \) ....................B

Determine the values of:

(α) p and q in equation A;

222pRn → 21884Po + 92He

p = 84 + 2 = 86

218 + q = 222

q = 222 - 218 = 4

(β) m and n in equation B and identify X.

21483Bi → 21483Po + mnX

m = 214 - 214 = 0; m = 0
n = 83 - 84 = -1;

n = -1;

o1X

 

(ii) Give a reason why it is important to dispose of radioactive waste safely.

Answer

-Toxic
- Emits harmful radiations

(d)(i) A certain atom emits ultra-violet photon of wavelength 2.4 x 10 -7m

Calculate the energy of the photon:

[h = 6.6 x 10-34Js; c = 3.0 x 108ms-1]

Answer

E = \( \frac{hc}{\lambda} \)

E = \( \frac{6.6 \: \times \: 10^{-34} \: \times \: 3.0 \: \times \: 10^{8}}{2.4 \: \times \: 10^{-7} } \)

= \( \scriptsize 8.25 \: \times \: 10^{-19} J \)

 

E4_______________ -6.0 x 10-19J

E3_______________ -8.2 x 10-19J

E2_______________ -8.8 x 10-19J

E1_______________ -16.7 x 10-19J

(ii) The figure above illustrates the energy levels of the atom. Copy the figure in your answer booklet and indicate on it, the energy level transitions which cause the emission of the photon in d(i) above.

Answer

The enrgy level transitions which cause the emission of the photon in d(i) is E3

Question 1

A particle is projected horizontally at 10ms-1 from the top of a tower 20m high. Calculate the horizontal distance travelled by the particle when it hits the level ground. [g = 10ms-2]

Solution

R = \( \scriptsize U \normalsize \sqrt{ \frac{2H}{g}} \)

R = \( \scriptsize 10 \normalsize \sqrt{ \frac{2 \: \times \: 20}{10}} \)

R = \( \scriptsize 10\: \times \: 2\)

= 20m

Question 2

A tennis ball projected at an angle θ attains a range R = 78m. If the velocity imparted to the ball by the racket is 30ms-1, calculate θ. [g = 10ms-2]

Solution 

Recall that:-

R = \( \frac{u^2sin2 \theta}{g} \)

Information given:- Range, R= 78m, acceleration due to gravity,g = 10ms-2,

θ = ?, U = 30ms-1

R = \( \frac{30^2 sin2 \theta}{10} \)

sin2θ =  \( \frac{780}{900} \)

sin2θ= 0.8667

2θ = sin-10.8667 = 60°

θ = \( \frac{60^o}{2} \)

θ = 30°

Question 3

An electron moves with a speed of 2.00 x 107ms-1in an orbit in a uniform magnetic field of 1.2 x 10-3 Calculate the radius of the orbit.
[Mass of an electron = 9.11 x 10-31kg, charge on an electron = 1.61 x 10-19C]

Solution

Remember that \(\frac{Mv^2}{r} \scriptsize = qvB \)

That is, centripetal force = magnetic field force

Making r the subject of the formula

r = \( \frac{mV^2}{Bvq} \)

∴ r = \( \frac{mV}{Bq} \)

r = \( \frac{9.11 \: \times \: 10^{-31} \: \times \: 2 \: \times \: 10^7}{1.2 \: \times \: 10^{-3} \: \times \: 1.61 \: \times 10^{-19}} \)

r = \( \scriptsize 9.43 \: \times \: 10^{-2}m\)

Question 4

A metallic bar 50cm long has a uniform cross-sectional area of 4.0cm2. If a tensile force of 35KN produces an extension of 0.25mm, calculate the value of young's modulus.

Answer

Remember that:

Young's modulus ( E ) =  \( \frac{\frac{F}{A}}{\frac{e}{l}} \)

F = 35KN, A = 4cm2 = 0.04m2, e = 0.25mm = 0.00025m

l = \( \frac{50}{100} \scriptsize = 0.5m \)

Area, A = 4.0cm2 = \( \frac{4}{100^2} \scriptsize = 0.0004m \)

Tensile Force, F = 35N

extension, e = 0.25mm = \( \frac{0.25}{1000m} \scriptsize = 0.00025m\)

Young modulus ( E ) =  \( \frac{\frac{F}{A}}{\frac{e}{l}}  \\ = \frac{F \: \times \: l}{A \: \times \:  e} \\ = \frac{35 \: \times \: 0.5}{0.0004 \: \times \:  0.00025} \\ \scriptsize = 1.75 \: \times \: 10^{11} Nm^{-2} \)

Question 5

(a) Explain how a gas can be made to conduct electricity.

