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Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 8 Rectification of Errors Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 8 Rectification of Errors

1. Answer in One Sentence:

Question 1.
What is an Arithmetical error?
Answer:
An error committed to totaling the amount columns of journal and ledger is called arithmetical error.

Question 2.
What do you mean by one-sided errors?
Answer:
One-sided errors are those errors that either affect the debit or credit aspect of the transaction. It affects the agreement of the trial balance.

Question 3.
What do you mean by two-sided errors?
Answer:
Two-sided errors are those errors that affect both the aspects of a transaction i.e. debit and credit. It does not affect the agreement of the trial balance.

Question 4.
What is a suspense account?
Answer:
An account that is opened to transfer the difference in the totals of the trial balance is known as a suspense account.

Question 5.
What are accounting errors?
Answer:
Mistakes or errors committed while writing the books of accounts are known as accounting errors.

2. Give one word/term or phrase for each of the following statements.

Question 1.
Transactions remained to be recorded at all in the books of account.
Answer:
Errors of omission

Question 2.
Errors are always rectified by passing rectification entries.
Answer:
Two-sided errors

Question 3.
Errors that affect the debit and credit side unequally.
Answer:
One-sided errors

Question 4.
Errors that affect the debit and credit side equally.
Answer:
Two-sided errors

Question 5.
An error in which the transaction is entered in the original book but not posted into the ledger.
Answer:
The error of posting/partial omission

Question 6.
Error in the process of transferring the entry from original books into the ledger.
Answer:
Error of posting

Question 7.
Errors can be rectified without passing rectification entries.
Answer:
One-sided errors

3. Select the most appropriate alternative from those given below and rewrite the sentence.

Question 1.
Wages paid for the installation of machinery wrongly debited to wages account is an error of ___________
(a) omission
(b) principle
(c) commission
(d) duplication
Answer:
(b) principle

Question 2.
If the trial balance shows a short credit the suspense account will have a ___________ balance.
(a) debit
(b) zero
(c) credit
(d) nil
Answer:
(c) credit

Question 3.
If the trial balance does not agree the difference of the trial balance is placed in ___________ account.
(a) Personal
(b) Suspense
(c) Rectification
(d) Real
Answer:
(b) Suspense

Question 4.
Errors which compensate the effect of each other are called ___________ errors.
(a) compensating
(b) one-sided
(c) two sided
(d) clerical
Answer:
(a) compensating

Question 5.
One sided errors are disclosed by ___________
(a) Trial Balance
(b) Suspense Account
(c) Journal
(d) Ledger Account
Answer:
(a) Trial Balance

4. State whether the following statements are True or False with reasons.

Question 1.
Errors of principle are not disclosed by the Trial Balance.
Answer:
This statement is True.
The error of Principle is those where some basic principles of bookkeeping and accountancy are not Properly Followed while recording a business transaction. Such errors can be rectified by passing journal entries.
Eg.: Capital expender showed as revenue expenditure or vice versa. It won’t affect the Trial balance. If will agreed.

Question 2.
Transaction not recorded in the books is an error of principle.
Answer:
This statement is False.
Transaction not recorded in the books is an error of omission and not an error of principle.

Question 3.
If the purchase hook is undermasted, the purchase account is debited in rectification.
Answer:
This statement is True.
Purchases are expenses. All expenses are debited as per the rule of nominal account. When purchases are underacted they can be corrected by debiting in rectification.

Question 4.
When a transaction is not recorded according to the principles of book-keeping the error is said to be an error of principle.
Answer:
This statement is True.
The error of Principle is those where some basic principles of bookkeeping and accountancy are not properly followed while recording a business transaction.
Eg.: Capital expenditure is shown as revenue expenditure or Vice-a-Varsa and they are called an error of principle.

Question 5.
The error of omission is disclosed by the Trial Balance.
Answer:
This statement is False.
The complete omission of a transaction will bot disclosed by a trial balance. Trial balance, balances will be agreed with such errors. So the omission of transaction or an error of omission will not be disclosed by Trial Balance.

5. Do you agree or disagree with the following statements.

Question 1.
A temporary account opened to rectify the entry is known as suspense A/c.
Answer:
Agree

Question 2.
Rectified entries are passed in Ledger.
Answer:
Disagree

Question 3.
Compensating errors affect the agreement of Trial Balance.
Answer:
Disagree

Question 4.
There is only one type of error.
Answer:
Disagree

Question 5.
Transaction recorded without following the accounting principles and rules are known as Errors of Principle.
Answer:
Agree

6. Complete the following sentences.

Question 1.
___________ sided errors affect the total of Trial Balance.
Answer:
One

Question 2.
___________ sided errors do not affect the Trial Balance.
Answer:
Two

Question 3.
One sided error do not require ___________ entry.
Answer:
Rectification

Question 4.
Errors which are committed in writing the accounts are called error of ___________
Answer:
Posting

Question 5.
Under casting of Sales book is corrected by ___________ Sales Account.
Answer:
Crediting

Question 6.
The disagreement of Trial Balance indicates that an ___________ has been committed.
Answer:
Error

Question 7.
An asset has been purchased for the firm. However, the amount was debited to the purchase account. It is an error of ___________
Answer:
Principle

Question 8.
___________ account is necessary for rectification of one-sided error.
Answer:
Suspense A/c

Question 9.
An item of ₹ 95 has been debited to a personal account as ₹ 59. It is an error of ___________
Answer:
Commission

Question 10.
Rectification entries are passed in ___________
Answer:
Journal Proper

Question 11.
Two sided errors are rectified by passing ___________ entry.
Answer:
Rectification

Question 12.
If the Trial Balance does not tally, its difference is transferred to ___________ Account.
Answer:
Suspense

Question 13.
Under casting of an account is ___________ sided error.
Answer:
One

Question 14.
If the transaction is not at all recorded in the books, it is called an error of ___________
Answer:
Complete Omission

Question 15.
If the total balance shows short a credit, the suspense account will have a ___________ balance.
Answer:
Credit

Question 16.
An error that affects debit as well as credit side, it is called as ___________ errors.
Answer:
Two-Sided

Question 17.
Wages paid for installation of machinery debited to Wages Account is an error of ___________
Answer:
Principle

Question 18.
Errors which cancel out the effect of one another is called ___________ errors.
Answer:
Compensating

Question 19.
One sided errors are disclosed by ___________
Answer:
Trial Balance

Question 20.
In an error of omission, the debit and credit are ___________
Answer:
Equal

Practical Problems

Question 1.
The trial balance of Sagar did not agree. It showed an excess credit of ₹ 7,550. Sagar put the difference to the suspense account. He located the following errors:
1. Sales return book was overcast by ₹ 1,200.
2. Purchase book was undercast by ₹ 750.
3. Goods returned to Mahesh ₹ 1,000 were recorded through the sales book.
4. Credit purchases from Mahadev ₹ 6,000 were recorded through the sales book.
5. Credit purchases from Damodhar ₹ 4,000 were recorded through the sales book. However, Damodhar’s account was correctly credited.
6. Salary paid ₹ 3,500 was debited to the employee’s personal account.
Give journal entries to rectify the above errors and prepare Suspense Account.
Solution:
Journal Proper

Question 2.
The trial balance of Radhika did not agree. Radhika put the difference to the suspense account. Subsequently, she located the following errors.
1. Furniture purchased for ₹ 6,000 was posted to the purchases account as ₹ 600.
2. Repairs to Machinery ₹ 500 were debited to the Machinery account.
3. Wages paid for the installation of Machinery ₹ 750 was posted to wages account.
4. Purchased material ₹ 8,000 and Wages ₹ 2,000 were used for construction of the building. No adjustment was made in the books.
5. Total of sales returns book ₹ 2,000 was not posted to the ledger.
6. Old Furniture sold to Dinesh at its book value of ₹ 2,500 was recorded through sales book.
Give journal entries and prepare Suspense Account.
Solution:
Journal Proper

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 3 Basic Analytical Techniques Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 3 Basic Analytical Techniques

Question 1.
Give reason: Purification of a chemical substance is important before investigating its composition and properties.
Answer:

• Chemical substances occur in nature in impure stage.
• Also, chemical substances synthesized in the laboratory are obtained in crude and impure form.
• Impurities present in the chemical substances may interfere with the properties to be determined (e.g. melting point or boiling point).
• Therefore, before investigating composition and properties of a given chemical substance, it is important to obtain it in the pure form.

Question 2.
What are the different types of impurities that a solid may contain?
Answer:
A solid substance may contain two types of impurities:

• Impurities which are soluble in the same solvent as the main substance.
• Impurities which are not soluble in the same solvent as the main substance.

Question 3.
For which of the following cases, is the process of filtration feasible? Why?
Case 1: A solid substance containing impurities which are soluble in the same solvent as the main substance.
Case 2: A solid substance containing impurities which are not soluble in the same solvent as the main substance.
Answer:
Impurities which are not soluble in the same solvent as the main compound can be separated by a simple process called filtration. Hence, for ‘Case 2’, filtration is more feasible.

Question 4.
Describe the process of filtration with a neat and labelled diagram.
Answer:
i. Impurities which are not soluble in the same solvent as the main compound can be separated by a simple process called filtration.
ii. Procedure:
a. A circular piece of filter paper is folded to form a cone and fitted in the funnel.
b. The funnel is fixed on a stand and a beaker is kept below.
c. The mixture which has to be purified is added to a suitable solvent in which the main compound dissolves.
d. The paper is made moist, and the solution to be filtered is poured on the filter paper.
e. Diagram:

iii. The insoluble part remaining on the filter paper is called residue and the liquid which pass through the filter paper and collected in the beaker is called filtrate.
iv. This process is similar to separating tea leaves from decoction of tea or sand from mixture of sand and water.

Question 5.
Why is safety bottle used when filtration is carried out under suction?
Answer:
The safety bottle is used to prevent sucking of the filtrate into suction pump.

Question 6.
Name the steps involved in the process of crystallization.
Answer:
Steps involved in the process of crystallization:

• Preparation of a saturated solution
• Hot filtration
• Cooling of the filtrate
• Filtration

Question 7.
How is saturated solution of the crude solid prepared?
Answer:

• A saturated solution of the crude solid is prepared by boiling it in a small but sufficient quantity of a suitable solvent.
• The main solute from the sample of the crude solid dissolves to form a saturated solution on boiling.

[Note: The solution is not saturated with respect to the soluble impurities, as they are in small proportion.]

Question 8.
Explain the following steps with respect to the process of crystallization.
i. Preparation of a saturated solution
ii. Hot filtration
iii. Cooling of the filtrate
iv. Filtration
Answer:
i. Preparation of a saturated solution:

• A saturated solution of the crude solid is prepared by boiling it in a small but sufficient quantity of a suitable solvent.
• On doing so the main solute forms an almost saturated solution, but the solution is not saturated with respect to the soluble impurities, as they are in small proportion.

ii. Hot filtration: The hot saturated solution is quickly filtered to remove undissolved impurities as residue. Filtration under suction can be employed for rapid filtration.

iii. Cooling of the filtrate:

• The hot filtrate is allowed to cool.
• On cooling, the filtrate becomes supersaturated with respect to the main dissolved solute because solubility of a substance decreases with lowering of temperature.
• The excess quantity of the dissolved solute comes out of the solution in the form of crystals.
• The dissolved impurities, however, do not supersaturate the solution, as their quantity is small.
• These continue to stay in the solution in dissolved state even on cooling. Therefore, the separated crystals are free from soluble impurities.

iv. Filtration:

• The crystals obtained on cooling are further purified by filtration to remove insoluble impurities.
• The filtrate obtained is called as mother liquor.
• The crystals obtained after filtration are free from soluble as well as insoluble impurities.

Question 9.
Name the common solvents used in the process of crystallization.
Answer:
The commonly used solvents are water, ethyl alcohol, methyl alcohol, acetone, ether or their combinations.

Question 10.
Describe the process of crystallization of common salt from impure sample with the help of a diagram.
Answer:

• Impure sample of a common salt is added to the required quantity of water and stirred with a glass rod.
• More amount of salt is added and the solution is heated till no more salt dissolves.
• The hot saturated solution is filtered off to remove insoluble impurities while the filtrate is collected in an evaporating dish.
• The filtrate is allowed to cool which results in the formation crystals of pure salt (NaCl) leaving behind the soluble impurities.
• The crystals are filtered and dried.

The diagram is as follows:

Question 11.
Which solvent is used for the purification of copper sulphate and benzoic acid?
Answer:
The solvent used for the purification of copper sulphate and benzoic acid is water.

Question 12.
Define: Fractional crystallization
Answer:
Fractional crystallization is a process wherein two or more soluble substances having widely different solubilities in the same solvent at same temperature are separated by crystallization.

Question 13.
Give a brief description of the principle of fractional crystallization.
Answer:
Fractional crystallization is based on the differences in solubilities of two or more compounds in the same solvent at the same temperature. That is, the substance which is least soluble crystallizes out first and the most soluble substance crystallizes out last.
e.g. Mixture of two solutes A and B can be purified by fractional crystallization as follows:

• Preparation of a saturated solution: Mixture of two solutes A and B are dissolved in a suitable hot solvent to prepare a saturated solution.
• Hot filtration: The hot saturated solution is filtered to remove insoluble impurities.
• Cooling of the filtrate: Hot filtrate is allowed to cool. On cooling, the solute which is least soluble crystallizes out first leaving behind the most soluble substance in the mother liquor.
• Filtration: The crystals formed are filtered, washed with solvent and dried. Crystals obtained will be of a solute which is least soluble in a given solvent.
• Concentration of a mother liquor: The mother liquor is concentrated by evaporating the solvent. These crystals are filtered and dried to obtain the second purified component (which was more soluble in given solvent).

