Emulsions

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Emulsions

Emulsions are colloidal solution in which a liquid is dispersed in an another liquid. Generally there are two types of emulsions.

  1. Oil in water (O/W)
  2. Water in oil (W/O)

Example:

  • Milk is example of the oil in water type emulsion.
  • Stif greases are emulsion of water in oil i.e. water dispersed in lubricating oil.
  • The process of preparation of emulsion by the dispersal of one liquid in another liquid is called Emulsification.
  • A colloid mill can be used as a homogeniser to mix the two liquid. To have a stable emulsion a small amount of emulsifier or emulsification agent is added.

Several Types of Emulsifiers are known.

  1. Most of the lyophillic colloids also act as emulsifiers. Example: glue, gelatine.
  2. Long chain compounds with polar groups like soap and sulphonic acids.
  3. Insoluble powders like clay and lamp black also act as emulsifiers.

Identification of Types of Emulsion

The two types of emulsions can be identified by the following tests.

(i) Dye Test

A small amount of dye soluble in oil is added to the emulsion. The emulsion is shaken well. The aqueous emulsion will not take the colour whereas oily emulsion will take up the colour of the dye.

(ii) Viscosity Test

Viscosity of the emulsion is determined by experiments. Oily emulsions will have higher value than aqueous emulsion.

(iii) Conductivity Test

Conductivity of aqueous emulsions are always higher than oily emulsions.

(iv) Spreading Test

Oily emulsions spread readily than aqueous emulsion when spread on an oily surface.

Deemulsification:

Emulsion can be separated into two separate layers. The process is called Deemulsification.

Various Deemulsification Techniques are Given Below

  1. Distilling of one component
  2. Adding an electrolyte to destroy the charge
  3. Destroying the emulsifir using chemical methods
  4. Using solvent extraction to remove one component
  5. By freezing one of the components
  6. By applying centrifugal force
  7. Adding dehydrating agents for water in oil (W/O) type
  8. Using ultrasonic waves.
  9. Heating at high pressures.

Inversion of Phase:

The change of W/O emulsion into O/W emulsion is called inversion of phases.

For example:

An oil in water emulsion containing potassium soap as emulsifying agent can be converted into water in oil emulsion by adding CaCl2 or AlCl3. The mechanism of inversion is in the recent developments of research.

Colloid, Dispersion Phase and Dispersion Medium

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Colloid, Dispersion Phase and Dispersion Medium

Origin of study of colloid starts with Thomas Graham who observed diffusion of that a solution of sugar, urea or sodium chloride through a membrane but not glue, gelatine or gum. He called the former substances as crystalloids and the latter as colloids (In Greek, kola as gum, eidos-like).

Later it was realised that any substance can be converted into a colloid by reducing its particle size to 1-200nm.

Hence, colloid is a homogeneous mixture of two substances in which one substance (smaller proportion) is dispersed in another substance (large proportion).

In a colloid, the substance present in larger amount is called dispersing medium and the substance present in less amount is called dispersed phase.

Classifications of Colloidal Solution

Probably the most important colloidal systems have dispersed phase as solid and the dispersion medium as liquid. If the dispersion medium considered is water, then the colloids are referred as hydrosols or aquasols.

If the dispersion medium is an alcohol, the colloid is termed as alcosol, and if benzene is the dispersion medium, it is called as benzosol.

One more type of classification is based on the forces acting between the dispersal phase and dispersion medium.

In lyophillic colloids definite attractive force or affinity exists between dispersion medium and dispersed phase. Examples: sols of protein and starch. They are more stable and will not get precipitated easily. They can be brought back to colloidal solution even after the precipitation by addition of the dispersion medium.

In a lyophobic colloids, no attractive force exists between the dispersed phase and dispersion medium. They are less stable and precipitated readily, but can not be produced again by just adding the dispersion medium. They themselves undergo coagulation after a span of characteristic life time.

They are called irreversible sols
Examples: sols of gold, silver, platinum and copper.

The following table lists the types of colloids based on the physical states of dispersed phase and dispersion medium.

Classification of Colloids Based on the Physical State of Dispersed Phase and Dispersion Medium.

Dipersion Medium

Dispersed Phase Name of the Colloid

Examples

1. Gas Liquid Liquid Aerosol Fog Aerosol spray
2. Gas Solid Solid Aerosol Smoke, Air pollutants likes fumes, dust
3. Liquid Gas Foam Whipped cream, Shaving cream, Soda water, Froth
4. Liquid Liquid Emulsion Milk, Cream, Mayonnaise
5. Liquid Solid Sol Inks, Paints, Collodial gold
6. Solid Gas Solid foam Pumice stone, Foam rubber bread
7. Solid Liquid Gel Butter, Cheese
8. Solid Solid Solid sol Pearls, opals, coloured glass alloys colloidal dispersed eutics

Preparation of Colloids

Many lyophillic substances are made in their colloidal form by warming with water. Rubber forms colloidal solution with benzene. Soap spontaneously forms a colloidal solution by just mixing with water.

In general, colloidal are prepared by the following methods.

1. Dispersion Methods:
In this method larger particles are broken to colloidal dimension.

2. Condensation Methods:
In this method, smaller atom or molecules are converted into larger colloidal sized particles.

1. Dispersion Methods

(i) Mechanical Dispersion:

Using a colloid mill, the solid is ground to colloidal dimension. The colloid mill consists of two metal plates rotating in opposite direction at very high speed of nearly 7000 revolution/minute.

Colloid, Dispersion Phase and Dispersion Medium img 1

The colloidal particles of required colloidal size is obtained by adjusting the distance between two plates. By this method, colloidal solutions of ink and graphite are prepared.

(ii) Electro Dispersion:

A brown colloidal solution of platinum was first prepared by George Bredig in 1898. An electrical arc is struck between electrodes dispersed in water surrounded by ice. When a current of 1 amp/100 V is passed an arc produced forms vapours of metal which immediately condense to form colloidal solution.

By this method colloidal solution of many metals like copper, silver, gold, platinum, etc. can be prepared Alkali hydroxide is added as an stabilising agent for the colloidal solution.

Colloid, Dispersion Phase and Dispersion Medium img 2

Svedberg modified this method for the preparation of non aqueous inflammable liquids like pentane, ether and benzene, etc using high frequency alternating current which prevents the decomposition of liquid.

(iii) Ultrasonic Dispersion

Sound waves of frequency more than 20kHz (audible limit) could cause transformation of coarse suspension to colloidal dimensions.

Colloid, Dispersion Phase and Dispersion Medium img 3

Claus obtained mercury sol by subjecting mercury to sufficiently high frequency ultrasonic vibrations.

The ultrasonic vibrations produced by generator spread the oil and transfer the vibration to the vessel with mercury in water.

(iv) Peptisation:

By addition of suitable electrolytes, precipitated particles can be brought into colloidal state. The process is termed as peptisation and the electrolyte added is called peptising or dispersing agent

2. Condensation Methods:

When the substance for colloidal particle is present as small sized particle, molecule or ion, they are brought to the colloidal dimension by condensation methods. Here care should be taken to produce the particle with colloidal size otherwise precipitation will occur. Various chemical methods for the formation of colloidal particles.

(i) Oxidation

Sols of some non metals are prepared by this method.

(a) When hydroiodic acid is treated with iodic acid, I2 sol is obtained.
HIO3 + 5HI → 3H2O + I2 (Sol)

(b) When O2 is passed through H2Se, a sol of selenium is obtained.
H2Se + O2 → 2H2O + Se (sol)

(ii) Reduction

Many organic reagents like phenyl hydrazine, formaldehyde, etc are used for the formation of sols. For example: Gold sol is prepared by reduction of auric chloride using formaldehyde.

2 AuCl3 + 3HCHO + 3H2O → Au(sol) + 6HCl + 3HCOOH

(iii) Hydrolysis

Sols of hydroxides of metals like chromium and aluminium can be produced by this method.

For Example,
FeCl3+3H2O → Fe(OH)3+3HCl

(iv) Double Decomposition

For the preparation of water insoluble sols this method can be used. When hydrogen sulphide gas is passed through a solution of arsenic oxide, a yellow coloured arsenic sulphide is obtained as a colloidal solution.
As2O3 +3H2S → As2S3 + 3H2O

(v) Decomposition

When few drops of an acid is added to a dilute solution of sodium thiosulphate, the insoluble free sulphur produced by decomposition of sodium thiosulphate accumulates into small, clusters which impart various colours blue, yellow and even red to the system depending on their growth within the size of colloidal dimensions.

Colloid, Dispersion Phase and Dispersion Medium img 4

3. By Exchange of Solvent:

Colloidal solution of few substances like phosphorous or sulphur is obtained by preparing the solutions in alcohol and pouring them into water. As they are insoluble in water, they form colloidal solution.

P in alcohol + water → Psol.

Purification of Colloids

The colloidal solutions due to their different methods of preparation may contain impurities. If they are not removed, they may destablise and precipitate the colloidal solution. This is called coagulation. Hence the impurities mainly electrolytes should be removed to increase the stabilisation of colloid. Purification of colloidal solution can be done by the following methods.

  1. Dialysis
  2. Electrodialysis
  3. Ultrafilteration.

1. Dialysis

In 1861, T. Graham separated the electrolyte from a colloid using a semipermeable membrane (dialyser). In this method, the colloidal solution is taken in a bag made up of semipermeable membrane. It is suspended in a trough of flowing water, the electrolytes diffuse out of the membrane and they are carried away by water.

2. Electrodialysis

The presence of electric field increases the speed of removal of electrolytes from colloidal solution. The colloidal solution containing an electrolyte as impurity is placed between two dialysing membranes enclosed into two compartments filled with water.

When current is passed, the impurities pass into water compartment and get removed periodically. This process is faster than dialysis, as the rate of diffusion of electrolytes is increased by the application of electricity.

Colloid, Dispersion Phase and Dispersion Medium img 5

3. Ultrafiltration

The pores of ordinary filter papers permit the passage of colloidal solutions. In ultra filtrations, the membranes are made by using collodion cellophane or visiking. When a colloidal solution is filtered using such a filter, colloidal particles are separated on the filter and the impurities are removed as washings.

This process is quickened by application of pressure. The separation of sol particles from electrolyte by filteration through an ultrafilter is called ultrafiltration. Collodion is 4% solution of nitrocellulose in a mixture of alcohol and water.

Properties of Colloids

1. Colour

The colour of a sol is not always the same as the colour of the substance in the bulk. For example bluish tinge is given by diluted milk in reflected light and reddish tinge in transmitted light.

Colour of the sol, generally depends on the following factors.

  • Method of preparation
  • Wavelength of source of light.
  • Size and shape of colloidal particle
  • Whether the observer views the reflected light or transmitted light.

2. Size

The size of colloidal particles ranges from 1nm (10-9m) to 1000 nm (10-6m) diameter.

3. Colloidal Solutions are Heterogeneous in Nature Having two Distinct Phases

Though experiments like dialysis, ultrafiltration and ultracentrifuging clearly show the heterogeneous nature in the recent times colloidal solution are considered as border line cases.

4. Filtrability

As the size of pores in ordinary filter paper are large the colloidal particles easily pass through the ordinary filter papers.

5. Non-Setting Nature

Colloidal solutions are quite stable i.e. they are not affcted by gravity.

6. Concentration and Density

When the colloidal solution is dilute, it is stable. When the volume of medium is decreased coagulation occurs. Generally, density of sol decreases with decrease in the concentration.

7. Diffusability

Unlike true solution, colloids diffuse less readily through membranes.

8. Colligative Properties

The colloidal solutions show colligative properties i.e. elevation of boiling point, depression in freezing point and osmotic pressure. Measurements of osmotic pressure is used to find molecular weight of colloidal particle.

9. Shape of Colloidal Particles

It is very interesting to know the various shapes of colloidal particles. Here are some examples

Colloidal Particles

Shapes

As2S3 Spherical
Fe(OH)3sol (blue gold sol) Disc or plate like
W2O5sol (tungstic acid sol) Rod like

10. Optical Property

Colloids have optical property. When a homogeneous solution is seen in the direction of light, it appears clear but it appears dark, in a perpendicular direction.

Colloid, Dispersion Phase and Dispersion Medium img 6

But when light passes through colloidal solution, it is scattered in all directions. This effect was first observed by Faraday, but investigations are made by Tyndall in detail, hence called as Tyndall effect.

The colloidal particles absorb a portion of light and the remaining portion is scattered from the surface of the colloid. Hence the path of light is made clear.

11. Kinetic Property

Robert Brown observed that when the pollen grains suspended in water were viewed through ultra microscope, they showed a random, zigzag ceaseless motion.

This is called Brownian movement of colloidal particles.

This can be explained as follows

The colloidal sol particles are continuously bombarded with the molecules of the dispersion medium and hence they follow a zigzag, random, continuous movement.

Brownian Movement Enables Us,

I. To calculate Avogadro number.

II. To confirm kinetic theory which considers the ceaseless rapid movement of molecules that increases with increase in temperature.

III. To understand the stability of colloids:

As the particles in continuous rapid movement they do not come close and hence not get condensed. That is Brownian movement does not allow the particles to be acted on by force of gravity.

Colloid, Dispersion Phase and Dispersion Medium img 7
Colloid, Dispersion Phase and Dispersion Medium img 8

12. Electrical Property

(i) Helmholtz Double Layer

The surface of colloidal particle adsorbs one type of ion due to preferential adsorption. This layer attracts the oppositely charged ions in the medium and hence at the boundary separating the two electrical double layers are setup. This is called as Helmholtz electrical double layer.

As the particles nearby are having similar charges, they cannot come close and condense. Hence this helps to explain the stability of a colloid.

(ii) Electrophoresis:

When electric potential is applied across two platinum electrodes dipped in a hydrophilic sol, the dispersed particles move toward one or other electrode. This migration of sol particles under the influence of electric field is called electrophoresis or cataphoresis.

If the sol particles migrate to the cathode, then they posses positive (+) charges, and if the sol particles migrate to the anode then they have negative charges(-). This from the direction of migration of sol particles we can determine the charge of the sol particles. Hence electrophoresis is used for detection of presence of charges on the sol particles.

Colloid, Dispersion Phase and Dispersion Medium img 9

Few Examples of Charges of Sols Detected by Electrophoresis are Given Below:

Positively charge colloids Negatively charge colloids
Ferric hydroxide Ag, Au & Pt
Aluminium hydroxide Arsenic sulphide
Basic dyes Clay
Haemoglobin Starch

(iii) Electro Osmosis

A sol is electrically neutral. Hence the medium carries an equal but opposite charge to that of dispersed particles. When sol particles are prevented from moving, under the influence of electric field the medium moves in a direction opposite to that of the sol particles. This movement of dispersion medium under the influence of electric potential is called electro osmosis.

Colloid, Dispersion Phase and Dispersion Medium img 10

13. Coagulation or Precipitation

The flocculation and settling down of the sol particles is called coagulation.
Various method of coagulation are given below:

  • Addition of electrolytes
  • Electrophoresis
  • Mixing oppositively charged sols.
  • Boiling

Addition of Electrolytes

A negative ion causes the precipitation of positively charged sol and vice versa. When the valency of ion is high, the precipitation power is increased. For example, the precipitation power of some cations and anions varies in the following order

Al3+ > Ba2+ > Na+, Similarly [Fe(CN)6]3- > SO42- > Cl

The precipitation power of electrolyte is determined by finding the minimum concentration (millimoles/lit) required to cause precipitation of a sol in 2 hours. This value is called flocculation value. The smaller the flocculation value greater will be precipitation.

Electrophoresis

In the electrophoresis, charged particles migrate to the electrode of opposite sign. It is due to neutralization of the charge of the colloids. The particles are discharged and so they get precipitated.

By Mixing two Oppositively Charged Sols

When colloidal sols with opposite charges are mixed mutual coagulation takes place. It is due to migration of ions from the surface of the particles.

By Boiling

When boiled due to increased collisions, the sol particles combine and settle down.

14. Protective Action

Generally, lyophobic sols are precipitated readily even with small amount of electrolytes. But they are stabilised by addition of a small amount of lyophillic colloid.

A small amount of gelatine sol is added to gold sol to protect the gold sol.

Zsigmondy introduced the term ‘gold number’ as a measure of protecting power of a colloid. Gold number is defined as the number of milligrams of hydrophilic colloid that will just prevent the precipitation of 10ml of gold sol on the addition of 1ml of 10% NaCl solution. Smaller the gold number greater the protective power.

Colloid

Gold number

Gelatin 0.005-0.01
Egg albumin 0.08-0.10
Gum Arabic 0.1-0.15
Potato starch 25

Zeolite Catalysis

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Zeolite Catalysis

The details of heterogeneous catalysis will be in complete, if zeolites are not discussed. Zeolites are microporous, crystalline, hydrated, alumino silicates, made of silicon and aluminium tetrahedron. There are about 50 natural zeolites and 150 synthetic zeolites. As silicon is tetravalent and aluminium is trivalent, the zeolite matrix carries extra negative charge.