Answer

The molecules /atom of gas must be ionized before the gas can conduct electricity. Ionisation of the gas requires that the gas pressure is very low within the closure and a very high voltage is applied to the enclosed gas.

 

(b)Name the electric charge carriers in gases.

Answer

  The electric charge carriers in gases are the

- electrons
- ions

Question 6

The diagram above represents the graph of electron energy against the frequency of the radiation incident on a metal surface. Interpret the:

(a) slope of the graph;

Answer

The slope of the graph is the Planck's constant, (h)

(b) intercept, OC;

Answer

Intercept, OC represents the work function (f) of the metal

(c) intercept, OK.

Answer

Intercept OK, depicts the threshold frequency, fo. of the incident's radiation

Question 7

(a) State two conditions under which photo-electrons can be emitted from the surface of a metal.

Answers

- The wavelength of the incidents radiation must be greater than the threshold wavelength for metal surfaces
- The frequency of the incidents radiation must be greater than the threshold frequency
- The energy of the incident radiation must be greater than the work function of the metal

 

(b) List two-particle characteristics of electromagnetic waves.

Answers

- Photoelectric effect
- Thermionic emission
- Radiations from the heated black bodies
- Emission and absorption of light [incandescence]
- Compton effects

Question 8

(a) (i) Give two examples of rotational motion

Answer

- The rotation of the earth about its axis.
- A cricket ball spinning about its axis.
- The wheel of a moving car.
- The rotation of the blades of an electric fan.

(ii) Give two examples of linear motion

Answer

- An aeroplane moving on the runway.
- A bee flying from one flower to another.
- A ball rolling.

(iii) Describe a laboratory experiment to determine the density of an irregularly shaped solid.

Answer

The mass m of the solid is determined using a chemical/beam balance, a graduated measuring cylinder is partially filled with water and the initial volume V1 is recorded the solid is completely immersed in the water and the final volume V2 is recorded.

Volume of the solid = V2 - V1

Density of the solid \( \scriptsize \rho = \normalsize \frac{m}{V_2 \: - \: V_1} \)

 

(c) State Newton's second law of motion

Answer

Newton's second law of motion states that the rate of change of momentum is directly proportional to the impressed force and takes place in the direction of that force.

 

(d) Explain the term inertia

Answer

Inertia is the term given to the tendency of an object in uniform motion to remain in this state of motion, or an object at rest to remain at rest.

- the more mass a body has ,the greater is its inertia

(e) (i)

The diagram above illustrates a body of mass 5.0kg being pulled by a horizontal force F. If the body accelerates at 2.0ms-2 and experiences a frictional force of 5N, calculate the

[g = 10ms-2]

(i)  net force on it

Answer

Information given; Mass = 5kg, Acceleration, a = 20ms-2, Force, F1 = 5N, F = ?

Net force = F - F1 = Ma

= 5 x 2 = 10N

 

(ii) magnitude of F

Answer

Magnitude of F = F1 + net force (ma)
= 5 + 10 = 15N

 

(iii) coefficient of kinetic friction.

Answer

Coefficient of kinetic friction
from the formula
F1 = μR
∴   μ =  \(\frac{F_1}{R} \)

μ = coefficient of kinetic friction

F1 = Frictional force

R = reaction = mg = 5 x 10 = 50N

μ  = \( \frac{5}{50} \)

∴  μ = 0.1

Question 9

(a) Define heat capacity and state its unit.

Answer

Heat capacity of a substance is the quantity of heat required to raise the temperature of unit mass of the substance by 1oC (or 1K). The unit is Joules per kilogram per Kelvin (J/kg.K)

 

(b) List two effects of heat on a substance.

Answer

- It leads to expansion of a substance /contraction

- It leads to change of state

- It leads to a change in colour

- It leads to a change in electrical resistance

 

(c) Explain how a tightly fitted glass stopper could be removed from a reagent bottle.