Question 14.
Which type of impure liquids can be purified by the process of distillation?
Answer:
Distillation technique can be employed for the purification of

• volatile liquids from non-volatile impurities.
• liquids having sufficient difference in their boiling point.

Question 15.
Explain the construction of simple distillation unit using neat labelled diagram.
Answer:
i. The apparatus used for simple distillation is shown in the figure below:

ii. It consists of round bottom flask fitted with a cork having a thermometer.
iii. The flask has a sidearm through which it is connected to a condenser.
iv. The condenser has a jacket with two outlets through which water is circulated.
v. The liquid to be distilled is taken in the round bottom flask fixed by clamp.
vi. The flask is placed in a water bath or oil bath or sometimes wire gauze is kept on a stand as shown in the figure.

Question 16.
State the principle involved and describe the process to separate acetone and water from their mixture.
Answer:
i. Acetone and water can be separated from their mixture by simple distillation.

ii. Principle: Acetone and water are two miscible liquids having an appreciable difference (more than 30 K) in their boiling points. Acetone boils at 56 °C while boiling point of water is 100 °C. When the mixture of acetone and water is heated and temperature of the mixture reaches 56 °C acetone will distil out first. Once all acetone distils out, and when the temperature rises to 100 °C water will distil out.

iii. Process to separate acetone and water from their mixture:

• Take the mixture of water and acetone in the distillation flask.
• Heat the flask on a water bath carefully. At 56 °C acetone will distil out, collect it in receiver.
• After all acetone distilled, change the receiver. Discard a few mL of the liquid. As the temperature reaches 100 °C water will begin to distil. Collect this in another receiver.

Question 17.
What is the advantage of fractional distillation over simple distillation?
Answer:
If in a mixture, the difference in boiling points of two liquids is not appreciable/large, they cannot be separated using simple distillation. To separate such liquids, fractional distillation is used.

Question 18.
Label the following diagram and explain the process by giving example.

Answer:
The labelled diagram is as follows:

i. In fractional distillation, vapours first pass through the fractionating column.
ii. Vapours of more volatile liquid with lower boiling point rise up more than the vapours of liquid having higher boiling point.
e.g.

• Suppose we have a mixture of two liquid ‘A’ and ‘B’ having boiling points 363 K and 373 K respectively.
• ‘A’ is more volatile and ‘B’ is less volatile. As the mixture is heated, vapours of ‘A’ along with a little vapours of ‘B’ rise up and come in contact with the large surface of the fractionating column.
• Vapours of ‘B’ condense rapidly into the distillation flask. While passing through the fractionating column, there is an exchange between the ascending vapours and descending liquid. The vapours of ‘B’ are scrubbed off by the descending liquid, this makes the vapours richer in ‘A’.
• This process is repeated each time the vapours and liquid come in contact with the surface in the fractionating column.
• Rising vapours become richer in ‘A’ and escape through the fractionating column and reach the condenser while the liquid in the distillation flask is richer in ‘B.
• The separated components are further purified by repeating the process.

Question 19.
Give two examples of a mixture that can be separated by fractional distillation.
Answer:

1. Mixture of acetone (b.p. 329 K) and methyl alcohol (b.p. 337.7 K)
2. Mixture of acetone (b.p. 329 K) and benzene (b.p. 353 K)

Question 20.
Give one industrial application of fractional distillation.
Answer:
Fractional distillation is used in petroleum industry to separate different fractions of crude oil.

Question 21.
Write a short note on distillation under reduced pressure.
Answer:

• Liquids having very high boiling points or which decompose on heating are purified by the method of distillation under reduced pressure.
• In this method, the liquid is made to boil at a temperature lower than its normal boiling point by reducing the pressure on its surface.
• The external pressure is reduced using a water pump or vacuum pump, e.g. Glycerol can be separated from soap by using this method.

Question 22.
Write the principle of solvent extraction and explain the process with labelled diagram.
Answer:
Principle: Extraction of compound takes place based on the difference in solubility of compound in two liquids.

• In this process, the solute distributes itself between two immiscible liquids. From the aqueous phase the solute gets extracted in the organic phase.
• On shaking for a few times with small volumes of organic phase, most of the solute gets extracted into the organic phase.
• Then solute is then recovered from organic solvent either by evaporation of organic solvent or distillation.

Diagram:

Question 23.
Write a short note on continuous extraction method.
Answer:

• During solvent extraction, if the solute is found to be less soluble in organic phase, then continuous extraction method is employed.
• In this method, the same amount of organic solvent is used repeatedly for extraction.
• This ensures that the most of the solute gets extracted in the organic phase.
• This technique involves continuous distillation of the solvent within the same assembly. Hence, the use of large quantity of organic solvent is avoided.

Question 24.
Match the following:

	Process		Used in the purification/separation of
i.	Crystallization	a.	Acetone and benzene
ii.	Simple distillation	b.	Benzoic acid and water
iii.	Fractional distillation	c.	Impure copper sulphate
iv.	Solvent extraction	d.	Acetone and water

Answer:
i – c,
ii – d,
iii – a,
iv – b

Question 25.
What is chromatography? Explain the principle behind it.
Answer:
Chromatography is a technique used to separate components of a mixture, and also purify compounds.
Principle: The principle of separation of substances in chromatography is based on the distribution of the solutes in two phases, i.e., stationary phase and mobile phase.

• Chromatography uses two phases for separation.
• This technique is based on the difference in rates at which components in the mixture move through the stationary phase under the influence of the mobile phase.
• In this technique, first the mixture of components is loaded at one end of the stationary phase and then the mobile phase is allowed to move over the stationary phase. The mobile phase can be a pure solvent or a mixture of solvents.
• Depending on the relative affinity of the components toward the stationary phase and mobile phase, they remain on the surface of the stationary phase or move along with the mobile phase, and gradually get separated.

Question 26.
Give a brief description of column chromatography with an illustration.
Answer:
Column chromatography involves the separation of components over a column of stationary phase. The stationary phase material can be alumina, silica gel.
Procedure:

• A slurry of the stationary phase material is filled in a long glass tube provided with a stopcock at the bottom and a glass wool plug at the lower end.
• The mixture to be separated is dissolved in a suitable solvent and then it is loaded on top of adsorbent column.
• A suitable mobile phase which could be a single solvent or a mixture of solvents is then poured over the adsorbent column.
• The mixture along with the mobile phase slowly moves down the column.
• The solutes get adsorbed on the stationary phase and depending on the degree to which they are adsorbed, they get separated from each other.
• The component which is readily adsorbed are retained on the column and others move down the column to various distances forming distinct bands.
• The component which is less strongly adsorbed is desorbed first and leaves the column first, while the strongly adsorbed component is eluted later.
• The solutions of these components are collected separately.
• These different components can be recovered by evaporating the solvent.

Question 27.
How is TLC plate or chromplate prepared?
Answer:
TLC plate or chromplate is prepared by applying a thin layer (0.2 mm thick) of adsorbent silica gel or alumina spread over a glass plate.

Question 28.
Describe the process of thin layer chromatography (TLC) and separation of components in it.
Answer:
i. Process:

• A thin layer (about 0.2 mm thick) of an adsorbent like silica gel or alumina is spread over a thin glass plate (called chromplate or TLC plate). This plate acts as a stationary phase.
• With the help of a capillary tube, the solution of the mixture to be separated is spotted at above 2 cm (on base line) from one end of the TLC plate.
• The TLC plate is then placed in a closed jar containing a suitable solvent (mobile phase or eluant).
• As the mobile phase rises up the plate, the components of the mixture move up along with the mobile phase to different distances depending upon their degree of adsorption, thus resulting in complete separation.

ii. Separation of components:

• If the components are coloured, they appear as separated coloured spots on the plate.
• If the components are not coloured but have property of fluorescence, they can be visualised under UV light, or the plate can be kept in a chamber containing a few iodine crystals. The Iodine vapours are adsorbed by the components and the spots appear brown.
• Amino acids are visualised by spraying the plate with a solution of ninhydrin. This is known as spraying agent.

Question 29.
Name the physical state each of stationary phase and mobile phase in partition chromatography.
Answer:
In partition chromatography, both stationary and mobile phases are in liquid state.

Question 30.
State the principle of partition chromatography.
Answer:
Partition chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases.

Question 31.
Describe the process of paper chromatography.
Answer:
Process of paper chromatography:

• The mixture of the compound to be analysed is dissolved in a suitable solvent and spotted on the chromatography paper about 2 cm from one end of the paper using a glass capillary.
• The paper is then suspended in a chamber containing the mobile phase.
• The mobile phase rises up the paper and flows over the spot, due to capillary action.
• Different solutes are retained differently on the paper depending on their selective partitioning between the two phases. The paper strip so developed, is known as chromatogram.

Question 32.
Name the following:
i. A glass plate coated with a thin layer of silica gel.
ii. A spraying agent used for the visualization of amino acids.
Answer:
i. Chromplate/TLC plate
ii. Ninhydrin

Question 33.
Write a short note on R_f value.
Answer:
i. In chromatography, migration of the solute relative to the solvent front gives an idea about the relative retention of the solutes (or components of t the mixture) on the stationary phase.
ii. The relative adsorption of solutes is expressed in terms of its R_f value.
The symbol R_f stands for Retardation Factor.

Question 34.
In a chemical laboratory, Priyal was asked to isolate an organic compound from its aqueous solution. She added ethyl acetate to the given sample, separated the organic layer and kept it for evaporation. At the end of her practical, Priyal found few crystals in the beaker which she kept for evaporation. Answer the following questions:
i. In the above passage, which method was used by Priyal for separation? State its principle.
ii. Why do you think the organic compound dissolved in ethyl acetate?
iii. Illustrate the method of separation used in the passage with an example.
Answer:
i. Method used: Solvent extraction method.
Principle: Extraction of compound takes place based on the difference in solubility of compound in two liquids,

• In this process, the solute distributes itself between two immiscible liquids. From the aqueous phase the solute gets extracted in the organic phase.
• On shaking for a few times with small volumes of organic phase, most of the solute gets extracted into the organic phase.
• Then solute is then recovered from organic solvent either by evaporation of organic solvent or distillation.

ii. An organic compound (non-polar) dissolves in organic solvents (non-polar) because of the dipole-dipole interactions in between them (like dissolves like). Water is a polar solvent and it is unlikely that the covalent constituents of the organic substance is strong enough to break the ionic bonds. Any substance dissolves in other because it is able to break the bonds between the solvent molecules and form weak bonds with the solvent molecules. Hence, the organic compound will be more soluble in ethyl acetate as compared to water and this helps in its isolation from aqueous solution.

iii. An example for the separation of organic compound using solvents extraction method is: Benzoic acid in water can be extracted from its aqueous solution by using benzene.

Question 35.

Based on the above diagram, answer the following questions:
i. Name the chromatographic technique involved.
ii. From the developed chromatogram, state which has the highest and which has the lowest R_f value?
iii. Based on the TLC, which component would elute out at the end of a column chromatography?
iv. Mention two applications of TLC method.
Answer:
i. Thin layer chromatography
ii. Based on the developed chromatogram, spot ‘x’ has the highest Rf value while spot ‘z’ the lowest Rf value.
iii. Based on the TLC, spot ‘z’ being strongly adsorbed will elute at the end of a column chromatography.
iv. Applications of TLC are:

• Separation of plant pigments from its mixture.
• Separation of impurities from a given organic compound.
• Separation of different amino acid.