To balance the negative charge, there are extra framework cations for example, H+ or Na+ ions. Zeolites carrying protons are used as solid acid catalysts and they are extensively used in the petrochemical industry for cracking heavy hydrocarbon fractions into gasoline, diesel, etc., Zeolites carrying Na+ ions are used as basic catalysts.

One of the most important applications of zeolites is their shape selectivity. In zeolites, the active sites namely protons are lying inside their pores. So, reactions occur only inside the pores of zeolites.

Reactant Selectivity:

When bulkier molecules in a reactant mixture are prevented from reaching the active sites within the zeolite crystal, this selectivity is called reactant shape selectivity.

Transition State Selectivity:

If the transition state of a reaction is large compared to the pore size of the zeolite, then no product will be formed.

Product Selectivity:

It is encountered when certain product molecules are too big to diffuse out of the zeolite pores.

Phase Transfer Catalysis:

Suppose the reactant of a reaction is present in one solvent and the other reactant is present in an another solvent. The reaction between them is very slow, if the solvents are immiscible. As the solvents form separate phases, the reactants have to migrate across the boundary to react. But migration of reactants across the boundary is not easy.

For such situations a third solvent is added which is miscible with both. So, the phase boundary is eliminated, reactants freely mix and react fast. But for large scale production of any product, use of a third solvent is not convenient as it may be expensive.

For such problems phase transfer catalysis provides a simple solution, which avoids the use of solvents. It directs the use a phase transfer catalyst (a phase transfer reagent) to facilitate transport of a reactant in one solvent to the other solvent where the second reactant is present. As the reactants are now brought together, they rapidly react and form the product.

Example:

Substitution of Cl and CN in the following reaction.
R-Cl + NaCN → R-CN + NaCl

organic phase aqueous phase organic phase aqueous phase

R – C l = 1 – chlorooctane
R – C N = 1 – cyanooctane

By direct heating of two phase mixture of organic 1-chlorooctane with aqueous sodium cyanide for several days, 1-cyanooctane is not obtained. However, if a small amount of quaternary ammonium salt like tetraalkylammoniumchloride is added, a rapid transition of 1-cyanooctane occurs in about 100% yield after 1 or 2 hours.

In this reaction, the tetraalkylammonium cation, which has hydrophobic and hydrophilic ends, transports CN from the aqueous phase to the organic phase using its hydrophilic end and facilitates the reaction with 1-chloroocatne as shown below:

Zeolite Catalysis img 1

So phase transfer catalyst, speeds up the reaction by transporting one reactant from one phase to another.

Nano Catalysis:

Nano materials such a metallic nano particles, metal oxides, etc., are used as catalyst in many chemical transformation, Nanocatalysts carry the advantages of both homogeneous and heterogeneous catalyses.

Like homogeneous catalysts, the nanocatalysts give 100% selective transformations and excellent yield and show extremely high activity. Like the heterogeneous catalysts, nanocatalysts can be recovered and recycled. Nanocatalysts are actually soluble heterogeneous catalysts. An example for nanoparticles catalysed reaction is given below.

Zeolite Catalysis img 2

Enzyme Catalysis

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Enzyme Catalysis

Enzymes are complex protein molecules with three dimensional structures. They catalyse the chemical reaction in living organism. They are often present in colloidal state and extremely specific in catalytic action. Each enzyme produced in a particular living cell can catalyse a particular reaction in the cell.

Some common examples for enzyme catalysis

1. The peptide glycyl L-glutamyl L-tyrosin is hydrolysed by an enzyme called pepsin.

2. The enzyme diastase hydrolyses starch into maltose
2(C6H10O5)n + nH2O → nC12H22O11

3. The yeast contains the enzyme zymase which converts glucose into ethanol.
C6H12O6 → 2C2H5OH + 2CO2

4. The enzyme micoderma aceti oxidises alcohol into acetic acid.
C2H5OH + O2 → CH3COOH + H2O

5. The enzyme urease present in soya beans hydrolyses the urea.
NH2 – CO – NH2 + H2O → 2NH3 + CO2

Mechanism of Enzyme Catalysed Reaction

The following mechanism is proposed for the enzyme catalysis
E + S ⇄ ES → P + E

Where E is the enzyme, S the substrate (reactant), Es represents activated complex and P the products.

Enzyme Catalysis img 1

Enzyme Catalysed Reaction show Certain General Special Characteristics.

(i) Effective and efficient conversion is the special characteristic of enzyme catalysed reactions. An enzyme may transform a million molecules of reactant into product in a minute.
For eg. 2H2O2 → 2H2O + O2

For this reaction, the activation energy is 18k cal/mole without a catalyst. With colloidal platinum as a catalyst the activation energy is 11.7 kcal/mole.

But with the enzyme catalyst the activation energy of this reaction is less than 2kcal/mole.

(ii) Enzyme catalysis is highly specific in nature.

H2N-CO-NH2 + H2O → 2NH3 + CO2

The enzyme urease which catalyses the reaction of urea does not catalyse the following reaction of methyl urea

H2N-CO-NH-CH3 + H2O → No reaction

(iii) Enzyme catalysed reaction has maximum rate at optimum temperature. At first rate of reaction increases with the increase of temperature, but above a particular temperature the activity of enzyme is destroyed. The rate may even drop to zero. The temperature at which enzymic activity is high or maximum is called as optimum temperature.

Enzyme Catalysis img 2

For Example:

  • Enzymes involved in human body have an optimum temperature 37°C/98°F
  • During high fever, as body temperature rises the enzymatic activity may collapse and lead to danger.

(iv) The rate of enzyme catalysed reactions varies with the pH of the system. The rate is maximum at a pH called optimum pH.

Enzyme Catalysis img 3

(v) Enzymes can be inhibited i.e. poisoned. Activity of an enzyme is decreased and destroyed by a poison. The physiological action of drugs is related to their inhibiting action.

Example: Sulpha drugs. Penicillin inhibits the action of bacteria and used for curing diseases like pneumonia, dysentery, cholera and other infectious diseases.

(vi) Catalytic activity of enzymes is increased by coenzymes or activators. A small non protein (vitamin) called a coenzyme promotes the catalytic activity of enzyme.

Catalysis

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Catalysis

In 1836 Berzelius identified certain substances loosen the bond in the reacting molecules and increased the rate of the reaction. But he also found these substances didn’t undergo any change chemically. In order to indicate the property, he gave them the name catalyst. (In greek, kata-wholly, lein-to loosen).

Later it was identified that there were many substances which retarded the speed of a reaction.

Hence a catalyst is defined as a substance which alters the rate of chemical reaction without itself undergoing chemical change. The phenomenon which involves the action of a catalyst is called catalysis.

Positive and Negative Catalysis:

In positive catalysis, the rate of a reaction is increased by the presence of catalyst but in negative catalysis, the rate of reaction is decreased by the presence of a catalyst. The two main types of catalysis

  1. Homogeneous Catalysis and
  2. Heterogeneous Catalysis

Homogeneous Catalysis

In a homogeneous catalysed reaction, the reactants, products and catalyst are present in the same phase.

Illustration (1):

Catalysis img 1

In this reaction the catalyst NO, reactants, SO2 and O2, and product, SO3 are present in the gaseous form.

Illustration (2):

In the decomposition of acetaldehyde by I2 catalyst, the reactants and products are all present in the vapour phase.

CH3CHO(g) + [I2](g) → CH4(g) + CO(g) + [I2](g)

Let us consider some examples in which the reactants, products and catalyst are present in aqueous solution.

(1) Hydrolysis of cane sugar with a mineral acid as catalyst

Catalysis img 2

(2) Ester hydrolysis with acid or alkali as catalyst

Catalysis img 3

Heterogeneous Catalysis

In a reaction, the catalyst is present in a different phase i.e. it is not present in the same phase as that of reactants or products. This is generally referred as contact catalysis and the catalyst present is in the form of finely divided metal or as gauze.

Illustration

(i) In the manufacture of sulphuric acid by contact process SO3 is prepared by the action of SO2 and O2
in the presence of Pt or V2O5 as a catalyst.

Catalysis img 4

(ii) In the Haber’s process for the manufacture of ammonia, iron is used as a catalyst for the reaction between Hydrogen and Nitrogen.

Catalysis img 5

(iii) Oxidation of ammonia is carried out in presence of platinum gauze

Catalysis img 6

(iv) The hydrogenation of unsaturated organic compounds is carried out using finely divided nickel as a catalyst.

Catalysis img 7

(v) Decomposition of H2O2 occurs in the presence of the Pt catalyst

Catalysis img 8

(vi) In the presence of anhydrous AlCl3, benzene reacts with ethanoyl chloride to produce acetophenone

Catalysis img 9

Characteristics of Catalysts

  1. For a chemical reaction, catalyst is needed in very small quantity. Generally, a pinch of catalyst is enough for a reaction in bulk.
  2. There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.
  3. A catalyst itself cannot initiate a reaction. It means it can not start a reaction which is not taking place. But, if the reaction is taking place in a slow rate it can increase its rate.
  4. A solid catalyst will be more effective if it is taken in a finely divided form.
  5. A catalyst can catalyse a particular type of reaction, hence they are said to be specific in nature.
  6. In an equilibrium reaction, presence of catalyst reduces the time for attainment of equilibrium and hence it does not affect the position of equilibrium and the value of equilibrium constant.
  7. A catalyst is highly effective at a particular temperature called as optimum temperature.
  8. Presence of a catalyst generally does not change the nature of products.

For example: 2SO2 + O2 → SO3
This reaction is slow in the absence of a catalyst, but fast in the presence of Pt catalyst

Promoters and Catalyst Poison

1. In a catalysed reaction the presence of a certain substance increases the activity of a catalyst. Such a substance is called a promoter.
2. For example in the Haber’s process of manufacture of ammonia, the activity of the iron catalyst is increased by the presence of molybdenum.
3. Hence molybdenum is called a promoter. In the same way Al2O3 can also be used as a promoter to increase the activity of the iron catalyst.

On the other hand, certain substances when added to a catalysed reaction decreases or completely destroys the activity of catalyst and they are often known as catalytic poisons.

Few examples,

In the reaction, 2SO2 + O2 → 2SO3 with a Pt catalyst, the poison is As2O3
blocks the activity of the catalyst. So, the activity is lost.

In the Haber’s process of the manufacture of ammonia, the Fe catalyst is poisoned by the presence of H2S.

In the reaction, 2H2 + O2 → 2H2O,
CO acts as a catalytic poison for Pt – catalyst

Auto Catalysis

In certain reactions one of the products formed acts as a catalyst to the reaction. Initially the rate of reaction will be very slow but with the increase in time the rate of reaction increases.

Auto catalysis is observed in the following reactions.

CH3COOC2H5 + H2O → CH3COOH + C2H5OH

Acetic acid acts as the autocatalyst

2AsH3 → 2As + 3H2

Arsenic acts as an autocatalyst

Negative Catalysis

In certain reactions, presence of certain substances, decreases the rate of the reaction. Ethanol is a negative catalyst for the following reaction.

(i) 4CHCl3 + 3O2 → 4COCl2 + 2H2O + 2Cl2

Ethanol decreases the rate of the reaction

(ii) 2H2O2 → 2H2O + O2

In the decomposition of hydrogen peroxide, dilute acid or glycerol acts as a negative catalyst.

Theories of Catalysis

For a chemical reaction to occur, the reactants are to be activated to form the activated complex. The energy required for the reactants to reach the activated complex is called the activation energy. The activation energy can be decreased by increasing the reaction temperature. In the presence of a catalyst, the reactants are activated at reduced temperatures in otherwords, the activation energy is lowered.

The catalyst adsorbs the reactants activates them by weakening the bonds and allows them to react to form the products. As activation energy is lowered in presence of a catalyst, more molecules take part in the reaction and hence the rate of the reaction increases.

The action of catalysis in chemical reactions is explained mainly by two important theories. They are

  • The intermediate compound formation theory
  • The adsorption theory.

The Intermediate Compound Formation Theory

A catalyst acts by providing a new path with low energy of activation. In homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the catalyst is regenerated.

Consider the reactions:

A + B → AB (1)
A + C → AC (intermediate) (2)
C is the catalyst
AC + B → AB + C (3)

Activation energies for the reactions (2) and (3) are lowered compared to that of (1). Hence the formation and decomposition of the intermediate accelerate the rate of the reaction.

Example 1:

The mechanism of Fridel craft reaction is given below

Catalysis img 10

The action of catalyst is explained as follows

CH3Cl + AlCl3 → [CH3]+[AlCl4]

It is an intermediate.

C6H6 + [CH3+] [AlCl4] → C6H5CH3 + AlCl3 + HCl

Example 2:

Thermal decomposition of KClO3 in presence of MnO2 proceeds as follows.

Steps in the reaction
2KClO3 → 2KCl + 3O2 can be given as
It is an intermediate.
6MnO3 → 6MnO2 + 3O2

Example 3:

Formation of water due to the reaction of H2 and O2 in the presence of Cu proceeds as follows.
Steps in the reaction H2 + \(\frac{1}{2}\)O2 → H2O can be given as

2Cu + \(\frac{1}{2}\)O2 → Cu2O

It is an intermediate.

Cu2O + H2 → H2O + 2Cu

Example 4:

Oxidation of HCl by air in presence of CuCl2 proceeds as follows. Steps in the reaction

4HCl + O2 → 2H2O + 2Cl2 can be given as

2CuCl2 → Cl2 + Cu2Cl2
2Cu2Cl2 + O2 → 2Cu2OCl2

It is an intermediate.

2Cu2OCl2 + 4HCl → 2H2O + 4CuCl2

This theory describes

  • The specificity of a catalyst and
  • The increase in the rate of the reaction with increase in the concentration of a catalyst.

Limitations

  • The intermediate compound theory fails to explain the action of catalytic poison and activators (promoters).
  • This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Adsorption Theory

Langmuir explained the action of catalyst in heterogeneous catalysed reactions based on adsorption. The reactant molecules are adsorbed on the catalyst surfaces, so this can also be called as contact catalysis. According to this theory, the reactants are adsorbed on the catalyst surface to form an activated complex which subsequently decomposes and gives the product.

The various steps involved in a heterogeneous catalysed reaction are given as follows:

  1. Reactant molecules diffuse from bulk to the catalyst surface.
  2. The reactant molecules are adsorbed on the surface of the catalyst.
  3. The adsorbed reactant molecules are activated and form activated complex which is decomposed to form the products.
  4. The product molecules are desorbed.
  5. The product diffuse away from the surface of the catalyst.

Catalysis img 11

Active Centres

The surface of a catalyst is not smooth. It bears steps, cracks and corners. Hence the atoms on such locations of the surface are co-ordinatively unsaturated. So, they have much residual force of attraction. Such sites are called active centres. So, the surface carries high surface free energy.

The presence of such active centres increases the rate of reaction by adsorbing and activating the reactants. The adsorption theory explains the following

1. Increase in the surface area of metals and metal oxides by reducing the particle size increases acting of the catalyst and hence the rate of the reaction.

Catalysis img 12

2. The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.

3. A promoter or activator increases the number of active centres on the surfaces.

Adsorption and Absorption

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Adsorption and Absorption

Solid surfaces have the ability to attract the contacting species due to free valency or residual force on them.

For example:
charcoal adsorbs ammonia, silica gel adsorbs water., charcoal adsorbs colorants from sugar.

These examples prove that adsorption is a surface phenomenon. In contrast to adsorption, absorption is a bulk phenomenon i.e. the adsorbate molecules are distributed throughout the adsorbent.

  1. Adsorbent is the material on which adsorption takes place.
  2. Adsorbed substance is called an adsorbate.
  3. The surface of separation of the two phases where the concentration of adsorbed molecule is high is known as interface.
  4. In adsorption, if the concentration of a substance in the interface is high, then it is called positive adsorption.
  5. If it is less, then it is called negative adsorption.
  6. The process of removing an adsorbed substance from the surface is called desorption.
  7. The gaseous molecules like He, Ne, O2, SO2 and NH3 and solutions of NaCl or KCl can be adsorbed by suitable adsorbents. These are referred as adsorbates.
  8. Silica gel and metals like Ni, Cu, Pt, Ag and Pd and certain colloids can act as adsorbents.

Characteristics of Adsorption

1. Adsorption can occur in all interfacial surfaces i.e. the adsorption can occur in between gas-solid, liquid solid, liquid-liquid, solid-solid and gas-liquid.

2. Adsorption is a spontaneous process and it is always accompanied by decrease in free energy. When ∆G reaches zero, the equilibrium is attained. We know,

∆G = ∆H – T ∆S where ∆G is Change in Free energy.
∆H is Change in enthalpy and ∆S = Change in entropy.

3. When molecules are adsorbed, there is always a decrease in randomness of the molecules. ie., ∆S < 0, and T∆S is negative. Hence, adsorption is exothermic. Adsorption is a quick process whereas absorption is a slow process.

Types of Adsorption

Adsorption is classifid as physical adsorption and chemical adsorption, depending on the nature of forces acting between adsorbent and adsorbate. In chemical adsorption, gas molecules are held to the surface by formation of chemical bonds. Since strong bond is formed, nearly 400 KJ / mole is given out as heat of adsorption.