Answer

The neck of the glass bottle should be warmed gradually from outside using warm cloth this makes the glass to expand creating space for the cock to be removed.

 

(d) A quantity of pepper soup of mass 800g poured into a plastic container with a tight-fitting lid has a temperature of 30oC. The container is then placed in a microwave oven rated 1200W and operated for 3 minutes.

(i) Calculate the final temperature attained by the soup. (Assuming no heat losses)

Answer

Information given; mass = 800g = 0.8kg, θ1 = 30oc, θ2 = ?
θ = θ2 - θ1
Power = 1200W,  c = 4000JKg-1K-1
time = 3 mins = (3 x 60) = 180sec
It should be noted that; Heat gained by the pepper soup = mcθ

power = Heat/time = energy/time

∴ Heat = power x time

= 1200 x 180 = 216,000 = mcθ

∴ θ = \( \frac{216000}{mc} \)

θ = \( \frac{216000}{0.8 \: \times \: 4000} \scriptsize = 67.5^0C \)

Final temperature = 67.5 + 30 = 97.5oC

(ii) Explain why containers with tight-fitting lids are not suitable for use in microwave cooking.

Answer

The molecules of the the steam from the content are confined within the container as the temperature increases the rate of collision of the molecules with walls of the container increases thus the pressure inside the container increases If it is tightly fitted ,the pressure can can make the container burst thus releasing its hot content.

(iii) When the soup is brought out and allowed to cool, a dent is observed on the container. Explain

Answer

The container will be observed with a dent because the condensation of the steam in the container will lead to a decrease in pressure in the container making the atmospheric pressure greater than the pressure in the container

Question 10

(a) State the three characteristics of sound and the factor on which each of them depends.

Answer

- Pitch depends on the frequency of the soundwave
- Loudness depends on the mass/amplitude of the air which can be set into vibration
- Quality/timbre depends on the overtones/harmonics present in the note.

(b) Explain resonance as applied to sound

Answer

Resonance is a phenomenon which occurs whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses or signals, received from some other system or body which is vibrating with the same frequency.
When the imposed frequency is the same as the natural frequency, the vibration build-up to a large amplitude. Therefore, Resonance is an effect caused by a vibrating body setting another body vibrating both having the same natural frequency. Resonance is what happens when a book placed on a surface is blown by a rotating fan.

(c) What role does echo play in the construction of a concert hall?

Answer

Echo sensitizes the construction engineers on the need to line the interior wall with appropriate sound-absorbing materials.

(d) The surface of an ear drum (assumed circular) has a radius 2.1 mm. It resonates with an amplitude of 0.8 x 10-7m as a result of impulses received from an external body vibrating at 2400 Hz. If the resulting pressure change on the ear drum is 3.6 x 10-5NM-2,calculate the:

(i) period of oscillation;

Answer

Period of oscillation T = \( \frac{1}{F} \)

T = \( \frac{1}{2400} \)

= 0.00042s

(ii) velocity;

Answer

Velocity = V = fλ = 2400 x 0.8 x 10-7
= 2.4 x 10³ x 0.8 x 10 -7
= 1.2 x 104-7
= 1.2 x 10 -3m/s

(iii) acceleration;

Answer

Acceleration
a =w2 r = (2πf)2r , here r = Amplitude
= (2 x 3.14 x  2400)2 x 0.8 x 10-7

= 18.2 ms -2

(iv) force

Answer

To calculate force

Force = Pressure x Area

radius = 2.1mm = 0.0021m

Area of circle = 2πr = 3.14 x (0.0021)2

Area = 1.38 x 10-5m 2

Pressure = 3.6 x 10-5NM-2

Force = 3.6 x 10-5 x 1.38 x 10-5

= 4.968 x 10(-5 - 5)

= 4.968 x 10-10N

Question 11

(a) Define electromotive force.

Answer

Electromotive force of a cell is the energy needed to take a unit positive charge around a circuit in which the cell is connected

- the p.d across the terminal of the cell when the cell/battery/generator is on open circuit

-the energy per unit charge transformed in the cell.

 

(b) (i) State the principle of operation of a potentiometer.