Multiple Choice Questions

1. If a crude solid is made of mainly one substance and has some impurities then it is purified by ……………..
(A) crystallization
(B) distillation
(C) extraction
(D) sublimation
Answer:
(A) crystallization

2. Impure common salt can be purified by ……………
(A) crystallization
(B) distillation
(C) extraction
(D) sublimation
Answer:
(A) crystallization

3. Which of the following solvents is most commonly used for the crystallization of copper sulphate?
(A) Water
(B) Acetone
(C) Ether
(D) Methanol
Answer:
(A) Water

4. In distillation of liquid, water condenser is used ……………
(A) to boil the liquid
(B) to collect the liquid
(C) to condense hot vapours of the liquid
(D) to adsorb the liquid
Answer:
(C) to condense hot vapours of the liquid

5. Separation of binary mixture of acetone and methyl alcohol is done by ……………
(A) simple distillation
(B) fractional distillation
(C) fractional crystallization
(D) re-crystallization
Answer:
(B) fractional distillation

6. Which of the following method is used to separate different fractions of crude oil?
(A) Solvent extraction
(B) Simple distillation
(C) Fractional distillation
(D) TLC
Answer:
(C) Fractional distillation

7. The method used to separate a given organic compound present in aqueous solution by shaking with a suitable solvent in which the compound is more soluble than water is called ……………….
(A) simple distillation
(B) fractional distillation
(C) solvent extraction
(D) crystallization
Answer:
(C) solvent extraction

8. Adsorption chromatography is a chromatographic technique based on the principle of ……………
(A) differential adsorption
(B) differential solubility
(C) differential extraction
(D) all of these
Answer:
(A) differential adsorption

9. The stationary phase and mobile phase in TLC are ……………. respectively.
(A) solid and liquid
(B) solid and gas
(C) liquid and solid
(D) liquid and liquid
Answer:
(A) solid and liquid

10. Which of.the following is most commonly used for the visualization of amino acids in chromatography?
(A) Ultraviolet light
(B) Spraying agent
(C) Sunlight
(D) X-rays
Answer:
(B) Spraying agent

11. The stationary phase and mobile phase in partition chromatography are ………….. respectively.
(A) solid and liquid
(B) solid and gas
(C) liquid and solid
(D) liquid and liquid
Answer:
(D) liquid and liquid

12. Paper chromatography is based on the principle of …………….
(A) adsorption
(B) partition
(C) solubility
(D) volatility
Answer:
(B) partition

13. In paper chromatography, the mobile phase rises up the chromatography paper due to ………………
(A) evaporation of volatile solvent
(B) capillary action
(C) gravitational force
(D) differential adsorption
Answer:
(B) capillary action

14. Which of the following is a type of partition chromatography?
(A) Column chromatography
(B) Thin layer chromatography
(C) Paper chromatography
(D) Both (B) and (C)
Answer:
(C) Paper chromatography

15. The principle of differential adsorption is applicable for which of the following chromatographic technique?
(A) Column chromatography
(B) Thin layer chromatography
(C) Paper chromatography
(D) Both (A) and (B)
Answer:
(D) Both (A) and (B)

16. Which of the following method will give clean separation of sample of chloroform (organic liquid) and water in short time span?
(A) TLC
(B) Distillation under reduced pressure
(C) Solvent extraction
(D) Simple distillation
Answer:
(C) Solvent extraction

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 9 Morphology of Flowering Plants Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 9 Morphology of Flowering Plants

Question 1.
Explain how angiosperms are classified into different types based on habitat.
Answer:
Angiosperms can be classified into following types based on habitat:

1. Hydrophytes – Growing in aquatic habitat e.g. Hydrilla
2. Xerophytes – Growing in regions with scanty or no rainfall like desert e.g. Opimtia
3. Psammophytes – Growing in sandy soil e.g. Elymus
4. Lithophytes – Growing on rock e.g. Couchidium, Cladopus, Dalzellia, Paphiopedilum orchids, rock felt fem.
5. Halophytes – Growing in saline soil e.g. Mangrove plants like Rhizophora

Question 2.
Draw neat and labelled diagram of a typical angiospermic plant. Classify its vegetative and reproductive structures.
Answer:

Vegetative structures in angiospermic plant	Reproductive structures in angiospermic plant
Root, Stem, Leaf	Flowers, Fruits, Seeds

Question 3.
Label the various regions of a typical root in the given figure and explain them in detail.
Answer:

A typical root possesses the following regions:
1. Root cap:
a. A parenchymatous multicellular structure in the form of cap, present over young growing root apex is known as root cap.
b. Cell of root cap secrete mucilage for lubricating passage of root through the soil.
c. Cells of root cap show presence of starch granules which help in graviperception and geotropic movement of root.
d. Usually single root cap is present in plants. But in plants like Pandanus or screw pine multiple root caps are present.
e. In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.
f. Due to presence of root cap the growing apex of root.

2. Meristematic region or region of cell division:
a. The apex of the root is a growing point about 1 mm in length protected by root cap. This region is called as region of cell division or meristematic region.
b. The structure is developed by compactly arranged thin walled actively dividing meristematic cells.
c. These cells bring about longitudinal growth of root.

3. Region of elongation:
a. This region of cells is present just above zone of cell division.
b. The cells are newly formed and show rapid elongation to bring about increase in length of the root.
c. The cells help in absorption of mineral salts.

4. Region of root hair or region of absorption:
a. A region of root hair / absorption/piliferous zone is made up of numerous hair like outgrowths.
b. The epiblema or piliferous layer produces tubular elongated unicellular structures known as root hair.
c. They are in close contact with soil particles and increase surface area for absorption of water.
d. Root hair are short lived or ephimeral and are replaced after every 10 to 15 days.

5. Region of maturation/region of differentiation:
a. It is the uppermost major part of the root.
b. The cells of this region are quite impermeable to water due to thick wall.
c. The cells show differentiation and form different types of tissues.
d. This region helps in fixation of plant and conduction of absorbed substances.
e. Development of lateral roots also takes place from this region.

Question 4.
What are the primary functions of root?
Answer:
Primary functions of root are, fixation or anchorage of plant body in the soil, absorption of water and minerals from soil and conduction of absorbed materials up to the stem base, etc.

Question 5.
Which type of root system is found in plants like maize, wheat and sugarcane? Explain in detail.
Answer:

1. Adventitious root system is found in plants like maize, wheat and sugarcane.
2. Adventitious root develops from any part other than radicle.
3. Such roots may develop from the base of the stem, nodes or from leaves.
4. In monocots, radicle is short lived.
5. A thick cluster of equal sized roots arise from the base of a stem. It is also known as fibrous root system as they look like fibre. The growth of roots is superficial.
6. Adventitious roots in some plants are used for vegetative propagation. E.g. Euphorbia, Carapichea ipecacuanha (Ipecac) etc.

Question 6.
What are metamorphosed roots?
Answer:
When roots have to perform some special type of function in addition to or instead of their normal function they develop some structural changes. Such roots are called as metamorphosed roots.

Question 7.
Complete the given chart and explain the modification of tap root for storage of food.
Answer:

1. Modification of tap root for storage of food:
a. WTien tap root stores food it becomes swollen, fleshy and also develops definite shape.
b. Main or primary root is the main storage organ but sometimes hypocotyl part of embryo axis also joins the main root. Secondary roots remain thin.
c. On the basis of shape, swollen tap roots are classified as Fusiform, Conical and Napiform.
2. Fusiform root:
The fusiform root is swollen in the middle and tapering towards both ends forming spindle shaped structure, e.g. Radish (Raphanus sativus)
3. Conical root:
The conical root is broad at its morphological base and narrows down towards its apex. e.g. Carrot (Daucus car ota)
4. Napiform root:
In napiform root, base of root is highly swollen, almost spherical in shape and abruptly narrows down towards its apex. e.g. Beet (Beta vulgaris)

Question 8.
Identify the type of swollen tap root in the figures given below.

Answer:
Figure ‘a’: Conical root;
Figure ‘b’: Fusiform root;
Figure ‘c’: Napiform root

Question 9.
Answer the following questions:
1. Identify the label ‘X’ in the given figure of respiratory roots. Give its function.

2. Give any tw o examples of plants in which respiratory roots are present.
Answer:
1. X: Lenticels
In respiratory roots or pneumatophores, gaseous exchange occurs through lenticels.
2. Examples of plants in which respiratory roots are present:
Rhizophora, Avicennia, Sormeratia, Heritiera fames (sundri), etc.

Question 10.
Identify the types of modified adventitious roots in the figures given below and explain in detail.

Answer:
1. Figure ‘a’ indicates simple tuberous root.
a. Simple tuberous roots become swollen and do not show definite shape.
b. They are produced singly.
c. The roots arise from nodes over the stem and penetrate into the soil, e.g. sweet potato or shakarkand (Ipomoea batatas).

2. Figure ‘b’ indicates fasciculated tuberous roots.
a. A cluster of roots arising from one point which becomes thick and fleshy due to storage of food is known as fasciculated tuberous root.
b. These clusters are seen at the base of the stem, e.g. Dahlia, Asparagus, etc.

3. Figure ‘c’ indicates Moniliform roots.
a. Some adventitious roots get swollen at regular intervals.
b. These gives them the appearance of beads of a necklace. Such roots are called as Moniliform roots, e.g. Spinacia oleracea (Indian Spinach).

4. Figure ‘d’ indicates Nodulose roots.
The cluster of long slender roots become enlarged at the tips forming nodules is known as nodulose roots, e.g. Arrow root (Maranta), Amhaldi or mango ginger (Curcuma amada).

Question 11.
Explain various types of adventitious roots which are modified for mechanical support.
Answer:
1. Prop roots / Columnar roots:
a. These roots arise from horizontal branches of tree like Banyan tree (Ficus benghalensis) and grow vertically downwards till they penetrate the soil.
b. These prop roots show secondary growth, become thick, act like pillars to provide mechanical support to the heavy branches.

2. Stilt roots:
a. These roots normally arise from a few lower nodes of a weak stem in some monocots, shrubs and small trees.
b. They show obliquely downward growth penetrating soil and provide mechanical support to the plant.
c. In the members of family Poaceae, the plants like Maize, Jowar, Sugarcane etc. produce stilt root in whorl around the node.
d. These roots provide additional support to the plant body.
e. In Screw pine or Pandanus (Kewada), stilt roots arise only from the lower surface of obliquely growing stem for additional support. These roots show multiple root caps.

3. Climbing roots:
Different climbers with weak stem produce roots at their nodes by means of which they attach themselves to support and thereby raise themselves above the ground.
e.g. Betel leaf or Pan, black pepper or Piper nigrum (Kali Mirch), Pothos or money plant.

4. Clinging roots:
a. These tiny roots develop along intemodes, show disc at tips, which exude sticky substance.
b. This substance enables plant to get attached with walls of buildings.
c. They do not damage substratum, e.g. English Ivy (Hedera helix).

5. Plank roots/Buttress:
a. These roots often develop at the base of large trees and form plank like extensions around stem.
b. These roots provide additional support, e.g. Silk cotton, Peepal, etc.

6. Buoyant roots:
Roots developed at the nodes of aquatic herbs like (Jussiaea repens), become highly inflated and spongy providing buoyancy and helping the plant to float.

Question 12.
What are sucking roots? Explain with the help of examples.
Answer:
Sucking roots or Haustoria:
1. These are the specialised microscopic sucking roots developed by parasitic plants to absorb nourishment from the host.
2. Viscum album is a partial parasite. It develops haustoria which penetrate into xylem of host plant for absorption of food.
3. In Cuscuta reflexa or Dodder (Amarvel) haustoria penetrates vascular strand and suck food from phloem, water and minerals from xylem. Cuscuta is leafless plant with yellow stem. It is a total parasite.

Question 13.
Enlist the important characteristics of stem.
Answer:
Characteristics of stem:

1. Stem is the ascending part of the plant body which develops from plumule and reproductive units.
2. It is usually positively phototropic, negatively geotropic and negatively hydrotropic.
3. It shows different types of buds (axillary, apical, accessory, etc.).
4. It is differentiated into nodes and intemodes.
5. At nodes it produces dissimilar organs such as leaves and flowers and similar organs such as branches.
6. Young stem is green and capable of photosynthesis.

Question 14.
Sketch and label a typical stem structure and write its primary functions.
Answer:

The primary functions of the stem are to produce and support branches, leaves, flowers and fruits; conduction of water and minerals and transportation of food to plant parts.

Question 15.
What is underground stem? When does it produce aerial shoots?
Answer:
In some herbaceous plants the stem which develops below soil surface is called underground stem. The underground stem remains dormant during unfavourable condition and on the advent of favourable condition produces aerial shoots.

Question 16.
Draw neat and labelled diagram of rhizome of ginger.
Answer:

Question 17.
Potato which we eat is an underground part of a plant, however it can not be considered as root. Justify the given statement.
Answer:

1. Potato is a stem tuber.
2. It is an underground stem, modified for storage of food material.
3. Special underground branches of stem at their tips becomes swollen due to storage of food which is mostly starch.
4. Stem tuber shows distinct nodes, but not intemodes hence it is classified as stem.
5. At nodal part, it shows scale leaves with axillary buds, which are commonly called as ‘eyes’.
6. Under favourable conditions, ‘eyes’ can produce aerial shoots.
7. Potato tuber can be propagated vegetatively. [Note: In stem tuber, internodes are present but they are not very distinct.]

Question 18.
What are tunicated and compound tunicated bulbs?
Answer:
1. Tunicated bulb:
When fleshy scale leaves are arranged on stem in concentric manner, bulb is called as tunicated bulb or layered bulb. E.g. Onion
2. Compound tunicated bulb:
When fleshy scale leaves arranged on stem, partially overlap each other by their margins only, such bulb is called compound tunicated or scaly bulb. e.g. Garlic

Question 19.
Which type of modified stem is present in Colocasia and Amorphophallus? Explain with the help of neat and labelled diagram.
Answer:

1. In Colocasia and Amorphophallus corm is present, which is an underground stem modified for storage of food.
2. Corm is swollen underground spherical or subspherical vertically growing stem.
3. It is condensed structure with circular or ring like nodes.
4. It shows presence of axillary buds and scales.
5. Adventitious buds are produced which help in vegetative propagation.
6. Adventitious roots are produced at lower part of the stem.

Question 20.
1. What are sub aerial stems?
2. Explain the different types of sub aerial stems. Give atleast one example of each.
Answer:
1. Subaerial stems:
a. These are generally weak or straggling stems growing over the ground and need support for perpetuation.
b. Sometimes these stems are found to grow beneath the soil surface also. Thus, they show contact with both air and soil.
c. Subaerial stems are meant for perennation and vegetative propagation.
d. Scale leaves and axillary buds are present over stem surface. Axillary buds develop into aerial shoots.

