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]]>

In a circuit, current may divide up where circuit splits into two separate branches. This doesn’t mean that current will disappear or an extra current will show up from nowhere. This is the basis for Kirchhoff’s first law.

Kirchhoff’s first law states that the current entering any point in a circuit is equal to the sum of the currents leaving the same point. Said differently, the algebraic sum of the currents into any junction is zero.

This law is also called Kirchhoff’s current law. It is an expression of conservation of charge i.e. the amount of charge entering a point must leave the same point. Kirchhoff’s first law obeys principle of conservation of charge.

This law deals with electromotive force (e.m.f.s) and the voltages in a circuit.

Kirchhoff’s second law states that the sum of e.m.f.s around any loop in a circuit is equal to the sum of the p.d.s (potential difference) around the loop.

Kirchhoff’s second law is a consequence of the principle of conservation of energy i.e. energy gain passing through sources of e.m.f = energy lost passing through components with p.d.s. Kirchhoff’s second law obeys principle of conservation of energy.

Ex 1

The diagram shows a junction in a circuit. What is the current Y ?

A +0.2 A

B −0.2 A

C +1.8 A

D −1.8 A

Solution

Current flowing to th point = Current leaving the point

3 + 0.4 = 2.2 + Y + 1.4

3.4 = 3.6 + Y

Y = 3.4 – 3.6 =- 0.2 A

Ex 2

Use Kirchhoff’s laws to determine the currents I1, I2 and I3 in the circuit on the right.

Solution

I2 + I3 = I1 (obeying Kirchhoff’s first law)

Using Kirchhoff’s second law to solve the loop

Ist loop (Clockwise direction)

3 = 10 * I1 +20*I2 (eq 1)

2nd loop (anti-clockwise direction)

1.5 = 20* I2 (eq 2)

I2 = 1.5/20 = 0.075 A

subtituting I2 in eq 1

3 = 10*I1 + 1.5

1.5 = 10* I1

I1 = 0.15 A

I2 + I3 = I1

0.075 + I3 = 0.15

I3 = 0.15 – 0.075 = 0.075 A

Recommended: Note on Direct Current

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]]>The post What is temperature and temperature measurement in Physics appeared first on ServantBoy.

]]>We are used to the idea that a thermometer shows the temperature of an object with which is in contact with.

What I will be discussing in this tutorial is listed below

- What is temperature or how temperature is defined in physics?
- What are the different units of temperature?
- Conversion of different temperature scales
- Temperature measurement
- Temperature examples
- Types of thermometer

Temperature is a measure of degree of hotness or coldness of an object on some scale. You can also say it tells us the direction in which energy flows (energy flow from a region of higher temperature to the region of lower temperature).

When two objects in contact have the same temperature, there will be no transfer of thermal energy between them. This is referred to as thermal equilibrium.

The unit of temperature includes Celsius, Kelvin, and Fahrenheit. However, the S.I base unit of temperature is Kelvin. Measuring temperature in Kelvin scale is better than that of Celsius scale in that one of its fixed points, absolute zero has much significance than the higher or lower fixed points of Celsius scale.

Note: it is impossible to have a temperature lower than zero Kelvin (0 K is absolute zero).

To convert from Celsius to Kelvin = temperature in Celsius + 273.15

To convert from Kelvin to Celsius = temperature in Kelvin – 273.15

Example

Convert -23 degree Celsius to Kelvin

Solution

-23 + 273.15 = 250.15 K

Note: the triple point of water is the temperature at which ice, water, and water vapour can co-exist. And it is defined as 273.16 K (273.16 – 273.15 = 0.01 C)

To convert from Celsius to Fahrenheit or vice-versa

C/5 = (F-32)/9

C is Celsius scale while F is Fahrenheit scale

Example

Convert 50 degree Celsius to Fahrenheit

50/5 = (F-32)/9

10 = (F-32)/9

10 X 9 = F-32

90 = F -32

F = 90 + 32

F = 122

Temperature of an object is measured with a thermometer.

Types of thermometer

- Resistance thermometer
- Thermocouples
- Mercury-in-glass thermometer
- Clinical thermometer

**Differences between Resistance thermometer and Thermocouple**

Resistance thermometer is larger than thermocouple, has greater thermal capacity therefore slower acting while thermocouple is smaller, has smaller capacity, therefore quicker acting and can measure temperature at a point.

Recommended

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]]>The post Heat Capacity and Specific Heat Capacaity | Introduction appeared first on ServantBoy.

]]>Specific heat capacity of a substance is the quantity of heat required to raise the temperature of unit mass (1kg) of the substance by 1K or 1^{o}C. The S.I unit is J/kgk or J/kgC

The law of conservation of energy shows that the heat we supply does not just disappear. It is usually transformed into some other kind of energy. In this case, the heat supplied to the water is transformed to the internal energy of the water molecules. I shall focus on conservation principle of heat energy, heat capacity, specific heat capacity, latent heat.