Answer

- The p.d of a length of wire is directly proportional to the length of the wire

-when a steady current flows through a uniform wire, equal lengths of the wire will have equal potentials

(ii) State two advantages that a potentiometer has over a voltmeter in measuring the potential difference.

Answer

- It is very accurate/has no errors
- It is more sensitive
- It eliminates errors due to the internal internal resistance of the cell

 

(c) (i) Sketch and label a diagram of a gold-leaf electroscope

Answer

Gold leaf electroscope

(ii) Give one use of a gold-leaf electroscope.

Answer

-Detects the [electrostatic] charge
-Detects the nature of [electrostatic] charge
-Determines the magnitude of the induced charges

 

(d) (i) Explain the action of a magnetic relay.

Answer

The mode of action of magnetic relay relies on the principle of electromagnetic induction, a small current passing through one circuit controls another circuit containing devices which require larger current.

(ii) List two factors which determine the magnitude of an induced emf in a coil.

Answer

- Speed of rotation of the coil
- Cross section area of the coil
- Number of turns

(iii) A current of 5 A passes through a straight wire in a uniform magnetic field of flux density 2.0 x 10-3 T. Calculate the force per unit length exerted on the wire when it is inclined at 30° to the field.

Answer

Information provided

B = \( \scriptsize 2.0 \: \times \: 10^{-3} \)

I = 5A

θ = 30º

L = ?

F = ?

From the formula

F = BIL sinθ

To get force per unit length, divide through by L

= \( \frac{F}{L} = \frac{BILsin \theta}{L}  \)

= \( \scriptsize BIsin \theta \)

Force per unit length =  \( \frac{F}{L}  \)

= \( \scriptsize 2.0 \: \times \: 10^{-3} \: \times \: 5sin30^o \)

= \( \scriptsize 10 \: \times \: 10^{-3} \: \times \: 0.5\)

= \( \scriptsize 5 \: \times \: 10^{-3} Nm^{-1}\)

Question 12

(a) Write Einstein's photoelectric equation and identify each component of the equation.

Answer

Einsten photoelectric equation is given as: E = W0  +  KEmax
E = Energy of radiation incidence
W0 = Work function
K.Emax = maximum kinetic energy of photoelectron

(b) For a photocell, state, one factor each that is responsible for the:

(i) emission

Answer

Emission - frequency/wavelength of incident radiation

(ii) rate of emission;

Rate of emission - intensity of incident radiation /number of protons

Answer

(iii) energy of photoelectrons.

Answer

Energy of photoelectrons - frequency/wavelength of incident radiation

 

(c)(i) Two nuclear equations are given below:

\( \scriptsize ^{222}_{p}RN \to \:\: ^{218}_{84}Po + \:\: ^q_2He\)  ...................A

\(\scriptsize  ^{214}_{83}RN \to\:\: ^{214}_{84}Po + \:\: ^m_nX \) ....................B

Determine the values of:

(α) p and q in equation A;

222pRn → 21884Po + 92He

p = 84 + 2 = 86

218 + q = 222

q = 222 - 218 = 4

(β) m and n in equation B and identify X.

21483Bi → 21483Po + mnX

m = 214 - 214 = 0; m = 0
n = 83 - 84 = -1;

n = -1;

o1X

 

(ii) Give a reason why it is important to dispose of radioactive waste safely.

Answer

-Toxic
- Emits harmful radiations

(d)(i) A certain atom emits ultra-violet photon of wavelength 2.4 x 10 -7m

Calculate the energy of the photon:

[h = 6.6 x 10-34Js; c = 3.0 x 108ms-1]

Answer

E = \( \frac{hc}{\lambda} \)

E = \( \frac{6.6 \: \times \: 10^{-34} \: \times \: 3.0 \: \times \: 10^{8}}{2.4 \: \times \: 10^{-7} } \)

= \( \scriptsize 8.25 \: \times \: 10^{-19} J \)

 

E4_______________ -6.0 x 10-19J

E3_______________ -8.2 x 10-19J

E2_______________ -8.8 x 10-19J

E1_______________ -16.7 x 10-19J

(ii) The figure above illustrates the energy levels of the atom. Copy the figure in your answer booklet and indicate on it, the energy level transitions which cause the emission of the photon in d(i) above.

Answer

The enrgy level transitions which cause the emission of the photon in d(i) is E3