Examples

(a) Adsorption of O2 on tungsten, Adsorption of H2 on nickel, Adsorption of ethyl alcohol vapours on nickel.

In physical adsorption, physical forces such as van der waals force of attraction, dipole – dipole interaction, dispersion forces etc., exist between adsorbent and adsorbate. As these forces are weak, heat of adsorption is low, hence physical adsorption occurs at low temperatures.

Examples

(a) Adsorption of N2 on mica.
(b) Adsorption of gases on charcoal.

The following table 10.1 illustrates the distinction between chemical and physical adsorption.

Distinction Between Chemical and Physical Adsorption

Chemical adsorption or Chemisorption or Activated adsorption Physical adsorption or van der waals
adsorption or Physisorption
1. It is very slow 1. It is instantaneous
2. It is very specific depends on nature of
adsorbent and adsorbate.
2. It is non-specific
3. Chemical adsorption is fast with increase pressure, it can not alter the amount 3. In Physisorption, when pressure increases the extent of adsorption increases
4. When temperature is raised chemisorption first increases and then
decreases
4. Physisorption decreases with increase in
temperature
5. Chemisorption involves transfer of electrons between the adsorbent and adsorbate 5. No transfer of electrons
6. Heat of adsorption is high i.e., from 40-
400kJ/mole
6. Heat of adsorption is low in the order of
40kJ/mole.
7. Monolayer of the adsorbate is formed 7. Multilayer of the adsorbate is formed on
the adsorbent
8. Adsorption occurs at fixed sites called
active centres. It depends on surface area
8. It occurs on all sides
9. Chemisorption involves the formation
of activated complex with appreciable
activation energy
9. Activation energy is insignifiant

Factors Affecting Adsorption

The adsorption is well understood by considering the various factors affecting it. Qualitatively, the extent of surface adsorption depends on the following factors.

  1. Nature of adsorbent
  2. Nature of adsorbate
  3. Pressure
  4. Concentration at a given temperature.

1. Surface Area of Adsorbent:

As the adsorption is a surface phenomenon it depends on the surface area of adsorbent. i.e., higher the surface area, higher is the amount adsorbed.

2. Nature of Adsorbate

The nature of adsorbate can inflence the adsorption. Gases like SO2, NH3, HCl and CO2 are easily liquefiable as they have greater van der waal’s force of attraction. On the other hand, permanent gases like H2, N2 and O2 can not be liquefied easily. These permanent gases are having low critical temperature and adsorbed slowly, while gases with high critical temperature are adsorbed readily.

3. Effect of Temperature

When temperature is raised chemisorption fist increases and then decreases. whereas physisorption decreases with increase in temperature.

4. Effect of Pressure:

chemical adsorption is fast with increase in pressure, it can not alter the amount of adsorption. In Physisorption the extent of adsorption increases with increase in pressure.

Adsorption Isotherms and Isobars

Adsorption isotherms represents the variation of adsorption at constant temperature. When amount of adsorption is plotted versus temperature at constant pressure it is called adsorption isobar. Adsorption isobars of physisorption and chemisorption are different as represented in the graphs.

Adsorption and Absorption img 1

In physical adsorption, \(\frac{x}{m}\) decreases with increase in Temprature, But in chemical adsorption, \(\frac{x}{m}\) increases with rise in temperature and then decreases.

The increase illustrates the requirement of activation of the surface for adsorption is due to fact that formation of activated complex requires certain energy. The decrease at high temperature is due to desorption, as the kinetic energy of the adsorbate increases.

Adsorption Isotherms

Adsorption isotherm can be studied quantitatively. A plot between the amount of adsorbate adsorbed and pressure (or concentration of adsorbate) at constant temperature is called adsorption isotherms. In order to explain these isotherms various equations were suggested as follows:

(i) Freundlich adsorption isotherm.
According to Freundlich,
\(\frac{x}{m}\) = kp1/n

where x is the amount of adsorbate, adsorbed on ‘m’ gm of adsorbent at a pressure of p. K and n are constant introduced by freundlich. Value n is always less than unity.

This equation is applicable for adsorption of gases on solid surfaces. The same equation becomes \(\frac{x}{m}\) = K c1/n, when used for adsorption in solutions with c as concentration.

This equation quantitively predicts the effect of pressure(or concentration) on the adsorption of gases(or adsorbates) at constant temperature.

Taking log on both sides of equation \(\frac{x}{m}\) = Kp1/n
log \(\frac{x}{m}\) = logK + \(\frac{1}{n}\)log P

Adsorption and Absorption img 2

Hence the intercept represents the value of log k and the slope gives \(\frac{1}{n}\).

This equation explains the increase of \(\frac{x}{m}\) with increase in pressure. But experimental values show the deviation at low pressure.

Limitations

This equation is purely empirical and valid over a limited pressure range. The values of constants k and n also found vary with temperature. No theoretical explanations were given.

Applications of Adsorption

Though we have innumerable applications for adsorption, we consider few of them

1. Gas Masks:

During world war I charcoal gas mask was employed by both the British and American. Activated charcoal was found to be one of the best adsorbents.

2. To create high vacuum in vessels, Tail and Dewar used activated charcoal. For dehydration and also purification of gases like CO2, N2, Cl2, O2 and He, alumina and silica are employed. In the blast furnace silica gel is also used for drying air.

3. One of the highly important use of adsorption is the softening of hardwater. Permutit is employed for this process which adsorbs Ca2+ and Mg2+ ions in its surface, there is an ion exchange as shown below it occurs on the surface.

Na2Al2Si4O12 + CaCl2 → CaAl2Si4O12 + 2NaCl

Exhausted permutit is regenerated by adding a solution of common salt.

CaAl2Si4O12 + 2NaCl → Na2Al2Si4O12 + CaCl2

4. Ion Exchange Resins

Ion exchange resins are working only based on the process of adsorption. Ion exchange resins are used to demineralise water. This process is carried out by passing water through two columns of cation and anion exchange resins.

Adsorption and Absorption img 3

5. Petroleum Refining and Refining of Vegetable Oil

Fuller’s earth and silica gel are used for refining process.

6. Decolourisation of Sugar:

Sugar prepared from molasses is decolourised to remove coloured impurities by adding animal charcoal which acts as decolourising material.

7. Chromatography

The chromatographic technique is applied for separation of components in a mixture. It is mainly based on adsorption of components on the surface of adsorbents. This method is very effective and used for identification, detection and estimation of many substances even if they are contained in micro quantities.

8. Catalysed Reaction

Catalysis is an important branch of surface chemistry which is based on the phenomenon of adsorption of materials on the catalyst surface.

Examples:

In the Haber’s process, ammonia is manufactured from N2 and H2 as shown by the following reactions.

In this process, Fe is the catalyst and Mo is a promoter. The surface of the Fe catalyses the reaction.

In the hydrogenation of oils to obtain vanaspathi, Nickel is used as a catalyst. Nickel surface catalyses the reaction.

Adsorption and Absorption img 4

9. Qualitative Analysis

When blue litmus solution is added to Al3+ ion, a red coloration is seen due to the acidic nature of the solution. Addition of ammonium hydroxide to it gives a blue lake. This is due to the adsorption of blue colour litmus compound on the surface of Al (OH)3 Which is formed during the addition of NH4OH.

10. Medicine:

Drugs cure diseases by adsorption on body tissues.

11. Concentration of Ores of Metals

Sulphides ores are concentrated by a process called froth fltation in which light ore particles are wetted by pine oil.

12. Mordants and Dyes

Most of the dyes are adsorbed on the surface of the fabrics. Mordants are the substances used for fining dyes onto the fabric.

13. Adsorption Indicators

In the precipitation titrations, the end point is indicated by an external indicator which changes its colour after getting adsorbed on precipitate. It is used to indicate the end point of the titration.

Thermodynamics of Cell Reactions

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Thermodynamics of Cell Reactions

We have just learnt that in a galvanic cell, the chemical energy is converted into electrical energy. The electrical energy produced by the cell is equal to the product of the total charge of electrons and the emf of the cell which drives these electrons between the electrodes.

If ‘n’ is the number of moles of electrons exchanged between the oxidising and reducing agent in the overall cell reaction, then the electrical energy produced by the cell is given as below.

Electrical energy = Charge of ‘n’ mole of electrons × Ecell ……. (9.20)
Charge of 1 mole of electrons = one Faraday (1F)
∴ Charge of ‘n’ mole of electrons = nF
Equation (9.20) ⇒ Electrical energy = nFEcell ……. (9.21)

∴ Charge of one elctron = 1.602 × 10-19C
∴ Charge of one mole of elctron = 6.023 × 1023 × 1.602 × 10-19C
= 96488 C
i.e., 1F = 96500 C

This energy is used to do the electric work. Therefore the maximum work that can be obtained from a galvanic cell is

(Wmax)cell = – nFEcell …………… (9.22)

Here the (-) sign is introduced to indicate that the work is done by the system on the surroundings.

We know from the Second Law of thermodynamics that the maximum work done by the system is equal to the change in the Gibbs free energy of the system.

i.e., Wmax = ∆G …………… (9.23)

From (9.22) and (9.23),

∆G = – nFEcell ……………. (9.24)

For a spontaneous cell reaction, the DG should be negative. The above expression (9.24) indicates that Ecell should be positive to get a negative ∆G value.

When all the cell components are in their standard state, the equation (9.24) becomes

∆G° = – nFE°cell ………….. (9.25)

We know that the standard free energy change is related to the equilibrium constant as per the following expression.

∆G° = – RTlnKeq …………… (9.26)
Comparing (9.25) and (9.26),
⇒ E°cell = \(\frac{2.303RT}{nF}\)log Keq ………….. (9.27)

Nernst Equation

Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction. Let us consider an electrochemical cell for which the overall redox reaction is,
xA + yB ⇄ lC + mD. The reaction quotient Q for the above reaction is given below

Thermodynamics of Cell Reactions img 1 ……….. (9.28)

We have already learnt that,
∆G = ∆G° + RT lnQ

The Gibbs free energy can be related to the cell emf as follows [∴equation (9.24) and (9.25)]
∆G = – nFE cell; ∆G° = – nFE°cell
Substitute these values and Q from (9.28) in the equation (9.29)

Thermodynamics of Cell Reactions img 2 ………… (9.30)

Divide the whole equation (9.30) by (-nF)

Thermodynamics of Cell Reactions img 3 ………… (9.31)

The above equation (9.31) is called the Nernst equation. At 25°C (298K), the above equation (9.31) becomes,

Thermodynamics of Cell Reactions img 4 ……………… (9.32)

Let us calculate the emf of the following cell at 25°C using Nernst equation

Cu (s)|Cu2+(0.25 aq, M)|Fe2+(0.1 aq M)|Pt(s)
Given: (E°)Fe2+|Fe2+ = 0.77V and (E°)Cu2+|Cu = 0.34 V

Half reactions are
Cu (s) → Cu2+(aq) + 2e …………… (1)
2 Fe3+(aq) + 2e → 2Fe2+(aq) …………. (2)

the overall reaction is

Cu (s) + 2Fe3+(aq) → Cu2+(aq) + 2Fe2+ (aq), and n = 2
Apply Nernst equation at 25°C.

Thermodynamics of Cell Reactions img 5

Given standard reduction potential of Cu2+|Cu is 0.34V

Thermodynamics of Cell Reactions img 6

Electrolytic Cell and Electrolysis

Electrolysis is a process in which the electrical energy is used to cause a non-spontaneous chemical reaction to occur; the energy is often used to decompose a compound into its elements. The device which is used to carry out the electrolysis is called the electrolytic cell.

The electrochemical process occurring in the electrolytic cell and galvanic cell are the reverse of each other. Let us understand the function of a electrolytic cell by considering the electrolysis of molten sodium chloride.

The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they are connected to an external DC power supply via a key as shown in the figure (9.8). The electrode which is attached to the negative end of the power supply is called the cathode, and the one which attached to the positive end is called the anode. Once the key is closed, the external DC power supply drives the electrons to the cathode and at the same time pull the electrons from the anode.

Thermodynamics of Cell Reactions img 7

Cell Reactions

Na+ ions are attracted towards cathode, where they combine with the electrons and reduced to liquid sodium.

Cathode (Reduction)

Na+(l) + e → Na(l) E° = – 2.71
Similarly, Cl ions are attracted towards anode where they lose their electrons and oxidised to chlorine gas.

Anode (Oxidation)

2Cl(l) → Cl2(g) + 2e E° = – 1.36V

The overall reaction is,
2Na+(l) + 2Cl(l) → 2Na(l) + Cl2(g) E° = – 4.07V

The negative E° value shows that the above reaction is a non spontaneous one. Hence, we have to supply a voltage greater than 4.07V to cause the electrolysis of molten NaCl. In electrolytic cell, oxidation occurs at the anode and reduction occur at the cathode as in a galvanic cell, but the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode is – ve and anode is + ve.

Faraday’s Laws of Electrolysis
First Law

The mass of the substance (m) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell.

i.e m α Q

We know that the charge is related to the current by the equation I = \(\frac{Q}{t}\) ⇒ Q = It
∴ m α It
(or)
m = Z It …………. (9.33)

Where is Z is known as the electro chemical equivalent of the substance produced of the electrode. When, I = 1A and t = 1sec, Q = 1C, in such case the equation (9.32) becomes, (9.33)
⇒ m = Z …………….. (9.34)

Thus, the electrochemical equivalent is defined as the amount of substance deposited or liberated at the electrode by a charge of 1 coulomb.

Electro Chemical Equivalent and Molar Mass

Consider the following general electrochemical redox reaction
Mn+(aq) + ne → M(s)
We can infer from the above equation that ‘n’ moles of electrons are required to precipitate 1 mole of Mn+ as M(s).

The quantity of charge required to precipitate one mole of Mn+ = Charge of ‘n’ moles of electrons
= nF

In other words, the mass of substance deposited by one coulomb of charge

Thermodynamics of Cell Reactions img 8 …………… (9.35)

Second Law

Thermodynamics of Cell Reactions img 9

When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents.

Let us consider three electrolytic cells connected in series to the same DC electrical source as shown in the figure 9.9. Each cell is filled with a different electrolytes namely NiSO4, CuSO4 and CoSO4, respectively.

When Q coulomb charge is passed through the electrolytic cells the masses of Nickel, copper and cobalt deposited at the respective electrodes be mNi, mCu and mCo, respectively.

According to Faraday’s second Law,

Thermodynamics of Cell Reactions img 10 ………….. (9.36)

Batteries

Batteries are indispensable in the modern electronic world. For example, Li – ion batteries are used in cell phones, dry cell in flashlight etc. These batteries are used as a source of direct current at a constant voltage. We can classify them into primary batteries (non – rechargeable) and secondary batteries (rechargeable). In this section, we will briefly discuss the electrochemistry of some batteries.

Leclanche Cell

Anode: Zinc container
Cathode: Graphite rod in contact with MnO2
Electrolyte: ammonium chloride and zinc chloride in water

Emf of the cell is about 1.5V
Cell reaction

Oxidation at Anode
Zn (s) → Zn2+(aq) + 2e ……… (1)

Reduction at Cathode
2 NH4+(aq) + 2e → 2NH3(aq) + H2(g) ………… (2)

The hydrogen gas is oxidised to water by MnO2
H2 (g) + 2 MnO2 (s) → Mn2O3(s) + H2O(l)
Equation (1) + (2)+(3) gives the overall redox reaction

Thermodynamics of Cell Reactions img 11

Zn (s) + 2NH4+(aq) + 2 MnO2(s) → Zn2+(aq) + Mn2O3 (s) + H2O(l) + 2NH3 ………… (4)

Ammonia produced at the cathode combines with Zn2+ to form a complex ion [Zn(NH3)4]2+(aq). As the reaction proceeds, the concentration of NH4+ will decrease and the aqueous NH3 will increase which lead to the decrease in the emf of cell.

Mercury Button Cell

Anode: zinc amalgamated with mercury
Cathode: HgO mixed with graphite
Electrolyte: Paste of KOH and ZnO

Thermodynamics of Cell Reactions img 18

Overall reaction: Zn (s) + HgO (s) → ZnO (s) + Hg (l)
Cell emf: about 1.35V.
Uses: It has higher capacity and longer life. Used in pacemakers, electronic watches, cameras etc.

Thermodynamics of Cell Reactions img 12

Secondary Batteries

We have already learnt that the electrochemical reactions which take place in a galvanic cell may be reversed by applying a potential slightly greater than the emf generated by the cell. This principle is used in secondary batteries to regenerate the original reactants. Let us understand the function of secondary cell by considering the lead storage battery as an example.

Lead Storage Battery

Anode: Spongy lead
Cathode: lead plate bearing PbO2
Electrolyte: 38% by mass of H2SO4 with density 1.2g / mL.