Heat Capacity = mass of substance x specific heat capacity

S.H.C =

From the definition of S.H.C, it follows that

Heat capacity of 5kg of aluminium = Mass (5kg) x specific heat capacity of aluminium (910 J/kg.k) = 4550 J/K

or

Heat capacity of 2kg of water = mass (2kg) x S.H.C of water (4200 J/kg.k) = 8400 J/K

**Factors that does not affect specific heat capacity of a material**

- It doesn’t depend on how much of the material is present.
- It’s also independent of the temperature interval i.e. if you double or tripple the teperature S.H.C is not affected

Example1 Calculate the quantity of heat required to raise the temperature of 4kg of copper from 25^{0}C to 95^{0}C [Take specific heat capacity of copper = 390 J/Kgk]

Quantity of heat

M = 4kg, C = 390J/Kgk, ( = 95 – 25 =70

Q = 4 x 390×70

Q = 109200J

**Determination of specific heat capacity of a solid **

1. Method of Mixture

2. Electrical

Example 2: A piece of copper block of specific heat capacity 400J/kgk falls through a vertical distance of 20m from rest, calculate the rise in temperature of the copper block on hitting the ground when all its energies are converted into heat.

This potential energy is converted into kinetic energy as it falls and the kinetic energy to heat energy as it hits the ground.

Example 3: A heating coil is rated 75W, calculate the time it will take this coil to heat 1.4kg of water at 30^{0}C to 100^{0}C (specific heat capacity of water =4200J/KgK)

Heat energy supplied by the heater = Heat absorbed by water

Ivt = Mcq

But power = IV,

Energy = Pt = Mcq

75 x t = 1.4 x 4200 x (100 – 30)

t = 54885 s

**Change of state**

Substance can exist in any three state of matter namely solid, liquid or gas. The state in which a substance exists depends on the temperature. Solid when heated change to liquid and when the liquid is further heated, it changes to gas.

Note: during change of state temperature remain constant.

The heat supplied to a solid (e. g ice) during change of state is used to overcome the attractive forces that hold the solid molecule together. It does not make the substance warmer and so it is not detected by a thermometer. The heat is used mainly in making the solid molecule move freely as in a liquid. Also the heats supply to a liquid at its boiling point is used to overcome the attractive forces that hold the liquid molecule together and push back the surrounding air molecule.

Q = ML (Joule)

Where L is called the specific latent heat

L = Q/M = J/kg

Example 4: Calculate the amount of heat energy required to change 20kg of ice water at 0^{0}C. Specific content heat of ice = 336 x 10^{3} J/kg

Q = 20 x 366 x

Q= 732 x J

Q= 7.32 x J

Click for solutions for UTME past questions on heat energy

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]]>The post Understanding Centre of gravity of an Object appeared first on ServantBoy.

]]>Every particle or object has weight because of the force of gravity on them, said differently; every object is attracted towards the centre of the earth by the gravitational force. The point where all the weight of the object may appear to act is referred to as centre of gravity.

Stability of an object (stable, unstable, and neutral) depends on the CG of that object. There are factors that affect the C.G. of an object, they are:

- The distribution of the masses
- The gravitational free strength which is the same as acceleration of free fall g

It is easier to find the centre of gravity of a regular object (uniform object) than irregular object. The C.G of a uniform object is at its mid-point.

- The C.G for a person standing upright is in the middle of the body behind the navel
- The C.G of a lamina or thin sheet or triangular plate is two-thirds of the distance of the median from the corresponding point. And this can be found by suspending any of the objects freely from two to three points.

For instance, the C.G of a rigid body in the figure below is the single point through which the weight of the body is considered to act.

The center of mass of an object is sometimes called its center of gravity. However, the center of mass is defined independently of any gravitational effect.

Centre of mass of an object is the point where the total mass appears to act.

If the distribution of masses and gravitational field strength (g) remains constant on all part of the object, the CG will coincide with the CM of the object.

Example on centre of gravity

A man of weight 600 N stands at the end of a uniform wooden plank, which is pivoted as shown in the diagram. What is the weight of the wooden plank?

Solution

The CG of the plank is roughly at the midle since it is a uniform wooden plank. Therefore, the distance of the weight of the plank is 1.5m from the man.

Sum of clockwise momnet = sum of anticlockwise moment

w x 1 = 600 x 0.5 (the moment is taking from the pivot)

w = 300N

To view more example, click here

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]]>The post Why Ammeter is Connected in Series And Voltmeter connected in Parallel appeared first on ServantBoy.

]]>**Function of Ammeter**

Ammeter is used to measure the value of flowing current in a circuit.

**Function of Voltmeter**

Voltmeter is used to measure the value of the potential difference or voltage across the load (resistor). The reading on the voltmeter indicates the energy transferred to the component by each unit of the charge.