2. Types of subaerial stems:
a. Trailer:
1. The shoot spreads over the ground without striking adventitious roots.
The branches are either flat i.e. procumbent or partly vertical i.e. decumbent.
e. g. Euphorbia, Tridax etc. [Any one example]

b. Runner:
1. They are special narrow, prostrate or horizontal green branches which develop at the base of erect shoots known as crown.
2. Runners spread in all directions to produce new crowns with bunch of adventitious roots.
3. Presence of nodes with scale leaves and axillary buds is observed.
e.g. Cynodon (Lawn grass) Centella (Hydrocotyl / Brahmi), Oxalis etc. [Any one example]

c. Stolons:
1. The slender lateral branch arising from the base of main axis is known as stolon.
2. In some plants it is above ground (wild strawberry).
3. Primarily stolon shows upward growth in the form of ordinary branch, but when it bends and touches the ground terminal bud grows into new shoot and develops adventitious roots.
e.g. Wild Strawberry, Jasmine, Mentha, etc. [Any one example]

d. Sucker:
1. It is non-green, runner like branch of stem.
2. It grows horizontally below soil initially and then comes above the soil surface obliquely to produce a new plant.
3. Sucker can be termed as underground runner.
e. g. Chrysanthemum, Banana etc. [Any one example]

e. Offset:
1. These are one intemode long runners in rosette plants at ground or water level.
2. Offset helps in vegetative propagation.
e.g. Water hyacinth or Jal kumbhi (Eichhornia) and Pistia. [Any one example]

Question 21.
Observe the given figures and identify P, Q, R and S representing the different types of sub aerial shoot.
Answer:
‘P’: Trailer ‘Q’: Runner ‘R’: Stolon ‘S’: Offset

Question 22.
Draw neat and labelled diagram of Eichhornia showing offset.
Answer:

Question 23.
What are metamorphosed stems?
Answer:
Stem or its vegetative part develops various modifications to carry out specialized functions. Such modified stems are called as metamorphosed stems.

Question 24.
Describe the aerial modifications of stem.
Answer:
Different aerial modifications shown by stem are as follows:
1. Stem tendrils:
a. Tendrils are thin, wiry, photosynthetic, leafless coiled structures.
b. They give additional support to developing plant.
c. Tendrils have adhesive glands for fixation.
d. Apical bud in Vitis quadrangularis gets modified into tendril. The further growth is carried out by axillary bud.
e. In Passiflora axillary bud gets modified in tendril.
f. Extra axillary bud is the one which grows outside the axil. This bud in cucurbita gets modified into tendril.
g. Normally floral buds are destined to produce flowers. But in plants like Antigonon they produce tendrils.

2. Thorn:
a. It is modification of apical or axillary bud.
b. Thom is hard pointed and mostly straight structure (except Bougainvillea where it is curved and useful for climbing).
c. It provides protection against browsing animals and also helps in reducing transpiration.
d. Apical bud develops into thorn in Carrisa whereas axillary bud develops into thorn in Duranta, Citrus, Bougainvillea, etc.

3. Phylloclade:
a. Modification of stem into leaf like photosynthetic organ is known as phylloclade.
b. Being stem it possesses nodes and internodes.
c. It is thick, fleshy and succulent, contains mucilage for retaining water e.g. Opuntia, Casuarina (Cylindrical shaped phylloclade) and Muehlenbeckia (ribbon like phylloclade).

4. Cladodes:
a. The branches of limited growth i.e. one intemode long and performing photosynthetic function are called as cladodes.
b. True leaves are reduced to spine or scales to reduce rate of transpiration, e.g. Asparagus.

5. Cladophylls:
These are leaf like structures bore in the axil of scale leaf. It has floral bud and scale leaf in the middle i.e. upper half is leaf and lower half is stem. e.g. Ruscus.

6. Bulbils:
a. In plants like Dioscorea, etc. axillary bud becomes fleshy and rounded due to storage of food called as bulbil.
b. When it falls off it produces new plant and help in vegetative propagation.

Question 25.
Observe the given figures oi Asparagus, Vitis quadrangularis and Passiflora. Which of the following is labelled incorrectly?
Answer:
Figure ‘a’ is incorrectly labelled.
It represents Cladode oi Asparagus.

Question 26.
How does thorn in Carissa differ from that of Duranta?
Answer:
In Carrisa, apical bud develops into thorn, whereas in Duranta, axillary bud develops into thorn.

Question 27.
Enlist the general characteristics of a leaf.
Answer:
General characteristics of a leaf:

1. Leaves are the most important appendages as they carry out photosynthesis and also help to remove excess amount of water from plant body through transpiration.
2. Leaves are exogenous in origin and develops from leaf primordium.
3. Leaf is dorsiventrally flattened lateral appendage of stem, produced at nodal region.
4. Leaf is thin, expanded and green due to presence of photosynthetic pigments, i.e. chlorophyll.
5. Axil of leaf shows presence of axillary bud.
6. Leaf shows limited growth, does not show apical bud or a growing point.

Question 28.
Give an account of various parts of a typical dicot leaf.
Answer:
A typical dicot leaf shows presence of three main parts Leaf base or Hypopodium, Petiole or Mesopodium and Leaf lamina/blade or Epipodium.
1. Leaf base or Hypopodium:
a. The point by which leaf remains attached to the stem is known as leaf base.
b. The nature of leaf base varies in different plants. It may be pulvinus (swollen), sheathing or ligulate, etc.

Question 29.
How simple leaf differs from a compound leaf?
Answer:
The leaf with entire lamina is called simple leaf, whereas leaf in which leaf lamina is divided into many leaflets is called as compound leaf.

Question 30.
Identify the type of pinnately compound leaves in the figures given below and give one example of each.
Answer:

Figure ‘a’: Paripinnately compound leaves (e.g. Cassia)
Figure ‘b’: Imparipinnately compound leaves (e.g. Rosa)
Figure ‘c’: Bipinnately compound leaves (e.g. Caesalpinia)
Figure ‘d’: Tripinnately compound leaves (e.g. Moringa)
Figure ‘e’: Decompound leaves (e.g. Coriandrum)

Question 31.
Explain how leaves modify to perform different functions other than photosynthesis and gaseous exchange.
Answer:
Leaves show different types of modification as follows:
1. Leaf spines:
Sometimes entire leaf is modified into spines (Opuntia) or margin of leaf becomes spiny (Agave) or stipule modifies into spine (Acacia, Zizyphus) to check the rate of transpiration and to protect plant from grazing.

2. Leaf tendril:
In some weak stems, leaf, leaflet or other part modifies to produce thin, green, wiry, coiled structure called as leaf tendril. It helps in climbing and provides additional support.

3. Leaf hooks:
In plants like Bignonia unguis-cati (Cat’s nail) the terminal three leaflet get modified into three! stiff curve and pointed hooks used to cling over the bark of tree.

4. Phyllode:
When petiole of leaf becomes flat, green and leaf like it is called as phyllode. In Acacia auriculoformis the normal leaf is bipinnately compound and falls off soon. The petiole modifies itself into phyllode. It is a xerophytic adaptation.

Question 32.
Identify different types of tendril in the figures given below and give one example of each.

Answer:
Figure ‘a’: Whole leaf tendril (e.g. Lathyrus)
Figure ‘b’: Leaflet tendril (e.g. Pisum sativum)
Figure ‘c’: Leaf tip tendril (e.g. Gloriosa)
Figure ‘d’: Stipular tendril (e.g. Smilax)

Question 33.
Define phyllotaxy. Why do leaves show phyllotaxy?
Answer:
Phyllotaxy:
1. Arrangement of leaves on the stem and branches in a specific manner is known as phyllotaxy.
2. It enables leaf to get sufficient light for photosynthesis.

Question 34.
Which type of phyllotaxy is show n by leaves of Mango, Nerium and Jamun?
Answer:

1. Phyllotaxy shown by leaves of Mango is alternate phyllotaxy. In this type, single leaf from each node.
2. Phyllotaxy shown by leaves of Nerium is whorled phyllotaxy. In this type, many leaves arise from each node
   and form a whorl.
3. Phyllotaxy shown by leaves of Jamun is opposite superposed phyllotaxy. In this type, a pair of opposite leaves are arranged one above the other in the same plane.

Question 35.
What are pinnately compound and palmately compound leaves? Give any two examples of each.
Answer:
1. Pinnately compound: Leaflets are present laterally on a common axis called rachis, which represents the midrib of the leaf. e.g. Cassia, Rose, Caesalpinia, Moringa, Coriandrum, etc. [Any two examples]
2. Palmately compound: In this all the leaflets are attached at the tip of petiole.
e.g. Citrus, Zorina, Oxalis, Marsilea, Bombax [Any two examples] [Note: Another example of palmately compound leaf (Bifoliate) is Balanites roxburghii.]

Question 36.
Define inflorescence.
Answer:
A specialised axis or branch over which flowers are produced or borne in definite manner is known inflorescence.

Question 37.
Write significance of inflorescence.
Answer:
Significance of inflorescence:

1. Inflorescence makes a flower more conspicuous to attract the insects and birds for pollination.
2. It provides more chances for cross pollination.
3. An insect can pollinate many flowers in inflorescence in a single visit.
4. In an inflorescence, flowers open successively and not simultaneously. This improves chances of pollination as flowering period is longer.

Question 38.
Define flower.
Answer:
Flower is highly modified and condensed shoot meant for sexual reproduction.

Question 39.
Complete the table by giving the meaning of following terminologies related to flower.
Answer:

1. Complete flower: Presence of all four floral whorls	Incomplete flower: Absence of any one of the floral whorls.
2. Pedicellate flower: Flower with pedicel	Sessile flower: Flower without pedicel
3. Bracteate flower: Flower with bract at the base of pedicel	Ebracteate flower: Flower without bract
4. Perfect flower: Both androecium and gynoecium are present, also called as dicliny or bisexual flower.	Imperfect flower: Any one reproductive whorl is present also called as monocliny or unisexual flower.
5. Actinomorphic flower: The flower can be cut in any plane passing through the centre in order to obtain two identical halves. Flowers show radial symmetry, e.g. Sunflower	Zygomorphic flower: The flower can be cut only along one plane passing through the centre in order to obtain two identical halves. Flowers show bilateral symmetry e.g. Sweet Pea flower.
6. Unisexual flower: It can be either staminate (male)/ pistillate (female) flower	Neuter flower: When both reproductive whorls are absent, it is said to be neuter flower
7. Monoecious plant: Male and female reproductive flowers are borne on same plant, e.g. Maize	Dioecious plant: Only one type of unisexual flowers are present on plant, e.g. Ray floret of sunflower

Question 40.
Mango is called as polygamous plant. Why?
Answer:
Mango produces all types of flowers like staminate, bisexual and neuter, hence it is called as polygamous plant.

Question 41.
What is insertion of floral whorls?
Answer:
The position and arrangement of rest of the floral whorls (calyx, corolla, androecium) with respect to gynoecium on the thalamus is known as insertion of floral whorls. .

Question 42.
With help of neat and labelled diagrams explain the classification of flowers based on the position of ovary on the thalamus.
Answer:
Classification of flowers based on the position of ovary on the thalamus.

1. Hypogynous flower:
When the convex or conical thalamus is present in flower, ovary occupies the highest position while other floral parts are below ovary. Ovary is said to be superior and flower is called as hypogynous flower, e.g. Brinjal, Mustard, China rose etc. It is denoted as G in floral formula.

2. Perigynous flower:
When cup shaped or saucer shaped thalamus is present in a flower, ovary and other floral parts occupy about same position. Such an ovary is said to be semi- superior or semi-inferior. All floral whorls are at the rim of thalamus. Flower is called as perigynous. e.g. Rose, etc. It is denoted as G- in floral formula.

3. Epigynous flower:
When thalamus completely encloses ovary and may show fusion with wall; the other floral parts occupy superior position and ovary becomes inferior. Such flower is called as epigynous flower, e.g. Ray florets of Sunflower, Guava, Cucumber etc. It is denoted as G in floral formula.

Question 43.
What is thalamus?
Answer:
Thalamus:
1. The upper, swollen, condensed, knob-like part of the pedicel is called thalamus. It is also called receptacle or torus.
2. In a typical flower, the thalamus consists of four compactly arranged nodes and three highly condensed intemodes.
3. From each node of thalamus, whorl of modified leaves is produced.

Question 44.
With the help of a diagram explain the floral parts of a typical flower.
Answer:

Floral parts of a typical flower:
1. Calyx (K):
a. It is outermost floral whorl and individual members are known as
sepals.
b. Sepals are usually green in colour and perform photosynthesis.
c. If all the sepals are united, the condition is gamosepalous and if they are free, the condition is called as polysepalous.
d. Gamosepalous calyx is found in China rose and polysepalous calyx is found in Brassica.
e. The main function of sepals is to protect inner floral parts in bud condition.
f. Sometimes sepals become brightly coloured (petaloid sepals) and attract insects for pollination,
e.g. Mussaenda etc.
g. Sepals modify into hairy structures called as pappus. Such calyx helps in dispersal of fruit, e.g. Tridex.

2. Corolla (C):
a. It is second floral whorl from outer side and variously coloured.
b. The individual member is called as petal.
c. Petals may be sweet to taste, possess scent, odour, aroma or fragrance etc.
d. The condition in which petals are free is said to be polypetalous (e.g. Rose) and if they are fused it is called as gamopetalous (e.g. Datura).
e. The main function of corolla is to attract different agencies for pollination.