Oxidation occurs at the anode

Pb(s) → Pb2+(aq) + 2e ……….. (1)
The Pb2+ ions combine with SO42-(aq) → PbSO4 (s) …………. (2)

Reduction occurs at the cathode

PbO2 (s) + 4 H+(aq) + 2e → Pb2+(aq) + 2H2O(l) ………. (3)
The Pb2+ ions also combine with SO42- → Pb2+(aq) + 2H2O(l)

The emf of a single cell is about 2V. Usually six such cells are combined in series to produce 12 volt

The emf of the cell depends on the concentration of H2SO4. As the cell reaction uses SO42-
ions, the concentration H2SO4 decreases. When the cell potential falls to about 1.8V, the cell has to be recharged.

Recharge of the Cell

As said earlier, a potential greater than 2V is applied across the electrodes, the cell reactions that take place during the discharge process are reversed. During recharge process, the role of anode and cathode is reversed and H2SO4 is regenerated.

Oxidation occurs at the cathode (now act as anode)

image

Reduction occurs at the anode (now act as cathode) PbSO4(s) + 2e → Pb(s) + SO42-(aq)
Overall reaction
2PbSO4(s) + 2H2O(l) → Pb(s) + PbO2(s) + 4H+(aq) + 2SO42-(aq)

Thus, the overall cell reaction is exactly the reverse of the redox reaction which takes place while discharging.

Uses:
Used in automobiles, trains, inverters etc.

The lithium – ion Battery

Anode: Porus graphite
Cathode: Transition metal oxide such as CoO2.

Thermodynamics of Cell Reactions img 13

Electrolyte: Lithium salt in an organic solvent
At the anode oxidation occurs
Li (s) → Li+(aq) + e

At the cathode reduction occurs
Li+ + CoO2(s) + e → LiCoO2(s)

Overall reactions
Li (s) + CoO2 → LiCoO2 (s)

Both electrodes allow Li+ ions to move in and out of their structures.

During discharge, the Li+ ions produced at the anode move towards cathode through the non – aqueous electrolyte. When a potential greater than the emf produced by the cell is applied across the electrode, the cell reaction is reversed and now the Li+ ions move from cathode to anode where they become embedded on the porous graphite electrode. This is known as intercalation.

Thermodynamics of Cell Reactions img 14

Uses:

Used in cellular phones, laptops, computers, digital cameras, etc.

Fuel Cell:

The galvanic cell in which the energy of combustion of fuels is directly converted into electrical energy is called the fuel cell. It requires a continuous supply of reactant to keep functioning. The general representation of a fuel cell is follows.

Thermodynamics of Cell Reactions img 15

Fuel | Electrode | Electrolyte | Electrode | Oxidant

Let us understand the function of fuel cell by considering hydrogen – oxygen fuel cell. In this case, hydrogen act as a fuel and oxygen as an oxidant and the electrolyte is aqueous KOH maintained at 200°C and 20 – 40 atm. Porous graphite electrode containing Ni and NiO serves as the inert electrodes.

Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively.

Oxidation occurs at the anode:
2H2g + 4 OH(aq) → 4H2O(l) + 4e
Reduction occurs at the cathode O2(g) + 2H2O(l) + 4e → 4 OH(aq)

The overall reaction is 2H2 (g) + O2 (g) → 2H2O(l)

The above reaction is the same as the hydrogen combustion reaction, however, they do not react directly ie., the oxidation and reduction reactions take place separately at the anode and cathode respectively like H2 – O2 fuel cell. Other fuel cells like propane – O2 and methane O2 have also been developed.

Corrosion

We are familiar with the rusting of iron. Have you ever noticed a green film formed on copper and brass vessels?. In both, the metal is oxidised by oxygen in presence of moisture. This redox process which causes the deterioration of metal is called corrosion.

As the corrosion of iron causes damages to our buildings, bridges etc… it is important to know the chemistry of rusting and how to prevent it. Rusting of iron is an electrochemical process.

Electrochemical Mechanism of Corrosion

The formation of rust requires both oxygen and water. Since it is an electrochemical redox process, it requires an anode and cathode in different places on the surface of iron. The iron surface and a droplet of water on the surface as shown in figure (9.15) form a tiny galvanic cell. The region enclosed by water is exposed to low amount of oxygen and it acts as the anode.

The remaining area has high amount of oxygen and it acts as cathode. So based on the oxygen content, an electro chemical cell is formed corrosion occurs at the anode i,e,. in the region enclosed by the water as discussed below.

Thermodynamics of Cell Reactions img 16

At Anode (Oxidation):

Iron dissolves in the anode region

2Fe (s) → 2Fe2+ (aq) + 4e E° = 1.23V

The electrons move through the iron metal from the anode to the cathode area where the oxygen dissolved in water, is reduced to water.

At Cathode (Reduction):

The reaction of atmospheric carbon dioxide with water gives carbonic acid which furnishes the H+ ions for reduction.

O2(g) + 4H+(aq) + 4e → 2H2O(l) E° = 0.44 + 1.23 = 1.67V

The positive emf value indicates that the reaction is spontaneous.

Fe2+(aq) + O2(g) + 4H+(aq) → 4Fe3+(aq) + 2H2O(l)
4Fe3+(aq) + 4H2O(l) → Fe2O3.H2O(s) + 6H+(aq)

Other metals such as aluminium, copper and silver also undergo corrosion, but at a slower rate than iron. For example, let us consider the reduction of aluminium,
Al(s) → Al3+(aq) + 3e

Al3+, which reacts with oxygen in air to forms a protective coating of Al2O3. This coating act as a protective film for the inner surface. So, further corrosion is prevented.

Protection of Metals form Corrosion

This can be achieved by the following methods.

  • Coating metal surface by paint.
  • Galvanizing – by coating with another metal such as zinc. zinc is stronger reducing agent than iron and hence it can be more easily corroded than iron. i.e., instead of iron, the zinc is oxidised.
  • Cathodic protection – In this technique, unlike galvanising the entire surface of the metal to be protected need not be covered with a protecting metal.

Instead, metals such as Mg or zinc which is corroded more easily than iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but Mg or Zn is corroded.

Passivation:

The metal is treated with strong oxidising agents such as concentrated HNO3. As a result, a protective oxide layer is formed on the surface of metal.

Alloy Formation:

The oxidising tendency of iron can be reduced by forming its alloy with other more anodic metals.
Example, stainless steel – an alloy of Fe and Cr.

Electrochemical Series

We have already learnt that the standard single electrode potentials are measured using standard hydrogen electrode. The standard electrode potential at 298K for various metal metal ion electrodes are arranged in the decreasing order of their standard reduction potential values as shown in the figure. This series is called electrochemical series.

The standard reduction potential (E°) is a measure of the oxidising tendency of the species. The greater the E° value, greater is the tendency shown by the species to accept electrons and undergo reduction. So higher the (E°) Value, lesser is the tendency to undergo corrosion.

Thermodynamics of Cell Reactions img 17

Electrochemical Cell

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Electrochemical Cell

Electrochemical cell is a device which converts chemical energy into electrical energy and vice versa. It consists of two separate electrodes which are in contact with an electrolyte solution. Electrochemical cells are mainly classified into the following two types.

1. Galvanic Cell (Voltaic Cell):

It is a device in which a spontaneous chemical reaction generates an electric current i.e., it converts chemical energy into electrical energy. It is commonly known as a battery.

2. Electrolytic Cell:

It is a device in which an electric current from an external source drives a nonspontaneous reaction i.e., it converts electrical energy into chemical energy.

Galvanic Cell

We have already learnt in XI standard that when a zinc metal strip is placed in a copper sulphate solution, the blue colour of the solution fades and the copper is deposited on the zinc strip as red – brown crust due to the following spontaneous chemical reaction.

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

The energy produced in the above reaction is lost to the surroundings as heat.

In the above redox reaction, Zinc is oxidised to Zn2+ ions and the Cu2+ ions are reduced to metallic copper.
The half reactions are represented as below.

Zn(s) → Zn2+(aq) + 2e (oxidation)
Cu2+(aq) + 2e → Cu(s) (reduction)

If we perform the above two half reactions separately in an apparatus as shown in figure 9.5, some of the energy produced in the reaction will be converted into electrical energy. Let us understand the function of a galvanic cell by considering Daniel cell as an example. It uses the above reaction for generation of electrical energy. The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.

Oxidation Half Cell

A metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker, as shown in Figure 9.5.

Electrochemical Cell img 1

Reduction Half Cell

A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker, as shown in Figure 9.5.

Electrochemical Cell img 2

Joining the Half Cells

The zinc and copper strips are externally connected using a wire through a switch (k) and a load (example: volt meter). The electrolytic solution present in the cathodic and anodic compartment are connected using an inverted U tube containing a agar-agar gel mixed with an inert electrolytes such as KCl, Na2SO4 etc., The ions of inert electrolyte do not react with other ions present in the half cells and they are not either oxidised (or) reduced at the electrodes. The solution in the salt bridge cannot get poured out, but through which the ions can move into (or) out of the half cells.

When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip. Ths is due to the following redox reactions which are taking place at the respective electrodes.

Anodic Oxidation

The electrode at which the oxidation occurs is called the anode. In Daniel cell, the oxidation take place at zinc electrode, i.e., zinc is oxidised to Zn2+ ions by loosing its electrons. The Zn2+ ions enter the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip. Electrons are liberated at zinc electrode and hence it is negative (- ve).

Zn(s) → Zn2+(aq) + 2e (loss of electron-oxidation)

Cathodic Reduction

As discussed earlier, the electrons flow through the circuit from zinc to copper, where the Cu2+ ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode. Here, the electrons are consumed and hence it is positive (+ ve).

Cu2+ + 2e → Cu (s) (gain of electron-reduction)

Salt Bridge

The electrolytes present in two half cells are connected using a salt bridge. We have learnt that the anodic oxidation of zinc electrodes results in the increase in concentration of Zn2+ in solution. i.e., the solution contains more number of Zn2+ ions as compared to SO42- and hence the solution in the anodic compartment would become positively charged.

Similarly, the solution in the cathodic compartment would become negatively charged as the Cu2+ ions are reduced to copper i.e., the cathodic solution contain more number of SO42- ions compared to Cu2+.

To maintain the electrical neutrality in both the compartments, the non reactive anions Cl (from KCl taken in the salt bridge) move from the salt bridge and enter into the anodic compartment, at the same time some of the K+ ions move from the salt bridge into the cathodic compartment.

Completion of Circuit

Electrons flow from the negatively charged zinc anode into the positively charged copper cathode through the external wire, at the same time, anions move towards anode and cations are move towards the cathode compartment. This completes the circuit.

Consumption of Electrodes

As the Daniel cell operates, the mass of zinc electrode gradually decreases while the mass of the copper electrode increases and hence the cell will function until the entire metallic zinc electrode is converted in to Zn2+ or the entire Cu2+ ions are converted in to metallic copper.

Unlike Daniel cell, in certain cases, the reactants (or) products cannot serve as electrodes and in such cases insert electrode such as graphite (or) platinum is used which conducts current in the external circuit.

Galvanic Cell Notation

The galvanic cell is represented by a cell diagram, for example, Daniel cell is represented as

Zn (s)|Zn2+(aq)||Cu2+(aq)|Cu (s)

In the above notation, a single vertical bar (|) represents a phase boundary and the double vertical bar (||) represents the salt bridge. The anode half cell is written on the left side of the salt bridge and the cathode half cell on the right side.

The anode and cathode are written on the extreme left and extreme right, respectively. The emf of the cell is written on the right side after cell diagram.

Electrochemical Cell img 3

Example

The net redox reaction of a galvanic cell is given below
2 Cr (s) + 3Cu2+(aq) → 2Cr3+ (aq) + 3Cu (s)

Write the half reactions and describe the cell using cell notation.
Anodic oxidation 2Cr (s) → 2Cr3+ (aq) + 6e ….. (1)
Cathodic reduction : 3Cu2+ (aq) + 6e → 3 Cu (s) ….. (2)

Cell Notation is
Cr (s)|Cr 3+(aq)||Cu2+(aq)|Cu(s)

Emf of a Cell

We have learnt that when two half cells of a Daniel cell are connected, a spontaneous redox reaction will take place which results in the flow of electrons from anode to cathode. The force that pushes the electrons away from the anode and pulls them towards cathode is called the electromotive force (emf) (or) the cell potential. The SI unit of cell potential is the volt (v).

When there is one volt difference in electrical potential between the anode and cathode, one joule of energy is released for each columb of charge that moves between them.

i.e., 1J = 1C × 1V ….. (9.18)

The cell voltage depends on the nature of the electrodes, the concentration of the electrolytes and the temperature at which the cell is operated. For example

At, 25°C , The emf of the below mentioned Daniel cell is 1.107 Volts

Zn (s)|Zn2+(aq, 1M)||Cu2+(aq, 1M)|Cu (s) E° = 1.107 V0lts

Measurement of Electrode Potential

The overall redox reaction can be considered as the sum of two half reactions i.e., oxidation and reduction. Similarly, the emf of a cell can be considered as the sum of the electrode potentials at the cathode and anode,

Ecell = (Eox)anode + (Ered)cathode ….. (9.19)

Here, (Eox)anode represents the oxidation potential at anode and (Ered)cathode represents the reduction potential at cathode. It is impossible to measure the emf of a single electrode, but we can measure the potential difference between the two electrodes (Ecell) using a voltmeter.

If we know the emf of any one of the electrodes which constitute the cell, we can calculate the emf of the other electrode from the measured emf of the cell using the expression (9.19). Hence, we need a reference
electrode whose emf is known.

For that purpose, Standard Hydrogen Electrode (SHE) is used as the reference electrode. It has been assigned an arbitrary emf of exactly zero volt. It consists of a platinum electrode in contact with 1M HCl solution and 1 atm hydrogen gas. The hydrogen gas is bubbled through the solution at 25°C as shown in the figure 9.6. SHE can act as a cathode as well as an anode.

The Half cell reactions are given below.

If SHE is used as a cathode, the reduction reaction is
2H+(aq, 1M) + 2e → H2(g, 1 atm) E° = 0 volt

If SHE is used as an anode, the oxidation reaction is
H2(g, 1 atm) → 2H+(aq, 1M) + 2e E° = 0 volt

Electrochemical Cell img 4

Illustration

Let us calculate the reduction potential of zinc electrode dipped in zinc sulphate solution using SHE.

Step: 1

The following galvanic cell is constructed using SHE
Zn (s) | Zn2+(aq, 1M)|| H+(aq, 1M)| H2 (g, 1atm)| Pt (s)

Step: 2

The emf of the above galvanic cell is measured using a volt meter. In this case, the measured emf of the above galvanic cell is 0.76V.

Calculation
We know that,

cell = (E°ox)zn|zn2++ (E°red)SHE [From equation (9.19)]
cell = 0.76 and (E°red)SHE = 0V . Substitute these values in the above equation

Electrochemical Cell img 5

⇒ 0.76V = (E°ox)zn|zn2+ + 0V
⇒ (E°ox)zn|zn2+ = 0.76V

This oxidation potential corresponds to the below mentioned half cell reaction which takes place at the cathode.

Zn → Zn2++ 2e (Oxidation)

The emf for the reverse reaction will give the reduction potential
Zn2++2e → Z ; E° = – 0.76V
∴ (Eo°red)zn2+|zn = – 0.76V.

Variation of Molar Conductivity With Concentration

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Variation of Molar Conductivity With Concentration

Friedrich Kohlraush studied the molar conductance of different electrolytes at different concentrations. He observed that, increase of the molar conductance of an electrolytic solution with the increase in the dilution. One such experimental results is given in the following table for better understanding.

Variation of Molar Conductivity With Concentration img 1

Based on the above such results, Kohlraush deduced the following empirical relationship between the molar conductance (Λm) and the concentration of the electrolyte (C).

Λm = ° – k\(\sqrt{C}\) ………… (9.11)

The above equation represents a straight line of the form y = mx + c. Hence, the plot of Λm Vs \(\sqrt{C}\) gives a straight line with a negative slope of – k and the y intercept. Λ°m. Where Λ°m is called the limiting molar value in very dilute solutions.

For strong electrolytes such as KCl, NaCl etc., the plot, Λm Vs \(\sqrt{C}\), gives a straight line as shown in the graph (9.4). It is also observed that the plot is not a linear one for weak electrolytes.

For a strong electrolyte, at high concentration, the number of constituent ions of the electrolyte in a given volume is high and hence the attractive force between the oppositely charged ions is also high. Moreover the ions also experience a viscous drag due to greater solvation.

These factors attribute for the low molar conductivity at high concentration. When the dilution increases, the ions are far apart and the attractive forces decrease. At infinite dilution the ions are so far apart, the interaction between them becomes insignificant and hence, the molar conductivity increases and reaches a maximum value at infinite dilution.

Variation of Molar Conductivity With Concentration img 2

For a weak electrolyte, at high concentration, the plot is almost parallel to concentration axis with slight increase in conductivity as the dilution increases. When the concentration approaches zero, there is a sudden increase in the molar conductance and the curve is almost parallel to Λm axis.