Now you need to understand series and parallel connection;

In series connection,

- Current are the same through the loads
- In a closed circuit only ONE path for current to flow
- Total potential difference = sum of individual potential difference across the loads
- Total resistance = sum of individual resistances

In parallel connection,

- Potential difference across each resistor
- Total current = sum of individual currents
- Effective resistance less than individual resistance

The ammeter is connected in series with the load so that there is the same current in both. This can only be possible if the resistance on the ammeter is very low so as to ensure correct measurement of current in the circuit.

For an ideal ammeter, the value of the resistance should be zero. This means that all the voltage should appear across the load, so that current can be measured accurately.

From, V =IR

Voltage is directly proportional to current flow. If there is no voltage on the ammeter, then current won’t flow. And this can only be possible if the resistance on the ammeter is zero.

A voltmeter is designed to have a high resistance and should be connected in parallel in order to measure accurately.

By this, no current will flow through the voltmeter i.e. it won’t disrupt the current that should flow across a resistor or load.

For current not to flow through a voltmeter, the resistance of an ideal voltmeter should be infinity.

Kindly note current in a circuit will choose to flow through a less resistive path. This is also a reason why the resistance of a voltmeter must be high to avoid current flow across it.

Mathematically,

If the load is r and the voltmeter resistance is infinity connected in parallel to each other, the effective resistance will be

1/r + 1/infinity = 1/R

1/infinity = 0

1/r + 0 = 1/R

R = r (i.e. the effective resistance is the same as the resistance on the load)

This shows clearly that voltmeter will not alter the flow of current in the circuit.

Recommended: Note on Direct Current

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]]>The post How to solve questions on dynamics for Cambridge A level appeared first on ServantBoy.

]]>This article is intended to help students who finds it difficult to solve questions on dynamics. Below are some questions on dynamics and steps on how to answer them.

Question 1

A brick weighing 20 N rests on an inclined plane. The weight of the brick has a component of 10 N

parallel with the plane. The brick also experiences a frictional force of 4 N.

What is the acceleration of the brick down the plane? Assume that the acceleration of free fall g is

equal to 10 m s–2.

Question 2

The diagram shows a barrel suspended from a frictionless pulley on a building. The rope

supporting the barrel goes over the pulley and is secured to a stake at the bottom of the building

A man stands close to the stake. The bottom of the barrel is 18 m above the man’s head. The

mass of the barrel is 120 kg and the mass of the man is 80 kg.

The man keeps hold of the rope after untying it from the stake and is lifted upwards as the barrel

falls.

What is the man’s upward speed when his head is level with the bottom of the barrel? (Use

g = 10 m s–2.)

Question 3

A box of mass 8.0 kg rests on a horizontal, rough surface. A string attached to the box passes

over a smooth pulley and supports a 2.0 kg mass at its other end

When the box is released, a frictional force of 6.0 N acts on it.

What is the acceleration of the box?

Question 4

The momentum of an object changes from 160 kg m s–1 to 240 kg m s–1 in 2 s.

What is the mean resultant force on the object during the change?

Question 5

Two spheres approach each other along the same straight line. Their speeds are u1 and u2

before collision. After the collision, the spheres separate with speeds v1 and v2 in the directions

shown below.

Which equation must be correct if the collision is perfectly elastic?

A u1 – u2 = v2 + v1

B u1 – u2 = v2 – v1

C u1 + u2 = v2 + v1

D u1 + u2 = v2 – v1

Solutions

Solution Q1

weight of the box is 20N

weight = mg

20 = m * 10

m = 2kg

since the box slide down the plane, it means there is a resultant force or net force

F – Fr = ma

F = force parallel to the plane = 10N

10 – 4 = 2 * a

6 = 2a

a = 3 ms-2

Solution Q2

The force acting in this questions are the weight of the man, the weight of the barrel and the tension on the rope.

As the man moves upward, this is what happens. The tension on the rope acting upward is greater than the weight of the man acting downward. So this gives a resultant force

mathematically,

T – wm = m1a1……..iing eqn i

wm represents the weight of the man

m1 is the mass of the man

a1 is the acceleration of the man

As the barrel moves downward. The tension on the rope acting upward is less than the weight on the barrel acting downwards. So this gives a resultant force

dont forget that weight(gravitational force) is always acting downwards to the centre of the earth.