3. Perianth (P):
a. Many times, calyx and corolla remain undifferentiated. Such member is known as tepal.
b. The whorl of tepals is known as Perianth.
c. It protects other floral whorls.
d. If all the tepals are free the condition is called as polyphyllous and if they are fused the condition is called as gamophyllous.
e. Sepaloid perianth shows green tepals, while petaloid perianth shows brightly coloured tepals. e.g. Lily, Amaranthus, Celosia, etc.
f. Petaloid tepal helps in pollination and sepaloid tepals can perform photosynthesis.

4. Androecium (A):
a. It is third floral whorl from outer side.
b. Androecium is male reproductive part of a flower.
c. The individual member is known as stamen.
d. If all the stamens are free the condition is polyandrous and synandrous if they are fused.
e. Typical stamen shows three different parts:
1. Anther: It is terminal in position. Anther produces pollen grains. It is usually dithecous (two anther lobes), tetralocular/tetra sporangiate (four pollen sacs) structure, e.g. Datura.
In some plants it is monothecous (single lobed), bilocular or bisporangiate structure e.g. Hibiscus.
2. Filament: It is a stalk of stamen and bears anther at its tip. It raises anther to a proper height for easy dispersal of pollen grains.
3. Connective: It is in continuation with the filament. It is similar to mid rib and connects two anther lobes together and also with the filament.

5. Gynoecium (G):
a. It is the female reproductive part of a flower and innermost in position.
b. It is also known as pistil.
c. The individual member of gynoecium is known as carpel.
d. The number of carpels may be one to many.
e. If all the carpels are fused the condition is described as syncarpous and if they are free the condition is described as apocarpous.
f. The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.
g. A typical carpel consists of three parts stigma, style and ovary.
1. Stigma is a terminal part of carpel which receives pollen grains during pollination. It helps in
germination of pollen grain. Stigma shows variation in structure to suit the pollinating agent.
2. Style is narrow thread like structure that connects ovary with stigma.
3. Ovary is basal swollen fertile part of the carpel. Ovules are produced in ovary on a soft fertile tissue called placenta.

Question 45.
Explain the term Epicalyx.
Answer:
Epicalyx:

1. Epicalyx is an additional whorl of sepal like structures formed by bracteole which occurs on the outside of calyx.
2. These are 5-8 in number.
3. It is a characteristic feature of family Malvaceae.
4. They are protective in function, e.g. Ladies finger

Question 46.
Match the columns.

Column I (Tvpe of calyx)	Column II (Nature of sepals)	Column III (Example)
1. Caducous	(a) Sepals remain even after fruit formation	1. Brinjal, Pea
2. Deciduous	(b) Sepals fall off as soon as the flower bud opens	2. Lotus, Mustard
3. Persistent	(c) Sepals survive till (withering of petals) fruit formation	3. Argemone (Poppy)

Answer:

Column I (Type of calyx)	Column II (Nature of sepals)	Column III (Example)
1. Caducous	(b) Sepals fall off as soon as the flower bud open	3. Argemone (Poppy)
2. Deciduous	(c) Sepals survive till (withering of petals) fruit formation	2. Lotus, Mustard
3. Persistent	(a) Sepals remain even after fruit formation	1. Brinjal, Pea

Question 47.
Define aestivation and explain its different types.
Answer:
Aestivation:
1. The mode of arrangement of sepals, petals or tepals in a flower with respect to the members of same whorl is known as aestivation.
2. Different types of aestivation are as follows:
a. Valvate: Margins of sepals or petals remain either in contact or lie close to each other but do not overlap, e.g. Calyx of Datura, Calotropis.
b. Twisted: Margins of each sepal or petal is directed inwards and is overlapped. While the other margin is directed outwards and overlap the margin of adjacent, e.g. Corolla of China rose, Cotton etc.
c. Imbricate: One of the sepals or petals is internal and is overlapped at both the margins. One is external i.e. both of its margin overlap adjacent member. Rest of the sepals / petals have one inner or overlapped margin and outer or overlapping margin, e.g. Cassia, Bauhinia, etc.
d. Vexillary: Corolla is butterfly shaped and consists of five petals. Outermost and largest is known as standard or vexillum, two lateral petals are wings and two smaller fused forming boat shaped structures keel. e.g. Pisum sativum

Question 48.
Answer the following:
1. Define the term Adelphy.
2. Give three types of stamens based on adelphy. Draw a diagram of each type.
Answer:
Adelphy:
1. When stamens are united by filaments and anthers are free, the condition is called adelphy.
2. Three types of stamens based on adelphy:
Monadelphous stamens, Diadelphous stamens, Polyadelphous stamens.

Question 49.
Define the following terms and give one example of each:

1. Epipetalous stamens
2. Epiphyllous stamens
3. Syngenesious stamens
4. Synandrous stamens

Answer:

1. Epipetalous stamens: When the stamens are united to petals they are described as epipetalous stamen. E.g. Datura
2. Epiphyllous stamens: When the stamens are united to tepals they are described as epiphyllous stamens. E.g. Lily
3. Syngenesious stamens: When anthers are united and filaments are free, such stamens are called as syngenesious stamens. E.g. Sunflower
4. Synandrous stamens: When both anthers and filaments are fused, such stamens are called as synandrous stamens. E.g. Cucurbita

Question 50.
Define placentation. Explain its types.
Answer:
1. Placentation: The mode of arrangement of ovules on the placenta within the ovary is called placentation.
2. Types of placentation:
a. Marginal: Ovules are placed on the fused margins of unilocular ovary, e.g. Pea, Bean etc.
b. Axile: Ovules are placed on the central axis of a multilocular ovary, e.g. China rose, Cotton, etc.
c. Parietal: Ovules are placed on the inner wall of unilocular ovary of multicarpellary syncarpus ovary,
e. g. Papaya, Cucumber, etc.
d. Basal: Single ovule is present at the base of unilocular ovary, e.g. Sunflower, Rice, Wheat.
e. Free central: Ovules are borne on central axis which is not attached to ovary wall, e.g.Argemone, Dianthus.

Question 51.
What are parthenocarpic fruits? Give any two examples.
Answer:
Fruits which are produced from ovary without fertilization are called as parthenocarpic fruits, e.g. Cultivated Banana and Grapes.

Question 52.
How true fruit differs from false fruit or pseudofruit? Give one example of each.
Answer:
The fruit which develops only from ovary is called as true fruit or eucarp. e.g. Mango. The fruit which develops from other than ovary floral part is called as false fruit or pseudocarp

Question 53.
Identify the parts of mango and apple fruit in the given figures.
Answer:
1. parts of mango fruit:
X: Epicarp
Y: Mesocarp
Z: Endocarp

2. Part of apple fruit:
P: Thalamus
Q: Seed
R: Endocarp
S: Mesocarp

Question 54.
Identify terms a,b,and c in the given chart.

Answer:
a: Epicarp
b: Mesocarp
c: Endocarp

Question 55.
What are simple fruits? Explain its different types.
Answer:
1. Fruits which develop from one ovary of one flower are called as simple fruits.
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
1. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
2. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)

b. Fleshy fruits: These fruits have thick pericarp. It is further divided into two types based on nature endocarp:
1. Drupe: In these fruits, endocarp is hard and stony, e.g. Mango
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato

Question 56.
What are aggregate fruits? Enlist the types of aggregate fruits.
Answer:
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)

Question 57.
Write a short note on composite fruits.
Answer:
1. The fruits which develop from many ovaries of many flowers of a complete inflorescence are called composite fruits, (e.g. fig) and from catkin inflorescence are called sorosis (e.g. Pineapple).
2. Fruits which develops from hypanthodium inflorescence are called syconus.

Question 58.
Match the columns with column II.

Column I	Column II
1. Composite fruit	(a) Custard apple
2. Parthenocarpic fruit	(b) Cultivated Banana
3. Aggregate fruit	(c) Mango
	(d) Pineapple

Answer:

Column I	Column II
1. Composite fruit	(d) Pineapple
2. Parthenocarpic fruit	(b) Cultivated Banana
3. Aggregate fruit	(a) Custard apple

Question 59.
Describe the structure of a seed.
Answer:
Structure of a seed:

1. Seed is a reproductive unit that developed from fertilized mature ovule.
2. The seed is made up of seed coat, embryo with or without endosperm and one or two cotyledons.
3. Outer most covering of a seed is called seed coat.
4. It shows outer layer called as testa and inner layer called as tegmen.
5. Hilum is a scar on the seed coat through which seed attach to the fruit.
6. Embryo of a seed is enclosed within seed coat.
7. Embryonal axis consists of radicle and plumule.
8. The part of embryonal axis between cotyledon and plumule is epicotyl, while the part between cotyledons and radicle is hypocotyl.
9. The nutritive tissue in a seed called endosperm.

[Note: Dicotyledonous seed is a non-endospermic or exalbuminous, as it lacks endosperm at maturity.]
[Note: Students can scan the given Q.R code to study the structure of dicotyledonous and monocotyledonous seed.]

Question 60.
Describe the family Fabaceae with suitable floral diagram.
Answer:

• Example: Pea plant (Pisum sativum)
• Habit: Tree, shrubs, herbs.
• Root: Root with root nodules.
• Stem: Erect or climber.
• Leaves: Alternate phyllotaxy, Pinnately compound leaves.
• Inflorescence: Racemose
• Flower: Zygomorphic, bisexual, complete.
• Calyx: Sepals five, gamosepalous, imbricate aestivation.
• Corolla: Petals five, polypetalous, consisting of a larger posterior petal vexillum, two lateral petals wings and two anterior ones forming a keel, vexillary aestivation.
• Androecium: Stamens ten, diadelphous.
• Gynoecium: Ovary superior, monocarpellary, unilocular with many ovules, marginal placentation. Fruit: Legume.
• Seed: Non-endospermic

Question 61.
Apply your knowledge

Question 1.
Tendrils are seen in the following plants. Identify whether they are stem tendrils or leaf tendrils,

1. Vitis
2. Smilax
3. Lathyrus
4. Passiflora
5. Cucurbita
6. Gloriosa

Answer:
Stem tendrils → Vitis, Passiflora, Cucurbita
Leaf tendrils → Smilax, Lathyrus, Gloriosa

Question 2.
Which type of roots are shown by halophytes growing in Sundarbans in West Bengal?
Answer:
Pneumatophores

Question 3.
Find out from internet the state flower of Sikkim. Write about the type of roots shown by this plant.
Answer:
Dendrobium is the state flower of Sikkim. It shows epiphytic roots.
Epiphytic roots:

1. Epiphytic plants like Vanda, Dendrobium grow on branches of trees in dense rain forests and are unable to obtain moisture from soil.
2. Such plants produce epiphytic roots which hang in the air.
3. The roots are provided with a spongy membranous absorbent covering of the velamen tissue.
4. The cells of velamen tissue are hygroscopic and have porous walls, thus they can absorb moisture from air.
5. Epiphytic roots can be silvery white or green and are without root cap.

Question 62.
Quick Review:

Question 63.
Exercise:

Question 1.
Draw neat and labelled diagram of tap root showing different regions.
Answer:
A typical root possesses the following regions:
1. Root cap:
a. A parenchymatous multicellular structure in the form of cap, present over young growing root apex is known as root cap.
b. Cell of root cap secrete mucilage for lubricating passage of root through the soil.
c. Cells of root cap show presence of starch granules which help in graviperception and geotropic movement of root.
d. Usually single root cap is present in plants. But in plants like Pandanus or screw pine multiple root caps are present.
e. In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.
f. Due to presence of root cap the growing apex of root is

2. Meristematic region or region of cell division:
a. The apex of the root is a growing point about 1 mm in length protected by root cap. This region is called as region of cell division or meristematic region.
b. The structure is developed by compactly arranged thin walled actively dividing meristematic cells.
c. These cells bring about longitudinal growth of root.

3. Region of elongation:
a. This region of cells is present just above zone of cell division.
b. The cells are newly formed and show rapid elongation to bring about increase in length of the root.
c. The cells help in absorption of mineral salts.

4. Region of root hair or region of absorption:
a. A region of root hair / absorption/piliferous zone is made up of numerous hair like outgrowths.
b. The epiblema or piliferous layer produces tubular elongated unicellular structures known as root hair.
c. They are in close contact with soil particles and increase surface area for absorption of water.
d. Root hair are short lived or ephimeral and are replaced after every 10 to 15 days.

5. Region of maturation/region of differentiation:
a. It is the uppermost major part of the root.
b. The cells of this region are quite impermeable to water due to thick wall.
c. The cells show differentiation and form different types of tissues.
d. This region helps in fixation of plant and conduction of absorbed substances.
e. Development of lateral roots also takes place from this region.

Question 2.
Write a short note on root cap.
Answer:
Root cap:
a. A parenchymatous multicellular structure in the form of cap, present over young growing root apex is known as root cap.
b. Cell of root cap secrete mucilage for lubricating passage of root through the soil.
c. Cells of root cap show presence of starch granules which help in graviperception and geotropic movement of root.
d. Usually single root cap is present in plants. But in plants like Pandanus or screw pine multiple root caps are present.
e. In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.
f. Due to presence of root cap the growing apex of root.

Question 3.
Name the region of a root which is located just above the zone of a cell division.
Answer:
Region of elongation:
a. This region of cells is present just above zone of cell division.
b. The cells are newly formed and show rapid elongation to bring about increase in length of the root.
c. The cells help in absorption of mineral salts.

Question 4.
Name any two plants in which root caps are replaced by root pockets.
Answer:
In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.