This is due to the fact that the dissociation of the weak electrolyte increases with the increase in dilution (Ostwald dilution law). Λ°m values for strong electrolytes can be obtained by extrapolating the straight line, as shown in figure (9.4). But the same procedure is not applicable for weak electrolytes, as the plot is not a linear one, Λ°m values of the weak electrolytes can be determined using Kohlraush’s law.

Debye – Huckel and Onsager Equation

We have learnt that at infinite dilution, the interaction between the ions in the electrolyte solution is negligible. Except this condition, electrostatic interaction between the ions alters the properties of the solution from those expected from the free – ions value. The influence of ion-ion interactions on the conductivity of strong electrolytes was studied by Debye and Huckel.

They considered that each ion is surrounded by an ionic atmosphere of opposite sign, and derived an expression relating the molar conductance of strong electrolytes with the concentration by assuming complete dissociation. Later, the equation was further developed by Onsager. For a uni – univalent electrolyte the Debye Huckel and Onsager equation is given below.

Λm = Λ°m – (A + BΛ°m)\(\sqrt{C}\) ………. (9.12)

Where A and B are the constants which depend only on the nature of the solvent and temperature. The expression for A and B are

Variation of Molar Conductivity With Concentration img 3

Here, D is the dielectric constant of the medium, η the viscosity of the medium and T the temperature in Kelvin.

Kohlraush’s Law

The limiting molar conductance Λ°m is the basis for kohlraush law. At infinite dilution, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions. i.e., the molar conductivity is due to the independent migration of cations in one direction and anions in the opposite direction.

°m)NaCl = (λ°m)Na + (λ°m)Cl

In general, according to Kohlraush’s law, the molar conductivity at infinite dilution for a electrolyte represented by the formula AxBy, is given below.

°m)AxBy = x(λ°m)Ay+ + y(λ°m)Bx- ………… (9.13)

Kohlraush arrived the above mentioned relationship based on the experimental observations such as the one as shown in the table. These result show that at infinite dilution each constituent ion of the electrolyte makes a definite contribution towards the molar conductance of the electrolyte irrespective of nature of other ion with which it is associated i.e;

°m)KCl – (Λ°m)NaCl = 149.86 – 126.45
{(λ°K)K + (λ°m)Cl} – {(λ°m)Cl} = 23.41
°m)K+ – (λ°m)Na+ = 23.41

Similarly, we can conclude that (λ°m)Br – (λ°m)Cl = 2.06

Applications of Kohlrausch’s Law

1. Calculation of Molar Conductance at Infinite Dilution of a Weak Electrolyte.

It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally. However, the same can be calculated using Kohlraush’s Law.

For example, the molar conductance of CH3COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCl, NaCl and CH3COONa.

Variation of Molar Conductivity With Concentration img 4
Variation of Molar Conductivity With Concentration img 5

Equation (1) + Equation (2) – Equation (3) gives,

Variation of Molar Conductivity With Concentration img 6

2. Calculation of Degree of Dissociation of Weak Electrolytes

The degree of dissociation of weak electrolyte can be calculated from the molar conductivity at a given concentration and the molar conductivity at infinite dilution using the following expression

Variation of Molar Conductivity With Concentration img 7 ………… (9.14)

Calculation of dissociation constant using Λm values. According to Ostwald dilution Law,

Ka = \(\frac{α2C}{(1-α}\) ……….. (9.15)

Substitute α value in the above expression (9.15)

Variation of Molar Conductivity With Concentration img 8

3. Calculation of Solubility of Sparingly Soluble Salts

Substances like AgCl, PbSO4etc., are sparingly soluble in water. The solubility product of such substances can be determined using conductivity measurements.

Let us consider AgCl as an example

AgCl(s) ⇄ Ag+ + Cl
Ksp = [Ag+] [Cl]

Let the concentration of [Ag+] be ‘C’ molL-1.
As per the stoichiometry, if [Ag+] = C, then [Cl] also equal to ‘C’ mol L-1.

Ksp = C.C
⇒ Ksp = C2
We know that the concentration (in mol dm-3) is related to the molar and specific conductance by the following expressions

Variation of Molar Conductivity With Concentration img 9

Substitute the concentration value in the relation Ksp = C2

Variation of Molar Conductivity With Concentration img 10 ………… (9.17)

Motion in a Straight Line Class 11 Important Extra Questions Physics Chapter 3

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 3 Motion in a Straight Line. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

Class 11 Physics Chapter 3 Important Extra Questions Motion in a Straight Line

Motion in a Straight Line Important Extra Questions Very Short Answer Type

The Velocity Calculator is a handy physics tool to quickly get your answer.

Question 1.
Can a moving body have relative velocity zero with respect to another body? Give an example.
Answer:
Yes, two trains running on two parallel tracks with the same velocity in the same direction.

Free Projectile Motion Calculator – calculate projectile motion step by step.

Question 2.
Can there be motion in two dimensions with acceleration in only one dimension?
Answer:
Yes, projectile motion.

Question 3.
Is it true that a body is always at rest in a frame that is fixed to the body itself?
Answer:
Yes.

Question 4.
Tell under what condition a body moving with uniform velocity can be in equilibrium?
Answer:
When the net force on the body is zero.

Question 5.
What does the speedometer records: the average speed or the instantaneous speed?
Answer:
It records (or measures) the instantaneous speed.

Question 6.
Can an object be accelerated without speeding up or slowing down? Give examples,
Answer:
Yes, circular motion.

Question 7.
Is it possible to have the rate of change of velocity constant while the velocity itself changes both in magnitude and direction? Give an example.
Answer:
Yes, in projectile motion.

Question 8.
Which motion is exactly represented by Δs = vΔt?
Answer:
It Represents motion with uniform velocity.

Question 9.
In which frame of reference is the body always at rest?
Answer:
The body is always at rest in the frame attached to it i. e. inertial frame of reference.

Question 10.
What is common between the two graphs shown in figs, (a) and (b)?
Answer:
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 1
Both these graphs represent that the velocity is negative.

Question 11.
What is common between the two graphs shown in figs, (a) and (b)?
Answer:
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 2
Both these graphs represent that velocity is positive.

Question 12.
What is meant by a point object in Physics?
Answer:
An object is said to be a point object if its dimensions are very small as compared to the distance covered by it.

Question 13.
The displacement of a body is zero. Is the distance covered by it is necessarily zero?
Answer:
No.

Question 14.
Which of the velocity or speed is measured by the speedometer of a vehicle?
Answer:
Speed.

Question 15.
Can you think of a situation where a body falling under gravity has constant velocity? Give example.
Answer:
Yes, the terminal velocity of a body.

Question 16.
Give an example of a motion which even though is accelerated motion yet it is called uniform motion.
Answer:
Uniform circular motion.

Question 17.
How many-dimensional motion does the following have?
(a) Train moving fast on its track.
Answer:
One dimensional motion.

(b) A lizard moving on a wall in a room.
Answer:
Two-dimensional motion.

(c) Kite flying in the sky.
Answer:
and

(d) Bee flying in a closed room.
Answer:
Three-dimensional motion.

Question 18.
When is the average velocity over an interval of time becomes equal to instantaneous velocity?
Answer:
When the velocity is constant.

Question 19.
A coolie carries a bag of luggage from one side of a platform to another side on the same platform. How far vertically the load is shifted?
Answer:
Zero.

Question 20.
The displacement of a body is proportional to the square of time along a straight line. Is the body moving with constant velocity or constant acceleration?
Answer:
It is moving with constant acceleration.

Question 21.
When the train in which you are sitting starts moving by the side of another train without jerks, you find that the other train is moving but when you look to the platform you find that your train is moving. Name the phenomenon responsible for such a motion.
Answer:
Relative velocity is the phenomenon responsible for such a motion.

Question 22.
Under what condition the magnitude of the average velocity of a particle is equal to the average speed?
Answer:
The magnitude of the average velocity of a particle is equal to the average speed if it moves with constant velocity.

Question 23.
Two particles A and B are moving along the same straight line with B being ahead of A. Velocities remaining unchanged, what would be the effect on the magnitude of relative velocity if A is ahead of B? ’
Answer:
The magnitude of the relative velocity will remain the same i.e. no effect on its magnitude.

Question 24.
Define the speed of the object.
Answer:
The speed of an object is defined as the distance covered by it per unit of time.

Question 25.
Why the speed of an object cannot be negative?
Answer:
The speed of an object cannot be negative because the distance can never be negative.

Question 26.
Can a body have zero velocity and still accelerating?
Answer:
Yes.

Question 27.
Can the direction of the velocity of a body change, when acceleration is constant?
Answer:
Yes.

Question 28.
Is the acceleration of a car is greater when the accelerator is pushed to the floor or when the brake pedal is pushed hard?
Answer:
The acceleration of the car is greater when the brake pedal is pushed hard because the car comes to rest suddenly i. e. the rate of change of velocity of the car is large in this case, so the acceleration.

Question 29.
The displacement is given by x = 2 + 4t + 5t2. Find the value of instantaneous acceleration.
Answer:
a = \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = 10

Question 30.
A stone is thrown vertically upwards from the surface of Earth. What is the direction of the velocity and acceleration of the stone?
(a) on its upward motion
Answer:
Velocity is vertically upward and acceleration is vertically downward.

(b) on its downward motion?
Answer:
Both velocity and acceleration are vertically downward.

Question 31.
Can Earth be regarded as a point object if only the orbital motion of Earth around the Sun is considered? Why?
Answer:
Yes. This is because the size of Earth is very small as compared to the size of the orbit of the Earth around the Sun.

Question 32.
The motion of two persons is shown by two straight lines on a displacement time graph intersecting each other at a certain point. What information do you get from the point of intersection?
Answer:
This means that the two persons cross each other at a certain place at a given instant of time.

Question 33.
Following two equations represents the x – t relation for the motion of an objects.
x (t) = x(0) + v(0)t + \(\frac{1}{2}\) at2
and x(t) = v(0)t + \(\frac{1}{2}\) at2
What is the difference between them?
Answer:
The first equation is a more general form of motion as it contains information regarding the initial position of the object.

Question 34.
Can the speed of a body change if its velocity is constant? Why?
Answer:
No, the speed of a body cannot change if its velocity is constant which means that both the magnitude and direction of velocity do not change. The magnitude of velocity is speed, so speed cannot change.

Question 35.
If the instantaneous velocity of a particle is zero, will its instantaneous acceleration be necessarily zero?
Answer:
No.

Question 36.
What is the shape of the displacement time graph of a particle having an average velocity equal to its instantaneous velocity?
Answer:
In this case, the velocity is uniform, so the x – t graph is a straight line.

Question 37.
Can there be a two-dimensional motion with acceleration in one dimension only? Give example.
Answer:
Yes, a projectile motion which is two-dimensional one has acceleration only in one dimension i.e. vertically downward.

Question 38.
Under what condition will the distance and displacement of a moving object will have the same magnitude?
Answer:
The distance and displacement of a moving object will have the same magnitude when it is moving with uniform velocity along a straight line.

Question 39.
Under what condition an object in motion cannot be considered a point object?
Answer:
A moving object cannot be considered as a point object if its size is not negligible as compared to the distance travelled by it.

Question 40.
Define a point object.
Answer:
It is defined as an object having dimensions (length, breadth, thickness etc.) very small as compared to the distance covered by it.

Question 41.
Is the following graph possible for the motion of a particle moving along a straight line?
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 3
Answer:
No.

Question 42.
Explain why the graph in the above question is not possible?
Answer:
This is because the speed for a given time is negative and speed is always positive.

Question 43.
Why the following graph is not possible for the motion of a particle moving along a straight line?
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 4
Answer:
This is because here the path length decreases with time while it must either increase or must remain constant.

Question 44.
What happens to kinematic equations under time reversal?
Answer:
The kinematic equations of motion don’t change in the form under time reversal i.e. if t is replaced by -t.

Question 45.
What happens to the uniform motion of a body when it is given an acceleration at right angles to its motion?
Answer:
The body will come in a circular motion when it is given an acceleration at right angles to its motion.

Question 46.
To deal with physical phenomena, we consider objects even as big as Sun a point objects. Can you name physical phenomena in which Earth cannot be taken as a point object?
Answer:
The occurrence of solar or lunar eclipse does not allow Earth to be taken as a point object otherwise the phenomena cannot be explained.

Question 47.
The average velocity of a body moving with uniform acceleration is given by \(\frac{1}{2}\) (u + v). Ii the acceleration changes from point to point can the average velocity be still given by this expression? Give reason.
Answer:
No, the average velocity cannot be given by \(\frac{1}{2}\) (u + v) in case the acceleration varies from point to point i.e. if it is not uniform. This is because the slope of the v-t graph does not remain the same at all points.

Question 48.
Acceleration is defined as the rate of change of velocity. Suppose we call the rate of change of acceleration SLAP. Then (i) What is the unit of SLAP.
Answer:
SLAP = Acceleration/time (By definition).

∴ Its unit will be = \(\frac{\mathrm{ms}^{2}}{\mathrm{~s}}\) = ms-3.

(ii) How can you calculate instantaneous SLAP?
Answer:
Average SLAP = \(\frac{\Delta \mathrm{a}}{\Delta \mathrm{t}}\)

∴ Instantaneous SLAP = Limiting value of average SLAP
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 5

Question 49.
Why is the time stated twice in stating acceleration?
Answer:
Since acceleration is the double rate of change of displacement

i. e. a = \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\), so time is stated twice in stating acceleration.

Question 50.
Separate the following in one, two and three-dimensional motion :
(a) a kite flying on a windy day.
(b) an insect crawling on a globe.
(c) a carom coin rebounding from the side of the board,
(d) a planet revolving around its star.
(e) the motion of a boat.
(f) the motion of a dropped body.
(g) the motion of a tennis ball.
(h) a charged particle moving under an electric field.
(i) movement of a saw while cutting wood.
(j) molecular motion.
(k) a charged particle moving under a magnetic field.
Answer:

  • One dimensional motion : (e), (f), (i)
  • Two dimensional motion : (b), (c), (d), (g), (h), (k).
  • Three dimensional motion : (a), (j).

Motion in a Straight Line Important Extra Questions Short Answer Type

Question 1.
Prove that the average velocity of a particle over an interval of time is either smaller than or equal to the average speed of the particle over the same interval.
Answer:
Average velocity is defined as the ratio of the total displacement to the total time. Average speed is defined as the ratio of the total distance to the total time. Since displacement is less than or equal to the distance, therefore the average velocity is less than or equal to the average speed.

Question 2.
Two trains each of the length 109 m and 91 m are moving in opposite directions with velocities 34 km h-1 and 38 km h-1 respectively. At what time the two trains will completely cross each other?
Answer:
Let l1, l2 be the lengths of the two trains.
v1, v2 be their velocities respectively.

∴ l1 = 109m, l2 = 91 m, v1 = 34kmh-1, v2 = 38kmh-1.
As the trains are moving in opposite directions so relative velocity of the trains is given by
v1 – (- v2) = v1 + v2
= 34 + 38 = 72 kmh-1
= 72 × \(\frac{5}{18}\) = 20 ms-1

Total distance to be covered by the two trains in crossing each other
= l1 + l2= 109 + 91 = 200 m

If t be the time taken in crossing, then t can be calculated using the relation
x = vt
or
t = \(\frac{200}{20}\) = 10s

Question 3.
Ambala is at a distance of 200 km from Delhi. Ram sets out from Ambala at a speed of 60 km h-1 and Sham set out at the same time from Delhi at a speed of 40 km h-1. When will they meet?
Answer:
S = 200 km. Let VR and vs be the speeds of Ram and Sham respectively moving in opposite directions.
∴ vR = 60 kmh-1, vS = 40 kmh-1.

∴ Relative velocity of Ram w.r.t. Sham is
VRS = VR – (- VS)
= VR + VS
= 60 + 40 = 100 kmh-1

If t = time after which they will meet, then
t = time taken in covering 200 km distance with VRS
i.e. t = \(\frac{200}{v_{\mathrm{RS}}}=\frac{200 \mathrm{~km}}{100 \mathrm{kmh}^{-1}}\) = 2h.

∴ Time after which they meet = 2h.

Question 4.
A car travelling at a speed of 60 km h-1 on a straight road is ahead of a scooter travelling at a speed of 40 km h-1. How would the relative velocity be altered if the scooter is ahead of the car?
Answer:
vc = speed of car = 60 kmh-1
vs = speed of scooter = 40 kmh-1
vcs = relative velocity of car w.r.t. scooter
= vc – vs
= 60 – 40
= 20 kmh-1

Similarly vsc = relative velocity of scooter w.r.t. car
= vs – vc
= 40 – 60
= – 20 kmh--1

Thus we conclude that the magnitude of the relative velocity is the same in both cases but the direction of relative velocity is reversed if the scooter is ahead of the car.

Question 5.
Draw the position-time graphs for two objects initially occupying different positions but having zero relative velocity.
Answer:
The positive T time graphs for two objects initially occupying different positions but having zero relative velocity are parallel to each other as shown in Fig.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 6

Question 6.
A ball is thrown vertically upward with a velocity of 20 ms-1. It takes 4 seconds to return to its original position. Draw a velocity-time graph for the motion of the ball and answer the following questions:
At which point P, Q, R, the stone has :
(a) reached its maximum height.
(b) stopped moving?
Answer:
Let P represent the initial position at the time when the ball is thrown vertically upward.
Q represents the highest point reached by the ball.
R represents the original position of the ball after 4 seconds.

Thus the velocity-time graph for the motion of the ball is as shown in Fig.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 7
(a) We know that at the highest point, the velocity of the object is zero. So stone will reach its maximum height corresponding to point Q.

(b) The stone has stopped moving at point Q because at Q, v = 0.

Question 7.
“It is the velocity and not the acceleration which decides the direction of motion of a body.” Justify this statement with the help of a suitable example.
Answer:
The direction of velocity is always in the direction of motion of the body whereas the direction of acceleration may or may not be in the direction of motion of the body. Thus we conclude that it is the velocity that decides the direction of motion of the body.

Example: When a ball is thrown vertically upwards, the direction of motion of the ball and velocity is the same i.e. vertically upwards. On the other hand, the acceleration due to gravity on the ball acts vertically downwards i.e. opposite to the direction of motion of the ball.

Question 8.
Two buses A and B starting from the same point move in a mutually perpendicular direction with speeds uA km h-1 and uB km h-1 respectively. Calculate the relative velocity of A w.r.t B.
Answer:
Since uA and uB are in mutually perpendicular directions, they will cover uA and uB km in one hour respectively. Thus if v km be the separation between them in one hour,
Then v = \(\sqrt{\mathrm{u}_{\mathrm{A}}^{2}+\mathrm{u}_{\mathrm{B}}^{2}}\)
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 8
Thus if vAB be the relative speed of A w.r.t. B, then
vAB = \(\sqrt{\mathrm{u}_{\mathrm{A}}^{2}+\mathrm{u}_{\mathrm{B}}^{2}}\) kmh …(1)

If θ be the direction of vAB w.r.t. uA, Then
tan θ = \(\frac{u_{A}}{v_{A B}}=\frac{u_{A}}{\sqrt{u_{A}^{2}+u_{B}^{2}}}\) …(2)

Thus, equations (1) and (2) give the magnitude and direction of relative velocity of A w.r.t. B.

Question 9.
A draw velocity-time graph for a body which
(i) accelerates uniformly from rest,
Answer:
The required velocity-time graph is shown in Fig. here

Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 9
OA part of the graph represents the uniformly accelerated motion of the body.

(ii) then moves with a uniform velocity and
Answer:
AB part of the graph represents motion with uniform velocity.

(iii) finally retarded uniformly.
Answer:
BC part of the graph represents motion with uniform retardation of the body.

Question 10.
From a velocity-time graph, how do you calculate the average acceleration of a moving body?
Answer:
The slope of the velocity-time curve gives the acceleration of the body. For this purpose, we take a small interval of time Δt and a corresponding change in velocity Δv such that Δt = t2 – t1 and Δv = v2 – v1. Over a very small interval of time Δt, arc AB may be considered as a chord AB, then.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 10
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 11

Question 11.
State whether the following two graphs in Fig. here represent the same type of motion or not. Name the motion of the particle.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 12
Answer:
Yes. Both of these graphs represent the same type of motion. The motion of the particle represented by these graphs is non-uniform motion i.e. the particle is moving with variable velocity.

Question 12.
Draw the velocity-time graph for an object moving with uniform velocity. What does it show for t < 0?
Answer:
As the object moves with uniform velocity, the magnitude and direction of its velocity remain the same at all points of its path. Thus v – t graph’ is a straight line parallel to the time axis as shown in Fig. here.

For t < 0, the v – t graph shows that the object is at rest till t = 0.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 13

Question 13.
The displacement time graph of a body is shown in the figure below. What does the curve for t < 0 and t > 0 show?
Answer:
The line ∥ to t – axis in fig. shows that the object is at rest t ∥ t = 0. For t > 0, it shows that it is still at rest but at some another poi it at a distance XQ from the point for t < 0.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 14

Question 14.
Distinguish between one, two and three-dimensional motion.
Answer:
The motion of an object along a straight line in a fixed direction is called one-dimensional motion.

  • The motion of an object in a plane is called two-dimensional motion
  • The motion of an object in space is called three-dimensional motion

Question 15.
A ball thrown upward reaches a height and comes bad downward. Out of the following statements, which one is true for displacement, velocity and acceleration.
(a) It varies continuously but never changes the sign.
(b) It varies continuously with the maximum, in the beginning, being zero at the top.
(c) It remains constant throughout the course of the journey.
(d) It only changes the sign when the ball is at the top.
Answer:

  • Statement (a) is true for displacement as it varies from starting
  • Statement (b) is true for velocity as it is maximum at the time o projecting the ball and becomes zero at the highest point i.e. top.
  • Statement (c) is true for acceleration as it is always constant throughout the course of the journey and is equal to ‘g’.

Question 16.
Derive the expression for the time taken by a body dropped from a height h to reach at Earth.
Answer:
Here,
initial velocity, u = 0
acceleration, a = g
distance covered, S = h

Let t = time taken
Using the relation
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 15

Question 17.
In which of the following cases, the body may be considered a point object:
(a) A railway carriage moving without jerks between two stations.
(b) A monkey sitting on the shoulder of a cyclist moving smoothly in a circular track.
1 A beaker tumbling down the edge of a table, A spinning cricket ball that turns sharply on hitting the ground.
Answer:
(a) and (b) as the distance moved by the bodies is much larger; the size of the body.

Question 18.
What do you understand by positive and negative time?
Answer:
The origin of time is called zero time. The instant of time which is after the origin of time is called positive time and the instant of, which is taken before the origin of time is called negative time.

Question 19.
If the displacement time graph of a particle is parallel to the displacement axis (b) the time axis, what will be the velocity particle? Why?
Answer:
(a) When the displacement-time graph is parallel to the displacement axis, the velocity of the particle is infinity.
We know that velocity = \(\frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}\)
when the graph is parallel to the displacement axis, then Δt = 0, so velocity = infinity.

(b) When the displacement-time graph is parallel to the time axis, the velocity of the particle is zero as in this case Δx = 0.

Question 20.
An object is in uniform motion along a straight line. What will be its position-time graph if
(a) x0 = +ve, v = +ve,
(b) x0 = +ve, v = – ve,
(c) x0 = – ve, v = +ve,
(d) x0 = – ve, v = – ve.
The letters x0 and v represent the position of the object at time t = 0 and the uniform velocity of the object respectively.
Answer:
The position of an object at any time t moving with a uniform velocity along a straight line is given by
x = x0 + vt ….(1)
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 16
(a) If X0 > 0, v > 0 i.e. both positive, then the position-time graph is as shown in Fig. (a).
(b) If X0 = +ve, v = – ve, then the x – t graph is as shown in Fig. (b).
(c) If X0 = – ve, v = +ve, then the x – t graph is as shown in Fig. (c).
(d) If both x0 and v are – ve, then the x -t graph is as shown in Fig. (d).

Question 21.
Define displacement. What are its characteristics?
Answer:
It is defined as the change in the position of an object in a particular direction.

Characteristics of displacement:

  1. It is a vector quantity.
  2. It has units of length.
  3. The magnitude of displacement is called distance.
  4. The diode valve curve between voltage and current is quite like curve ABCD.

Question 23.
What are the important points about the uniform motion?
Answer:
The following are some important points about the uniform motion:

  1. The velocity in uniform motion does not depend upon the time interval (t2 – 1,).
  2. The velocity in uniform motion is independent of the choice of origin.
  3. The average and the instantaneous velocities have the same value in uniform motion.
  4. No force acts on the object having uniform motion.
  5. Velocity is taken to be positive when the object moves toward the right of the origin and it is taken -ve if an object moves toward the left of the origin.

Question 24.
A car is being driven at a uniform velocity u. The driver suddenly puts his foot on the accelerator and the speed increases to v. Unfortunately after that his brakes failed. Show his velocity-time graph.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 17
Answer:
Initially, the graph will be parallel to the time axis and is represented by AB when the velocity is uniform. Thereafter it is accelerated and its velocity becomes v. The car will then move with this velocity represented by CD.

Question 25.
Draw position-time graphs of two objects moving along a straight line when their relative velocity is (a) zero and (b) non-zero.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 18
Answer:
The motion of two objects moving along a straight line can be represented by two parallel lines on the position-time graph.

In case of zero relative velocity the two vectors \(\vec{A}\) and \(\vec{B}\) will have the same magnitude as shown in fig. (a) and when the relative velocity is non-zero, the length of the two vectors i.e. the magnitude of vectors is different even though they are parallel and is shown in fig.(b).

Question 26.
Prove that the distance travelled by an object in nth second is given by
Snth = u + \(\frac{a}{2}\) (2n – 1)
Answer:
Derivation: Let Sn and Sn-1 be the distances covered by an object in n and n-1 seconds respectively.
Let u = its initial velocity
a = acceleration produced in the object

∴ Using the relation
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 19
If Snth is the distance covered by the object in nth second. Then
Snth = Sn – Sn-1
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 20
Hence proved.

Question 27.
Is it possible that the velocity of an object be in a direction other than the direction of acceleration? When?
Answer:
Yes, when a body moves in a circular path, then the direction of the velocity is along the tangent to the point on the circle and the acceleration is always towards its centre.

Question 28.
Is the rate of change of acceleration with the time important to describe the motion of a body? Why?
Answer:
No, because it is observed that only velocity and acceleration are sufficient to understand and explain the motion of a body.

Question 29.
Explaining with an example, why does a person sitting in a train think that the other train is at rest when both are moving on parallel tracks with the same speed and in the same direction?
Answer:
This is because the relative velocity of the train in which the person is sitting w.r.t. the other train is zero.
e.g. Let two trains A and B are moving along east with a velocity of 50 km/h i.e. vA = vB = 50 kmh1.

∴ relative velocity of A w.r.t. B is given by
VAB = vA – vB = 50 – 50 = 0.

Question 30.
Can a body be said to be at rest as well as in motion? Explain.
Answer:
Yes, both rest and motion are relative terms. A body at rest w.r.t. one object may be in motion w.r.t. another object, e.g. a person sitting in a moving train is at rest w.r.t. other passengers in the train but at the same time, he is in motion w.r.t. the surroundings (trees or buildings) on the side of the track.

Motion in a Straight Line Important Extra Questions Long Answer Type

Question 1.
Define the following terms :
(a) speed
Answer:
Speed : It is defined as the time rate of change of position
i. e. distance of an object.

i.e. Speed = \(\frac{\text { Distance travelled by the object }}{\text { Time taken }}\)

(b) uniform speed
Answer:
Uniform Speed: An object is said to be moving with uniform speed if it covers equal distances in equal small intervals of time.

(c) variable speed
Answer:
Variable Speed: An object is said to be moving with variable speed if it covers equal distances in unequal small intervals of time.

(d) average speed
Answer:
Average Speed: It is used to measure the variable speed of an object.

It is defined as the ratio of the total distance travelled by the object to the total time taken.
∴ vav = \(\frac{\text { Total }}{\text { Total time taken }}\)

(e) instantaneous speed
Answer:
Instantaneous Speed: It is defined as the speed of an object at a given instant of time. It is denoted by vins.
∴ If Δs be the distance covered by an object in a small time interval Δt s.t. Δt → 0,
Then
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 21
Thus in the case of the uniform motion of an object, the instantaneous speed is equal to its uniform speed.

(f) velocity
Answer:
Velocity: It is defined as the time rate of change of displacement of an object.

(g) uniform velocity
Answer:
Uniform Velocity: An object is said to be moving with uniform velocity if it undergoes equal displacements in equal intervals of time however small these intervals may be.

(h) variable velocity
Answer:
Variable Velocity: An object is said to be moving with variable velocity if either its magnitude (i.e. speed) or its direction or both change with time.

(i) uniform motion
Answer:
Uniform Motion: An object is said to be in uniform motion if it undergoes equal displacements in equal intervals of time which may be very small.

(j) average velocity in uniform
Answer:
Average Velocity in Uniform Motion: The velocity of an object in uniform motion may be defined as the ratio of the .displacement of the object to the total time interval for which the motion takes place.
i.e. v = \(\frac{x_{2}-x_{1}}{t_{2}-t_{1}}\)

(k) relative velocity motion
Answer:
Relative Velocity: The relative velocity of a moving object with respect to another object is defined as the rate of change of relative position of one object w.r.i. another object.
Or
It is the velocity with which one object moves with respect to another object.

(l) instantaneous velocity
Answer:
The instantaneous velocity of an object: It is defined as the velocity of an object at any instant of time or any point on its path.
Or
It is defined as the limiting value of the average velocity of the object as Δt → 0.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 22

(m) acceleration
Answer:
Acceleration: It is defined as the time rate of change of velocity of an object. It is a vector quantity.

(n) retardation
Answer:
Retardation: It is defined as the negative acceleration produced in the object.

(o) variable acceleration
Answer:
Variable Acceleration: An object is said to be moving with variable acceleration if its velocity changes by unequal magnitudes in equal intervals of time.

(p) average acceleration
Answer:
Average Acceleration: It is defined as the ratio of change in velocity in a given time interval to the total time taken.

(q) uniform acceleration
Answer:
Uniform Acceleration: An object is said to be moving with uniform acceleration if it undergoes equal changes in velocity in equal intervals of time.

(r) instantaneous acceleration.
Answer:
Instantaneous Acceleration: It is defined as the acceleration of an object at a particular instant of time or at a particular point on its path.
Or
It may be defined as the limiting value of the average acceleration in a small time interval around that instant when the time-interval tends to zero.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 23

Question 2.
Explain the importance of the position-time graph.
Answer:
1. The importance of a position-time graph is that its slope gives the velocity of the object in uniform motion.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 24
Let us consider the position-time graph of an object moving with uniform velocity represented by the line DB making angle 0 with the time axis. Let the coordinates of D and B be (x, t) and (x’, t’ ) respectively. Let BA and DC’ be perpendiculars drawn from B and D respectively on the time axis and BE and DC be perpendiculars on the y-axis from B and D.
Now BE’ = CE = x’ – x
and C’A = t’ – t = DE’

Then velocity = \(\frac{x^{\prime}-x}{t^{\prime}-t}=\frac{B E^{\prime}}{D E^{\prime}}\) = tanθ
So, velocity v = slope of position-time graph.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 25

2. The position-time graph for a stationary object is a straight line parallel to the time axis. Here the slope of the curve is zero, which means the object is stationary as v = 0.

3. In the case of variable velocity the position time curve is not a straight line. In this case, the slope of the curve gives the average velocity
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 26
= slope or chord AB when Δt → 0,

then the slope of curve gives the instantaneous velocity.
Instantaneous velocity =
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 27
Thus the position-time graph gives information about velocity.

Question 3.
Derive relations :
(i) v = u + at
Answer:
v = u + at:
Derivation: By def. of acceleration, we know that
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 28
where v1 and v2 are the velocities of an object at times t1 and t2 respectively.
If v1 = u (initial velocity of the object) at t1 = 0
v2 = v (final velocity of the object) at t2 = t
Then (1) reduces to v = u + at
Hence derived.

(ii) v2 – u2 = 2as
Answer:
Derivation : We know that acceleration is given by a =\(\frac{v_{2}-v_{1}}{t_{2}-t_{1}}\) , where v1 and v2, t1 and t2 are as in (1).
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 29
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 30
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 31

(iii) s = ut + \(\frac{1}{2}\) at2.
Answer:
Derivation:
Let x1, V1 = position and velocity of the object at time t1.
x2, v2 = position and velocity of the object at time t2.
a = uniform acceleration of the object.

Also Let vav = average velocity in t2 – t1 interval
∴ By definition
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 32
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 33
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 34
Hence derived.

Numerical Problems:

Question 1.
In a car race, car A takes a time t s less than car B and finishes the finishing point with a velocity v more than that of car B. Assuming that the cars start from rest and travel with constant acceleration a, and a respectively show that
v = t\(\sqrt{\mathbf{a}_{1} \mathbf{a}_{2}}\)
Answer:
Let v1, v2 be the finishing velocities of car A and car B respectively. Let t1, t2 be the finishing time-intervals for car A and car B respectively.
d = distance travelled by both cars.
∴ according to the statement
V = v1 – v2 ….(i)
t = t2 – t1 …(ii)

Let d = distance covered during a race by each car
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 35
Also using equation
s = ut + \(\frac{1}{2}\) at2, we get
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 36
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 37

Question 2.
A car accelerates from rest at a constant rate a, for sometime after which it decelerates at a constant rate B to come to rest. If the total time elapsed is t s, then calculate
(a) maximum velocity reached.
(b) total distance travelled.
Answer:
The given situation is shown in fig. below:
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 38
Let t1 and t2 be the times of acceleration and deceleration respectively of the car.

If t = total time of journey,
then t = t1 + t2 … (i)

Also let v = maximum velocity reached
(1) For accelerated motion, using equations, v = u + at, we get
v = 0 + αt1
or
v = αt1 … (ii)

(2) For decelerated motion,
here, u = v, final velocity is zero.
∴ 0 = v + (- β)t2
v = βt2 …(iii)

∴ From (ii) and (iii), we get
αt1 = βt2
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 39
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 40

Question 3.
The driver of a train moving at a speed v1 sights another train at a distance d ahead of hint moving in the same direction with a slower speed v2. He applies brakes and gives a constant deacceleration ‘a’ to his train. Show that there will be no collision if
d > \(\frac{\left(\mathbf{v}_{1}-\mathbf{v}_{2}\right)^{2}}{2 \mathbf{a}}\)
Answer:
Let the collision be just avoided i.e. the rear train comes to rest after travelling a distance d with a declaration ‘a’.
the initial speed of 1 st train = v1
the initial speed of 2nd train = v2
If u = initial relative speed of 1 st w.r.t. 2nd train, then
u = v1 – v2

Final velocity of rear train v = 0
∴ using equations, v2 – u2 = 2as, we get
0 – (v1 – v2)2 = 2(- a) d (Here a = -a, s = d.)
d = \(\frac{\left(\mathbf{v}_{1}-\mathbf{v}_{2}\right)^{2}}{2 \mathbf{a}}\)
Thus there will be no collision, if
d > \(\frac{\left(\mathbf{v}_{1}-\mathbf{v}_{2}\right)^{2}}{2 \mathbf{a}}\)

Question 4.
A stone is dropped from the top of a tower and after one second, another stone is dropped from a point 20 m below the top. If both the stones reach the ground at the same time, calculate the height of the tower (take g = 10 mf2).
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 73
Answer:
The given situation is shown in fig.
Let t = time taken by the first stone to reach the ground
h = height of the tower
u1 = initial velocity O
∴ using equation
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 41
∴ t – 1 = Time taken by the second stone to reach the ground and
h – 20 = height of the point from where it is dropped l.
∴ u2 = initial velocity = 0

Thus using equation (1), we get
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 42
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 43

Question 5.
A body travels half of its total path in the last second of its fall from rest. Calculate the time and height of fall.
Answer:
Let h = height of fall
t = time of fall
a = g = 9.8ms-2
u = 0

∴ using equation,
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 44

Also according to the statement first half i.e. \(\frac{h}{2}\) is covered in (t – 1)s.
∴ using (i), we get
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 45
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 46
As last half of the journey takes 1 s, so t cannot be less than 1 s.
∴ t = 3.41s
∴ From(ii), h = 4.9 × (3.414)2
= 57.11 m
∴ h = 57.11, t = 3.414s.

Question 6.
A train moves from one station to another in 2h time. Its speed during the motion is shown in fig. Calculate
(a) the maximum acceleration during the journey.
(b) the distance covered during 0.75 to 1.00 h interval.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 47
Answer:
(a) Maximum acceleration during the journey
= slope of AB part of the graph
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 48
∴ acceleration = 160 kmh-2.

(b) Distance covered during the interval 0.75 to 1.00 h
= Area under the v -1 curve from 0.75 to 1.00
= Area of trapezium CABD
= Area of rectangle AEDC + area of AAEB
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 49

Question 7.
When the speed of the car is y, the minimum distance over which it can be stopped is z. If the speed of the car becomes v, then what will be the minimum distance over which it can be stopped?
Answer:
Case I:
Here, the initial velocity of car = v
The final velocity of car = 0

Let a = acceleration of car Distances covered before stopping = x
∴ Using the equation,
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 50
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 51

Question 8.
A body describes 10m in the third second and 12m in the Sth second with uniform acceleration. Find the distance covered by it in :
(a) next 3′ seconds and
(b) 8th second of its motion.
Answer:
Let the initial velocity of body = v
and let acceleration of body = a
Also, we know that distance covered during the nth second is given by
Snth =u + \(\frac{a}{2}\)(2n – 1)

∴ Distance travelled in 3rd and 5th second is:
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 52
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 53
(a) using the equation, S = ut + at2, Let us find the distance covered by the body in 5s and 8 seconds.
Take u = 7.5,ms-1, a= I ms2

If x1 and x2 be the distances covered in 5s and 8s respectively.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 54
= 60 + 32 = 92 m

∴ Distance travelled by the body in next 3 seconds is given by
= x2 – x1
= 92 – 50 = 32 m

(b) The distance trave1ìdby the body in 8th second (x8) is given by
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 55

Question 9.
The velocity of an object (ms-1) moving along a straight line is given by
v = 10 + 3t2 …. (1)
Find (a) change in velocity in the interval when t = 2s and 5s.
(b) average acceleration in the same interval.
(c) instantaneous acceleration at t = 4s.
Answer:
Comparing equation (1), with the equation v = u + at, we find that u = 10ms-1,
(a) Now velocity of the object after 2 and 5 seconds is calculated as :
v = 10 + 3t2
putting t = 2, v = 10 + 3 × 22 = 10 + 3 × 4 = 22ms-1
putting t = 5, v = 10 + 3 × 52 = 10 + 3 × 25 = 85 ms-1.
∴ Change in velocity from 2s to 5 s
= 85 – 22 = 63 ms-1.

(b) We know that average acceleration in a given time interval is given by
aav = \(\frac{\Delta \mathrm{v}}{\Delta \mathrm{t}}\)

Thus aav in the interval 2s to 5 is given by
aav = \(\frac{85-22}{5-2}=\frac{63}{3}\) = 21 ms-2.

(c) We know that instantaneous acceleration is given by
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 56

Question 10.
A car travels at a rate of 30 km h-1 for 5 minutes and then at 40 km h-1 for 10 minutes.
Find (i) the total distance covered by the car
(ii) the average speed of the car.
Answer:
v1 = 30 kmh-1 t1 = 5 min = \(\frac{5}{60}\) h = \(\frac{1}{12}\) h.
If S1 be the distance covered in 5 min., then
S1 = v1 t1 = 30 × \(\frac{1}{12}\) = \(\frac{5}{2}\) km = 2.5 km

Now v2 = 40 kmh-1, t2 = 10 minutes = \(\frac{1}{6}\) h

If S2 = distance covered in 10 min., then
S2 = v2 t2 = 40 × \(\frac{1}{6}\) = \(\frac{20}{3}\) km = 6.67 km.

(i) If S be the total distance, then
S = S1 + S2 = 2.5+ 6.67 = 9.17 km.

(ii) If vav be the average speed of the car,
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 57

Question 11.
A particle covered half the distance with a velocity v0. The remaining half distance was covered with a velocity v1 for half the time and with velocity v2 for another half of the time. Find the mean speed of the particle averaged over the whole time of motion.
Answer:
Let S = total distance covered by the particle.
t1 = time taken to cover half the distance i.e. \(\frac{S}{2}\)

v0 = velocity of the particle
\(\frac{S}{2}\) = Distance covered with v0.

∴ using equation x = ut, we get
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 58

Now let 2t = time to cover the remaining distance \(\frac{S}{2}\).
If S1 = distance covered with v1 in time t.
S2 = distance covered with v2 in time t,
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 59
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 60

Question 12.
In a circus, a motorcyclist takes 4 rounds on the same track in the globe of radius r with a velocity 5r.
Find (i) total displacement
Answer:
radius of track = r, velocity v = 5r.
As the starting and endpoint in a circular track are the same, so the net displacement = 0

(ii) total distance covered
Answer:
Distance covered in one round = perimeter of the circular track of radius r
= 2π x r
∴ total distance covered in 4 rounds = 4 × 2πr = 8πr.

(iii) total time is taken by him.
Answer:
Let t = time taken by the motorcyclist in completing 4 rounds.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 61

Question 13.
The displacement of a particle moving in one dimension under the action of a constant force is related to the time t by the equation t = \(\sqrt{x}\) + 3, where x is in metre and t is in second. Find the displacement of the particle when its velocity is zero.
Answer:
t = \(\sqrt{x}\) + 3

\(\sqrt{x}\) = t – 3 ….(1)

Squaring on both sides of equation (1), we get
x = (t – 3)2
= t2 + 9 – 6t …..(2)

If v be the velocity of the particle,
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 62
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 63

Question 14.
A car moving along a straight road with a speed of 144 km h-1 is brought to a stop within a distance of 200 m. Calculate its retardation and the time it takes to stop.
Answer:
Here, S = distance covered before topping = 200 m.
u = initial speed of car = 144 km h-1 = 144 × \(\frac{5}{18}\) ms-1 = 40 ms-1
v = its final speed = 0

Let a = retardation of the car =?
using the relation,
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 64

Also Let t = time taken ?
∴ using the relation, v = u + at, we get
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 65

Question 15.
A stone falls from the top of the tower in 8s. How much time will it take to cover the first quarter of the distance starting from the top?
Answer:
Here u = initial speed oftcie = 0
h = height of tower
t = time taken in falling by h = 8s

Let t1 = time taken to cover first quarter i.e. \(\frac{h}{4}\) = ?
a = g = acce. due to gravity

using the relation, S = ut + \(\frac{1}{2}\)at2
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 66

Question 16.
A ball is thrown vertically upwards with a velocity of 30 ms-1. If access. due to gravity is 10 ms-2, what will be the distance travelled by ¡tin the first and last second of upward motion?
Answer:
Here, u = initial velocity in the upward direction 30ms-1
Let after a time t, the ball stops i.e. v = 0
a = acce.= g = 10 ms-2

Let Snth be the distance covered by the ball in the last second i.e. nth second of its motion =7
∴ using the relation, v = u + at,
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 67

Let x1 be the distance covered by it in first second of its motion,
∴ using the relation, S = ut + \(\frac{1}{2}\) gt2, we get
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 68
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 69

Question 17.
A rocket is fired vertically from the ground. It moves upwards with a constant acceleration of 10 ms-2 for the 30s after which the fuel is consumed. After what time from the instant of firing, the rocket will attain the maximum height? Take g = 10ms-2.
Answer:
Here, a = g = 10ms-2 for t = 30s
∴ u = 0 for first part of the motion
v = velocity attained by the rocket after 30s = ?

∴ using the relation,
v = u + at …(1)
we get v = 0 + 10 × 30
= 300 ms-1

for second part of motion,
u = v = 300 ms-1
v1 = final velocity = 0

Let t1 be the time of journey for this part of the motion,
∴ from equation (1), we get
0 = 300 – 10 × t1

Let T = total time of flight in which the rocket attains maximum height,
T = t + t1
= 30 + 30
= 60 second.

Question 18.
On a foggy day, two car drivers spot each other when they are just 80 m apart. They were travelling at 72 km h-1 and 60 km h-1 respectively. Both of them simultaneously apply brakes which retard both the cars at 5ms-2. Tell whether they will avert a collision or not?
Answer:
Here, let u, and u2 be the initial velocities of the two cars.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 70

If S1 and S2 be the distances covered by the two cars before coming to rest (i.e. v = 0), then
using the relation,
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 71

If S be the total distance covered by two cars before coming to rest, then
S = S1 + S2 = 40 + 27.7 = 67.7 m.

Also S1 = 80 m.
Now clearly the total distance covered by the cars before coming to rest is less than 80 m, so the collision will be averted.

Question 19.
A stone is dropped from a rising balloon at a height of 76 m above the ground and reaches the ground in 6s. What was the velocity of the balloon when the stone was dropped?
Answer:
Let u be the velocity of the balloon in an upward direction at the point
A = initial velocity of stone in an upward direction.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 72
The stone rises to the top, comes to rest and then it starts coming back to the ground to hit B.

Let us take A as origin :
∴ Net vertical distance covered by the stone, y = – 76 m.
a = – g = – 9.8 ms-2
(Here y and acc. are taken as -ve because they are in a downward direction w.r.t. A)
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 74

Question 20.
A car moves at a velocity of 2.24 km h-1 in the first minute, at 3.60 km h-1 in the second minute and at 5.18 km h-1 in the third minute. Calculate the average velocity in these three minutes.
Answer:
Let x1, x2 and x3 be the displacements of car in 1 st, 2nd and 3rd minutes respectively and let v1, v2 and v3 be the velocities of these time intervals,
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 75

∴ v1 = 2.24 kmh-1, v2 = 3.60 kmh-1, v3 = 5.18 kmh-1
time intervals = t1 = t2 = t3 = 1 minute = \(\frac{1}{60}\) h
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 76
If x be the total displacement, then
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 77
If v be the average velocity in these 3 minutes, then
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 78

Question 21.
A man is walking due East at the rate of 3 km h-1. Rain appears to fall down vertically at the rate of 3 km h-1. Find the actual velocity and direction of the rainfall.
Answer:
Let a = angle made by the rainfall with the vertical Velocity of man represented by OA due East = 3 km h-1
As rain appears to fall vertically downward, so OB = 3 km h-1 represents the velocity of rain w.r.t. man. The actual velocity of rainfall is represented by OC and is given by
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 79
Thus the rain is actually falling at 3\(\sqrt{2}\) km h-1 at an angle of 45° west of the vertical.
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 80

Question 22.
A car starting from rest and moving with uniform acceleration possesses average velocities 5 ms-1, 10 ms-1 and 15 ms-1 in the first, second and third seconds. What is the total r distance covered by the car in these three seconds?
Answer:
Here, let v1, v2 and v3 be the average velocities of 1st, 2nd and 3rd secs, respectively.
∴ v1 = 5 ms-1, v2 = 10 ms-1, v3 = 15 ms-1
t1 = t2 = t3 = 1 s = time intervals

If x1, x2 and x3 be the distances covered in these seconds resp., then
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 81

Question 23.
A person is running at a maximum speed of 3 ms-1 along the length of a train to catch hold of the door of a compartment. When he is just 2.5 m from the door, the train steams off with an acceleration of 1 ms-2. Find how long it takes him to catch the door?
Answer:
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 82
Here, let u1 = 3 m s-1, be the initial speed of man at point A.
For train x1 = 2.5 m = distance of the train from man.
u = 0, a = acceleration of train = 1 ms-2

Let C be their meeting point after a time t.
If BC = x, then usirig the relation,
S = ut + \(\frac{1}{2}\) at2, we get

For train, x = 0 + \(\frac{1}{2}\) × 1 × t2 = 0.5t2 … (1)
For man, using the relation,
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 83
∴ t = 1, 5 i.e. he will catch the door first after Is and then after 5s.
t = 1s.

Question 24.
The displacement of a body along the x-axis changes according to the relation: x = 20 – 15t + 4t2, where x is in metres, t in seconds. Determine its position, velocity and acceleration at t = O.
Answer:
Here, x = 4t2 – 1 5t + 20 … (1)
v = \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\) = 4.2t – 15 = 8t – 15 …(2)

Also a = \(\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}\) = 8
At t = 0, the position (x), velocity (y) and acceleration (a) are given by
x = 4.0 – 15 × 0 + 20 = 20 m
v = 8 × 0 – 15 = – 15ms-1
and a = 8 ms-2.

Question 25.
A body undergoes a uniformly accelerated motion. Its velocity after 5 seconds is 25 ems and after 8 seconds it is 34 cms’. Calculate the distance covered in the 12th second.
Answer:
Let u and a be the initial speed and acceleration of the body.
Let v1 and v2 be its velocities after 5s and 8s respectively
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 84
Let Snth be the distance covered in the 12th second.

Using the relation,
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 85

Value-Based Type:

Question 1.
On a two-lane road, there are hoardings hanged on the electric poles “Save Energy”.
If car a is travelling at a speed of 54 km h1 and car ‘B’ moves with 90 km h-1 from the opposite direction.
Now, answer these questions:
(a) Which values are depicted in the above problem? Write down the four ways by which the energy can be saved.
Answer:
Save energy to save our environment.

The four ways to save energy are as under :

  1. By using CFLs, switching off appliances when not in use.
  2. Saving water by using it efficiently.
  3. Using solar energy.
  4. By using public transport.

(b) Write a slogan on “save energy”.
Answer:
Try your self

(c) Find the velocity of B with respect to A? Also, find the velocity of the ground with respect to B?
Solution:
VA = + 54 km h-1 = 15 ms-1
VB = 90 km h-1 = – 25 ms-1

(Takingthe velocity of car A positive N v and car B negative)
∴ Relative velocity of B with respect to A
= VB – VA = – 25 – 15 = – 40 ms-1

i.e the car Bappers to A to move with a speed 40 ms-1 from opposite direction.
Relative velocity of ground w.r.t B = 0 – VB = 0 – (- 25)
= 25 ms-1

Question 2.
A candle march was organised by an NGO. The theme of the candle march was “No use of non-biodegradable products”. The organizer made a semicircular track of radius R on which peoples have to run taking a poster in their hands. If Mahesh travelled from A to B in time t with constant speed, then answer the following:
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 86
(i) Why should non-biodegradable products not be used? Justify your answer.
Answer:
Biodegradable products are reusable and cause less pollution. So, non-biodegradable products must be avoided.