mathematically,

wb – T = m2a2……..ii

wb = weight of the barrel

m2 = mass of the barrel

a2 = acceleration of the barrel

combining eqn i and ii and eliminate T

you will have

wb – wm =m2a2 + m1a1

note the two objects will have the same acceleration

a2 = a1

wb – wm = (m1 + m2) a1

wb = 120*10 = 1200N

wm = 80 *10 = 800N

1200 – 800 = (120 + 80) a1

400 = 200*a1

a1 = 2 ms-2

recall,

*v ^{2}=u^{2}+2as*

the man and the barrel would have covered half of the distance when his head is level with the bottom of the barrel

i.e s = 9m

** v^{2}** = 0 + 2 * 2 * 9

** v^{2}**= 36

v = 6 ms-1

Solution Q3

The same approach we use in Q2 is applicable here

wb – fr = (m1 + m2)a1

wb = is the weight of the box

m1 is the mass of the box, m2 is the mass at the other end

20 – 6 = 10 * a

14 = 10a

a = 1.4 ms-2

Solution to Q4

change in momentum = p1 – p2

change in momentum = 240 – 160 = 80kgms-1

resultant force = change in momentum / time = 80 / 2 = 40N

Recommended: Short note on dynamics

Solution to Q5

Before collision, the balls move in opposite direction to each other

After collision, the balls move in the same direction

recall that,

relative speed of approach = relative speed of separation

u1 – u2 = v2 – v1 ( when they are moving in the same direction both before and after collision)

But in this case, they move in opposite direction before collision

in this case

u1 + u2 = v2 – v1

The answer is D

OR

If you look closely, u1 and v1 are in the same direction, that means

u1 + v1

u2 and v2 are in opposite direction, that means

v2 – u2

using conservation of linear momentum

u1 + v1 = v2 – u2

rearrange

u1 + u2 = v2 – v1

D is the correct answer

For additional solved solutions on dynamic click here

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]]>The post Solutions to Cambridge A level June 2015 and 2016 physics appeared first on ServantBoy.

]]>I took some questions out of 2015 and 2016 May/June Cambridge A level physics paper 11 and provid solutions to them based on their difficulty level. However, if you have any questions that you need additional explanation in 2015 and 2016 past papers, kindly comment below.

Question 1

The average kinetic energy E of a gas molecule is given by the equation E = 3/2 kT where T is the absolute (kelvin) temperature.

What are the SI base units of k ?

A kg–1m–1s2K B kg–1m–2s2K C kg m s–2K–1 D kg m2s–2K–1

Solution

The S.I base unit of Kinetic energy = Kgm2s-2

For temperature = K

K = E/T = kgm2s-2 / k

K = kgm2s-2k-1

D is the correct option

Question 2

A uranium-238 nucleus, 238U92 , undergoes nuclear decays to form uranium-234, 234U92 .Which series of decays could give this result?

A emission of four β-particles

B emission of four γ-rays

C emission of one α-particle and two β-particles

D emission of two α-particles and eight β-particles

Solution

alpha emission = 4He2

beta emission = 0e-1

For nucleon number and proton number to be conserved Uranium-238 must undergo one α-particle and two β-particles

C is the correct option

Recommended: Solutions to Cambridge A level Oct/Nov Paper 1 Physics questions

Question 3

A cell of e.m.f. 2.0 V and negligible internal resistance is connected to a network of resistors as shown.

What is the current I ?

A 0.25 A B 0.33 A C 0.50 A D 1.5 A

Solutiothe two 4 ohms are in parallel

effective resistance is 1/R = 1/4 + 1/4

1/R = 2/4

R = 2 ohms

This 2 ohms is in series with the second 2 ohms

R = 2 + 2 = 4 ohms

This 4 ohms is in parallel with the 2 ohms above, therefore the same E.m.f flows through them

For the effective resistance of 4 ohms

E = IR

2 = I*4

I = 1/2 A

2 ohms and the two 4 ohms are in series, the same current will flow through them

The current that will flow through the 4 ohms = half of 1/2 = 1/4 = 0.25 A

A is the correct option

Question 4

A charged oil drop of mass m, with n excess electrons, is held stationary in the uniform electric field between two horizontal plates separated by a distance d.

The voltage between the plates is V, the elementary charge is e and the acceleration of free fall is g.

What is the value of n ?

A eV/mgd B mgd/ev C meV/gd D gd/mev

Solution

Electric field intensity = F/ne = V/d

F = mg

mg/ne = V/d

mgd = neV

n = mgd /eV

B is the correct answer

Question 5

A sound wave has a speed of 330 m s–1 and a frequency of 50 Hz. What is a possible distance between two points on the wave that have a phase difference of 60°?

A 0.03 m B 1.1 m C 2.2 m D 6.6 m

Solution

V = fλ

330 = 50*λ

λ = 330/50 = 6.6 m

phase angle = 2πx/λ

π in rad = 180

60 = π/3

π/3 = 2πx/λ

λ = 6x

6x = 6.6

x = 1.1 m

B is the correct answer

Question 6

A fisherman lifts a fish of mass 250 g from rest through a vertical height of 1.8 m. The fish gains a speed of 1.1 m s–1.

What is the energy gained by the fish?