Question 5.
Write a short note on pneumatophores.
Answer:
1. a. Plants growing in marshy region (halophytes) produce upwardly growing roots called as
pneumatophores or respiratory roots.
b. The main root system of these plants does not get sufficient air for respiration as soil is water logged.
c. Due to this, mineral absorption of plant also gets affected.
d. To overcome this problem underground roots, develop special roots which are negatively geotropic; growing vertically upward.
e. These roots are conical projections present around main trunk of plant.
f. Respiratory roots show presence of lenticels which helps in gaseous exchange.
2. Examples of plants in which respiratory roots are present:
Rhizophora, Avicennia, Sormeratia, Heritiera fames (sundri), etc.

Question 6.
Name the four types of adventitious roots modified for food storage.
Answer:
1. Figure ‘a’ indicates simple tuberous root.
a. Simple tuberous roots become swollen and do not show definite shape.
b. They are produced singly.
c. The roots arise from nodes over the stem and penetrate into the soil, e.g. sweet potato or shakarkand (Ipomoea batatas).

2. Figure ‘b’ indicates fasciculated tuberous roots.
a. A cluster of roots arising from one point which becomes thick and fleshy due to storage of food is known as fasciculated tuberous root.
b. These clusters are seen at the base of the stem, e.g. Dahlia, Asparagus, etc.

3. Figure ‘c’ indicates Moniliform roots.
a. Some adventitious roots get swollen at regular intervals.
b. These gives them the appearance of beads of a necklace. Such roots are called as Moniliform roots, e.g. Spinacia oleracea (Indian Spinach).

4. Figure ‘d’ indicates Nodulose roots.
The cluster of long slender roots become enlarged at the tips forming nodules is known as nodulose roots, e.g. Arrow root (Maranta), Amhaldi or mango ginger (Curcuma amada).

Question 7.
Explain in detail modification of tap roots for food storage.
Answer:
1. Modification of tap root for storage of food:
a. WTien tap root stores food it becomes swollen, fleshy and also develops definite shape.
b. Main or primary root is the main storage organ but sometimes hypocotyl part of embryo axis also joins the main root. Secondary roots remain thin.
c. On the basis of shape, swollen tap roots are classified as Fusiform, Conical and Napiform.
2. Fusiform root:
The fusiform root is swollen in the middle and tapering towards both ends forming spindle shaped structure, e.g. Radish (Raphanus sativus)
3. Conical root:
The conical root is broad at its morphological base and narrows down towards its apex. e.g. Carrot (Daucus car ota)
4. Napiform root:
In napiform root, base of root is highly swollen, almost spherical in shape and abruptly narrows down towards its apex. e.g. Beet (Beta vulgaris)

Question 8.
Write a short note on:
1. Stilt root
2. Climbing roots
Answer:
1. Stilt roots:
a. These roots normally arise from a few lower nodes of a weak stem in some monocots, shrubs and small trees.
b. They show obliquely downward growth penetrating soil and provide mechanical support to the plant.
c. In the members of family Poaceae, the plants like Maize, Jowar, Sugarcane etc. produce stilt root in whorl around the node.
d. These roots provide additional support to the plant body.
e. In Screw pine or Pandanus (Kewada), stilt roots arise only from the lower surface of obliquely growing stem for additional support. These roots show multiple root caps.

2. Climbing roots:
Different climbers with weak stem produce roots at their nodes by means of which they attach themselves to support and thereby raise themselves above the ground.
e.g. Betel leaf or Pan, black pepper or Piper nigrum (Kali Mirch), Pothos or money plant.

Question 9.
What is the function of prop roots in banyan tree?
Answer:
These prop roots show secondary growth, become thick, act like pillars to provide mechanical support to the heavy branches.

Question 10.
Name any two plants which show climbing roots.
Answer:
Climbing roots:
Different climbers with weak stem produce roots at their nodes by means of which they attach themselves to support and thereby raise themselves above the ground.
e.g. Betel leaf or Pan, black pepper or Piper nigrum (Kali Mirch), Pothos or money plant.

Question 11.
Identify the type of modified roots shown in the picture given below.

Answer:
Plank roots/Buttress:
a. These roots often develop at the base of large trees and form plank like extensions around stem.
b. These roots provide additional support, e.g. Silk cotton, Peepal, etc.

Question 12.
Identify the type of root marked as ‘X’ in the given picture and write a short note on it.

Answer:
Epiphytic roots:
1. Epiphytic plants like Vanda, Dendrobium grow on branches of trees in dense rain forests and are unable to obtain moisture from soil.
2. Such plants produce epiphytic roots which hang in the air.
3. The roots are provided with a spongy membranous absorbent covering of the velamen tissue.
4. The cells of velamen tissue are hygroscopic and have porous walls, thus they can absorb moisture from air.
5. Epiphytic roots can be silvery white or green and are without root cap.

Question 13.
Identify the type of aerial modification of stem in the given figure.

Answer:
Cladophylls:
These are leaf like structures bore in the axil of scale leaf. It has floral bud and scale leaf in the middle i.e. upper half is leaf and lower half is stem. e.g. Ruscus.

Question 14.
Write a short note on corm.
Answer:

1. In Colocasia and Amorphophallus corm is present, which is an underground stem modified for storage of food.
2. Corm is swollen underground spherical or subspherical vertically growing stem.
3. It is condensed structure with circular or ring like nodes.
4. It shows presence of axillary buds and scales.
5. Adventitious buds are produced which help in vegetative propagation.
6. Adventitious roots are produced at lower part of the stem.

Question 15.
Observe the type of sub aerial stem in the given picture of Eichhornia and explain it.

Answer:
Offset:
1. These are one intemode long runners in rosette plants at ground or water level.
2. Offset helps in vegetative propagation.
e.g. Water hyacinth or Jal kumbhi (Eichhornia) and Pistia. [Any one example]

Question 16.
Explain the type of stem modification in Opuntia.
Answer:
Phylloclade:
a. Modification of stem into leaf like photosynthetic organ is known as phylloclade.
b. Being stem it possesses nodes and internodes.
c. It is thick, fleshy and succulent, contains mucilage for retaining water e.g. Opuntia, Casuarina (Cylindrical shaped phylloclade) and Muehlenbeckia (ribbon like phylloclade).

Question 17.
Identify the structure ‘X’ in the given figure and explain it.

Answer:
Bulbils:
a. In plants like Dioscorea, etc. axillary bud becomes fleshy and rounded due to storage of food called as bulbil.
b. When it falls off it produces new plant and help in vegetative propagation.

Question 18.
Describe the types of stem tendrils present in different plants.
Answer:
1. Stem tendrils:
a. Tendrils are thin, wiry, photosynthetic, leafless coiled structures.
b. They give additional support to developing plant.
c. Tendrils have adhesive glands for fixation.
d. Apical bud in Vitis quadrangularis gets modified into tendril. The further growth is carried out by axillary bud.
e. In Passiflora axillary bud gets modified in tendril.
f. Extra axillary bud is the one which grows outside the axil. This bud in cucurbita gets modified into tendril.
g. Normally floral buds are destined to produce flowers. But in plants like Antigonon they produce tendrils.

Question 19.
What are cladodes?
Answer:
Cladodes:
a. The branches of limited growth i.e. one intemode long and performing photosynthetic function are called as cladodes.
b. True leaves are reduced to spine or scales to reduce rate of transpiration, e.g. Asparagus.

Question 20.
What are tendrils?
Answer:
Tendrils are thin, wiry, photosynthetic, leafless coiled structures.

Question 21.
What is sympodial growth in ginger rhizome and monopodial growth in lotus rhizome?
Answer:
1. Growth of rhizome takes place with lateral buds, such growth is known as sympodial growth, e.g. Ginger (Zingiber officinale), Turmeric (Curcuma domestica), Canna etc.
2. In plants where rhizomes grow obliquely, terminal bud brings about growth of rhizomes. This is known as monopodial growth, e.g. Nymphea, Nelumbo (Lotus), Pteris (Fern) etc.

Question 22.
In xerophytic plant like Opuntia, the stem becomes photosynthetic. Give reason.
Answer:
1. Xerophytes are the plants which grow in regions with scanty or no rainfall like desert.
2. In Xerophytes, leaves get modified into spines or get reduced in size to check the loss of water due to transpiration.
3. As the leaves are modified into spines, the stem becomes green in colour to do the function of photosynthesis.

Question 23.
Define compound leaves. Explain its two types.
Answer:
The leaf with entire lamina is called simple leaf, whereas leaf in which leaf lamina is divided into many leaflets is called as compound leaf.
1. Pinnately compound: Leaflets are present laterally on a common axis called rachis, which represents the midrib of the leaf. e.g. Cassia, Rose, Caesalpinia, Moringa, Coriandrum, etc. [Any two examples]
2. Palmately compound: In this all the leaflets are attached at the tip of petiole.
e.g. Citrus, Zorina, Oxalis, Marsilea, Bombax [Any two examples] [Note: Another example of palmately compound leaf (Bifoliate) is Balanites roxburghii.]

Question 24.
Define leaf venation. What are its two types?
Answer:
Leaf venation:
1. Arrangement of veins and veinlets in leaf lamina is known as venation.
2. There are two types of leaf venation: parallel venation which is found in monocot leaves and reticulate venation which is found in dicot leaves.

Question 25.
Draw neat and labelled diagram of structure of a typical leaf.
Answer:
A typical dicot leaf shows presence of three main parts Leaf base or Hypopodium, Petiole or Mesopodium and Leaf lamina/blade or Epipodium.
1. Leaf base or Hypopodium:
a. The point by which leaf remains attached to the stem is known as leaf base.
b. The nature of leaf base varies in different plants. It may be pulvinus (swollen), sheathing or ligulate, etc.

Question 26.
Which type of venation can be observed in monocot leaf?
Answer:
There are two types of leaf venation: parallel venation which is found in monocot leaves and reticulate venation which is found in dicot leaves.

Question 27.
Enlist the types of pinnately compound leaves and give one example of each.
Answer:
Figure ‘a’: Paripinnately compound leaves (e.g. Cassia)
Figure ‘b’: Imparipinnately compound leaves (e.g. Rosa)
Figure ‘c’: Bipinnately compound leaves (e.g. Caesalpinia)
Figure ‘d’: Tripinnately compound leaves (e.g. Moringa)
Figure ‘e’: Decompound leaves (e.g. Coriandrum)

Question 28.
What is opposite decussate phyllotaxy? Give one example.
Answer:
Figure ‘c’ represents opposite decussate phyllotaxy. In this type of phyllotaxy, a pair of leaf arise from each node and the consecutive pair at right angle to the previous one. e.g. Calotropis.

Question 29.
Explain in detail androecium of an angiospermic flower.
Answer:
Androecium (A):
a. It is third floral whorl from outer side.
b. Androecium is male reproductive part of a flower.
c. The individual member is known as stamen.
d. If all the stamens are free the condition is polyandrous and synandrous if they are fused.
e. Typical stamen shows three different parts:
1. Anther: It is terminal in position. Anther produces pollen grains. It is usually dithecous (two anther lobes), tetralocular/tetra sporangiate (four pollen sacs) structure, e.g. Datura.
In some plants it is monothecous (single lobed), bilocular or bisporangiate structure e.g. Hibiscus.

2. Filament: It is a stalk of stamen and bears anther at its tip. It raises anther to a proper height for easy dispersal of pollen grains.

3. Connective: It is in continuation with the filament. It is similar to mid rib and connects two anther lobes together and also with the filament.

4. Gynoecium (G):
a. It is the female reproductive part of a flower and innermost in position.
b. It is also known as pistil.
c. The individual member of gynoecium is known as carpel.
d. The number of carpels may be one to many.
e. If all the carpels are fused the condition is described as syncarpous and if they are free the condition is described as apocarpous.
f. The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.
g. A typical carpel consists of three parts stigma, style and ovary.

1. Stigma is a terminal part of carpel which receives pollen grains during pollination. It helps in
germination of pollen grain. Stigma shows variation in structure to suit the pollinating agent.
2. Style is narrow thread like structure that connects ovary with stigma.
3. Ovary is basal swollen fertile part of the carpel. Ovules are produced in ovary on a soft fertile tissue called placenta.

Question 30.
Draw neat and labelled diagram of a typical flower.
Answer:
Floral parts of a typical flower:
1. Calyx (K):
a. It is outermost floral whorl and individual members are known as
sepals.
b. Sepals are usually green in colour and perform photosynthesis.
c. If all the sepals are united, the condition is gamosepalous and if they are free, the condition is called as polysepalous.
d. Gamosepalous calyx is found in China rose and polysepalous calyx is found in Brassica.
e. The main function of sepals is to protect inner floral parts in bud condition.
f. Sometimes sepals become brightly coloured (petaloid sepals) and attract insects for pollination,
e.g. Mussaenda etc.
g. Sepals modify into hairy structures called as pappus. Such calyx helps in dispersal of fruit, e.g. Tridex.

2. Corolla (C):
a. It is second floral whorl from outer side and variously coloured.
b. The individual member is called as petal.
c. Petals may be sweet to taste, possess scent, odour, aroma or fragrance etc.
d. The condition in which petals are free is said to be polypetalous (e.g. Rose) and if they are fused it is called as gamopetalous (e.g. Datura).
e. The main function of corolla is to attract different agencies for pollination.

