class-11 SUBJECT ENGLISHC CHAPTER-7 THE ADVENTURE

  EVENTS CONVENT HIGH SCHOOL

03/02/2022      CLASS- 11   SESSION 2021-22
SUBJECT : ENGLISH HORNBILL

CHAPTER-7

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CLASS-11 SUBJECT BIOLOGY CHAPTER-15PLANT GROWTH AND DEVELOPMENT,

 EVENTS CONVENT HIGH SCHOOL

26/10/2021      CLASS-11     SESSION 2021-22
SUBJECT :BIOLOGY

CHAPTER-15 PLANT GROWTH AND DEVELOPMENT

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1. Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
Solution: Growth is defined as a vital process which brings about an irreversible and permanent change in the shape, size, form, weight and volume of a cell, organ or whole organism, accompanied with increase in dry matter.
Differentiation is a localised qualitative change in size, biochemistry, structure and function of cells, tissues or organs, e.g., fibre, vessel, tracheid, sieve tube, mesophyll, leaf etc. Thus it is a change in form and physiological activity. It results in specialisation for particular functions.
Development may be defined as a process which includes growth, differentiation and maturation in a regular sequence in the life history of a cell, organ or organism viz., seed germination, growth, differentiation, flowering, seed formation and senescence. Dedifferentiation is the process by which the differentiated cells which have lost the ability to divide under certain circumstances, become meristematic and regain the divisibility. Redifferentiation is defined as maturation or differentiation of dedifferentiated cells to form cells which are unable to divide e.g., secondary xylem elements, cork cells etc., are formed by redifferentiation of secondary cambial cells.
Determinate growth is the ability of a cell, tissue or the organism to grow for a limited period of time. Meristem is a tissue consisting of unspecialised immature cells, possessing the power of continuous cell division and adding new cells to the body. Growth rate is defined as the increased growth per unit time.

2. Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?
Solution: A flowering plant consists of a number of organs viz., roots, stem, leaves, flowers, fruits etc. growing differently under different stages of life cycle. These plant organs require different parameters to demonstrate their growth. In plant organs like fruits, bulbs, corms etc. fresh weight is used for measuring their growth. In case of fruits, increase in volume, diameter etc., are also used as other parameters for the measurement of their growth. For flat organs like leaves, increase in surface area is used as the parameter. Stem and roots primarily grow in length and then in girth, thus increase in length and diameter are used for measuring their growth. Consequently, the flowering plants exhibit several parameters to demonstrate growth.

3. Describe briefly
(a) Arithmetic growth
(b) Geometric growth
(c) Sigmoid growth curve
(d) Absolute and relative growth rates
Solution: (a) Arithmetic growth: If the length of a plant organ is plotted against time it shows a linear curve, the growth is called arithmetic growth. In this growth, the rate of growth is constant and increase in growth occurs in arithmetic progression e.g., length of a plant is measured as 2,4, 6, 8,10,12 cms at a definite interval of 24 hrs. It is found in root or shoot elongating at constant rate. Arithmetic growth is expressed as Lt = L0 + rt Here, Lt = length after time t. L0 = length at the beginning, r = growth rate.

(b) Geometric growth: Geometric growth is the growth where both the progeny cells following mitosis retain the ability to divide and continue to do so. It occurs in many higher plants and in unicellular organisms when grown in nutrient rich medium. Number of cells is initially small so that initial growth is slow which is called lag phase. Later on, there is rapid growth at exponential rate. It is called log or exponential phase.

(c) Sigmoid growth curve: Geometric growth cannot be sustained for long. Some cells die. Limited nutrient availability causes slowing down of growth. It leads to stationary phase. There may be actually a decline. Plotting the growth against time will give a typical sigmoid or S-curve.

S-curve of growth is typical of most living organisms in their natural environment. It also occurs in cells, tissues and organs of plants.
(d) Absolute growth rate is the measurement of total growth per unit time. Relative growth rate is growth per unit time per unit initial growth.
Growth in given time period/ Measurement at start of time period
Suppose two leaves have grown by 5 cm2 in one day. Initial size of leaf A was 5 cm2 while that of leaf B was 50 cm2. Though their absolute growth is the same (5 cm2/day), relative rate of growth is faster in leaf A(5/5) because of initial small size than in leaf B(5/50).

4. List five main groups of natural plant growth regulators. Write a note on discovery, physiological functions and agricultural/ horticultural applications of any one of them.
Solution: There are five main groups of natural plant growth regulators which are very much recognised as natural hormones in plants. These are:

1.  Auxins
2. Gibberellins
3. Cytokinins
4. Abscisic acid
5. Ethylene

Discovery of auxin: In 1880, Charles Darwin and Francis Darwin worked with the coleoptile of canary grass (Phalaris sp.) and found the existence of a substance in coleoptile tip, which was able to recognise the light stimulus and leads to the bending of tip towards light. Boysen and Jensen (1910-1913) worked on Avena seedling and explained that the substances secreted in the tip are soluble in water (gelatin).
Paal (1919) reported that the substances secreted in the tip are translocated downwards and caused cell elongation in half portion which was on the dark side and hence bending was observed in opposite direction.
F.W. Went (1928) further refined this experiment and supported the observations of Paal. He was the first person to isolate and name these substances of tip as auxins (Greek Auxein – means ‘to grow’).
In 1931, Kogl and Haagen-Smith isolated
crystalline compounds from human urine.
These were named as auxin-a, auxin-b and
heteroauxin.

Physiological functions of auxins:

1. Auxins induce cambial cell divisions, shoot cell elongation and early differentiation of xylem and phloem in tissue culture experiments.
2. In general, auxins initiate rooting but inhibit the growth of roots. IBA is the most potent root initiator.
3. Auxins inhibit the growth of axillary buds (apical dominance) but enhance the size of carpel and hence earlier fruit formation.
4. Application of auxins retards the process of senescence (last degradative phase), the abscission of leaves, fruits, branches, etc.
5. Auxins induce feminisation, i.e., on male plant, female flowers are produced.

Agricultural/horticultural application of auxins:

1. Application of auxins like IAA, IBA, NAA induce rooting in stem cuttings of many plants. This method is widely used to multiply several economically useful plants.
2. Normally, auxins inhibit flowering however in litchi and pineapple, application of auxin promotes flowering thus used in orchards.
3. Auxin induces parthenocarpy in some plants including tomato, pepper, cucumber and Citrus, thus, produces seedless fruits of more economic value.
4. Auxins like 2, 4-D and 2, 4, 5-T are commercially used as weedicides, due to their low cost and greater chemical stability. They are selective herbicides (killing broad-leaved plants, but not grasses).
5. For checking premature fruit drop, auxins are applied which prevent the formation of abscission zone in the petiole or just below the fruit. Auxin regulates maturing fruit on the trees of apples, oranges and grape fruit. High doses of auxins can
   cause fruit drop. Thus, heavy applications of synthetic auxins are used commercially to promote a coordinated abscission of various fruits to facilitate harvesting.
6. Auxin, produced in the apical bud, suppresses the development of lateral buds, i.e., apical dominance. Thus practically used in prolonging the dormancy period of potato tubers.
7. Naphthalene acetamide is used to prevent the lodging (excessive elongation and development of weak plants, specially in gramineae) or falling of crops.
8. Auxin (2,4-D) promotes callus formation in tissue culture. Complete plantlets are regenerated from callus tissue, using auxins and cytokinin which are then transplanted into the soil. Now-a-days, this is a widely practised method of propagation in the field of agriculture and horticulture.

5. What do you understand by photoperiodism and vernalisation? Describe their significance.
Solution: The physiological mechanism for flower-ing is controlled by two factors: photoperiod or light period, i.e., photoperiodism and low temperature, i.e., vernalisation. Photoperiodism is defined as the flowering response of a plant to relative lengths of light/ dark period. Significance of photoperiodism is as follows:

1. Photoperiodism determines the season in which a particular plant shall flower. For example, short day plants develop flowers in autumn-spring period (e.g., Dahlia, Xanthium) while long day plants produce flowers in summer (e.g., Amaranthus).
2. Knowledge of photoperiodic effect is useful in keeping some plants in vegetative growth (many vegetables) to obtain higher yield of tubers, rhizomes etc. or keep the plant in reproductive stage to yield more flowers and fruits.
3. A plant can be made to flower throughout the year by providing favourable photoperiod.
4. Helps the plant breeders in effective cross-breeding in plants.
5.  Enable a plant to flower in different seasons.
   Vernalisation is promotion or induction of flowering by exposing a plant to low temperature for some time. Significance of vernalisation is as follows :
   (i) Crops can be grown earlier.
   (ii)Plants can be grown in such regions where normally they do not grow.
   (iii)Yield of the plant is increased.
   (iv)Resistance to cold and frost is increased.
   (v) Resistance to fungal diseases is increased.

6. Why is abscisic acid also known as stress hormone?
Solution: A fairly high concentration of abscisic acid (ABA) is found in leaves of plants growing under stress conditions, such as drought, flooding, injury, mineral deficiency etc. It is accompanied by loss of turgor and closure of stomata. When such plants are transferred to normal conditions, they regain normal turgor and ABA concentration decreases. Since the synthesis of ABA is accelerated under stress condition and the same is destroyed or inactivated when stress is relieved, it is also known as stress hormone.

7. ‘Both growth and differentiation in higher plants are open’. Comment.
Solution: Plant growth is generally indeterminate. Higher plants possess specific areas called meristems which take part in the formation of new cells. The body of plants is built on a modular fashion where structure is never complete because the tips (with apical meristem) “are open ended – always growing and forming new organs to replace the older or senescent ones. Growth is invariably associated with differentiation. The exact trigger for differentiation is also not known. Not only the growth of plants are open- ended, their differentiation is also open. The same apical meristem cells give rise to different types of cells at maturity, e.g., xylem, phloem, parenchyma, sclerenchyma fibres, collenchyma, etc. Thus, both the processes are indeterminate, unlimited and develop into
different structures at maturity i.e., both are open.

8. ‘Both a short day plant and a long day plant can produce flower simultaneously in a given place’. Explain.
Solution: A short day plant (SDP) flowers only when it receives a long dark period and short photoperiod, e.g., Xanthium, Dahlia etc. On the other hand, a long day plant (LDP) will flower only when it receives a long photoperiod and short dark period, e.g., wheat, oat etc. Thus critical photoperiod is that continuous duration of light which must not be exceeded in SDP and should always be exceeded in LDP in order to bring them to flower. Xanthium requires light for less than 15.6 hrs and Henbane requires light for more than 11 hrs. Xanthium (a SDP) and Henbane (DP) will flower simultaneously in light period between 11 to 15.6 hrs.

9. Which one of the plant growth regulators would you use if you are asked to
(a) induce rooting in a twig
(b) quickly ripen a fruit
(c) delay leaf senescence
(d) induce growth in axillary buds
(e) ‘bolt’ a rosette plant
(f) induce immediate stomatal closure in leaves.
Solution: (a) Auxins like IBA, NAA.
(b) Ethylene
(c) Cytokinins
(d) Cytokinins
(e) Gibberellins
(f) Abscisic acid (ABA)

10. Would a defoliated plant respond to photo- periodic cycle? Why?
Solution: No, a defoliated plant would not respond to photoperiodic cycle because photoperiodic stimulus is picked up by the leaves only. Even one leaf or a part of it is sufficient for this purpose. For perception of photoperiodic cycle, there must be the presence of leaves under inductive photoperiod, so that, the hormone responsible for flowering can be produced.

11. What would be expected to happen if:
(a) GA3 is applied to rice seedlings
(b) dividing cells stop differentiating
(c) a rotten fruit gets mixed with unripe fruits
(d) you forget to add cytokinin to the culture medium.
Solution:
(a) The coleoptile will elongate rapidly, as GA3 helps in cell growth.
(b) The development of callus (mass of undifferentiated cells) will take place.
(c) The unripe fruits will ripe quickly because of the increased rate of respiration due to emission of ethylene from rotten fruit.
(d) Cell division will retard and shoot will not initiate from the callus.

CLASS-11 SUBJECT BIOLOGY CHAPTER-9 EXCRETORY PRODUCTS AND THEIR ELIMINATION

 EVENTS CONVENT HIGH SCHOOL

09/10/2021      CLASS-11      SESSION 2021-22
SUBJECT :BIOLOGY

chapter-9
EXCRETORY PRODUCTS AND THEIR ELIMINATION

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1.Define Glomerular Filtration Rate (GFR).
Solution. The amount of filtrate formed by the kidneys per minute is called glomerular filtration rate (GFR). It is approximately 125 mL/min. in a healthy person.

2.Explain the autoregulatory mechanism of GFR.
Solution. The kidneys have built-in mechanisms for the regulation of glomerular filtration rate. One such efficient mechanism is carried out by juxta glomerular apparatus (JGA). JGA is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. A fall in GFR can activate the JG cells to release renin which can stimulate the glomerular blood flow and thereby the GFR back to normal.

3.Indicate whether the following statements are true or false.
(a) Micturition is carried out by a reflex.
(b) ADH helps in water elimination, making the urine hypotonic.
(c) Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
(d) Henle’s loop plays an important role in concentrating the urine.
(e) Glucose is actively reabsorbed in the proximal convoluted tubule.

Solution.(a) True (b) False (c) True (d) True (e) True

4.Give a brief account of the counter current mechanism.
Solution. The kidneys have a special mechanism for concentrating the urine, it is called counter current mechanism. The mechanism is said to be a counter current mechanism because the out flow (in the ascending limb) of Henle’s loop runs parallel to and in the opposite direction of the inflow (in the descending limb) and vasa recta. As the mechanism begins to function, the ascending limb of loop of Henle actively transports chloride and sodium ions out into the vasa recta from where it is secreted into the interstitial fluid. As a result the interstitial fluid around the loop of Henle contains large quantities of NaCl. The filtrate passes from the ascending limb of loop of Henle and enters a collecting duct. The collecting duct passes adjacent to the loop of Henle where the interstitial fluid contains large amounts of NaCl. The high osmotic pressure created by NaCl causes water to diffuse out of the collecting duct in the interstitial fluid and eventually to the blood of vasa recta. The filtrate becomes greatly concentrated and is now called urine. A similar counter current mechanism, operates between the interstitial fluid and blood passing through the vasa recta. As the blood capillary runs along the ascending limb of loop of Henle, NaCl diffuses out of the blood. The direction is reversed as the blood capillary passes along the descending limb of Henle. The blood flows in the vasa recta around the loop of Henle from ascending to the descending side while the fluid passing through the loop of Henle goes in the opposite direction. The arrangement helps to maintain the concentration gradient of NaCl.
The ‘overall function of counter current mechanism is to concentrate sodium chloride in the interstitial fluid and thereby cause water to diffuse out of the collecting ducts and concentrate the urine.

5.Describe the role of liver, lungs and skin in excretion.
Solution. Other than the kidneys, lungs, liver and skin also help in the elimination of excretory wastes. Lungs remove large amounts of C02 (18 litres/day) and also significant quantities of water every day. Liver secretes bile which contains substances like bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drugs. Most of these substances ultimately pass out along with digestive wastes. The sweat and sebaceous glands in the skin can eliminate certain substances through their secretions. Sweat produced by the sweat glands is a watery fluid containing NaCl, small amounts of urea, lactic acid etc. Sebaceous glands eliminate certain substances like sterols, hydrocarbons and waxes through sebum.

6.Explain micturition.
Solution. The process of passing out urine from the urinary bladder is called micturition. Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored. This causes stretching of the wall of bladder that leads to the stimulation of stretch receptors on the walls of the bladder. This sends signal to the CNS. The CNS passes on motor messages to initiate the contraction of smooth muscles of the bladder and simultaneous relaxation of the urethral sphincter causing the release of urine.

7.Match the items of column I with those of column II.
Column I                                     Column II
(a) Ammonotelism                   (i)Birds
(b) Bowman’s capsule             (ii)Water reabsorption
(c) Micturition                          (iii)Bony fish
(d) Uricotelism                         (iv)Urinary bladder
(e) ADH                                       (v)Renal tubule
Solution. (a) – (iii), (b) – (v), (c) – (iv), (d) – (i), (e) – (ii)

8.What is meant by the term osmoregulation?
Solution. The regulation of water and solute contents of the body fluids by the kidney is called osmoregualtion.

9.Terrestrialanimalsaregenerallyeitherureotelic or uricotelic, not ammonotelic, why?
Solution. Ammonotelic animals are aquatic animals that excrete ammonia which is highly soluble in water, thus large amount of water is also excreted. Terrestrial animals cannot afford to lose such large quantities of water from their bodies as they live in environment having water scarcity. They, therefore, excrete either urea (ureotelic) or uric acid (uricotelic) as these are less soluble in water.

10. What is the significance of juxta glomerular apparatus (JGA) in kidney function?
Solution. Juxta glomerular apparatus (JGA) is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. The JGA plays a complex regulatory role. A fall in glomerular blood flow/ glomerular blood pressure/GFR can activate the JG cells to release renin which converts angiotensinogen in blood to angiotensin I and further to angiotensin II. Angiotensin II, being a powerful vasoconstrictor, increases the glomerular blood pressure and thereby GFR. Angiotensin II also activates the adrenal cortex to release aldosterone. Aldosterone causes reabsorption of Na+ and water from the distal parts of the tubule. This also leads to an increase in blood pressure and GFR.

11 .Name the following.
(a) A chordate animal having flame cells as excretory structures.
(b) Cortical portions projecting between the medullary pyramids in the human kidney.
(c) A loop of capillary running parallel to the Henle’s loop.
Solution. (a) Cephalochordate – Amphioxus
(b) Columns of Bertini
(c) Vasa recta

12.Fill in the gaps.
(a) Ascending limb of Henle’s loop is________to water whereas the descending limb is________to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone________
(c) Dialysis fluid contains all the constituents as in plasma except________
(d) A healthy adult human excretes (on an average)________gm of urea/day.
Solution.
(a) Ascending limb of Henle’s loop is impermeable to water whereas the descending limb is permeable to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone ADH.
(c) Dialysis fluid contains all the constituents as in plasma except nitrogenous wastes.
(d) A healthy adult human excretes (on an average) 25 – 30 gm of urea/day.

CLASS-11 SUBJECT ENGLISH SNAPSHOT CHAPTER-2 THE ADDRESS

 EVENTS CONVENT HIGH SCHOOL

01/10/2021      CLASS-11     SESSION 2021-22
SUBJECT :ENGLISH SNAPSHOT

CHAPTER-2
THE ADDRESS

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A. Reading With Insight

Question 1:
‘Have you come back?’ said the woman, ‘I thought that no one had come back.’ Does this statement give some clue about the story? If yes, what is it?
Answer:
Yes, this statement gives some clue about the story. During the early part of the war Mrs Dorling had shifted the important belongings of her acquaintance Mrs S. from her house to 46, Marconi Street. These included table silver wares, antique plates and other nice things such as the iron anukkah candle-holder, woollen table cloth and green knitted cardigan with wooden buttons. Since Mrs S. had died during the war, Mrs Dorling did not expect anyone to come back and claim her costly belongings as she thought no one else knew her address.
The statement indicates the greedy and possessive nature of Mrs Dorling. She did not open the door to the daughter of her former acquaintance nor did she show any signs of recognition. She did not let the girl in. She refused to see her then saying it was not convenient for her to do. The narrator had gone to this address with a specific purpose—to see her mother’s belongings.
Even when she told Mrs Dorling that only she had come back, the woman with a broad back did not soften a bit. Thus the clash of interests is hinted at by the aforesaid statement.

Question 2:
The story is divided into pre-war and post-war times. What hardships do you think the girl underwent during these times?
Answer:
During the pre-war times, the narrator lived in some other city far away from home and she visited her mother only for a few days. During the first half of the war the narrator’s mother was always afraid that they might have to leave the place and lose all valuable belongings. The narrator lived in the city in a small rented room. Its windows were covered with blackout paper. She could not see the beauty of nature outside her room. The threat of death loomed large.
After the liberation, everything became normal again. Bread was getting to be a lighter colour. She could sleep in her bed without any fear of death. She could glance out of the window of her room each day. One day, she was eager to see all the possessions of her mother, which she knew were stored at number 46, Marconi Street. She went to that address. She felt disappointed when Mrs Dorling neither recognised her nor let her in. She asked her to come again someday. It was evident she wanted to put her off. She was eager to see, touch and remember her mother’s possessions. So, she had to take the trouble of visiting the place again.

Question 3:
Why did the narrator of the story want to forget the address?
Answer:
The narrator remembered the address her mother had told her only once. It was number 46, Marconi Street. Her mother’s acquaintance Mrs Dorling lived there. She had stored the valuable belongings of the narrator’s mother there. After her mother’s death, the narrator had an urge to visit the place. She wanted to see those things, touch them and remember. She went to the given address twice. She was successful in her second attempt to enter the living room. .
She found herself in the midst of things she wanted to see again. She felt oppressed in the strange atmosphere. Everything was arranged in a tasteless way. The ugly furniture and the muggy smell that hung there seemed quite unpleasant. These objects evoked the memory of the familiar life of former time. But they had lost their value since they had been separated from her mother and stored in strange surroundings. She no longer wanted to see, touch or remember these belongings. She resolved to forget the address. She wanted to leave the past behind and decided to move on.

Question 4:
‘The Address’ is a story of human predicament that follows war. Comment.
Answer:
The war creates many difficult and unpleasant situations for human beings. Sometime it becomes difficult to know what to do. The human predicament that follows war is amply illustrated through the experience of the narrator. The war had caused many physical difficulties as well as emotional sufferings to her. She had lost her dear mother. She went to 46, Marconi Street to see her mother’s valuable possessions. How greedy and callous human beings can become is exemplified by the behaviour of Mrs Dorling. She had stored all the valuable belongings of the narrator’s mother, but she refused to recognise the narrator. She did not even let her in. The presence of her mother’s possessions in strange atmosphere pained her. Now these valuables had lost all their importance for her as they had been separated from her mother. She could get no solace or comfort from them.

A. Short Answer Type Questions

Question 1:
How did Mrs Dorling react when the narrator said, m Mrs S’s daughter”?
Answer:
Mrs Dorling held her hand on the door as if she wanted to prevent it opening any
further. Her face showed no sign of recognition. She kept staring at the narrator without uttering a word.

Question 2:
What two reasons did the narrator give to explain that she was mistaken?
Answer:
She thought that perhaps the woman was not Mrs Dorling. She had seen her only once, for a brief interval and that too years ago. Secondly, it was probable that she had rung the wrong bell.

Question 3:
How did the narrator conclude that she was right?
Answer: The woman was wearing the green knitted cardigan of the narrator’s mother. The wooden buttons were rather pale from washing. She saw that the narrator was looking at the cardigan. She half hid herself again behind the door. Her reaction convinced the narrator that she was right.

Question 4:
What was the outcome of the interview between Mrs Dorling and the narrator?
Answer:
The interview was a flop as far as the narrator was concerned. Mrs Dorling refused
to see her and talk to her in spite of the narrator’s repeated requests.

Question 5:
Who had given the narrator the address, when and under what circumstances?
Answer:
The narrator’s mother had given her the address, years ago during the first half of
the war. The narrator came home for a few days and missed various things in the rooms. Then her mother told her about Mrs Dorling and gave her the address.

Question 6:
What did the narrator learn about Mrs Dorling from her mother?
Answer:
Mrs Dorling was an old acquaintance of the narrator’s mother. The latter had not seen her for several years. Then she suddenly turned up and renewed their contact.
Every time she left that place she took something with her—table silver, antique plates, etc. –

Question 7:
What reason did Mrs Dorling give for taking away the precious belongings of the narrator’s mother?
Answer:
Mrs Dorling suggested to the narrator’s mother that she should store her belongings at a safer place. She wanted to save all her nice things. She explained that they would lose everything if they had to leave the place.

Question 8:
What impression do you form of the narrator’s mother on the basis of her conversation with (i) Mrs Dorling and (ii) the narrator?
Answer:
The narrator’s mother was a kind-hearted, generous and liberal lady. She was fond of collecting valuable things. She is more worried about the physical risk to Mrs Dorling than losing them to her. She thought it an insult to tell her friends to keep those things for ever.

Question 9:
Did the narrator fee Up evinced about the views of her mother regarding Mrs Dorling? How do you know?
Answer:
The narrator did not feel convinced about her mother’s concern for Mrs Dorling.
The latter was keen on removing the precious possessions of the narrator’s mother to her own house. It seems that the narrator did not like Mrs Darling’s excessive interest in her mother’s belongings. It is evident from the questions she puts to her mother.

Question 10:
What does the narrator remember about Mrs Dorling as she saw her for the first time?
Answer:
Mrs Dorling was a woman with a broad back. She wore a brown coat and a shapeless hat. She picked up a heavy suitcase lying under the coat rack and left their house.
She lived at number 46, Marconi Street.

Question 11:
Why did the narrator wait a long time before going to the address number 46, Marconi Street?
Answer:
Initially, after the liberation, she was not at all interested in her mother’s belongings lying stored there. She was also afraid of being confronted with things that had belonged to her mother, who was now no more.

Question 12:
When did the narrator become curious about her mother’s possessions?
Answer:
The narrator became curious about her mother’s possessions as normalcy returned in the post-liberation period. She knew that those things must still be at the address her mother told her. She wanted to see them, touch and remember.

Question 13:
“I was in a room I knew and did not know,” says the narrator in the story ‘The Address’. What prompted her to make this observation?
Answer:
The narrator found herself in the midst of things she was familiar with and which she did want to see again. However, she found them in a strange atmosphere where everything was arranged in a tasteless way. They ugly furniture and the muggy smell created the feeling that she didn’t know the room.

Question 14:
“I just looked at the still life over the tea table,” says the narrator in the story ‘The Address’. What does she mean by ‘the still life? What prompted her to make this remark?
Answer:
By ‘the still life’, the narrator means the things over the tea table such as the table-cloth, tea pot, cups and spoons. The reference to antique box and silver spoons prompted her to make this remark.

Question 15:
How was the narrator able to recognise her own familiar woollen table- cloth?
Answer:
The narrator first stared at the woollen table-cloth. Then she followed the lines of the pattern. She remembered that somewhere there was a bum mark which had not been repaired. At last she found the bum mark on the table-cloth. This helped her to recognise her own familiar article.

Question 16:
“You only notice when something is missing.” What does the speaker exactly mean? What examples does she give?
Answer:
The speaker says that one gets used to touching one’s lovely things in the house. One hardly looks at them any more. It is only when something is missing that it is noticed either because it is to be repaired or it has been lent to someone.

Question 17:
How did narrator come to know that the cutlery they ate off every day was silver?
Answer:
Once the narrator’s mother asked her if she would help her polish the silver. The
narrator asked her which silver she meant. Her mother was surprised at her ignorance and replied that it was the spoons, forks and knives, i.e. the cutlery they ate off everyday.

Question 18:
Why did the narrator suddenly decide to leave?
Answer:
The narrator had visited 46, Marconi Street for a specific purpose—to see her mother’s belongings and touch them. However, these objects seemed to have lost their value in strange surroundings and on being severed from the life of former times.

Question 19:
How did the narrator reconcile herself to the loss of her mother’s precious belongings?
Answer:
The narrator felt that her mother had only lent them for safe custody and Mrs Dorling was not to keep everything. On seeing these objects, memories of her former life were aroused. She found no room for these precious belongings in her present life. So she reconciled to her fate.

Question 20:
“Of all the things I had to forget, that would be the easiest”. What does the speaker mean by ‘that’? What is its significance in the story?
Answer:
That’ here stands for the address. The words: number 46, Marconi Street, i.e. the address recur throughout the story. The address is important for the narrator at the beginning of the story. However, at the end of the story she resolves to forget it as she wants to break off with the past and move on with the present into the future.