A 0.15 J B 4.3 J C 4.4 J D 4.6 J

Solution

P.E = mgh = 0.25*1.8*9.81 = 4.415 J

K.E = 1/2 mv2 = 1/2 * 0.25 *1.1^2 = 0.15 J

Energy gained by the fish = P.E +K.E = 4.4 + 0.15 = 4.6 J

D is the correct answer

**Solutions to some questions in Cambridge A level May/June 2016 physics P11**

Question 1

A car accelerates uniformly from velocity u to velocity v in time t.

On the graph, which area equals the distance travelled by the car in time t ?

A NPTU + PQST B NPW V + VRSU C NPW V + WRST D PST + PQS

Solution

The distance covered by the car is NPSU which is a trapezium. The area under it will give the distance covered. In the trapezium there are three full boxes. The only option that gives three full boxes is option B.

Question 2

Two cars X and Y are positioned as shown at time t = 0. They are travelling in the same direction. X is 50 m behind Y and has a constant velocity of 30 m s–1. Y has a constant velocity of 20 m s–1.

What is the value of t when X is level with Y?

A 1.0 s B 1.7 s C 2.5 s D 5.0 s

Solution

The time X will be at the same dstance with car Y is t

the distance covered for car X is

50 + s = 30t

the distance covered for car Y is

s = 20t

50 + 20t = 30t

50 = 10t

t = 5s

D is the correct option

Question 3

Two spheres approach each other along the same straight line. Their speeds are u1 and u2 before they collide. After the collision, the spheres separate with speeds v1 and v2 in the directions shown below.

The collision is perfectly elastic. Which equation must be correct?

A u1 – u2 = v2 + v1

B u1 – u2 = v2 – v1

C u1 + u2 = v2 + v1

D u1 + u2 = v2 – v1

Solution

relative speed of approach = relative speed of separation for perfectly elastic collision

u1 + u2 = v2 – v1

D is the correct answer

Question 4

The diagram shows a man standing on a platform that is attached to a flexible pipe. Water is pumped through the pipe so that the man and platform remain at a constant height.

The resultant vertical force on the platform is zero. The combined mass of the man and platform is 96 kg. The mass of water that is discharged vertically downwards from the platform each second is 40 kg. What is the speed of the water leaving the platform?

A 2.4 m s–1 B 6.9 m s–1 C 24 m s–1 D 47 m s–1

Solution

The force exerted by the man and the platform = the vertical downward force of the water

96*9.81 = m * v/t

96 * 9.81 = m/t * v

v = 96*9.81 / 40 = 23.544 = 24ms-1

C is the correct option

Question 5

A solid metal cylinder stands on a horizontal surface, as shown.

The cylinder has length x and cross-sectional area A. The cylinder exerts a pressure p on the surface. The acceleration of free fall is g. Which expression gives the density of the metal of the cylinder?

A gx/p B p/gx C gx/pA D pA/gx

Solution

pressure = force / area

area = A

P = F / A

F = PA

mass = force/acceleration of free fall

mass = PA / g

density = mass / volume

volume = A *x = Ax

density =( PA/g)/Ax = P /gx

B is the correct option

Question 6

A trailer of weight 30 kN is attached to a cab at X, as shown in the diagram.

What is the upward force exerted at X by the cab on the trailer?

A 3 kN B 15 kN C 30 kN D 60 kN

Solution

F * 20 = 30 * 10

F = 300 / 20 = 15 KN

B is the correct answer

Question 7

Some gas in a cylinder is supplied with thermal energy q. The gas does useful work in expanding at constant pressure p from volume V0 to volume VF, as shown.

Which expression gives the efficiency of this change?

A pV0 / q B. Vf / Voq C. p(Vf – V0 ) / q D.(Vf – V0 )/V0q

Solution

Efficiency = workoutput / workinput

workinput = q

workoutput = p ( Vf – V0)

efficiency = p ( Vf – V0) /q

C is the correct option

Question 8

Two copper wires of equal length are connected in parallel. A potential difference is applied across the ends of this parallel arrangement. Wire S has a diameter of 3.0 mm. Wire T has a diameter of 1.5 mm.

What is the value of the ratio current in T/ current in S?

A. 1/4 B. 1/2 C. 2 D. 4

Solution

Resistance = (resistivity x length)/Area

Area = (pi x square of diameter)/4

length and resistivity is constant

meaning, Rs(ds)^2 = RT(dT)^2

RS/RT = (dS)^2/(dT)^2 = (3×3)/(1.5×1.5) = 4

Since the wires are connected in parallel, it means that the potential difference across each wire will be the same.

V = IR

IsRs = ItRt

It/Is = Rs / Rt = 4

i.e. the ratio current in T/ current in S = 4

D is the correct answer

Click here to download for free Preparatory guide Physics for Cambridge A level, Post UTME, and UTME

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]]>The post How to solve questions on linear expansivity appeared first on ServantBoy.

]]>It is defined as the increase in length, of the unit length of the material for one degree temperature rise.

In order to solve any question on linear expansity:

(i) Write down the formula

(ii) Write down the data given

(iii) Understand what you are asked to find

(iv) Make proper substitution

(v) Evaluate

**Advantages
**

- Used for making bimetallic strip which is used in thermostat
- Removal of tight glass stopper
- Red-hot rivets in ship building
- Expansion of metals is used in bimetallic thermometer

**Disadvantages**

- Expansion of railway lines
- Sagging of overhead wire
- Expansion of balance wheel or wrist-watch
- Cracking of glass cup when hot water is poured into the glass cup
- Expansion of metals or concrete bridges

Question 1

What is meant by the statement: The linear expansivity of a solid is 1.0 x 10-5 k-1

Solution

The statement means A unit length of the solid will expand in length by a fraction 1.o x 10-5 of the original per kelvin rise in temperature.

Question 2

A wire of length 100.0m at 300C has a linear expansivity of 2 x 10-5K-1. Calculate the length of the wire at a temperature of -100C{UTME 2013}

A. 99.92m B. 100.08m C. 100.04m D. 99.96m

Solution

l1 = new length

l0 = original length

Q = temperature change

0.00002 = change in length / 100(30-(-10))

0.00002 = change in length / 100*40

0.00002*4000 = change in length

change in length = 0.08

change in length = 100 – x

x = 100 – 0.08 = 99.92m

A is the correct option

Question 3

Two metals P and Q are heated through the same temperature difference. If the ratio of the linear expansivities of P to Q is 2: 3 and the ratio of their lengths is 3:4 respectively, the ratio of the increase in lengths of P to Q is{UTME 2012}

A. 1 : 2 B. 2 : 1 C. 8 : 9 D. 9 : 8

Solution

Temperature difference =( linear expansivity * length)/increase in length

For metal P,

Qp = ( linear expansivity * length)p/increase in lengthp

For metal Q,

Qq = ( linear expansivity * lengthq)/increase in lengthq

( linear expansivity * length)p/increase in lengthp = ( linear expansivity * lengthq)/increase in lengthq

increase in length of p:increase lengthq = ( L.E * length)p / ( L.E * lengthq)

increase in length of p:increase lengthq = 2/3 *3 /4 = 6 / 12 = 1:2

A is the correct option

Question 4

A metal of volume 40cm3 is heated from 300C to 900C, the increase in volume is{UTME 2011}

A. 0.40cm3 B. 0.14cm3 C 4.00cm3 D. 1.20cm3.

[L.E of the metal= 2.0 x 10-5K-1]

Solution

cubic expansivity = 3* linear expansivity = 3 * 0.00002 = 0.00006

0.00006 = increase in volume / 40*(90 – 30)

0.00006 = increase in volume / 40*60

increase in volume = 0.00006*2400 = 0.144cm3

B is the correct answer

Question 5

A blacksmith heated a metal whose cubic expansivity is 6.3 x 10-6K-1. The area expansivity is{UTME 2007}

A. 6.3 x 10-6K-1 B. 4.2 x 10-6K-1 C. 2.1 x 10-6K-1 D. 2.0 x 10-6K-1

Solution

cubic expansivity = 3* linear expansivity

0.0000063 = 3* L.E

linear expansivity = 0.0000063/3 = 0.0000021

Area expansivity = 2*0.0000021 = 0.0000042

B is the correct option

Question 6

Calculate the length which corresponds to a temperature of 20^0C if the ice and steam points of an ungraduated

thermometer are 400 mm apart

A. 80mm

B. 20mm

C. 30mm

D. 60mm

Solution

ice point = lower fixed point = 0^oC

steam point = upper fixed point = 100^oC

(100 -0)/(20 -0) = 400/(x – 0)

100/20 = 400 / x

20*400 = 100x

x = 8000/100

x = 80mm

A is the correct option

Question 7

A steel bridge is built in the summer when its temperature is 35.0°C. At the time of construction, its length is 80.00m. What is the length of the bridge on a cold winter day when its temperature is -12.0°C? (L.E of steel is 1.2 x10^-5)

Solution

initial length = 80 m

initial temperature = 35.0°C

final length = ?

final temperature = -12.0°C

Change in temperature = final – initial = -12-35 = -47.0°C

let change in length = x

o.oooo12 = x/80(-47)

x = 0.000012*80*(-47) = -0.04512 m

note

x = final length – initial length

-0.04512 = final length – 80

final length = 80 -0.04512 = 79.95488 m

The length of the bridge on a cold winter day = 79.95488 m

Question 8

A copper rod whose linear expansivity =1.70×10^-5°c is 20cm longer than an aluminum rod whose linear expansivity=2.20×10^-5°c. How long should the copper rod be if the difference in their length is to be independent of temperature.