3. Perianth (P):
a. Many times, calyx and corolla remain undifferentiated. Such member is known as tepal.
b. The whorl of tepals is known as Perianth.
c. It protects other floral whorls.
d. If all the tepals are free the condition is called as polyphyllous and if they are fused the condition is called as gamophyllous.
e. Sepaloid perianth shows green tepals, while petaloid perianth shows brightly coloured tepals. e.g. Lily, Amaranthus, Celosia, etc.
f. Petaloid tepal helps in pollination and sepaloid tepals can perform photosynthesis.

4. Androecium (A):
a. It is third floral whorl from outer side.
b. Androecium is male reproductive part of a flower.
c. The individual member is known as stamen.
d. If all the stamens are free the condition is polyandrous and synandrous if they are fused.
e. Typical stamen shows three different parts:
1. Anther: It is terminal in position. Anther produces pollen grains. It is usually dithecous (two anther lobes), tetralocular/tetra sporangiate (four pollen sacs) structure, e.g. Datura.
In some plants it is monothecous (single lobed), bilocular or bisporangiate structure e.g. Hibiscus.
2. Filament: It is a stalk of stamen and bears anther at its tip. It raises anther to a proper height for easy dispersal of pollen grains.
3. Connective: It is in continuation with the filament. It is similar to mid rib and connects two anther lobes together and also with the filament.
4. Gynoecium (G):
a. It is the female reproductive part of a flower and innermost in position.
b. It is also known as pistil.
c. The individual member of gynoecium is known as carpel.
d. The number of carpels may be one to many.
e. If all the carpels are fused the condition is described as syncarpous and if they are free the condition is described as apocarpous.
f. The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.
g. A typical carpel consists of three parts stigma, style and ovary.
1. Stigma is a terminal part of carpel which receives pollen grains during pollination. It helps in
germination of pollen grain. Stigma shows variation in structure to suit the pollinating agent.
2. Style is narrow thread like structure that connects ovary with stigma.
3. Ovary is basal swollen fertile part of the carpel. Ovules are produced in ovary on a soft fertile tissue called placenta.

Question 37.
How many carpels are present in gynoecium of Cucurbita and Hibiscus flower?
Answer:
The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.

Question 38.
Enlist the different types of aestivation and placentation in a flower.
Answer:
Different types of aestivation are as follows:
a. Valvate: Margins of sepals or petals remain either in contact or lie close to each other but do not overlap, e.g. Calyx of Datura, Calotropis.
b. Twisted: Margins of each sepal or petal is directed inwards and is overlapped. While the other margin is directed outwards and overlap the margin of adjacent, e.g. Corolla of China rose, Cotton etc.
c. Imbricate: One of the sepals or petals is internal and is overlapped at both the margins. One is external i.e. both of its margin overlap adjacent member. Rest of the sepals / petals have one inner or overlapped margin and outer or overlapping margin, e.g. Cassia, Bauhinia, etc.
d. Vexillary: Corolla is butterfly shaped and consists of five petals. Outermost and largest is known as standard or vexillum, two lateral petals are wings and two smaller fused forming boat shaped structures keel. e.g. Pisum sativum

2. Types of placentation:
a. Marginal: Ovules are placed on the fused margins of unilocular ovary, e.g. Pea, Bean etc.
b. Axile: Ovules are placed on the central axis of a multilocular ovary, e.g. China rose, Cotton, etc.
c. Parietal: Ovules are placed on the inner wall of unilocular ovary of multicarpellary syncarpus ovary,
e. g. Papaya, Cucumber, etc.
d. Basal: Single ovule is present at the base of unilocular ovary, e.g. Sunflower, Rice, Wheat.
e. Free central: Ovules are borne on central axis which is not attached to ovary wall, e.g.Argemone, Dianthus.

Question 39.
Explain in detail types of fruits.
Answer:
1. Fruits which develop from one ovary of one flower are called as simple fruits.
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
1. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
2. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)

b. Fleshy fruits: These fruits have thick pericarp. It is further divided into two types based on nature endocarp:
1. Drupe: In these fruits, endocarp is hard and stony, e.g. Mango
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)
1. The fruits which develop from many ovaries of many flowers of a complete inflorescence are called composite fruits, (e.g. fig) and from catkin inflorescence are called sorosis (e.g. Pineapple).
2. Fruits which develops from hypanthodium inflorescence are called syconus.

Question 40.
Give one example of each type of fruit given below:
1. Etario of berries
2. Berry
3. Etario of follicles
4. Cypsela
5. Sorosis
Answer:
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
3. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
4. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)
5. The fruits which develop from many ovaries of many flowers of a complete inflorescence are called composite fruits, (e.g. fig) and from catkin inflorescence are called sorosis (e.g. Pineapple).

Question 41.
Define parthenocarpic fruit and give one example.
Answer:
Fruits which are produced from ovary without fertilization are called as parthenocarpic fruits, e.g. Cultivated Banana and Grapes.

Question 42.
Write a short note on simple fruits.
Answer:
1. Fruits which develop from one ovary of one flower are called as simple fruits.
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
1. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
2. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)

b. Fleshy fruits: These fruits have thick pericarp. It is further divided into two types based on nature endocarp:
1. Drupe: In these fruits, endocarp is hard and stony, e.g. Mango
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato

Question 43.
Give examples of any two types of aggregate fruits.
Answer:
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)

Question 44.
Complete the given chart.

Answer:
It shows outer layer called as testa and inner layer called as tegmen.

Question 45.
Explain the family of pea plant in detail with suitable floral diagram.
Answer:

• Example: Pea plant (Pisum sativum)
• Habit: Tree, shrubs, herbs.
• Root: Root with root nodules.
• Stem: Erect or climber.
• Leaves: Alternate phyllotaxy, Pinnately compound leaves.
• Inflorescence: Racemose
• Flower: Zygomorphic, bisexual, complete.
• Calyx: Sepals five, gamosepalous, imbricate aestivation.
• Corolla: Petals five, polypetalous, consisting of a larger posterior petal vexillum, two lateral petals wings and two anterior ones forming a keel, vexillary aestivation.
• Androecium: Stamens ten, diadelphous.
• Gynoecium: Ovary superior, monocarpellary, unilocular with many ovules, marginal placentation. Fruit: Legume.
• Seed: Non-endospermic

Question 46.
Answer the following.
1. Explain the family Solanaceae with the help of floral diagram.
2. Give examples of economically important plants from family Liliaceae.
Answer:

• Example: Thom apple (Datura stramonium)
• Habit: Mostly herbs, shmbs and rarely small trees.
• Root: Tap root system
• Stem: Herbaceous, woody, erect, branched, hairy, sometimes it may be underground like in potato.
• Leaves: Alternate phyllotaxy, simple, reticulate venation.
• Inflorescence: Solitary, cymose.
• Flower: Actinomorphic, bisexual, complete.
• Calyx: Sepals five, gamosepalous, persistent, valvate aestivation.
• Corolla: Petals five, gamopetalous, valvate aestivation.
• Androecium: Stamens five, free epipetalous (adhesion).
• Gynoecium: Bicarpellary, syncarpous, superior ovary, bilocular, placenta swollen with many ovules, axile placentation.
• Fruits: Berry or capsule.
• Seeds: Many, endospermic.

2. Economically important plant from family Liliaceae:
Family Liliaceae includes many ornamental plants like tulip, Gloriosa, Medicinal plants like Aloe vera. Asparagus and source of colchicine, e.g. Colchicum autumnale.

Question 64.
Multiple Choice Questions:

Question 1.
Root is descending axis of plant body which is
(A) negatively geotropic
(B) hydrophobic
(C) negatively phototropic
(D) green and with intemodes
Answer:
(C) negatively phototropic

Question 2.
The root system grow out from the
(A) plumule of the embryo
(B) radicle of the embryo
(C) embryo of the seed
(D) all of these
Answer:
(B) radicle of the embryo

Question 3.
Adventitious roots develop from
(A) radicle
(B) any part of the plant body except the radicle
(C) flower
(D) embryo
Answer:
(B) any part of the plant body except the radicle

Question 4.
A fibrous root system is best adapted to perform which of the following functions?
(A) Storage of food
(B) Transport of water and organic food
(C) Absorption of water and minerals from the soil
(D) Anchorage of the plant into the soil
Answer:
(D) Anchorage of the plant into the soil

Question 5.
When the root is swollen in the middle and tapers at both ends, it will be called as root.
(A) tuberous
(B) fusiform
(C) conical
(D) napiform
Answer:
(B) fusiform

Question 6.
Pneumatophores are helpful in
(A) protein synthesis
(B) respiration
(C) transpiration
(D) carbohydrate metabolism
Answer:
(B) respiration

Question 7.
Sweet potato is a modification of
(A) leaf
(B) adventitious root
(C) tap root
(D) stem
Answer:
(B) adventitious root

Question 8.
Stilt roots are roots.
(A) primary
(B) adventitious
(C) secondary
(D) tap
Answer:
(B) adventitious

Question 9.
A spongy tissue called velamen is present in
(A) breathing roots
(B) parasitic roots
(C) tuberous roots
(D) epiphytic roots
Answer:
(D) epiphytic roots

Question 10.
Which of the following is NOT a type of adventitious root modified for storage of food?
(A) Fasciculated tuberous roots
(B) Simple tuberous root
(C) Napiform root
(D) Moniliform roots
Answer:
(C) Napiform root

Question 11.
In which of the following plants, root cap is replaced by root pocket?
(A) Pistia
(B) Pandanus
(C) Screw pine
(D) Hibiscus
Answer:
(A) Pistia

Question 12.
Which of the following is an example of stem tuber?
(A) Helianthus tuberosus
(B) Zingiber officinale
(C) Cynodon
(D) Chrysanthemum
Answer:
(A) Helianthus tuberosus

Question 13.
In stem tuber, the number of nodes and eyes is more towards
(A) rose end
(B) basal end
(C) heel
(D) both (B) and (C)
Answer:
(A) rose end

Question 14.
A rhizome differs from corm in its
(A) thickness
(B) basic organization
(C) direction of growth
(D) nature of leaves
Answer:
(C) direction of growth

Question 15.
The reduced stem of onion produces.
(A) Adventitious roots
(B) Prop roots
(C) Fusiform roots
(D) Fasciculated roots
Answer:
(A) Adventitious roots

Question 16.
Corm is ____________
(A) a horizontal underground stem.
(B) an underground root.
(C) an underground vertical stem.
(D) an aerial stem modification.
Answer:
(C) an underground vertical stem.

Question 17.
The stem modified to perform the function of leaf and with many internodes is called
(A) phylloclade
(B) cladode
(C) offset
(D) phyllode
Answer:
(A) phylloclade

Question 18.
_______ is a non-green runner like branch of stem, which develops from underground base of roots and found in Chrysanthemum.
(A) Corn
(B) Sucker
(C) Offset
(D) Tendril
Answer:
(B) Sucker

Question 19.
Ribbon shaped phylloclades are found in
(A) Ruscus
(B) Duranta
(C) Muehlenbeckia
(D) Bougainvillea
Answer:
(C) Muehlenbeckia

Question 20.
Axillary buds in Dioscorea becomes fleshy and rounded due to storage of food called as
(A) Stiphles
(B) Bulbils
(C) Offset
(D) Cladophylls
Answer:
(B) Bulbils

Question 21.
The leaves without petiole are called
(A) sessile
(B) petiolate
(C) rachis
(D) lamina
Answer:
(A) sessile

Question 22.
The type of leaves observed is mango plant is
(A) Compound leaves
(B) Bipinnately compound leaves
(C) Simple leaves with reticulate venation
(D) Simple leaves with parallel venation
Answer:
(C) Simple leaves with reticulate venation

Question 23.
In _________ , the terminal three leaflet get modified into three stiff leaf hooks.
(A) Lathyrus
(B) Pisum sativum
(C) Smilax
(D) Bignonia unguisi-cati
Answer:
(D) Bignonia unguisi-cati

Question 24.
Leaf apex is modified into tendril in
(A) Gloriosa
(B) Pea
(C) Smilax
(D) Lathyrus
Answer:
(A) Gloriosa

Question 25.
Modification of petiole into leaf-like structure is called ________ .
(A) cladode
(B) phylloclade
(C) phyllode
(D) pistillode
Answer:
(C) phyllode

Question 26.
The mode of arrangement of leaves on the stem and the branch is known as _______ .
(A) vernalization
(B) vernation
(C) venation
(D) phyllotaxy
Answer:
(D) phyllotaxy

Question 27.
Bipinnately compound leaves can be observed in
(A) Citrus
(B) Hibiscus
(C) Caesalpinia
(D) Coriandrum
Answer:
(C) Caesalpinia

Question 28.
The axis of the inflorescence is known as
(A) Thalamus
(B) Peduncle
(C) Pedicel
(D) Petiole
Answer:
(B) Peduncle

Question 29.
In racemose inflorescence
(A) growth of peduncle is infinite
(B) apical bud always terminates into flower
(C) growth of peduncle is finite.
(D) order of opening of flower is centrifugal.
Answer:
(A) growth of peduncle is infinite

Question 30.
In a typical flower thalamus consists of compactly arranged nodes and intemodes.
(A) three, two
(B) four, three
(C) three, four
(D) two, three
Answer:
(B) four, three

Question 31.
When the flower is epigynous, the ovary is said to be
(A) inferior
(B) superior
(C) semi-inferior
(D) semi-superior
Answer:
(A) inferior

Question 32.
When the gynoecium is present at the topmost position of the thalamus, the flower is known as
(A) inferior
(B) epigynous
(C) perigynous
(D) hypogynous
Answer:
(D) hypogynous

Question 33.
When all sepals are united, the condition is called as
(A) polysepalous
(B) gamosepalous
(C) polypetalous
(D) gamopetalous
Answer:
(B) gamosepalous

Question 34.
When sepals fall just after opening of the flower, they are termed as
(A) persistent
(B) caducous
(C) remnant
(D) deciduous
Answer:
(B) caducous

Question 35.
Fusion between members of a similar whorl is known as
(A) succession
(B) adhesion
(C) cohesion
(D) inflorescence
Answer:
(C) cohesion

Question 36.
Complete the analogy. Seed:ovule::Fruit:
(A) pericarp
(B) ovary
(C) embryo
(D) cotyledons
Answer:
(B) ovary

Question 37.
Which one of the following is not a fruit?
(A) Tomato
(B) Cucumber
(C) Pumpkin
(D) Potato
Answer:
(D) Potato

Question 38.
Pineapple is an example of _________ .
(A) simple dry fruit
(B) composite fruit
(C) aggregate fruit
(D) simple-fleshy fruit
Answer:
(B) composite fruit

Question 39.
Outer seed coat is called _______ .
(A) testa
(B) tegmen
(C) raphe
(D) micropyle
Answer:
(A) testa

Question 40.
Vexillum wings and keel corolla are found in family
(A) Solanaceae
(B) Fabaceae
(C) Liliaceae
(D) Malvaceae
Answer:
(B) Fabaceae

Question 65.
Competitive Corner:

Question 1.
Match the placental types (Column-I) with their examples (Column-II).