(ii) Find the
(a) Displacement
Answer:
Displacement = minimum distance between initial and final point = AB = 2 R

(b) Average speed
Answer:
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 87

(c) Average velocity
Answer:
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 88

(d) Average acceleration
Answer:
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 89

Question 3.
In sports day activities of a public school, the lines were drawn with chalk powder. Gagandcep an athlete runs a distance of 1500 m in the following manner, (i) Starting from rest, he accelerates himself uniformly at 2ms_1, till he covers a distance of 900 m. (ii) He then runs the remaining distance of 600 m at the uniform speed developed.
(i) Which value is depicted in the above problem?
Answer:
Importance of sports activities.

(ii) Do you think that sports activities are important in day-to-day life?
Answer:
Yes, sports activities are very much important in day-to-day life as
(a) It gives break from regular life
(b) It improves physical strength
(c) It relaxes our mind and body
(d) It helps to concentrate on other works
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 90im-90

(iii) Calculate the time taken by the athlete to cover the two parts of the distance covered.
Answer:
Let the athlete run from O to B Here, OB = 1500 m,
OA = 900 m, AB = 600 m
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 91
Case II : Taking motion of athelete from A to B u = 60 m/s, S = 600m, a = o (∵ Speed is uniform or constant)
t = t2 = ?
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 92

Question 4.
An old woman crossing the road was holding a money purse. She was not able to walk. A pickpocket snatches away her purse. A school student of class XI having seen this incident tries to help that old lady. He informs the polic^inspector who stands nearby. The inspector collects the money purse from the pickpocket and hands it over to the old lady.
(a) What value do you find in the school student?
Answer:
The student is sympathetic towards others, helping, and applies his presence of mind in solving the problems, knows how to use public services.

(b) Also the police inspector in a jeep is chasing the pickpocket on a straight road. The jeep is going at its maximum speed ‘v’. The pickpocket rides on the motorcycle of a waiting friend when the jeep is at a distance ‘d’ away. And the motorcycle starts with a constant acceleration ‘a’. Show that the pickpocket will be caught if v ≥ \(\sqrt{2ad}\).
Answer:
Let the police inspector catches the pickpocket after time t.
The distance covered by the pickpocket in this time interval is given by:
s = \(\frac{1}{2}\) at2

Distance travelled by the inspector (s1) = vt
the police inspector will catch the pickpocket if
Class 11 Physics Important Questions Chapter 3 Motion in a Straight Line 93

The police inspector will catch the bus if t is real i.e v2 ≥ 2ad or v ≥ \(\sqrt{2ad}\)
Hence, the minimum speed of the police inspector for catching the pickpocket is \(\sqrt{2ad}\) proved.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11

More Exercises

Midpoint Calculator is used to find the midpoint between 2 line segments using the midpoint formula. Use our calculator to find accurate midpoints step by step.

Question 1.
Find the co-ordinates of the mid-point of the line segments joining the following pairs of points:
(i) (2, – 3), ( – 6, 7)
(ii) (5, – 11), (4, 3)
(iii) (a + 3, 5b), (2a – 1, 3b + 4)
Solution:
(i) Co-ordinates of the mid-point of (2, -3), ( -6, 7)
\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) or \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q1.2

Question 2.
The co-ordinates of two points A and B are ( – 3, 3) and (12, – 7) respectively. P is a point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of P.
Solution:
Points are A (-3, 3), B (12, -7)
Let P (x1,  y1) be the point which divides AB in the ratio of m1 : m2 i.e. 2 : 3
then co-ordinates of P will be
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q2.1

Question 3.
P divides the distance between A ( – 2, 1) and B (1, 4) in the ratio of 2 : 1. Calculate the co-ordinates of the point P.
Solution:
Points are A (-2, 1) and B (1, 4) and
Let P (x, y) divides AB in the ratio of m1 : m2 i.e. 2 : 1
Co-ordinates of P will be
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q3.1

Question 4.
(i) Find the co-ordinates of the points of trisection of the line segment joining the point (3, – 3) and (6, 9).
(ii) The line segment joining the points (3, – 4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, – 2) and \(\left( \frac { 5 }{ 3 } ,q \right) \) respectively, find the values of p and q.
Solution:
(i) Let P (x1, y1) and Q (x2, y2) be the points
which trisect the line segment joining the points
A (3, -3) and B (6, 9)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q4.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q4.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q4.3

Question 5.
(i) The line segment joining the points A (3, 2) and B (5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x – 18y + k = 0. Find the value of k.
(ii) A point P divides the line segment joining the points A (3, – 5) and B ( – 4, 8) such that \(\frac { AP }{ PB } =\frac { k }{ 1 } \) If P lies on the line x + y = 0, then find the value of k.
Solution:
(i) The point P (x, y) divides the line segment joining the points
A (3, 2) and B (5, 1) in the ratio 1 : 2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q5.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q5.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q5.3

Question 6.
Find the coordinates of the point which is three-fourth of the way from A (3, 1) to B ( – 2, 5).
Solution:
Let P be the required point, then
\(\frac { AP }{ AB } =\frac { 3 }{ 4 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q6.2

Question 7.
Point P (3, – 5) is reflected to P’ in the x- axis. Also P on reflection in the y-axis is mapped as P”.
(i) Find the co-ordinates of P’ and P”.
(ii) Compute the distance P’ P”.
(iii) Find the middle point of the line segment P’ P”.
(iv) On which co-ordinate axis does the middle point of the line segment P P” lie ?
Solution:
(i) Co-ordinates of P’, the image of P (3, -5)
when reflected in x-axis will be (3, 5)
and co-ordinates of P”, the image of P (3, -5)
when reflected in y-axis will be (-3, -5)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q7.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q7.2

Question 8.
Use graph paper for this question. Take 1 cm = 1 unit on both axes. Plot the points A(3, 0) and B(0, 4).
(i) Write down the co-ordinates of A1, the reflection of A in the y-axis.
(ii) Write down the co-ordinates of B1, the reflection of B in the x-axis.
(iii) Assign.the special name to the quadrilateral ABA1B1.
(iv) If C is the mid point is AB. Write down the co-ordinates of the point C1, the reflection of C in the origin.
(v) Assign the special name to quadrilateral ABC1B1.
Solution:
Two points A (3, 0) and B (0,4) have been plotted on the graph.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q8.1
(i)∵ A1 is the reflection of A (3, 0) in the v-axis Its co-ordinates will be ( -3, 0)
(ii)∵ B1 is the reflection of B (0, 4) in the .x-axis co-ordinates of B, will be (0, -4)
(iii) The so formed figure ABA1B1 is a rhombus.
(iv) C is the mid point of AB co-ordinates of C” will be \(\frac { AP }{ AB } =\frac { 3 }{ 4 } \)
∵ C, is the reflection of C in the origin
co-ordinates of C, will be \(\left( \frac { -3 }{ 2 } ,-2 \right) \)
(v) The name of quadrilateral ABC1B1 is a trapezium because AB is parallel to B1C1.

Question 9.
The line segment joining A ( – 3, 1) and B (5, – 4) is a diameter of a circle whose centre is C. find the co-ordinates of the point C. (1990)
Solution:
∵ C is the centre of the circle and AB is the diameter
C is the midpoint of AB.
Let co-ordinates of C (x, y)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q9.1

Question 10.
The mid-point of the line segment joining the points (3m, 6) and ( – 4, 3n) is (1, 2m – 1). Find the values of m and n.
Solution:
Let the mid-point of the line segment joining two points
A(3m, 6) and (-4, 3n) is P( 1, 2m – 1)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q10.1

Question 11.
The co-ordinates of the mid-point of the line segment PQ are (1, – 2). The co-ordinates of P are ( – 3, 2). Find the co-ordinates of Q.(1992)
Solution:
Let the co-ordinates of Q be (x, y)
co-ordinates of P are (-3, 2) and mid-point of PQ are (1, -2) then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q11.1

Question 12.
AB is a diameter of a circle with centre C ( – 2, 5). If point A is (3, – 7). Find:
(i) the length of radius AC.
(ii) the coordinates of B.
Solution:
AC = \(\sqrt { { \left( 3+2 \right) }^{ 2 }+{ \left( -7-5 \right) }^{ 2 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q12.1

Question 13.
Find the reflection (image) of the point (5, – 3) in the point ( – 1, 3).
Solution:
Let the co-ordinates of the images of the point A (5, -3) be
A1 (x, y) in the point (-1, 3) then
the point (-1, 3) will be the midpoint of AA1.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q13.1

Question 14.
The line segment joining A \(\left( -1,\frac { 5 }{ 3 } \right) \) the points B (a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects y-axis. Calculate
(i) the value of a
(ii) the co-ordinates of P. (1994)
Solution:
Let P (x, y) divides the line segment joining
the points \(\left( -1,\frac { 5 }{ 3 } \right) \), B(a, 5) in the ratio 1 : 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q14.1

Question 15.
The point P ( – 4, 1) divides the line segment joining the points A (2, – 2) and B in the ratio of 3 : 5. Find the point B.
Solution:
Let the co-ordinates of B be (x, y)
Co-ordinates of A (2, -2) and point P (-4, 1)
divides AB in the ratio of 3 : 5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q15.1

Question 16.
(i) In what ratio does the point (5, 4) divide the line segment joining the points (2, 1) and (7 ,6) ?
(ii) In what ratio does the point ( – 4, b) divide the line segment joining the points P (2, – 2), Q ( – 14, 6) ? Hence find the value of b.
Solution:
(i) Let the ratio be m1 : m2 that the point (5, 4) divides
the line segment joining the points (2, 1), (7, 6).
\(5=\frac { { m }_{ 1 }\times 7+{ m }_{ 2 }\times 2 }{ { m }_{ 1 }+{ m }_{ 2 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q16.1

Question 17.
The line segment joining A (2, 3) and B (6, – 5) is intercepted by the x-axis at the point K. Write the ordinate of the point k. Hence, find the ratio in which K divides AB. Also, find the coordinates of the point K.
Solution:
Let the co-ordinates of K be (x, 0) as it intersects x-axis.
Let point K divides the line segment joining the points
A (2, 3) and B (6, -5) in the ratio m1 : m2.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q17.1

Question 18.
If A ( – 4, 3) and B (8, – 6), (i) find the length of AB.
(ii) in what ratio is the line joining AB, divided by the x-axis? (2008)
Solution:
Given A (-4, 3), B (8, -6)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q18.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q18.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q18.3

Question 19.
(i) Calculate the ratio in which the line segment joining (3, 4) and( – 2, 1) is divided by the y-axis.
(ii) In what ratio does the line x – y – 2 = 0 divide the line segment joining the points (3, – 1) and (8, 9)? Also, find the coordinates of the point of division.
Solution:
(i) Let the point P divides the line segment joining the points
A (3, 4) and B (-2, 3) in the ratio of m1 : m2 and
let the co-ordinates of P be (0, y) as it intersects the y-axis
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q19.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q19.2

Question 20.
Given a line segment AB joining the points A ( – 4, 6) and B (8, – 3). Find:
(i) the ratio in which AB is divided by the y-axis.
(ii) find the coordinates of the point of intersection.
(iii)the length of AB.
Solution:
(i) Let the y-axis divide AB in the ratio m : 1. So,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q20.1

Question 21.
(i) Write down the co-ordinates of the point P that divides the line joining A ( – 4, 1) and B (17,10) in the ratio 1 : 2.
(ii)Calculate the distance OP where O is the origin.
(iii)In what ratio does the y-axis divide the line AB ?
Solution:
(i) Let co-ordinate of P be (x, y) which divides the line segment joining the points
A ( -4, 1) and B(17, 10) in the ratio of 1 : 2.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q21.1

Question 22.
Calculate the length of the median through the vertex A of the triangle ABC with vertices A (7, – 3), B (5, 3) and C (3, – 1)
Solution:
Let D (x, y) be the median of ΔABC through A to BC.
∴ D will be the midpoint of BC
∴ Co-ordinates of D will be,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q22.1

Question 23.
Three consecutive vertices of a parallelogram ABCD are A (1, 2), B (1, 0) and C (4, 0). Find the fourth vertex D.
Solution:
Let O in the mid-point of AC the diagonal of ABCD
∴ Co-ordinates of O will be
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q23.1

Question 24.
If the points A ( – 2, – 1), B (1, 0), C (p, 3) and D (1, q) from a parallelogram ABCD, find the values of p and q.
Solution:
A (-2, -1), B (1, 0), C (p, 3) and D (1, q)
are the vertices of a parallelogram ABCD
∴ Diagonal AC and BD bisect each other at O
O is the midpoint of AC as well as BD
Let co-ordinates of O be (x, y)
When O is mid-point of AC, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q24.1

Question 25.
If two vertices of a parallelogram are (3, 2) ( – 1, 0) and its diagonals meet at (2, – 5), find the other two vertices of the parallelogram.
Solution:
Two vertices of a ||gm ABCD are A (3, 2), B (-1, 0)
and point of intersection of its diagonals is P (2, -5)
P is mid-point of AC and BD.
Let co-ordinates of C be (x, y), then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q25.1

Question 26.
Prove that the points A ( – 5, 4), B ( – 1, – 2) and C (5, 2) are the vertices of an isosceles right angled triangle. Find the co-ordinates of D so that ABCD is a square.
Solution:
Points A (-5, 4), B (-1, -2) and C (5, 2) are given.
If these are vertices of an isosceles triangle ABC then
AB = BC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q26.1

Question 27.
Find the third vertex of a triangle if its two vertices are ( – 1, 4) and (5, 2) and mid point of one sides is (0, 3).
Solution:
Let A (-1, 4) and B (5, 2) be the two points and let D (0, 3)
be its the midpoint of AC and co-ordinates of C be (x, y).
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q27.1

Question 28.
Find the coordinates of the vertices of the triangle the middle points of whose sides are \(\left( 0,\frac { 1 }{ 2 } \right) ,\left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right) and\left( \frac { 1 }{ 2 } ,0 \right) \)
Solution:
Let ABC be a ∆ in which \(D\left( 0,\frac { 1 }{ 2 } \right) ,E\left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right) andF\left( \frac { 1 }{ 2 } ,0 \right) \),
the mid-points of sides AB, BC and CA respectively.
Let co-ordinates of A be (x1, y1), B (x2, y2), C (x3, y3)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q28.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q28.2

Question 29.
Show by section formula that the points (3, – 2), (5, 2) and (8, 8) are collinear.
Solution:
Let the point (5, 2) divides the line joining the points (3, -2) and (8, 8)
in the ratio of m1 : m2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q29.1

Question 30.
Find the value of p for which the points ( – 5, 1), (1, p) and (4, – 2) are collinear.
Solution:
Let points A (-5, 1), B (1, p) and C (4, -2)
are collinear and let point A (-5, 1) divides
BC in the ratio in m1 : m2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q30.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q30.2

Question 31.
A (10, 5), B (6, – 3) and C (2, 1) are the vertices of triangle ABC. L is the mid point of AB, M is the mid-point of AC. Write down the co-ordinates of L and M. Show that LM = \(\\ \frac { 1 }{ 2 } \) BC.
Solution:
Co-ordinates of L will be
\(\left( \frac { 10+6 }{ 2 } ,\frac { 5-3 }{ 2 } \right) or\left( \frac { 16 }{ 2 } ,\frac { 2 }{ 2 } \right) or(8,1)\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q31.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q31.2

Question 32.
A (2, 5), B ( – 1, 2) and C (5, 8) are the vertices of a triangle ABC. P and.Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2.
(i) Find the co-ordinates of P and Q.
(ii) Show that PQ = \(\\ \frac { 1 }{ 3 } \) BC.
Solution:
A (2, 5), B (-1, 2) and C (5, 8) are the vertices of a ∆ABC,
P and Q are points on AB
and AC respectively such that \(\frac { AP }{ PB } =\frac { AQ }{ QC } =\frac { 1 }{ 2 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q32.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q32.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q32.3

Question 33.
The mid-point of the line segment AB shown in the adjoining diagram is (4, – 3). Write down die co-ordinates of A and B.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q33.1
Solution:
A lies on x-axis and B on the y-axis.
Let co-ordinates of A be (x, 0) and of B be (0, y)
P (4, -3) is the mid-point of AB
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q33.2

Question 34.
Find the co-ordinates of the centroid of a triangle whose vertices are A ( – 1, 3), B(1, – 1) and C (5, 1) (2006)
Solution:
Co-ordinates of the centroid of a triangle,
whose vertices are (x1, y1), (x2, y2) and
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q34.1

Question 35.
Two vertices of a triangle are (3, – 5) and ( – 7, 4). Find the third vertex given that the centroid is (2, – 1).
Solution:
Let the co-ordinates of third vertices be (x, y)
and other two vertices are (3, -5) and (-7, 4)
and centroid = (2, -1).
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q35.1

Question 36.
The vertices of a triangle are A ( – 5, 3), B (p – 1) and C (6, q). Find the values of p and q if the centroid of the triangle ABC is the point (1, – 1).
Solution:
The vertices of ∆ABC are A (-5, 3), B (p, -1), C (6, q)
and the centroid of ∆ABC is O (1, -1)
co-ordinates of the centroid of ∆ABC will be
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q36.1

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