Solution

For copper,

linear expansivity of copper = x

change in length of cooper = y

original length of copper = c

temperature change = t

x = y/(c*t)

For aluminium,

linear expansivity of aluminium = a

change in length of aluminium = b

original length of aluminium = d

temperature change = r

a = b/(d*r)

According to the question, copper is 20cm longer than an aluminum rod i.e. c = d + 0.2

y/t = x*c = x*(d+0.2) = 1.70×10^-5*(d+0.2)

b/r = a*d = 2.20×10^-5*d

note, y/t = b/r since the difference in length is independent on temperature

2.20×10^-5*d = 1.70×10^-5*(d+0.2)

d = 1.7/2.2 * (d+0.2)

d = 0.7727d + 0.1545

d-0.7727d = 0.15

0.2273d = 0.1545

d = 0.68m

c = d + 0.2 = 0.68 +0.2 = 0.88m

Recommended: Solved questions on other topics in physics

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]]>The post How to solve questions on D.C for UTME and Cambridge A level appeared first on ServantBoy.

]]>It is easy to solve any question on Direct Current (D.C) if you can follow the steps used to solve the questions below. Below are some of the past questions on Direct Current from Cambridge A level hysics and UTME.

Question 1

An electric generator has an e.m.f. of 240V and an internal resistance of 1Ω. If the current supplied by the generator is 20A when the terminal voltage is 220V, find the ratio of the power supplied to the power dissipated.{UTME 2008}

A. 11 : 1 B. 1 : 11 C. 12 : 11 D. 11 : 12

Solution

power supplied = IE

p= 20 * 240 = 4800

power dissipated = IV

p = 20 * 220= 4400 watt

ratio = 4800 / 4400

12 : 11

A is the correct answer

Question 2

Find the effective resistance in the diagram above.{UTME 2008}

A. 6Ω B. 12Ω C. 16Ω D. 24Ω

solution

the arrangement is series because there is no branching of current

Rt = R1 + R2 + R3 + R4 + R5 + R6

Rt = 4 + 4 + 4 + 4 + 4 + 4

Rt = 24Ω

D is the correct option

Question 3 and 4 are from cambridge may/june 2016 p11

Hurray! It is now free to download the small textbook in pdf that I made for student writing A level physics and UTME. Click to Visit the page to download

Question 3

In the circuit shown, X is a variable resistor whose resistance can be changed from 5.0 Ω to 500 Ω. The e.m.f. of the battery is 12.0 V. It has negligible internal resistance

What is the maximum range of values of potential difference across the output?

A 1.3 V to 11.1 V B 1.3 V to 12.0 V C 1.5 V to 11.1 V D 1.5 V to 12.0 V

Solution

Vout = (Rx / Rx + 40)* Vin

since Rx is a variable resistor, the vout will give a varying voltage

when Rx is 5Ω

Vout = (5 / 5 +40)* 12

Vout = 1.3 V

When Rx is 500 Ω

Vout =( 500 / 500 + 40)* 12

Vout = 11.1 V

so Vout will range from 1.3 v to 11.1 v

A is the correct option

Question 4

There is a current from P to R in the resistor network shown.

The potential difference (p.d.) between P and Q is 3 V.

The p.d. between Q and R is 6 V.

The p.d. between P and S is 5 V.

Which row in the table is correct?

Solution

potential difference between Q and S = p.d. between P and S is 5 V – potential difference (p.d.) between P and Q is 3 V.

potential difference between Q and S = PS – PQ

potential difference between Q and S = 5 – 3 = 2v

also,

potential difference between Q and S= The p.d. between Q and R is 6 V – potential difference between S and R

2 = 6 – SR

SR = 4 v

A is the correct option

Question 5

A battery of electromotive force (e.m.f.) 9.0 V and internal resistance 0.25 Ω is connected in series with two identical resistors X and a resistor Y, as shown in Fig below

The resistance of each resistor X is 0.15 Ω and the resistance of resistor Y is 2.7 Ω.

(i) Show that the current in the circuit is 2.8 A.

(ii) Calculate the potential difference across the battery.

(iii) Each resistor X connected in the circuit above is made from a wire with a cross-sectional area of 2.5 mm2. The number of free electrons per unit volume in the wire is 8.5 × 10^29 m–3. Calculate the average drift speed of the electrons in X.{cambridge may/ june 2016 p22}

Solution

the resistors are arranged in series

Rt = R1 + R2 + R3

Rt = 0.15 + 2.7 + 0.15

Rt = 3Ω

E = IR + Ir

9 = I (Rt + r)

9 = I ( 3 + 0.25)

9 = 3.25 * I

I = 9 / 3.25

I = 2.8 A

ii.

potential difference across the battery = IRt

p.d = 2.8 * 3

p.d = 8.3 v

iii

I = nevA

n is the number of charge

e is the electronic charge

v is the drift speed

A is the area

I is the current

A = 2.5 mm^2 which in meters will be 2.5 x 10^-6 m^2

v = I / evA

v = 2.77 / (8.5 × 10^29 × 1.6 × 10^–19 × 2.5 × 10^–6)

v = 8.147 x 10^-6 ms-1

Recommended: short note on direct current

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]]>The post How to solve questions on Work, Energy and Power appeared first on ServantBoy.