Column-I	Column-II
1. Basal	(p) Mustard
2. Axile	(q) China rose
3. Parietal	(r) Dianthus
4. Free central	(s) Sunflower

Choose the correct answer from the following option: [NEET (ODISHA) – 2019J
(A) i – r, ii – s, iii – p, iv – q
(B) i – q, ii – r, iii – s, iv – p
(C) i – p, ii – q, iii – r, iv – s
(D) i – s, ii – q, iii – p, iv – r
Answer:
(D) i – s, ii – q, iii – p, iv – r

Question 2.
Placentation in which ovules develop on the inner wall of the ovary or in peripheral part is:
(A) Parietal
(B) Free central
(C) Basal
(D) Axile
Answer:
(A) Parietal

Question 3.
Pneumatophores occur in
(A) Carnivorous plants
(B) Free-floating hydrophytes
(C) Halophytes
(D) Submerged hydrophytes
Answer:
(C) Halophytes

Question 4.
Sweet potato is a modified
(A) Tap root
(B) Adventitious root
(C) Stem
(D) Rhizome
Answer:
(B) Adventitious root

Question 5.
Root hairs develop from the region of
(A) Maturation
(B) Elongation
(C) Root cap
(D) Meristematic activity
Hint: Epidermal cells from the region of maturation form very fine and delicate, thread like structures called root hairs. These root hairs absorb water and minerals from the soil.
Answer:
(A) Maturation

Question 6.
Plants which produce characteristic pneumatophores and show vivipary belong to
(A) Mesophytes
(B) Halophytes
(C) Psammophytes
(D) Elydrophytes
Hint: Plants growing in swampy areas, marshy places and salt lakes are called halophytes. Many halophytes develop respiratory roots or pneumatophores. Pneumatophores are negatively geotropic and are provided with pores called lenticels. Since they grow in oxygen-deficient soil, their seeds germinate inside the fruit, when it is still attached with the parents, exhibiting vivipary.
Answer:
(B) Halophytes

Question 7.
In Bougainvillea thorns are the modifications of
(A) Stipules
(B) Adventitious root
(C) Stem
(D) Leaf
Hint: Thom is a hard, pointed, woody and usually straight structure produced by modification of axillary bud. It provides protection against browsing animals. In Bougainvillea, thorns are modified stems.
Answer:
(C) Stem

Question 8.
Coconut fruit is a
(A) Drupe
(B) Berry
(C) Nut
(D) Capsule
Hint: Dmpe is the simple, fleshy fruit developing from the monocarpellary superior ovary. It is generally one seeded. Coconut fruit is a dmpe with fibrous mesocarp and hard endocarp
Answer:
(A) Drupe

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 6 Bank Reconciliation Statement Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 6 Bank Reconciliation Statement

1. Answer in one sentence.

Question 1.
What is a Bank Reconciliation Statement?
Answer:
Bank Reconciliation Statement is a statement that shows the causes of disagreement between the balance shown by passbook and the balance shown by Cash Book under the column as on a particular date.

Question 2.
What is a Bank passbook?
Answer:
A Bank passbook is a copy of a Customer’s A/c in the Bank’s ledger.

Question 3.
What do you mean by the debit balance of Pass Book?
Answer:
Debit balance of passbook means overdraft as per passbook.

Question 4.
Which account is opened by a trader in Bank for his business operation?
Answer:
A current account is opened by a trader in Bank for his business operation.

Question 5.
On which side of the Cash Book interest on investment is to be shown?
Answer:
Interest on investment is to be shown on the debit/receipt side in the Cash Book.

Question 6.
On which side of the passbook, the direct deposit made by a customer is recorded?
Answer:
A direct deposit made by a customer is recorded on the credit side of the passbook.

Question 7.
What does the credit balance of Cash Book indicate?
Answer:
A credit balance of Cash Book indicates overdraft as per Cash Book.

2. Give one word/term/phrase which can substitute each of the following statements:

Question 1.
Excess of a total of debit side over a total of credit side of Cash Book, Bank column.
Answer:
Bank Balance as per Cash Book

Question 2.
An unfavourable balance is shown by the passbook.
Answer:
Overdraft

Question 3.
A copy of the customer’s account issued by the bank.
Answer:
Pass Book

Question 4.
Booklet or a statement that is used to record the banking transactions.
Answer:
Pass Book

Question 5.
The credit balance of the bank column of the Cash Book.
Answer:
Overdraft

Question 6.
Refusal by the bank to make payment of a cheque.
Answer:
Dishonour of Cheque

Question 7.
Document used to withdraw cash from the bank.
Answer:
Withdrawal Slip

Question 8.
Excess of a total of credit side over a total of debit side in the passbook.
Answer:
Balance as per Pass Book

Question 9.
Document used by the account holder to deposit cash and/or cheques into the bank.
Answer:
Pay-in-Slip

Question 10.
Document used by the account holder to withdraw cash from the bank and for making payment to outside parties through the bank.
Answer:
Cheque

3. Do you agree or disagree with the following statements:

Question 1.
Bank Reconciliation Statement is prepared by Bank.
Answer:
Disagree

Question 2.
Overdraft Facility is available to savings Bank Account.
Answer:
Disagree

Question 3.
The bank account holder can make payments to a third party by use of a pay-in slip.
Answer:
Disagree

Question 4.
Bank does not charge any interest on an overdraft balance.
Answer:
Disagree

Question 5.
Credit balance in Pass Book represents overdraft balance.
Answer:
Disagree

Question 6.
In the cash book of a trader Bank record the entries.
Answer:
Disagree

Question 7.
In the passbook, entries are made by the account holder.
Answer:
Disagree

Question 8.
Through mobile banking account holder can deposit physical cash into his account.
Answer:
Disagree

Question 9.
The right-hand side of the pay-in-slip is known as a counterfoil.
Answer:
Disagree

Question 10.
A cash withdrawal slip is used to deposit a cheque or cash into a bank account.
Answer:
Disagree

4. Select the most appropriate alternative from those given and rewrite the following statements:

Question 1.
Pass Book is _____________ of account holders transactions with bank.
(a) an extract
(b) balance sheet
(c) balance
(d) mode
Answer:
(a) an extract

Question 2.
When cheque is _____________ into bank Cash Book is debited.
(a) written
(b) issued
(c) deposited
(d) dishonoured
Answer:
(c) deposited

Question 3.
Overdraft facility is allowed to _____________ account.
(a) saving
(b) recurring
(c) current
(d) fixed deposit.
Answer:
(c) current

Question 4.
Overdraft means _____________ balance of Pass Book.
(a) opening
(b) debit
(c) credit
(d) closing
Answer:
(b) debit

Question 5.
Interest on bank overdraft is recorded on _____________ side of Pass Book.
(a) debit
(b) credit
(c) any
(d) both
Answer:
(a) debit

Question 6.
Credit balance in the Pass Book represents _____________
(a) overdraft
(b) bank balance
(c) loan borrowed
(d) negative
Answer:
(b) bank balance

Question 7.
Direct deposit by a customer will be recorded on _____________ side of Pass Book.
(a) debit
(b) credit
(c) left hand
(d) any
Answer:
(b) credit

Question 8.
Normally the Bank Reconciliation Statement is prepared at the end of a _____________
(a) day
(b) week
(c) year
(d) month
Answer:
(d) month

Question 9.
Bank charges charged by bank are recorded on _____________ side of Pass Book.
(a) debit
(b) credit
(c) any
(d) both
Answer:
(a) debit

5. Complete the following statements:

Question 1.
Bank issue _____________ to current account holders as record or summary for bank transactions.
Answer:
Bank statement

Question 2.
To deposit cash or cheque _____________ is used.
Answer:
Pay in Slip

Question 3.
Directly deposited by a customer in bank account appears an _____________ side of pass book.
Answer:
Credit

Question 4.
____________ is an unfavorable balance as per Pass Book.
Answer:
overdraft

Question 5.
Check deposited into Bank but not cleared is called _____________ cheque.
Answer:
Dishonoured

Question 6.
Expenses paid by the Bank as per standing instructions will be recorded at _____________ side of the passbook.
Answer:
Debit

Question 7.
Dividend collected by the bank will be recorded at _____________ side of cash book.
Answer:
Debit

Question 8.
_____________ type of bank account is open by trader.
Answer:
Current

6. State whether the following statements are True or False with reasons:

Question 1.
Bank Reconciliation Statement is prepared by the Bank.
Answer:
This statement is False.
A Bank Reconciliation statement shows the causes of disagreement between the balances shown by the bank passbook and the bank balance shown by the cash book, for a particular period of time generally a month. It is prepared by the trader as he has both the books to compare and find the differences.

Question 2.
Bank Reconciliation Statement is prepared at the end of every month.
Answer:
This statement is True.
Monthly preparation of Bank Reconciliation Statement assists in the regular monitoring of cash flows of a business and identification of accounting errors.

Question 3.
Overdraft facility is allowed to Proprietor’s Personal A/c.
Answer:
This statement is False.
Overdraft facility is allowed only to business current A/c and not to Proprietor’s Personal A/c.

Question 4.
The debit balance of Pass Book represents overdraft.
Answer:
This statement is True.
Debit Column of a passbook means withdrawals from the bank, When withdrawals are more than deposits it means the excess amount is withdrawn from the bank. It is a temporary loan payable by the trader to the bank. So the debit balance of Pass Book represents overdraft.

Question 6.
Bank charges debited by Bank increase bank balance as per Pass Book.
Answer:
This statement is False.
Bank charges are expenses for the business. Expenses decrease the bank balance as per Pass Book. Bank charges debited by the bank decrease the bank balance as per Pass Book.

Question 6.
Interest credited in PassBook is an income to the customer.
Answer:
This statement is True.
All incomes are shown on the credit side of the passbook. It is a deposit. So Interest Credited in Passbook is an income to the customer.

Question 7.
Bank Reconciliation Statement is prepared to detect the errors that take place in accounting.
Answer:
This statement is True.
A businessman maintains a cash book with a bank column to record his bank transactions whereas the bank also maintains a customer’s ledger account and issues him a Bank statement. There could be differences as per the bank balance in the cash book and bank balance in the passbook. To detect the errors that take place in accounting Bank Reconciliation Statement is prepared.

Question 8.
Overdraft as per Cash Book means debit balance as per Cash Book.
Answer:
This statement is False.
Overdraft as per Cash Book means credit balance as per cash book. Cashbook debit means deposits. When cash book debit balance is greater it means Bank Balance as per Cash Book.

Question 9.
Cheque deposited into Bank increases the Bank balance as per Cash Book.
Answer:
This statement is True.
The Debit side of the cash book means deposits. So cheque deposited into the Bank increases the Bank Balance as per Cash Book.

Question 10.
Payments made by the bank as per standing instructions are recorded on the Debit balance of the Pass Book.
Answer:
This statement is True.
The Debit side of the passbook represents payments. So any payments made by the bank, Bank debits the customer’s account and records on the Debit side of the passbook.

Question 11.
Bank column in Cash Book represents Proprietor’s Savings A/c.
Answer:
This statement is False.
Cashbook records only business transactions and not the personal A/c of the trader. The Bank column in Cash Book represents Business Current A/c and not the proprietor’s savings A/c.

7. Correct and rewrite the following statements.

Question 1.
Overdraft as per cash book means debit balance as per cash book.
Answer:
Normal bank balance as per cash book means debit balance as per cashback.

Question 2.
Bank column in cash book represents proprietors saving A/c.
Answer:
The Bank column in cashback represents the business’s current account.

Question 3.
Fixed deposit A/c is opened by traders for day-to-day business bank transactions.
Answer:
The current account is opened by traders for day-to-day business bank transactions.

Question 4.
The bank account is a real account.
Answer:
A bank account is a personal account

Question 5.
Interest charged by the bank on overdraft A/c is income for the business.
Answer:
Interest charged by the bank on overdraft A/c is an expense for the business.

8. Complete the following table.

Question 1.


Answer: