MCQ Questions for Class 8 Science Chapter 3 Synthetic Fibres and Plastics with Answers

Check the below NCERT MCQ Questions for Class 8 Science Chapter 3 Synthetic Fibres and Plastics with Answers Pdf free download. MCQ Questions for Class 8 Science with Answers were prepared based on the latest exam pattern. We have provided Synthetic Fibres and Plastics Class 8 Science MCQs Questions with Answers to help students understand the concept very well. https://mcqquestions.guru/mcq-questions-for-class-8-science-chapter-3/

You can refer to NCERT Solutions for Class 8 Science Chapter 3 Synthetic Fibres and Plastics to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Synthetic Fibres and Plastics Class 8 MCQs Questions with Answers

Choose the correct option in the following questions:

Class 8 Science Chapter 3 MCQ Question 1.
What is the other name for artificial silk?
(a) Nylon
(b) Rayon
(c) Acrylic
(d) Polyester

Answer

Answer: (b) Rayon


Synthetic Fibres And Plastics Class 8 MCQ Question 2.
Which fibre is used as artificial wool?
(a) Acrylic
(b) Rayon
(c) Nylon
(d) Cotton

Answer

Answer: (a) Acrylic


MCQ Questions For Class 8 Science Chapter 3 Question 3.
Polymers are made up of small units called
(a) layers
(b) molecules
(c) cells
(d) monomers

Answer

Answer: (d) monomers


Class 8 Science Ch 3 MCQ Question 4.
The strongest synthetic fibre is
(a) nylon
(b) rayon
(c) polyester
(d) acrylic

Answer

Answer: (a) nylon


Ncert Class 8 Science Chapter 3 MCQ Question 5.
Wood pulp is used to make
(a) plastic
(b) wool
(c) jute
(d) rayon

Answer

Answer: (d) rayon


MCQ For Class 8 Science Chapter 3 Question 6.
Which of the following is a natural fibre?
(a) Wool
(b) Nylon
(c) PVC
(d) Polythene

Answer

Answer: (a) Wool


Synthetic Fibres And Plastics Class 8 MCQ With Answers Question 7.
Melamine is
(a) thermoplastic polymer
(b) thermosetting polymer
(c) fibre
(d) elastomer

Answer

Answer: (b) thermosetting polymer


Class 8 Synthetic Fibres And Plastics MCQ Question 8.
Fibre produced in factories is called
(a) man-made fibre
(b) natural fibre
(c) synthetic fibre
(d) both (a) and (c)

Answer

Answer: (d) both (a) and (c)


Class 8 Chapter 3 Science MCQ Question 9.
PET is a
(a) polyester
(b) polyamide
(c) nylon
(d) thermosetting polymer

Answer

Answer: (a) polyester


Synthetic Fibres And Plastics MCQ Question 10.
Which of the following cannot be recycled?
(a) Toys
(b) Cooker handles
(c) Carry bags
(d) Plastic chair

Answer

Answer: (b) Cooker handles


Science Class 8 Chapter 3 MCQ Question 11.
The plastic which cannot be recycled is
(a) jute
(b) rayon
(c) petrochemicals
(d) bakelite

Answer

Answer: (d) bakelite


MCQ Questions For Class 8 Science With Answers Chapter 3 Question 12.
Bakelite is an example of
(a) fibre
(b) elastomer
(c) nylon
(d) thermosetting polymer

Answer

Answer: (d) thermosetting polymer


Ch 3 Science Class 8 MCQ Question 13.
Which term is used for polymers made up of a large number of glucose units?
(a) Protein
(b) Fructose
(c) Cellulose
(d) Polyester

Answer

Answer: (c) Cellulose


Chapter 3 Science Class 8 MCQ Question 14.
Which of the following is non-biodegradable?
(a) Paper
(b) Cotton cloth
(c) Wood
(d) Plastic

Answer

Answer: (d) Plastic


Class 8 Science Chapter 3 MCQ With Answers Question 15.
Which of the following is natural fibre obtained from plants?
(a) Cotton
(b) Wool
(c) Rayon
(d) Ketone

Answer

Answer: (a) Cotton


Match the following items given in Column ‘A with that in Column ‘B’.

Column ‘A’Column ‘B’
CottonThermoplastic
NylonPolyester
PVCBiodegradable
BakeliteArtificial fibres
PETArtificial wool
AcrylicChain of glucose units
cellouseNon-biodegradable
PaperNatural fibre
Plastic BagsCook wares
TeflonThermosetting
Answer

Answer:

Column ‘A’Column ‘B’
CottonNatural fibre
NylonArtificial fibres
PVCThermoplastic
BakeliteThermosetting
PETPolyester
AcrylicArtficial wool
cellouseChain of glucose units
PaperBiodegradable
Plastic BagsNon-biodegradable
TeflonCook wares

Fill in the blanks with appropriate words:

1. Synthetic fibres are made by ………….

Answer

Answer: human beings


2. Many small units combine to form a large single unit called a ………….

Answer

Answer: polymer


3. A nylon wire is …………. than steel wire.

Answer

Answer: stronger


4. All the synthetic fibres are prepared using raw materials of petroleum origin called ………….

Answer

Answer: petrochemicals


5. Plastics are …………. conductors of electricity.

Answer

Answer: poor


6. Nylon was prepared from …………. , …………. and ………….

Answer

Answer: coal, water, air.


State whether the statements given below are True or False’.

1. All fibres are made up of very large units which in turn are made up of many small units.

Answer

Answer: True


2. Nylon was also known as artificial silk in China.

Answer

Answer: False


3. Acrylic is an artificial wool.

Answer

Answer: True


4. Plastics are light, strong and durable.

Answer

Answer: True


5. It is not advised to store food items in plastic containers as it may react with the w ills of container.

Answer

Answer: False


6. Bake lite is poor conductor of electricity.

Answer

Answer: True


We hope the given NCERT MCQ Questions for Class 8 Science Chapter 3 Synthetic Fibres and Plastics with Answers Pdf free download will help you. If you have any queries regarding Synthetic Fibres and Plastics CBSE Class 8 Science MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Aldehydes, Ketones and Carboxylic Acids Class 12 MCQs Questions with Answers

Aldehyde Ketone And Carboxylic Acid MCQ Chapter 12 Question 1.

Which of the following compounds will give butanone on oxidation with alkaline KMnO4 solution ?

(A) Butan-l-ol
(B) Butan-2-ol
(C) Both of these
(D) None of these
Answer:
(B) Butan-2-ol

Explanation:
Butan-2-ol is secondary alcohol which on oxidation with alkaline KMnO4 solution gives butanone (ketone).
Aldehyde Ketone And Carboxylic Acid MCQ Chapter 12

Aldehydes And Ketones Class 12 MCQ Chapter 12 Question 2.

Write the IUPAC name of

Aldehydes And Ketones Class 12 MCQ Chapter 12
(A) 1-Aminopropanaldehyde
(B) 2-Aminopropanal
(C) 1-Aminoethan-l-al
(D) None of the above
Answer:
(B) 2-Aminopropanal

Explanation:
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 2
2-Aminopropanal

MCQ On Aldehydes Ketones And Carboxylic Acids Chapter 12 Question 3.

What kind of compounds undergo Cannizaro reactions ?

(A) Ketones with no a- hydrogen
(B) Aldehydes with a- hydrogen
(C) Carboxylic acids with a- hydrogen
(D) Aldehydes with no a- hydrogen
Answer:
(D) Aldehydes with no a- hydrogen

Explanation:
Aldehydes with no a-hydrogen undergo Canizzaro reaction.

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Class 12 Chemistry Chapter 12 MCQ Chapter 12 Question 4.

Write the product(s) in the following reactions:

MCQ On Aldehydes Ketones And Carboxylic Acids Chapter 12
(A) No product formed
Class 12 Chemistry Chapter 12 MCQ Chapter 12
Answer:
Option (B) is correct.

Explanation:
It is a nucleophilic addition reaction.
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 5

Aldehydes, Ketones And Carboxylic Acids MCQ With Answers Pdf Question 5.

Compounds A and C in the following reaction are

Aldehydes, Ketones And Carboxylic Acids MCQ With Answers Pdf
(A) identical
(B) positional isomers
(C) functional isomers
(D) optical isomers
Answer:
(B) positional isomers

Explanation:
MCQ Of Aldehydes And Ketones Class 12 Chapter 12
In compound A and C, position of -OH group is changed. So, these are positional isomers.

MCQ Of Aldehydes And Ketones Class 12 Chapter 12 Question 6.

In Clemmensen reduction carbonyl compound is treated with ………….

(A) zinc amalgam + HCl
(B) sodium amalgam + HCl
(C) zinc amalgam + nitric acid
(D) sodium amalgam + HNO3
Answer:
(A) zinc amalgam + HCl

Explanation:
Clemmensen reduction is used to convert carbonyl group to CH2 group as follows:
MCQ Of Aldehyde And Ketone Class 12 Chapter 12

MCQ Of Aldehyde And Ketone Class 12 Chapter 12 Question 7.

The reagent which does not react with both, acetaldehyde and benzaldehyde.

(A) Sodium hydrogen sulphite
(B) Phenyl hydrazine
(C) Fehling’s solution
(D) Grignard reagent
Answer:
(C) Fehling’s solution

Explanation:
Aliphatic aldehydes(acetaldehyde) reduce the Fehling’s solution to red cuprous oxide.
Aldehydes Ketones And Carboxylic Acids MCQ
Aromatic aldehydes (benzaldehyde) do not react with Fehling’s solution.

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Aldehydes Ketones And Carboxylic Acids MCQ Question 8.

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 34
(A) C6H5COOH + CH4
(B) C6H5COONa + CHI3
(C) C6H6 + CH3COONa + HI
(D) C6H5CH2COOH
Answer:
(B) C6H5COONa + CHI3

Explanation:
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 10

MCQs On Aldehydes And Ketones Class 12 Chapter 12 Question 9.

Predict the product of the following reaction:

MCQs On Aldehydes And Ketones Class 12 Chapter 12
(A)CH3CH2CH3
(B) CH3CHOHCH3
(C) CH3CH2CHO
(D) CH3CONHCH3
Answer:
(A)CH3CH2CH3

Explanation:
Aldehydes Ketones And Carboxylic Acids MCQ
It is a VVolff-Kishner reduction which converts MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 14 group into-CH2– group.

Aldehydes And Ketones MCQ Class 12 Chapter 12 Question 10.