]]>The solutions below gives an easy step to how to solve questions on work, Energy, and Power for students writing Cambridge A level and other exams like UTME or Post UTME.

Question 1

A hammer with 10 J of kinetic energy hits a nail and pushes it 5.0 mm into a plank. Both the hammer and nail come to rest after the collision. What is the approximate average force that acts on the nail while it moves through 5.0 mm?{ Cambridge A level may/june 2016 p11}

A 0.050 N B 2.0 N C 50 N D 2000 N

Solution

workdone by a net force = change in kinetic energy of a body

F x s = Ek

F x s = 10

F x 0.005 = 10

F = 10 / 0.005

F = 2000N

D is the correct option

Q2, 3 and 4 are from cambridge A level may/june 2016 p13

Question 2

An object of mass 0.30 kg is thrown vertically upwards from the ground with an initial velocity of 8.0 m s–1. The object reaches a maximum height of 1.9 m. How much work is done against air resistance as the object rises to its maximum height?

A 4.0 J B 5.6 J C 9.6 J D 15 J

Solution

workdone by a net force = change in kinetic energy of a body

workdone = 1/2 m v2

v2 = u2 – 2as

v2 = 64 – 2*9.81*1.9

v2 = 64 – 37.278

v2 = 26.722

workdone = 1/2 * 0.3 * 26.722

workdone = 4.0 J

A is the correct option

Recommended: Short note on work, energy and power

Question 3

A racing car has an output power of 300 kW when travelling at a constant speed of 60 m s–1. What is the total resistive force acting on the car?

A 5 kN B 10 kN C 50 kN D 100 kN

solution

power = force x velocity

300000 = force x 60

force = 300000 / 60

force = 5000 = 5KN

A is the correct option

Question 4

The diagram shows the design of a water wheel which drives a generator to produce electrical power. The flow rate of the water is 200 kg s–1. The generator supplies a current of 32 A at a voltage of 230 V.

Ignoring any changes in kinetic energy of the water, what is the efficiency of the system?

A 14% B 16% C 22% D 47%

Solution

efficiency = power output / power input

power output = IV

power output = 32*230 = 7360

power input = flow rate * a * h

a is the acceleration due to gravity

power input = 200*8*9.81 = 15696

efficiency = (7360 / 15696)*100%

efficiency = 47%

Q 5 and 6 are from cambridge A level may/june 2016 p12

Question 5

A boy on a bicycle starts from rest and rolls down a hill inclined at 30° to the horizontal. The boy and bicycle have a combined mass of 25 kg. There is a frictional force of 30 N, which is independent of the velocity of the bicycle.

What is the kinetic energy of the boy and the bicycle after rolling 20 m down the slope?

A 1850 J B 2450 J C 3050 J D 3640 J

Solution

mgsinθ – fr = ma

25 * 9.81*sin30 – 30 = ma

122.625 – 30 = ma

92.625 = ma

ma is the net force

the kinetic energy = net force X distance

kinetic energy = 92.625 * 20 = 1852 J = 1850J

A is the correct answer

Question 6

An escalator in an underground station has 250 people standing on it and is moving with a velocity of 4.3 m s–1. The average mass of a person is 78 kg and the angle of the escalator to the horizontal is 40°.

What is the minimum power required to lift these people?

A 54 kW B 64 kW C 530 kW D 630 kW

Solution

the vertical force = mgsinθ = 250*78*9.81*sin40 = 122962 N

minimum power = vertical force x velocity = 122962 x 4.3 = 530 Kw

C is the correct option

Question 7

A man has a mass of 80 kg. He ties himself to one end of a rope which passes over a single fixed pulley. He pulls on the other end of the rope to lift himself up at an average speed of 50 cm s–1.

What is the average useful power at which he is working? (Cambridge A level May/June 2017 p13 q17)

A 40 W B 0.39 kW C 4.0 kW D 39 kW

Solution

Power is the rate of energy expended per unit time i.e power = energy/time = force x velocity

force = mass x acceleration due to gravity. note g in CIE is always 9.81ms-2

f = 80 x 9.81 = 784.8N

velocity in ms-1 = 50/100 = 0.5ms-1

power = 784.8 x 0.5 = 392.4 watt = 0.39KW (B is the correct answer)

Question 8

Calculate the apparent weight loss of a man weighing 70kg in an elevator moving downwards with an acceleration of 1.5ms-2.{2013 UTME Physics – Type U}

A. 105N B. 686N C. 595N D. 581N

solution

when an elevator is moving down

net force = ma

net force = 70*1.5

net force = 105 N

the weight loss = net force

the weight loss = 105 N

A is the correct option

Click here to download for free Preparatory guide Physics for Cambridge A level, Post UTME, and UTME

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