Which of the following compounds is most reactive towards nucleophilic addition reactions ?

Aldehydes And Ketones MCQ Class 12 Chapter 12
Answer:
Option (A) is correct.

Explanation:
Methyl benzaldehyde < Benzal-dehyde < Propanone < Ethanal – reactivity to-wards nucleophilic substitution. Aldehydes are more reactive than aliphatic ketones. Aliphatic ketones are more reactive than aromatic ketones. The +1 effect is more in ketone than in aldehyde. Thus ketone will be least reactive in nucleophilic addition reactions. The presence of electron withdrawing group increases the reactivity towards the addition while the presence of electron donating group decreases the reactivity of compound towards nucleophilic addition. Benzaldehyde does not favour nucleophilic addition reaction due to resonance stabilisation.

MCQ Aldehydes And Ketones Class 12 Chapter 12 Question 11.

Formaldehyde reacts with methyl magnesium bromide followed by hydrolysis to form.

(A) Methanol
(B) Ethanol
(C) Propanol
(D) Butanol
Answer:
(B) Ethanol

Explanation:
Aldehyde And Ketone MCQ Class 12 Chapter 12

Aldehyde And Ketone MCQ Class 12 Chapter 12 Question 12.

Common name of Ethane-1,2-dioic acid is known as

(A) Oxalic acid
(B) Phthalic acid
(C) Adipic acid
(D) Acetic acid
Answer:
(A) Oxalic acid

Explanation:
Structural formula of Ethane-1, 2 dioicacidis
Aldehyde Ketone And Carboxylic Acid MCQ Questions
It is oxalic acid.

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Aldehyde Ketone And Carboxylic Acid MCQ Questions Question 13.

The carboxylic acid that does not undergo HVZ reaction is

(A) CH3COOH
(B) (CH3)2COOH
(C) CH3CH2CH2CH2COOH
(D) (CH3)3CCOOH
Answer:
(C) CH3CH2CH2CH2COOH

Explanation:
The carboxylic acids having a-hydrogen atom undergo HVZ reaction. Since (CH3)3C. COOH doesnot contain a-H-atom; so, it does not undergo HVZ reaction.

Multiple Choice Questions Aldehydes And Ketones Question 14.

Which of the following acids does not form anhydride ?

(A) Formic acid
(B) Acetic acid
(C) Propionic acid
(D) n-butyric acid
Answer:
(A) Formic acid

Explanation:
Formic acid(HCOOH) does not form anhydride because it does not contain u-C-atom.

Aldehydes And Ketones MCQs Class 12 Chapter 12 Question 15.

Which of the following is the strongest acid?

(A) Acetic acid
(B) Phenol
(C) Methyl alcohol
(D) Water
Answer:
(A) Acetic acid

Explanation:
Acetic acid is the strongest acid because it loses H+ ion to form carboxylic ion (CH3COO ) which gets stabilised by resonance:

Aldehydes Ketones And Carboxylic Acids Class 12 MCQ Question 16.

The reaction in which the aqueous solution of sodium salt of carboxylic acids on electrolysis give alkanes:

(A) Soda lime decarboxylation
(B) Kolbe’s electrolysis decarboxylation
(C) Dry distillation of calcium formate
(D) Reduction of carboxylic acid.
Answer:
(B) Kolbe’s electrolysis decarboxylation

Explanation:
It is Kolbe’s electrolytic decarboxylation.
RCOONa(aq) → RCOO + Na+
At anode, 2RCOO → R-R + 2CO2 + 2e Alkane
At cathode, 2H2 O + 2e + H2 + 2OH

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Aldehydes And Ketones Questions And Answers Pdf Class 12 Question 17.

Arrange the following compounds in increasing order of acid strength

Aldehydes And Ketones Questions And Answers Pdf Class 12
(A) (i) > (ii) > (iii)
(B) (ii) < (i) < (iii)
(C) (iii) < (i) < (ii)
(D) (iii) > (i) > (ii)
Answer:
(C) (iii) < (i) < (ii)

Explanation:
The electron with drawing group (-NO2) increases the acid strength of aromatic acids while electron releasing group (-CH3) decreases the acid strength of aromatic acids. Hence, the increasing order of add strength is given as
Aldehydes And Ketones Questions And Answers Pdf Class 12

Aldehydes And Ketones MCQ Pdf Class 12 Chapter 12 Question 18.

Complete the following reaction: CH3COONa + NaOH-

Aldehydes And Ketones MCQ Pdf Class 12 Chapter 12
Answer:
Option (C) is correct.

Explanation:
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 21

Question 19.

Identify the name of the given reaction:

Aldehydes Ketones And Carboxylic Acids Class 12 MCQ
(A) Etard reaction
(B) Hell-Volhard-Zelinsky reaction
(C) Stephen reaction
(D) None of the above
Answer:
(B) Hell-Volhard-Zelinsky reaction

Explanation:
Hell-Volhard-Zelinsky

Assertion And Reason Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is true

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 1.

Assertion (A): Oxidation of ketones is easier than aldehydes.
Reason (R): C-C bond of ketones is stronger than C-H bond of aldehydes.

Answer:
(D) A is false and R is true

Explanation:
Oxidation of aldehydes are easier than ketones.

Question 2.

Assertion (A): Benzaldehyde is less reactive than ethanal towards nucleophilic addition reactions.
Reason (R): Ethanal is more sterically hindered.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of carbonyl group present in elhanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance hence it is less reactive than ethanal towards nucleophilic addition reaction.

Question 3.

Assertion (A): Aromatic aldehydes and formaldehyde undergo Cannizzaro reaction.
Reason (R): Aromatic aldehydes are almost as reac¬tive as formaldehyde.

Answer:
(C) A is true but R is false

Explanation:
Aromatic aldehydes and formalde¬hyde do not contain a-hydrogen and thus undergo Cannizzaro reaction. Formaldehyde is more reactive than aromatic aldehydes.

Question 4.

Assertion (A): Aldehydes and ketones, both react with Tollen’s reagent to form silver mirror.
Reason (R): Both aldehydes and ketones contain a carbonyl group.

Answer:
(D) A is false and R is true

Explanation:
Both aldehydes and ketones have carbonyl group but only aldehydes react with Tollens’ reagent to give silver mirror.

Question 5.

Assertion (A): Benzoic acid does not undergo Friedel-craft’s reaction.
Reaction (R): The carboxyl group is activating and undergo electrophilic substitution reaction.

Answer:
(C) A is true but R is false

Explanation:
The carboxyl group (-COOH) is deactivating group because it is electron with drawing group. It decreases the electron density at benzene ring, hence deactivates it towards electrophilic substitution reactions.

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 6.

Assertion (A): Compounds containing -CHO group are easily oxidised to corresponding carboxylic acids.
Reason (R): Carboxylic acids can be reduced to alcohols by treatment with LiAlH4.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Compounds containing -CHO group are easily oxidised to corresponding* carboxylic acids.

Question 7.

Assertion (A): Aromatic carboxylic groups do not undergo Friedel- Crafts reaction.
Reason (R): Carboxyl group is deactivating and the catalyst aluminium chloride gets bonded to the carboxyl group.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Aromatic carboxylic groups do not urldergo Friedel-Crafts reaction because Carboxyl group is deactivating and the catalyst aluminium chloride gets bonded to the carboxyl group.

Question 8.

Assertion (A): Carboxylic acids are more acidic than phenols.
Reason (R): Phenols are ortho and para directing.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Carboxylic acids are more acidic than phenols as the carboxylate ion, the conjugate base of carboxylic acid is stabilized by two equivalent structures. Thus, the negative charge is delocalized effectively. However, in phenols, negative charge is less effectively delocalized over oxygen atom and carbon atoms in phenoxide ion.
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 23

Case-Based MCQs

I. Read the passage given below and answer the following questions:

Reduction of carboxylic acids and their derivatives plays an important role in organic synthesis, in both laboratory and industrial processes. Traditionally, the reduction is performed using stochiometric amounts of hydride reagents, generating stochiometric amounts of waste. A much more attractive, atom-economical approach is a catalytic reaction using H2; however, hydrogenation of carboxylic acid derivatives under mild conditions is a very challenging task, with amides presenting

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

the highest challenge among all classes of carbonyl compounds. Very few examples of the important hydrogenation of amides to amines, in which the C-O bond is cleaved with the liberation of water (Scheme 1), were reported. C-O cleavage of amides can also be affected with silanes as reducing agents. (Generation of amides to the with cleavage of the C-N products of C-O cleavage the case of anilides). The and neutral, homogeneous
Scheme 1. General Sche C-O cleavage
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 24

We have now prepared the new, dearomatized, bipyridine-based pincer complex 3, catalyst 3 (Here refered as Cat. 3). Remarkably, it efficiently catalyzes the selective hydrogenation of amides to form amines and alcohols (eq 1). The reaction proceeds under mild pressure and neutral conditions, with no additives being required.

Since the reaction proceeds well under anhydrous conditions, hydrolytic cleavage of the amide is not involved in this process. been reported.6 Amines and chemical, pharmaceutical and ch a reaction is conceptually step in amide hydrogenation bonvl group to form a very anhydrous condition involved in this pro
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 25

In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage.
(A) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(B) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(C) Assertion is correct statement but reason is wrong statement.
(D) Assertion is wrong statement but reason is correct statement.

Question 1.

Assertion (A): The use of catalyst 3 is an efficient method of preparation of primary amines
Reason (R): Use of catalyst 3 is a step down reaction.

Answer:
(B) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Question 2.

Assertion (A): Use of hydride catalyst or hydrogen brings about cleavage of C-O bond in amides.
Reason (R): Hydride catalyst or hydrogen cause to reduction of amides.

Answer:
(B) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 3.

Assertion (A): N-methyl ethanamide on reaction with catalyst 3 will yield ethanol and methanamine.
Reason (R): Use of Catalyst 3 brings about cleavage of C-N bond of amides

Answer:
(A) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Question 4.

Assertion (A): Aniline can be prepared from suitable amide using catalyst 3
Reason (R): The use of catalyst 3 is limited to aliphatic amides only.

Answer:
(C) Assertion is correct statement but reason is wrong statement.

II. Read the passage given below and answer the following questions:

Aldehydes, ketones and carboxylic acids are few of the major classes of organic compounds containing carbonyl group. Aldehydes are prepared by dehydrogenation or controlled oxidation of primary alcohols and controlled or selective reduction of acyl halides. Ketones are prepared by oxidation of secondary alcohols and hydration of alkynes. Carboxylic acids are prepared by the oxidation of primary alcohols, aldehydes and alkenes by hydrolysis of nitriles and by treatment of Grignard reagents with carbon dioxide.

Question 1.

Name a method by which both aldehydes and ketones can be prepared.

(A) Reduction of carboxylic acids
(B) Ozonolysis of alkenes
(C) Oxidation of alcohols
(D) All of the above
Answer:
(D) All of the above

Explanation:
Both aldehydes and ketones can be prepared by all these methods.

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 2.

How will you distinguish between aliphatic aldehydes and aromatic aldehydes ?

(A) Fehling’s test
(B) Benedict’s test
(C) Iodoform test
(D) Hinsberg reagent
Answer:
(A) Fehling’s test

Explanation:
On heating an aldehyde with Fehling’s reagent, a reddish brown precipitate is obtained. Aldehydes are oxidised to corresponding carboxvlate anion. Aromatic aldehydes do not respond to this test.
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 26

Question 3.

Name the main compounds A and B formed in the following reaction:

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 27
(A) CH3CH2COOH [A], CH3CH2CH3
(B) CH3CH2CHO [A], C2H4 [B]
(C) CH3COCH3 [A], CH3CH2CH3 [B]
(D) CH3COCH3 [A], C2H6 [B]
Answer:
(C) CH3COCH3 [A], CH3CH2CH3 [B]

Explanation:
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 28

Question 4.

The reagent which does not react with both, acetone and benzaldehyde.

(A) Sodium hydrogensulphite
(B) Phenyl hydrazine
(C) Fehlings’ solution
(D) Grignard reagent
Answer:
(C) Fehlings’ solution

Explanation:
Fehling’s solution does not react  with acetone and benzaldehyde as aromatic I aldehydes and ketones do not react with Fehling’s solution. :

OR

Through which of the following reactions number of carbon atoms can be increased in the chain?

(A) Grignard reaction
(B) Cannizzaro reaction
(C) Clemmenson reduction
(D) HVZ reaction
Answer:
(A) Grignard reaction

Explanation:
The number of C-atoms can be i increased in the chain by Grignard reaction.
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 29

III. Read the passage given below and answer the following questions:
Reductive alkylation is the term applied to the process of introducing alkyl groups into ammonia or a primary or secondary amine by means of an aldehyde or ketone in the presence of a reducing agent. The present discussion is limited to those reductive alkylations in which the reducing agent is hydrogen and a catalyst or “nascent” hydrogen, usually from a metalacid combination; most of these reductive alkylations have been carried out with hydrogen and a catalyst.

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

The principal variation excluded is that in which the reducing agent is formic acid or one of its derivatives; this modification is known as the Leuckart reaction. The process of reductive alkylation of ammonia consists in the addition of ammonia to a carbonyl compound and reduction of the addition compound or its dehydration product. The reaction usually is carried out in ethanol solution when the reduction is to be effected catalytically:
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 30
Since the primary amine is formed in the presence of the aldehyde it may react in the same way as ammonia, yielding an addition compound, a Schiff’s base (RCH= NCH2R) and finally, a secondary amine. Similarly, the primary amine may react with the imine, forming an addition product which also is reduced to a secondary amine Finally, the secondary amine may react with either the aldehyde or the imine to give products which are reduced to tertiary amines.
MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 31
Similar reactions may occur when the carbonyl compound employed is a ketone.

Question 1.

Ethanal on reaction with ammonia forms an imine
(X) which on reaction with nascent hydrogen gives
(Y) Identify ‘X’ and ‘Y’.

(A) X is CH3CH=NH and Y is CH3NH2
(B) X is CH3CHOHNH2 and Y is CH3CH2NH2
(C) X is CH3CHOHNH2and Y is CH3NH2
(D) X is CH3CH=NH and Y is CH3CH2NH2
Answer:
(D) X is CH3CH=NH and Y is CH3CH2NH2

Question 2.

Acetaldehyde is reacted with ammonia followed by reduction in presence of hydrogen as a catalyst. The primary amine so formed further reacts with ” acetaldehyde. The Schiff’s base formed during the reaction is:

(A) CH3CH=NHCH3
(B) CH3CH=NHCH2CH3
(C) CH3=NHCH2CH3
(D) CH3CH2CH=NHCH3
Answer:
(B) CH3CH=NHCH2CH3

Question 3.

The reaction of ammonia and its derivatives with aldehydes is called:

(A) Nucleophilic substitution reaction
(B) Electrophilic substitution reaction
(C) Nucleophilic addition reaction
(D) Electrophilic addition reaction
Answer:
(C) Nucleophilic addition reaction

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 4.

MCQ Questions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids - 32
The compound Q is:

(A) (CH3CH2CH2)3N
(B) (CH3CH2CH2)2N(CH2CH3)
(C) (CH3CH2)3N
(D) (CH3CH2)2NH
Answer:
(A) (CH3CH2CH2)3N

OR

Reductive alkylation of ammonia by means of an aldehyde in presence of hydrogen as reducing agents results in formation of:

(A) Primary amines
(B) Secondary amines
(C) Tertiary amines
(D) Mixture of all three amines
Answer:
(D) Mixture of all three amines

MCQ Questions for Class 12 Chemistry with Answers

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Sexual Reproduction in Flowering Plants Class 12 MCQ Questions With Answers

Class 12 Biology Chapter 2 MCQ Question 1.

Among the terms listed below, those that of are not technically correct names for a floral whorl are :

(i) Androecium
(ii) Carpel
(iii) Corolla
(iv) Sepal

(A) (i) and (iv)
(B) (iii) and (iv)
(C) (ii) and (iv)
(D) (i) and (ii)
Answer:
(C) (ii) and (iv)
Class 12 Biology Chapter 2 MCQ

Explanation :
All the four whorls of the plant with their relative position in flower can be indicated through following diagram. Sepals collectively form a whorl, called as calyx while technically the carpel is known as gynoecium. The floral whorls formed by petals and stamens are called as corolla and androecium, respectively.

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Sexual Reproduction In Flowering Plants Class 12 MCQ Question 2.

The embryo sac is related to ovule as is related to an anther.

(A) Stamen
(B) Filament
(C) Pollen grain
(D) Androecium
Answer:
(C) Pollen grain

Explanation :
The pollen grains represent the male gametophytes. As the anthers mature and dehydrate, the microspores dissociate from each other and develop into pollen grains. So, embryo sac is to ovule as pollen grain is to an ther

Chapter 2 Biology Class 12 MCQs Question 3.

In a typically complete, bisexual and hypogynous flower the arrangement of floral whorls on the thalamus from the outermost to the innermost is:

(A) Calyx, corolla, androecium and gynoecium
(B) Calyx, corolla, gynoecium and androecium
(C) Gynoedum, androecium, corolla and calyx
(D) Androecium, gynoecium, corolla and calyx
Answer:
(A) Calyx, corolla, androecium and gynoecium

Explanation :
In a typically complete, bisexual, and hypogynous flower the arrangement of floral whorls on the thalamus from the outer-most to the innermost is as follows :

(i) Calyx: It is the outermost whorl of sepals.
(ii) Corolla: It is a whorl of petals inside the calyx.
(iii) Androecium: It is a whorl of stamens inside the corolla.
(iv) Gynoecium: It is a whorl of pistils (in the center of the flower forming innermost whorls).

Chapter 2 Biology Class 12 MCQ Reproduction Question 4.

A dicotyledonous plant bears flowers but never produces fruits and seeds. The most probable cause for the above situation is:

(A) Plant is dioecious and bears only pistillate flowers
(B) Plant is dioecious and bears both pistillate and staminate flowers
(C) Plant is monoecious
(D) Plant is dioecious and bears only staminate flowers
Answer:
(D) Plant is dioecious and bears only staminate flowers

Explanation:
In dioecious plants, the unisexual male flower is staminate, that is, bearing stamens only,while the female is pistillate or bearing pistils only. For the production of fruits and seeds fertilization must take place, which is possible only in the presence of both male and female flowers. When the plant is dioecious, it will give rise to the following situations :
(i) If the plant is dioecious and bears only pistillate flowers, fertilization can take place with the help of pollinators.
(ii) If the plant is dioecious and bears only staminate flowers, fertilization cannot take place, because female gamete is non-motile which can’t reach the male gamete in order to fuse with it. When the plant is monoecious, that is, carrying both stamen and pistil together it may lead to self-fertilization and production of seed.

Chapter 2 Class 12 Biology MCQ Question 5.

The outermost and innermost wall layers of microsporangium in an anther are respectively:

(A) Endothecium and tapetum
(B) Epidermis and endodermis
(C) Epidermis and middle layer
(D) Epidermis and tapetum
Answer:
(D) Epidermis and tapetum

Explanation :
The outermost and innermost wall layers of microsporangium in an anther are respectively, epidermis and tapetum. A typical microsporangium is generally surrounded by four-wall layers, that is, the epidermis, (outermost protective layer), I endothecia, (middle fibrous layers) and the I tapetum (innermost nutritive layer).

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Biology Chapter 2 Class 12 MCQs Question 6.

During microsporogenesis, meiosis occurs in:

(A) Endothecium
(B) Microspore mother cells
(C) Microspore tetrads
(D) Pollen grains.
Answer:
(B) Microspore mother cells

Explanation :
During microsporogenesis, meiosis occurs in microspore mother cells. As the anther develops, the microspore mother cells of the sporogenous tissue undergo meiotic divisions to form microspore tetrads. The micro-spore tetrad after dehydration is separated into pollen grains.

Biology Class 12 Chapter 2 MCQs Question 7.

From among the sets of terms given below, identify those that are associated with the gynoecium.

(A) Stigma, ovule, embryo sac, placenta
(B) Thalamus, pistil, style, ovule
(C) Ovule, ovary, embryo sac, tapetum
(D) Ovule, stamen, ovary, embryo sac
Answer:
(A) Stigma, ovule, embryo sac, placenta

Explanation :
Gynoecium indicates the female reproductive part of the flower which consists of pistil. Each pistil has three parts, that is, stigma,style, and ovary. Inside the ovarian cavity, the placenta is located. Arising from the placenta there are the megasporangia, commonly called ovules. The functional megaspore undergoing the meiotic division develops into the female gametophyte or embryo sac. Thalamus, tapetum and stamen are not a part of gynoecium. Thalamus is the part of flower which forms the base on which all the floral whorls rest up on. Tapetum is the innermost nutritive layer or microsporangium and stamens are male reproductive part (androecium) of plant.

Class 12 Bio Ch 2 MCQ Question 8.

From the statements given below choose the option that are true for a typical female gametophyte of a flowering plant:

(i) It is 8-nucleate and 7-celled at maturity
(ii) It is free-nuclear during the development
(iii) It is situated inside the integument but outside the nucellus
(iv) It has an egg apparatus situated at the chalazal end

(A) i and iv,
(B) ii and iii
(C) i and ii
(D) ii and iv
Answer:
(C) i and ii

Explanation :
Statement (i) and (ii) are correct regarding female gametophyte of flowering plant. The female gametophyte or embryo sac is located inside the nucellus, enclosed within the integuments. In a majority of flowering plants, one of the megaspores is functional while the other three degenerates. Three repeated mitotic divisions of the functional megaspore result in the formation of seven-celled or eight-nucleate embryo sac. Six of the eight nuclei are organized at the two poles. Three cells grouped at micropylar end form egg apparatus and 3 at the chalazal end form antipodal cells. The large central cell at the center has two polar nuclei. The meiotic divisions in the formation of embryo sac are strictly free nuclear, that is nuclear divisions are not followed immediately by cell-wall formation. Gametophyte is situated at micropylar end, not at chalazal end.

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Chapter 2 Bio Class 12 MCQ Question 9.

Autogamy can occur in a chasmogamous flower if:

(A) Pollen matures before maturity of ovule
(B) Ovules mature before maturity of pollen
(C) Both pollen and ovules mature simultaneously
(D) Both anther and stigma are of equal lengths
Answer:
(C) Both pollen and ovules mature simultaneously

Explanation :
Autogamy is a method of self-pollination. It is a process in which the stigma of a flower receives pollen from the anther of same flower. For autogamy, both the sex organs of a chasmogamous flower should mature at the same time. As chasmogamous flowers open at maturity, pollen release and stigma receptivity should be synchronized for the process of autogamy. In such flowers, the length of anther and stigma plays a secondary role in autogamy, e.g., in case of protandry (in which pollens mature early) and protogyny (in which stigma matures early) leads to cross-pollination.

Reproduction In Flowering Plants Class 12 MCQ Question 10.

Choose the correct statement from the following:

(A) Cleistogamous flowers always exhibit autogamy
(B) Chasmogamous flowers always exhibit geitonogamy
(C) Cleistogamous flowers exhibit both autogamy and geitonogamy
(D) Chasmogamous flowers never exhibit autogamy
Answer:
(A) Cleistogamous flowers always exhibit autogamy

Explanation :
The pollination that occurs in opened flowers is called chasmogamy. It is of two types, that is, self-pollination (autogamy) and cross-pollination. Cross-pollination is of two types, that is, geitonogamy and xenogamy. So, we can say that chasmogamous flowers exhibit both autogamy (self-pollination) and allogamy (cross-pollination). While in cleistogamous flowers the anthers and stigma lie close to each other within the closed flowers. ‘ When anthers dehisce in [he flowers buds, pollen grains come in contact with the stigma or effective pollination. Thus, these flowers are invariably autogamous as there is no chance of cross-pollen landing on the stigma.

Class 12th Biology Chapter 2 MCQ Question 11.

A particular species of plant produces light, non-sticky pollen in large numbers and its stigmas are long and feathery. These modifications facilitate pollination by:

(A) Insects
(B) Water
(C) Wind
(D) Animals
Answer:
(C) Wind

Explanation :
Plants use two abiotic (wind and water) and one biotic (animals) agent to achieve pollination. Majority of plants use biotic agents for pollination. Pollination by wind is more common amongst abiotic pollination. It requires the light and non-sticky pollen grains so that, they can be transported in wind currents. They often possess well-exposed stamens (so that the pollens are easily dispersed into wind currents) and large often feathery stigma to easily trap air-borne pollen grains. Wind pollination is common in grasses. Pollination by water is called hydrophily which is quite rare in flowering plants but occurs in aquatic plants. Zoophily is pollination through the agency of animals. Entomophily (pollination by insects) is the most common type of zoophily which occurs through the agency of animals.

Biology Class 12 Chapter 2 MCQ Question 12.

From among the situations given below, choose the one that prevents both autogamy and geitonogamy

(A) Monoecious plant bearing unisexual flowers
(B)Dioecious plant bearing only male or female flowers
(C) Monoecious plant with bisexual flowers
(D) Dioecious plant with bisexual flowers
Answer:
Option (B) is correct

Explanation :
Dioecious plants (bearing only male or female flowers) prevent both autogamy and geitonogamy. Autogamy is a method of self-pollination in which the transfer of pollen grains from another to stigma of the same flower takes place. Geitonogamy is the transfer of pollen grains from another to stigma of another flower of the same plant. It is ecologically cross¬pollination which is supposed to be equivalent to self-pollination because all flowers on a plant are genetically identical.

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

MCQ Of Chapter 2 Biology Class 12 Question 13.

While planning for an artificial hybridization program involving dioecious plants, which of the following steps would not be relevant:

(A) Bagging of female flower
(B) Dusting of pollen on. stigma
(C) Emasculation
(D) Collection of pollen
Answer:
(C) Emasculation

Explanation :
Artificial hybridization is one of the major methods of crop improvement programs. This cross will make sure that only the desired pollen grains are used for pollination and the stigma is protected from contamination (from unwanted pollen). This is achieved by emasculation and bagging techniques. If the female parent produces unisexual flowers; there is no need for emasculation (a process of removal of anther). The female flower buds are bagged before the flowers open. When the stigma becomes receptive, pollination is carried out using the desired pollen and the flower rebagged. This protects them from contamination by unwanted pollen grains. When the female parent bears bisexual flowers, removal of anthers from the flower bud before the anther dehiscence is necessary,

Class 12 Biology Ch 2 MCQ Question 14.

In a flower, if the megaspore mother cell forms megaspores without undergoing meiosis and if one of the megaspores develops into an embryo sac, its nuclei would be:

(A) Haploid
(B) Diploid
(C) A few haploid and a few diploid
(D) With varying ploidy.
Answer:
(B) Diploid

Explanation :
In some species, the diploid egg cell is formed without reduction division and develops into an embryo without fertilization. IL is an asexual reproduction which occurs in the absence of pollinators or in extreme environments. In some species like citrus plants, nucellar cells surrounding ) the embryo sac start dividing and develop into embryos. It occurs in the megaspore mother cell without undergoing meiosis and produces diploid embryo sac through mitotic divisions. It helps in the preservation of desirable characters for indefinite period. Thus, it can be concluded that apomictic species produce diploid cells. Haploid cells will be formed during sexual reproduction when cells will undergo meiosis.

Class 12 Chapter 2 Biology MCQ Question 15.

Which one of the cell in an embryo-sac produces endosperm after double fertilization?

(A) Synergids cell
(B) Antipodal cell
(C) Central Cell
(D) Egg
Answer:
(C) Central Cell

Explanation :
In female gametophyte, central cell is involved in double fertilization that helps in the endosperm development While antipodal cells provide nourishment to the egg cell and synergid cell help in pollen tube growth.

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Ch 2 Biology Class 12 MCQ Question 16.

Starting from the innermost part, the correct sequence of parts in an ovule are,

(A) egg, nucellus, embryo sac, integument
(B) egg, embryo sac, nucellus, integument
(C) embryo sac, nucellus, integument, egg
(D) egg, integument, embryo sac, nucellus.
Answer:
(B) egg, embryo sac, nucellus, integument

Explanation:
Starting from the innermost part, the correct sequence of parts in an ovule is egg, embryo-sac, nucellus, and integument.

MCQ Biology Class 12 Chapter 2 Question 17.

In a fertilized embryo sac, the haploid, diploid and triploid structures are:

(A) Synergid, zygote and primary endosperm nucleus
(B) Synergid, antipodal and polar nuclei
(C) Antipodal, synergid and primary endosperm nucleus
(D) Synergid, polar nuclei and zygote.
Answer:
(A) Synergid, zygote and primary endosperm nucleus

Explanation :
In a fertilized embryo sac, the haploid, diploid and triploid structures are synergids, zygote and primary endosperm nucleus respectively.

Class 12 Bio Chapter 2 MCQ Question 18.

In an embryo sac, the cells that degenerate after fertilization are:

(A) Synergids and primary endosperm cell
(B) Synergids and antipodals
(C) Antipodals and primary endosperm cell
(D) Egg and antipodals.
Answer:
(B) Synergids and antipodals

Explanation :
In unfertilized embryo sac, the antipodal sand synergids are distinctly present at chalazal end and micropylar end respectively while, in fertilized embryo sac, antipodals and synergids gradually degenerate after the formation of zygote.

Biology Chapter 2 MCQ Class 12 Question 19.

In the embryos of a typical dicot and a grass, true homologous structures are:

(A) Coleorhiza and coleoptile
(B) Coleoptile and scutellum
(C) Cotyledons and scutellum
(D) Hypocotyl and radicle.
Answer:
(C) Cotyledons and scutellum

Sexual Reproduction In Flowering Plants Class 12 MCQ

Explanation:
A typical dicotyledonous embryo consists of two cotyledons. While embryos of monocotyledons possess only one cotyledon and it is called scutellum (in grass). Cotyledons of dicots are simple structures generally thick and swollen due to storage of food reserves (as in legumes) and embryo of monocots consists of one large and shield-shaped cotyledon known as scutellum situated towards one side (lateral) of the embryonal axis.

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

MCQ Of Biology Class 12 Chapter 2 Question 20.

The phenomenon observed in some plants wherein parts of the sexual apparatus is used for forming embryos without fertilization is called:

(A) Parthenocarpy
(B) Apombds
(C) Vegetative propagation
(D) Sexual reproduction.
Answer:
(B) Apombds

Explanation :
Apomixis refers to the formation of seeds without fertilization. The embryos are genetically identical to the parental plan

Question 21.

The phenomenon wherein, the ovary develops into a fruit without fertilization is called:

(A) Parthenocarpy
(B) Apomixis
(C) Asexual reproduction
(D) Sexual reproduction
Answer:
(A) Parthenocarpy

Explanation :
Parthenocarpy is the formation of seedless fruits without fertilization. The fruits developed from unfertilized ovaries are ‘ called parthenocarpic fruits.

Question 22.

Fragrant flowers with well-developed nectaries are an adaptation for

(A)hydrophily
(B) anemophily
(C) entomophily
(D) none of these
Answer:
(C) entomophily

Explanation :
Entomophily is a type of pollination which occur by the insects, butterfly, wasp, ants, beetles and mainly by bees which is most common, the flowers are colourful attract the insect. Nectar is given as reward to insects.

Question 23.

The total number of nuclei involved in double fertilization in angiosperm are

(A) two
(B) three
(C) four
(D) five
Answer:
(D) five

Explanation:
Double fertilization is the process in angiosperms. It involves fusion of one male gamete (haploid) with egg (haploid) to form v zygote (diploid) that gives rise to embryo accompanied with fusion of another male gamete (haploid) with two polar nuclei (secondary nucleus) to form primary endosperm nucleus (PEN) that gives rise to a nutritive tissue called endosperm.

Question 24.

Heterostyly as a contrivance for cross-pollination is found in

(A) Pennisetum
(B) Impatiens
(C) Primula vulgaris
(D) Oenothera
Answer:
(C) Primula vulgaris

Explanation :
Heterostyly is the presence s of 2—3 types of flower with different heights of styles and stamens are dimorphic heterostyly, there are two types of flower, pin eyed (long style and short stamens) and thrum eyed (short style and long stamens), Primula vulgaris (primrose), jasmine.

Assertion and Reason Based MCQs

Directions: In the following questions a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as :

(A) Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of assertion (A).
(B) Both assertion (A) and Reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(C) Assertion (A) is true but reason (R) is false.
(D) Assertion (A) is false but reason (R) is true.

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Question 1.

Assertion (A): Tapetum is a part of another wall that has 2-3 layers of cells.
Reason (R): Tapetum layers help in development and growth of pollen grain.

Answer:
(B) Both assertion (A) and Reason (R) are true but reason (R) is not the correct explanation of assertion (A).

Explanation :
In flowering plants, tapetum are the specialized cells that provide nutrition to the pollen grain within the anther.

Question 2.

Assertion (A): Pollen grains are best preserved as fossils.
Reason (R): The sporopollenin of exine is highly resistant to the action of strong acids and alkali and can withstand a high temperature.

Answer:
(A) Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of assertion (A).

Explanation:
Pollen grains are well preserved as fossils because of the presence of sporopollenin which is the most resistant organic material known.

Question 3.

Assertion (A): Tapetum is formed during the process of the formation of microsporangium.
Reason (R): The play an important role in guiding the pollen tubes into the synergid.

Answer:
(C) Assertion (A) is true but reason (R) is false.

Explanation :
Assertion is true, but the reason is wrong because tapetum plays an important role in nourishing pollen mother cells (PMCs)or microspores.

Question 4.

Assertion (A): Flowers are structure of sexual reproduction.
Reason (R): Different types of embryological process occur inside the flower.

Answer:
(A) Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of assertion (A).

Explanation:
Sexual reproduction involves the transfer fusion of male and female gametes known as pollination. The fertilized ovules: produce seeds that continue the next generation.

Question 5.

Assertion (A): Cleistogamous flowers can produce seeds without pollination.
Reason (R): Cleistogamous flowers have no chance of cross-pollination and they are invariably autogamous.

Answer:
(A) Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of assertion (A).

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Question 6.

Assertion (A): Entomophilous flowers are large, colorful, fragrant and rich in nectar.
Reason (R): If helps in attracting the pollinating agent.

Answer:
(A) Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of assertion (A).

Explanation :
Entomophily is a type of pollination, which is carried out by insects., Fragrance and color of the flowers attract insects.

Question 7.

Assertion (A): In Ophrys one petal of the flower bears on an uncanny resemblance to the female bee.
Reason (R): Two closely related species competing for the same resource can co-exist simultaneously.

Answer:
(A) Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of assertion (A).

Question 8.

Assertion (A): Perisperm is a haploid tissue.
Reason (R): Perisperm is the remains of nucellus which surround the embryo in certain seeds.

Answer:
(D) Assertion (A) is false but reason (R) is true.

Explanation:
Perisperm is a nutritive tissue of a seed derived from the nucellus and deposited | externally to the embryo sac. It is diploid.

Question 9.

Assertion (A): Pea, bean, mustard are non- albuminous seeds.
Reason (R): These seeds retain a part of endosperm as it is not completely used up during embryo development.

Answer:
(C) Assertion (A) is true but reason (R) is false.

Explanation :
Assertion is true but the reason is wrong because, in non-albuminous seeds, seeds does not retain any endosperm as it is completely used up during embryo development.

Question 10.

Assertion (A): Geitonogamous flowering plants are cross-pollinated plants.
Reason (R): In geitonogamous flowering plants the pollen is transferred to the stigma of another flower of another plant.

Answer:
(C) Assertion (A) is true but reason (R) is false.

Explanation:
A is true but R is wrong because in geitonogamy flower the pollen is transferred to the stigma of another flower of the same plant.

Question 11.

Assertion (A): Fertilization in flowers, produces fruits and seeds.
Reason (R): After fertilization, the ovary develops into fruits and ovule develops into seed.

Answer:
(A) Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of assertion (A).

Question 12.

Assertion (A): Seed is the final product of sexual reproduction in angiosperms.
Reason (R): A seed typically bears seed coat, cotyledons and an embryo axis.

Answer:
(B) Both assertion (A) and Reason (R) are true but reason (R) is not the correct explanation of assertion (A).

Explanation :
After fertilization, the ovary wall develops in fruits embryos captured in the seed as final product of sexual reproduction in plants. The seed bear protective seed coat, cotyledons, and embryo axis.

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Case-Based MCQs

Attempt any 4 sub-parts from each question. Each sub-part carries 1 mark.

I. Read the following text and answer the following questions on the basis of the same:
Study the given diagram and answer any of the four questions given below:

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants img 3

Question1.

This diagram represents which type of ovule

(A) Atropous
(B) Orthotropous
(C) Anatropous
(D) Amphitropous
Answer:
(C) Anatropous

Explanation:
Anatropous ovule is a completely inverted ovule turned back 180 degrees on its stalk. Ovule – a small body that contains the female germ cell of a plant; develops into a seed after fertilization.

Question 2.

A is the stalk of the ovule is called

(A) Hilum
(B) Pedicle
(C) Chalazal pole
(D) Funicle
Answer:
(D) Funicle

Explanation :
Funicle is Ihi’ stalk that attaches an ovule to the placenta in the ovary of a flowering plant. It contains a strand of conducting tissue leading from the placenta | into the chalaza.

Question 3.

The junction of attachment of funicle with the body of ovule at B is

(A) Funicle
(B) Hilum
(C) Nucellus
(D) Chalazal pole
Answer:
(B) Hilum

Explanation :
A scar on a seed (as a bean) marking the point of attachment of the ovule is called hilum. Ihere is small pore, called micropyle, which represent the micropyle of ovule.

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Question 4.

Tegmen develops from the part labelled C in the figure is called

(A) Inner integument
(B) Outer Integument
(C) Funicle
(D) Chalazal pole
Answer:
(A) Inner integument

Explanation :
The seed has two layers one is outer called the testa which is developed from the outer integument and another is inner layer called the legmen which is develops from the inner integument of the ovule.

Direction: In the following questions a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as :

(A) Both assertion (A) and reason (R) are true and (R) is correct explanation of assertion (A).
(B) Both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A).
(C) Assertion (A) is true, but reason (R) is false.
(D) Assertion (A) is false, but reason (R) is true.

Question 5.

Assertion (A): Most common type of ovule is anatrous.
Reason (R): Anatropous ovule is horse-shoe shaped.

Answer:
(C) Assertion (A) is true, but reason (R) is false.

Explanation :
Funiculus lies at the micropylar end and due to the unilateral growth of the ovule is called anatropous ovule. In angiosperms, when the curvature of the ovule affects the nucellus and later it becomes horseshoe-shaped. Such ovule is called amphitropous.

II. Read the following text and answer the following questions on the basis of the same:

Gynoecium is the female reproductive part of the flower. It may consist of a single or more than one pistil. This pistil may be free or fuse. Each pistils has three parts, stigma, style and ovary. Ovary has an ovarian cavity, which has one or many chambers or locules. The placenta is located inside the ovarian cavity. Megasporangia or ovules arise from the placenta.

Question 1.

In which of the following plants the number of ovules in an ovary is one?

(A) Mango
(B) Orchids
(C) Watermelon
(D) Papaya
Answer:
(A) Mango

Explanation :
Mango possess a single ovule in each ovary and orchids, watermelon, and papaya have multiple ovules present in each ovary.

Question 2.

A multicapellary, syncarpous gynoecium is found in:

(A) Papaver
(B) Brinjal
(C) Tomato
(D) All
Answer:
(D) All

Explanation :
Papaver, brinjal, and tomato all have multicar Bellary, syncarpous gynoecium. In this condition, carpels are more than one and fused.

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Question 3.

82% of ovules found in angiosperms are

(A) Anatropous
(B) Amphitropous
(C) Orthotropous
(D) Circinotropous
Answer:
(A) Anatropous

Explanation:
anatropous ovule is found in 82% angiosperm and it completely inverted ovule turned back 180 degrees on its stalk.

Question 4.

Which among the following cell is binucleate in an embryo sac?

(A) Antipodal cell
(B) Central cell
(C) Synergid
(D) Female gamete
Answer:
(B) Central cell

Explanation :
Central cell form binucleate endosperm mother cell upon fertilization with one of the two sperm cells, forms triploid endosperm to nourish embryo development.

Question 5.

Flowers with both androecium and gynoecium are called:

(A) Bisexual flowers
(B) Anther
(C) Unisexual flowers
(D) Androgynous
Answer:
(B) Anther

Explanation :
Androecium is the male part and gynoecium is the female part, and in those flowers have both of these they are called bisexual flowers.

III. Read the following text and answer the following questions on the basis of the same:

A typical anther is bilobed. It is a tetragonal structure consisting of four microsporangia. These microsporangia form pollen sac which on maturity
gets filled with a pollen grains. Pollen grains represent the male gametophytes, their cell wall is very hard. Pollen grains of many species cause severe allergies which cause various diseases in human beings.

Question 1.

Which among the following is a major cause of pollen allergy in India?

(A) Mirabilis
(B) Myosotis
(C) Parthenium
(D) Pistia
Answer:
(C) Parthenium

Explanation :
Parthenium is an invasive species in India, and its is also known as carrot grass or congress grass which is the major cause of allergy in India, the parthenium weed produces as much as 3,000 million pollen grains per square meter during the flowering season.

Question 2.

Select the odd one out with respect to wall layers of microsporangium in flowering plants.

(A) Integument
(B) Tapetum
(C) Endothecium
(D) Middle layers
Answer:
(A) Integument

Explanation :
The integuments arc the outer layer(s) of the ovule and develop into a seed 3 coat as the ovule matures following fertilisation.

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Question 3.

Study of pollen grains is called

(A) Bryology
(B) Mycology
(C) Algology
(D) Polynology
Answer:
(D) Polynology

Explanation :
Study of pollen grains is called polynology.

Question 4.

The prominent pollen grain aperture called germ pore is present in :

(A) Exine
(B) Intine
(C) Vegetative ceU
(D) Generative ceU
Answer:
(A) Exine

Explanation :
The prominent pollen grain aperture called germ pore is present in exine, it is decay-resistant outer coating of a pollen grain or spore.

Direction: In the following questions a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as :
(A) Both assertion (A) and Reason (R) are true and (R) is correct explanation of assertion (A).
(B) Both assertion (A) and Reason (R) are true, but reason (R) is not the correct explanation of assertion (A).
(C) Assertion (A) is true, but reason (R) is false.
(D) Assertion (A) is false, but reason (R) is true.

MCQ Questions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Question 5.

Assertion (A): The innermost layer of microsporangium is called tapetum.
Reason (R): Tapetum nourishes the development into pollen grains.

Answer:
(B) Both assertion (A) and Reason (R) are true, but reason (R) is not the correct explanation of assertion (A).

Explanation :
Cells of tapetum have dense cytoplasm and more than one nuclei, which help in nourishing the developing pollen grains.

MCQ Questions for Class 12 Biology with Answers

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

Magnetism and Matter Class 12 MCQs Questions with

Magnetism And Matter Class 12 MCQ Question 1.

A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as x-y plane. Its magnetic moment m

(A) is non-zero and points in the z-direction by symmetry.
(B) points along the axis of the toroid (m = mø ).
(C) is zero, otherwise there would be a field falling as \(\frac{1}{r^{3}}\) at large distances outside the toroid,
(D) is pointing radially outwards.
Answer:
(C) is zero, otherwise there would be a field falling as \(\frac{1}{r^{3}}\) at large distances outside the toroid,

Explanation:
As we know that a toroid can be considered as a ring shaped closed solenoid.

Magnetism And Matter Class 12 MCQ

So that it is like an endless cylindrical solenoid. So, the magnetic field is only confined inside the body of a toroid in the form of concentric magnetic lines of force. For any point inside, the empty space surrounded by toroid and outside the toroid, the magnetic field B is zero because the net current enclosed in these spaces is zero. So that, the magnetic moment of toroid is zero. In general, if we take r as a long distance outside the toroid, the m cc but this case is not possible here.

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

Class 12 Physics Chapter 5 MCQ Question 2.

Consider the two idealized systems : (i) a parallel plate capacitor with large plates and small separation and (ii) a long solenoid of length L, R, radius of cross-section. In (i), E is ideally treated as a constant between plates and zero outside. In (ii), magnetic field is constant inside the solenoid and zero outside. These idealized assumptions, however, contradict fundamental laws as below :

(A) Case (i) contradicts Gauss’s law for electrostatic fields.
(B) Case (ii) contradicts Gauss’s law for magnetic fields.
(C) Case (i) agrees with ∮E.dl = 0
(D) Case (ii) contradicts ∮H.dl = Im
Answer:
(B) Case (ii) contradicts Gauss’s law for magnetic fields.

Explanation:
According to Gauss’s law of electrostatic field,
∮E.ds = \(\frac{q}{\varepsilon_{0}}\)
So it does not contradict for electrostatic field as the electric field lines do not form continuous path.
According to Gauss’s law of magnetic field,
∮B.ds = 0
It is clear that it contradicts for magnetic field because there is magnetic field inside the solenoid, and no field outside the solenoid carrying current, but the magnetic field lines form the closed paths.

Magnetism And Matter MCQ Chapter 5 Question 3.

A rod of length L, along east-west direction is dropped from a height H. If B be the magnetic field due to Earth at that place and angle of dip is 0, then the magnitude of the induced e.m.f. across two ends of the rod when the rod reachs the Earth is –

(A) BLH cos θ
(B) BL cos θ x (2gH)1/2
(C) BL cos θ / (2gH)1/2
(D) None of the above
Answer:
(B) BL cos θ x (2gH)1/2

Explanation:
Horizontal component of magnetic field = B cos θ
Velocity of the rod = (2 gH)1/2
Induced e.m.f. = BLυ = BL cos θ x (2 gH)1/2

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

Chapter 5 Physics Class 12 MCQ Question 4.

A coil of N turns and radius R carries a current I. It is unwound and rewound to make a square coil of side a having same number of turns (N). Keeping the current I same, the ratio of the magnetic moments of the circular coil and the square coil is –

(A) \(\pi \frac{\mathrm{R}^{2}}{a^{2}}\)
(B) \(\pi \frac{a^{2}}{R^{2}}\)
(C) \(\frac{\mathrm{R}^{2}}{a^{2}}\)
(D) None of the above
Answer:
(A) \(\pi \frac{\mathrm{R}^{2}}{a^{2}}\)

Explanation:
Class 12 Physics Chapter 5 MCQ

Ch 5 Physics Class 12 MCQ Question 5.

A magnetic dipole moment is a vector quantity directed from:

(A) South to North
(B) North to South
(C) East to West
(D) West to East
Answer:
(A) South to North

Explanation:
Magnetic dipole moment vector is directed from South pole to north pole.

Magnetism MCQ Class 12 Question 6.

Time period of oscillation of a magnetic needle is –

(A) T = \(\sqrt{\frac{I}{M B}}\)
(B) T = \(2 \pi \sqrt{\frac{I}{M B}}\)
(C) T = \(2 \pi \sqrt{\frac{\mathrm{MB}}{\mathrm{I}}}\)
(D) T = \(\pi \sqrt{\frac{\mathrm{MB}}{\mathrm{I}}}\)
Answer:
(B) T = \(2 \pi \sqrt{\frac{I}{M B}}\)

Explanation:
Time period of oscillation of a magnetic needle is T = \(2 \pi \sqrt{\frac{I}{M B}}\)

Physics Class 12 Chapter 5 MCQ Question 7.

A magnetic needle is kept in a non-uniform magnetic field experiences

(A) a force as well as a torque
(B) a torque but not a force
(C) a force and a torque
(D) a force but not a torque
Answer:
(A) a force as well as a torque

Explanation:
Field being non-uniform the poles of the needle will experience non-uniform forces. Hence, the needle experiences a force as well as a torque.

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

MCQ On Magnetism Class 12 Question 8.

The magnetic field of Earth can be modelled by that of a point dipole placed at the centre of the Earth. The dipole axis makes an angle of 11.3° with the axis of Earth. At Mumbai, declination is nearly zero. Then,

(A) the declination varies between 11.3° W to 11.3° E.
(B) the least declination is 0°.
(C) the plane defined by dipole axis and Earth axis passes through Greenwich.
(D) declination averaged over Earth must be always negative.
Answer:
(A) the declination varies between 11.3° W to 11.3° E.

Explanation:
The magnetic field lines of the Earth resemble that of a hypothetical magnetic dipole located at the centre of the Earth.
The axis of the dipole does not coincide with the axis of rotation of the Earth and it is tilted at some angle (angle of declination).
In this situation, the angle of declination is approximately 11.3° with respect to the later. So, there is two possibilities arises as shown :
Magnetism And Matter MCQ Chapter 5
So that the declination varies between 11.3° W to 11.3° E.

MCQ Of Chapter 5 Physics Class 12 Question 9.

Let the magnetic field on Earth be modelled by that of a point magnetic dipole at the centre of Earth. The angle of dip at a point on the geographical equator

(A) is always zero.
(B) is always positive
(C) is always negative
(D) can be positive or negative or zero.
Answer:
(D) can be positive or negative or zero.

Explanation:
Angle of inclination or dip is the angle between Ihe direction of intensity of total magnetic field of the Earth and a horizontal line in the magnetic meridian. If the total magnetic field of the Earth is modelled by a point magnetic dipole at the centre, then it is in the same plane of geography al collator, line- the angle of dip on the geographical equator will he Jifferenl at dillerent point*. It m.iv be positive or negalive or may be zero at some points.

Magnetism Class 12 MCQ Chapter 5 Question 10.

Relative permeability of a magnetic material is 0.5. The material is –

(A) diamagnetic.
(B) ferromagnetic.
(C) paramagnetic.
(D) not a magnetic material.
Answer:
(A) diamagnetic.

Explanation:
Relative permeability of diamagnetic magnetic material is less than 1.

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

MCQ On Magnetism Class 12 Pdf Question 11.

Which of the following relation is correct?

(A) B = BV x BH
(B) B = BV / BH
(C) B = BV + BH
(D) B = \(\sqrt{B_{\mathrm{V}}^{2}+B_{\mathrm{H}}^{2}}\)
Answer:
(D) B = \(\sqrt{B_{\mathrm{V}}^{2}+B_{\mathrm{H}}^{2}}\)

Explanation:
BH – B cos θ
BV = B sin δ
B = \(\sqrt{B_{\mathrm{V}}^{2}+B_{\mathrm{H}}^{2}}\)

Class 12 Physics Ch 5 MCQ Question 12.

Ratio of total intensity of magnetic field at equator to poles is

(A) 1 : 1
(B) 1 : 2
(C) 2 : 1
(D) None of the above
Answer:
(A) 1 : 1

Explanation:
BH = B cos θ BV = BH sin δ
At equator, δ = 0°.
So, BH = B, BV = – 90°
At poles, δ = 90°.
So, BH = – B, BV = B
So, the ratio of total intensity of magnetic field at equator to poles is 1 :1.

MCQ On Magnetism And Matter Chapter 5 Question 13.

Which of the following is most suitable for the core of an electromagnet?

(A) Soft iron
(B) Steel
(C) Alnico
(D) Copper
Answer:
(A) Soft iron

Explanation:
Soft Iron gels magnetized faster but loses its magnetism as soon as the current slops flowing in solenoid. Icnce soft iron is said to have high susceptibility but low retentivily. This property of soft iron makes it suitable for core of electromagnets where we need strong but temporary magnetism as long as current is flowing.

Physics Chapter 5 Class 12 MCQ Question 14.

A ferromagnetic substance is heated above its curie temperature. Which of the following statements is correct?

(A) Ferromagnetic domains get perfectly arranged.
(B) Ferromagnetic domains get randomly arranged.
(C) Ferromagnetic domains are not at all influenced.
(D) Ferromagnetic material transforms into diamagnetic substance.
Answer:
(B) Ferromagnetic domains get randomly arranged.

Explanation:
On heating above Curie temperature, Ferromagnetic domains get randomly arranged and it transforms into paramagnetic substance.

Assertion And Reason Based MCQs (1 Mark each)

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is true

Class 12 Chapter 5 Physics MCQ Question 1.

Assertion (A): The magnetic field configuration with 3 poles is not possible.
Reason (R): No torque acts on a bar magnet itself due to its own field.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Magnetic poles exist in pairs. So assertion is true. The bar magnet does not exert a torque on itself in its own magnetic field. Torque is proportional to cross product of M and B . The angle between M and B being 0, the cross product is 0. So, there will be no torque. So reason is also true. But R cannot explain A.

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

Chapter 5 Class 12 Physics MCQ Question 2.

Assertion (A): Magnetic poles cannot be separated by breaking a bar magnet into two pieces.
Reason (R): When a magnet is broken into two pieces, the magnetic moment will be reduced to half.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation: Magnetic poles always exist in pairs even in atomic level. So assertion is true. When a magnet is broken into two pieces, the pole strength remains same; only the length becomes half. So, the magnetic moment becomes half. So, the reason is also true. But R is not the proper explanation of A.

MCQ On Magnetism And Matter Class 12 Question 3.

Assertion (A): The basic difference between magnetic lines of force and electric lines of force is electric lines of force are discontinuous and magnetic lines of force are continuous.
Reason (R): Magnetic lines of force exist in a magnet but no electric lines of force exist in a charged body.

Answer:
(A) Both A and R are true and R is the correct explanation of A.

Explanation:
Let us consider an electric dipole. The electric lines of force exist outside only and not inside the dipole.
Let us take a magnetic dipole. The magnetic lines of force exist outside as well as inside the dipole.
So, it can be said that magnetic lines of force are continuous and electric lines of force are discontinuous.
So assertion and reason both are true and reason explains the assertion too.

Chapter 5 MCQ Class 12 Physics Question 4.

Assertion (A): Gauss theorem is not applicable in magnetism.
Reason (R): Magnetic monopole does not exist.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Gauss’s theorem of magnetism is different from that for electrostatics because electric charges may not exist in pair but magnetic poles always exist in pair. So assertion is true. Magnetic monopole does not exist. Magnetic poles always exist in pair. So reason is also true and reason clearly explains the assertion.

Physics Chapter 5 MCQ Class 12 Question 5.

Assertion (A): A compass needle when placed at Earth’s magnetic pole rotates in vertical plane. Reason (R): The Earth has only horizontal component of its magnetic field at the poles.

Answer:
(D) A is false and R is true

Explanation:
Magnetic needle can rotate in horizontal plane only. But at poles, there is no horizontal component of Earth’s magnetic field.
So, the needle will remain horizontal and will point in any direction. Hence the assertion is false.
At poles, Earth has only vertical components of its magnetic field. Hence, the reason is also false.

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

Class 12th Physics Chapter 5 MCQ Question 6.

Assertion (A): Compass needle points the magnetic north-south direction.
Reason (R): The magnetic meridian of the earth merges with the axis of rotation of earth.

Answer:
(D) A is false and R is true

Explanation:
Compass needle points the magnetic north-south direction. So the assertion is true. Earth’s magnetic meridian is along its axis through magnetic north-south direction. Earth’s axis of rotation is along its geographic north-south direction. The angle between these two axes is 11.3°. Hence, the reason is also false.

Question 7.

Assertion (A): Ferromagnetic substances become paramagnetic beyond Curie temperature.
Reason (R): Domains are destroyed at high temperature.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
From Curie Weiss law,
x = \(\frac{C}{T-T_{C}}\)
As temperature increases beyond Curie temperature, susceptibility decreases and the ferromagnetic substances become paramagnetic. So, the assertion is true. Paramagnetic substance has no magnetic domain. At a very high temperature, the domains of ferromagnetic substance get destroyed and the substance transforms into paramagnetic substance. So, the reason is also true and properly explains the assertion.

Question 8.

Assertion (A): Gauss’s law of magnetism is different from Gauss’s law of electrostatics.
Reason (R): Isolated electric charge can exist but isolated magnetic pole cannot exist.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
in electrostatics, Gauss’s law:
= \(\int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d \mathrm{~A}}=\frac{q}{\varepsilon_{0}}\)
Gauss’s law of magnetism:
\(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d \mathrm{~A}}\) = 0

Gauss’s law of magnetism is different from Gauss’s law of electrostatics. Hence, the assertion is true. Electric charge may or may not exist in pair. But magnetic poles always exist in pair. No magnetic monopole exists. This is the reason why Gauss’s law of magnetism is different from Gauss’s law of electrostatics. So, the reason is also true and explains the assertion.

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

Case-Based MCQs

Attempt any 4 sub-parts out of 5. Each sub-part carries 1 mark.

I. Read the following text and answer the following questions on the basis of the same:
Earth’s magnetism: Earth’s magnetic field is caused by a dynamo effect. The effect works in the same way as a dynamo light on a bicycle. Magnets in the dynamo start spinning when the bicycle is pedaled, creating an electric current. The electricity is then used to turn on the light. This process also works in reverse.

If you have a rotating electric current, it will create a magnetic field. On Earth, flowing of liquid metal in the outer core of the planet generates electric currents. The rotation of Earth on its axis causes these electric currents to form a magnetic field which extends around the planet. The average magnetic field strength in the Earth’s outer core was measured to be 25 Gauss, 50 times stronger than the magnetic field at the surface. The magnetic field is extremely important for sustaining life on Earth.

Without it, we would be exposed to high amounts of radiation from the Sun and our atmosphere would be free to leak into space. This is likely what happened to the atmosphere on Mars. As Mars doesn’t have flowing liquid metal in its core, it doesn’t produce the same dynamo effect. This left the planet with a very weak magnetic field, allowing for its atmosphere to be stripped away by solar winds, leaving it uninhabitable. Based upon the study of lava flows throughout the world, it has been proposed that the Earth’s magnetic field reverses at an average interval of approximately 300,000 years. However, the last such event occurred some 780,000 years ago.

Question 1.

Which of the followings is the reason for Earth’s magnetism ?

(A) Rotation of electric current
(B) Rotation of Earth
(C) Attraction due to other celestial bodies
(D) Solar flares
Answer:
(A) Rotation of electric current

Explanation:
On Earth, flowing of liquid metal in the outer core of the planet generates electric currents. The rotation of Earth on its axis causes these electric currents to form a magnetic field which extends around the planet.

Question 2.

Electric current in the Earth’s body is generated due to:

(A) movement of charged particle in the atmosphere.
(B) flowing of liquid metal in the outer core.
(C) electric discharges during thunderstorm.
(D) its revolution round the Sun.
Answer:
(A) movement of charged particle in the atmosphere.

Explanation:
On Earth, flowing of liquid metal in the outer core of the planet generates electric currents.

Question 3.

Which planet has no own magnetic field ?

(A) Jupiter
(B) Neptune
(C) Mars
(D) Mercury
Answer:
(C) Mars

Explanation:
As Mars doesn’t have flowing liquid metal in its core, it doesn’t produce dynamo effect. So, it has very weak or almost no magnetic field.

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

Question 4.

Average magnetic field strength in the Earth’s outer core is:

(A) 5 Gauss
(B) 25 Gauss
(C) 500 Gauss
(D) Cannot be measured
Answer:
(B) 25 Gauss

Explanation:
The average magnetic field strength in the Earth’s outer core was measured to be 25 Gauss.

Question 5.

Which of the following statements is true ?

(A) Earth’s magnetic field is due to electric current induced in the ionosphere.
(B) The average magnetic field strength in the Earth’s outer core is equal to the magnetic field at the surface.
(C) Earth’s magnetic field reverses at an average interval of approximately 3,00,000 years.
(D) Angle of dip is same at every point of the surface of Earth.
Answer:
(C) Earth’s magnetic field reverses at an average interval of approximately 3,00,000 years.

Explanation:
Based upon the study of lava flows throughout the world, it has been proposed that the Earth’s magnetic field reverses at an average interval of approximately 300,000 years.

II. Read the following text and answer the following questions on the basis of the same:

If we move into space and study the Earth’s invisible magnetic field, it wouldn’t really look like a bar magnet at all. Earth’s magnetic field gets stretched out into a comet-like shape with a tail of magnetism that stretches millions of miles behind the earth, opposite to the Sun. The Sun has a wind of gas that pushes the earths field from the left to the right in the picture.

The core of the Earth is an electromagnet. Although the crust is solid, the core of the Earth is surrounded by a mixture of molten iron and nickle. The magnetic field of Earth is caused by currents of electricity that-flow in the molten core. These currents are hundreds of miles wide and flow at thousands of miles per hour as the Earth rotates.

Chapter 5 Physics Class 12 MCQ

The powerful magnetic field passes out through the core of the Earth, passes through the crust and enters space. This picture shows the solid inner core region (inner circle) surrounded by a molten outer core (the area between the two circles). The currents flow in the outer core, travel outwards through the rest of the earth’s interior. If the Earth rotated faster, it would have a stronger magnetic field.

By the time the field has reached the surface of Earth, it has weakened a lot, but it is still strong enough to keep your compass needles pointed towards one of its poles. All magnets have two poles: a North Pole and a South Pole. The magnetic poles of earth are not fixed on the surface, but wander quite a bit. The pole in the Northern Hemisphere seems to be moving northwards in geographic latitude by about 10 kilometres per year by an average.

Ch 5 Physics Class 12 MCQ

Question 1.

Earth’s magnetic field has a:

(A) shape of the magnetic field of a bar magnet.
(B) shape of the magnetic field of a horseshoe magnet.
(C) shape of a sphere.
(D) None of the above
Answer:
(D) None of the above

Explanation:
Earth’s magnetic field gets
stretched out into a comet-like shape with a tail of magnetism that stretches millions of miles < behind the Earth, opposite from the Sun.

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

Question 2.

Core of the Earth is:

(A) an electromagnet.
(B) a permanent magnet.
(C) a unipolar magnet.
(D) None of these
Answer:
(A) an electromagnet.

Explanation:
The core of the Earth is an electromagnet. Although the crust is solid, the core of the Earth is surrounded by a mixture of molten iron and nickle. The magnetic field of Earth is caused by currents of electricity that flow in the molten core.

Question 3.

The magnetic poles of Earth are:

(A) fixed on the surface
(B) wander throughout the Earth’s surface
(C) wander about 1000 kilometres per year on an average.
(D) wander about 10 kilometres per year on an average.
Answer:
(D) wander about 10 kilometres per year on an average.

Explanation:
The magnetic poles of Earth are not fixed on the surface, but wander quite a bit. The pole in the Northern Hemisphere seems to be moving northwards in geographic latitude by about 10 kms per year on an average.

Question 4.

Earth’s magnetic field may increase if:

(A) it rotates on its axis faster.
(B) its direction of rotation is changed.
(C) it revolves round the Sun faster.
(D) All of the above
Answer:
(A) it rotates on its axis faster.

Explanation:
If the Earth rotated faster, it would have a stronger magnetic field.

Question 5.

The Earth’s magnetism is due to:

(A) induction of Sun’s magnetism.
(B) current produced by the movement of molten metals.
(C) sea current.
(D) revolution of the Earth round the Sun.
Answer:
(B) current produced by the movement of molten metals.

Explanation:
The Earth crust is solid, the core of the Earth is surrounded by a mixture of molten iron and nickle. The magnetic field of Earth is caused by currents of electricity that flow in the molten core. These currents are hundreds of miles wide and flow at thousands of miles per horn as the Earth rotates.

III. Read the following text and answer the following questions on the basis of the same:

Super magnet:
The term super magnet is a broad term and encompasses several families of rare-earth magnets that include seventeen elements in the periodic table; namely scandium, yttrium, and the fifteen lanthanides. These elements can be magnetized, but have Curie temperatures below room temperature. This means that in their pure form, their magnetism only appears at low temperatures. However, when they form compounds with transition metals such as iron, nickel, cobalt, etc. Curie temperature rises well above room temperature and they can be used effectively at higher temperatures as well. The main advantage they have over conventional magnets is that their greater strength allows for smaller, lighter magnets to be used.

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

Super magnets are of two categories:
(i) Neodymium magnet:
These are made from an alloy of neodymium, iron, and boron. This material is currently the strongest known type of permanent magnet. It is typically used in the construction of head actuators in computer hard drives and has many electronic applications, such as electric motors, appliances, and magnetic resonance imaging (MRI).

(ii) Samarium-cobalt magnet:
These are made from an alloy of samarium and cobalt. This second- strongest type of rare Earth magnet is also used in electronic motors, turbo-machinery, and because of its high temperature range tolerance may also have many applications for space travel, such as cryogenics and heat resistant machinery.

Rare-earth magnets are extremely brittle and also vulnerable to corrosion, so they are usually plated or coated to protect them from breaking, chipping, or crumbling into powder. Since super magnets are about 10 times stronger than ordinary magnets, safe distance should be maintained otherwise these may damage mechanical watch, CRT monitor, pacemaker, credit cards, magnetically stored media etc.

These types of magnets are hazardous for health also. The greater force exerted by rare-earth magnets creates hazards that are not seen with other types of magnet. Magnets larger than a few centimeters are strong enough to cause injuries to body parts pinched between two magnets or a magnet and a metal surface, even causing broken bones. Neodymium permanent magnets lose their magnetism 5% every 100 years. So, in the truest sense Neodymium magnets may be considered as a permanent magnet.

Question 1.

Curie point of pure rare Earth elements is

(A) very high.
(B) below room temperature.
(C) 0 K.
(D) varies from element to element.
Answer:
(B) below room temperature.

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

Explanation:
Rare-Earth elements which can be magnetized have Curie temperatures below room temperature. This means that in their pure form, their magnetism only appears at low temperatures.

Question 2.

Neodymium and Samarium are

(A) diamagnetic.
(B) paramagnetic.
(C) ferromagnetic.
(D) not magnetic materials.
Answer:
(C) ferromagnetic.

Question 3.

Super magnets are about time stronger than ordinary magnets.

(A) 10
(B) 100
(C) 1000
(D)10000
Answer:
(A) 10

Question 4.

To raise the Curie point of rare Earth elements.

(A) they are coated with gold.
(B) compounds are formed with transition metals.
(C) they are oxidized.
(D) None of the above
Answer:
(B) compounds are formed with transition metals.

Explanation:
When rare-Earth elements form compounds with transition metals such as iron, nickel, cobalt, etc Curie temperatures thus rise well above room temperature.

MCQ Questions for Class 12 Physics Chapter 5 Magnetism and Matter

Question 5.

Neodymium permanent magnets lose their magnetism % every 100 years.

(A) 50
(B) 0.5
(C) 10
(D) None of the above
Answer:
(B) 0.5

Explanation:
Neodymium permanent magnets lose their magnetism 5% every 100 years. So, in the truest sense. Neodymium magnets may be considered as a permanent magnet.

MCQ Questions for Class 12 Physics with Answers

MCQ Questions for Class 12 Biology Chapter 1 Reproduction in Organisms with Answers

Check the below NCERT MCQ Questions for Class 12 Biology Chapter 1 Reproduction in Organisms with Answers Pdf free download. MCQ Questions for Class 12 Biology with Answers were prepared based on the latest exam pattern. We have provided Reproduction in Organisms Class 12 Biology MCQs Questions with Answers to help students understand the concept very well. https://mcqquestions.guru/mcq-questions-for-class-12-biology-chapter-1/

Reproduction in Organisms Class 12 MCQs Questions with Answers

Reproduction In Organisms Class 12 MCQ Question 1.
Pollination by Bat is called :
(a) Omithorphily
(b) Entemophily
(c) Cheropterophily
(d) Hydrophily

Answer

Answer: (c) Cheropterophily


Class 12 Biology Chapter 1 MCQ Question 2.
Anemophily takes place by:
(a) By Bird
(b) By Bat
(c) By Snail
(d) By Wind

Answer

Answer: (d) By Wind


MCQs Of Biology Class 12 Chapter 1 Question 3.
Entemophily take place by
(a) By Bird
(b) By Bat
(c) By Wind
(d) By Insect

Answer

Answer: (d) By Insect


Biology Class 12 Chapter 1 MCQ Question 4.
In higher plants, meiosis occurs?
(a) During formation of gametes
(b) During fertilization
(c) During embryogenesis
(d) In seeds

Answer

Answer: (a) During formation of gametes


Reproduction In Organisms MCQ Chapter 1 Question 5.
Earthworm is:
(a) Unisexual
(b) Bisexual
(c) Asexual
(d) Neutral

Answer

Answer: (b) Bisexual


Reproduction In Organisms MCQ With Answers Question 6.
An example of rhizome is
(a) Garlic
(b) Colocassia
(c) Ginger
(d) Onion

Answer

Answer: (c) Ginger


Class 12 Biology MCQ Chapter 1 Question 7.
Budding is the example of :
(a) Vegetative reproduction
(b) Tissue culture
(c) Sexual production
(d) Dispersal

Answer

Answer: (a) Vegetative reproduction


Reproduction In Organisms Class 12 MCQ Pdf Question 8.
What is hexapioid (on), in which total number, of chromosomes are 42. then number of chromosomes in its any gamete:
(a) 14
(b) 28
(c) 42
(d) 7

Answer

Answer: (d) 7


MCQ Reproduction In Organisms Class 12 Chapter 1 Question 9.
Asexual reproduction in Penicillium fungus usually takes place through?
(a) Budding
(b) Fission
(c) Conidia
(d) Gemmules

Answer

Answer: (c) Conidia


Biology Chapter 1 MCQ Class 12 Question 10.
When the gametes of two different strains fuse each other, then this process of reproduction is :
(a) Sexual reproduction
(b) Asexual, reproduction
(c) Vegetative reproduction
(d) None of these

Answer

Answer: (a) Sexual reproduction


MCQ On Reproduction In Organisms Class 12 Question 11.
Vegetative reproduction takes place in Bacteria :
(a) By binary fission
(b) By hybridization
(c) Byfiission
(d) By fission

Answer

Answer: (a) By binary fission


MCQs On Reproduction In Organisms Class 12 Question 12.
In potato, which structure takes part in vegetative reproduction?
(a) Adventitious root
(b) Apex bud of stem
(c) Germinating eye bud
(d) Arial shoot

Answer

Answer: (c) Germinating eye bud


MCQ On Reproduction In Organisms Pdf Chapter 1 Question 13.
Which structure takes part in gametogenesis :
(a) Male gamete)
(b) Female gamete
(c) Both (a) and (b)
(d) Male or female gamete

Answer

Answer: (c) Both (a) and (b)


Class 12 Biology Chapter 1 MCQs Question 14.
Which of the following is viviparous?
(a) Cockroach
(b) Scorpion
(c) Bee
(d) Butterfly

Answer

Answer: (b) Scorpion


Question 15.
In higher plants, fertilization is :
(a) External
(b) Internal
(c) In water
(d) In air

Answer

Answer: (b) Internal


Question 16.
The reproduction takes place by gemma :
(a) In higher plants
(b) In lower animals
(c) In some bryophytes
(d) In mammals

Answer

Answer: (c) In some bryophytes


Question 17.
The example of bulb is:
(a) Ginger
(b) Doob grass
(c) Onion
(d) Potato

Answer

Answer: (c) Onion


Question 18.
The organism similar to parent can be obtained by:
(a) Speed
(b) Zygote
(c) Gametes
(d) Binary fission

Answer

Answer: (d) Binary fission


Question 19.
The term scion is used in relation to :
(a) Embryology
(b) Sexual reproduction
(c) Pollen
(d) Grafting

Answer

Answer: (d) Grafting


Question 20.
Vegelative reproduction takes place in Bryophyllum :
(a) By buds
(b) By leaves
(c) By seeds
(d) By roots

Answer

Answer: (b) By leaves


Question 21.
Vegetative reproduction takes place in Agave:
(a) By bulbils
(b) By hybrid
(c) By stolon
(d) By rhizome

Answer

Answer: (a) By bulbils


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