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Here we are providing Online Education for Albert Einstein at School Important Extra Questions and Answers Class 11 English Snapshots, Extra Questions for Class 11 English was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-11-english/

Online Education for Albert Einstein at School Important Extra Questions and Answers Class 11 English Snapshots

Albert Einstein at School Extra Questions and Answers Short Answer Type

Albert Einstein At School Extra Questions Class 11 Question 1.
“I think it’s not facts that matter, but ideas.” To whom did Einstein say this and why?
Answer:
Einstein said it to Mr Braun, the history teacher, in his Munich school. He hated learning dates and facts by heart. He argued that ideas were more important than rote learning. He was not interested in knowing when the battles were fought but why they were fought.

Albert Einstein At School Question Answer Class 11 Question 2.
Do you think Albert is being impolite while answering the history teacher’s questions? Give your reasons.
Answer:
Though Albert addresses his history teacher politely, he is being impudent. If he did have a problem, he should have spoken to the teacher separately. Questioning the teacher’s views on education in a class was impolite and rude.
OR
Albert addresses his history teacher politely. His answers are straightforward and blunt but his opinions are strong. A firm conviction cannot be termed as rudeness.

Albert Einstein At School Questions And Answers Class 11 Question 3.
What characteristic of Einstein’s nature is highlighted by the exchanges between him and the teacher?
Answer:
The exchanges between Einstein and the teacher show him to be a person with firm convictions; he is frank and straightforward. Even at the cost of punishment he is not willing to compromise on his views. He comes across as one who will chart his own course rather than follow the beaten track.

Albert Einstein At School Important Questions Class 11 Question 4.
Why did Albert see no point in learning dates and facts?
Answer:
Einstein told Mr Braun, the history teacher, “I think it’s not facts that matter, but ideas.” He hated learning dates and facts by heart. He argued that ideas were more important than learning by rote. He was not interested in knowing when the battles were fought but why they were fought.

Albert Einstein Class 11 Extra Questions  Question 5.
What was Einstein’s reaction to the history teacher’s sarcasm?
Answer:
Mr Braun, his history teacher, was unhappy with Albert unwilling to leam facts or dates. When he admitted that he did not see any point in learning dates, the teacher taunted him by calling his views the ‘Einstein theory of education’. Albert argued that it was ideas that were important but nevertheless, felt miserable.

Class 11 Snapshot Chapter 4 Extra Questions Question 6.
Why did Albert feel miserable when he left school that day?
Answer:
When he left school that day, Albert felt miserable because his day at school, like most other days, had been bad and he had gotten into an argument with his teacher for which he had been punished. Secondly, he had to go back to the same school the next day. Moreover, the idea of going back to his lodgings with the atmosphere of domestic violence did not cheer him up.

Albert Einstein At School Questions Class 11 Question 7.
Why and what did his history teacher report to the head teacher?
Answer:
Mr Braun was shocked and furious with Albert’s candid reply. He believed that Albert had no desire to learn and was wasting his father’s money. He punished Albert by detaining him after school. He also reported to the head teacher that his work was horrible; he was a rebel and hindered teaching work in class.

Albert Einstein At School Extra Questions And Answers Class 11 Question 8.
Albert was equally unhappy at his lodging. Why?
Answer:
Albert was miserable at his lodging. His father was a man of modest means, had got him a room in an ugly area. Albert.had no comfort and did not like the food there. The atmosphere was bad as his landlady kept beating her children and her husband came drunk and beat her.

Albert Einstein At School Important Question Class 11 Question 9.
What did Yuri say to him about violence in the hostel?
Answer:
Yuri told Einstein that he was fortunate to have an accommodation to himself. The people around him were poor but not uncivilized like the ones with whom he shared his accommodation. Yuri also told him about the uncivilized students, sharing his accommodation, who fought. The authorities did not take action but merely told them not to do so.

Albert Einstein At School Class 11 Extra Questions Question 10.
Who was Elsa? What was her advice to Albert?
Answer:
Elsa was Einstein’s cousin who lived in Berlin. She visited him of and on. She encouraged him in his studies and tried to assure him that it was not difficult to pass his examination. All he had to do was learn like a parrot, like the stupid boys who did that and passed.

Extra Questions Of Albert Einstein At School Class 11 Question 11.
What book did Elsa see Albert carrying under his arm? What did she say about it? What does it reveal about him?
Answer:
Albert was interested in reading books on science. Elsa saw him with a geology book, a subject that was not taught in school. She pointed out to him that it would not help him pass his diploma. It however, showed that he was a learner and he liked reading what actually interested him.

Albert Einstein At School Extra Question Class 11 Question 12.
Why does the biographer refer to Albert’s interest in music as a comfort?
Answer:
Albert was having a bad time at school where he was supposed to cram facts in which he had no interest. The teachers would taunt him and punish him. Even at his lodging, he had no comfort. All this made him miserable. He turned to music as his only solace.

Albert Einstein At School Extra Question Answer Class 11 Question 13.
Why did the landlady ask Albert to stop playing music? How did he feel?
Answer:
Albert was miserable both in school and in his lodgings. He found comfort only in music. He liked playing the violin and played it till the landlady stopped him. She had no ear for music, and the wailing and howling of the children coupled with the sound of the music got on her nerves.

Albert Einstein Class 11 Important Question Question 14.
What kind of a certificate was Albert looking for? Why?
Answer:
Albert told Yuri that he wanted to discontinue with school. If, however, he went back to Milan he would be sent back by his father. He had a plan. If a doctor certified that he had a nervous breakdown and it would be bad for him to go to school, he could escape school.

Class 11 English Snapshot Chapter 4 Extra Questions Question 15.
Yuri calls Albert ‘the world’s worst liar’. Do you think this is an insult or a compliment?
Answer:
Yuri called Albert ‘the world’s worst liar’ and meant it as a compliment to him. Yuri implied that Albert was so honest and straightforward that he could not tell lies successfully. His voice or manner betrayed him when he lied.

Albert Einstein At School Question Answers Class 11 Question 16.
How did Yuri help him in his plan?
Answer:
Yuri knew that Albert was miserable and wanted to help him. He knew of no doctors but referred him to his friend, a medical student, Ernst Weil. This young doctor had a license to practise and Yuri felt he could be of help.

Question 17.
Why was Albert nervous when he met the doctor? What does this nervousness indicate about his nature?
Answer:
The whole day Albert had been wondering what to tell the doctor from whom he needed a certificate saying that he had had a nervous breakdown. He was in a nervous state, worrying about it, by the time he went to see the doctor. This showed his inherent truthfulness and honesty.

Question 18.
How did Albert hope to get admission to an Italian college without a diploma from the German school?
Answer:
Albert went to Mr Koch, his teacher of mathematics, to get a recommendation. Mr Koch admired Albert and acknowledged Einstein’s superiority of merit, over his own. He got a recommendation from Mr Koch that said that he was fit to join an institute for higher education in mathematics.

Question 19.
What reason did the head teacher give for expelling Albert from school?
Answer:
The head teacher expelled Albert from school because his work was ‘terrible’. Albert’s presence in the classroom also made it impossible for the teacher to teach and for other pupils to learn. He accused Albert of hindering serious work because he refused to learn and was in constant rebellion.

Question 20.
Describe Albert’s meeting with Ernst Weil.
Answer:
Albert related his problem honestly to Ernst Weil though Yuri had informed the doctor of everything beforehand. The doctor, a student till recently, understood his problem. He judged, had Albert not been close to a nervous breakdown, he would not have gone to a doctor, thereby certifying that he keep away from school for six months.

Albert Einstein at School Extra Questions and Answers Long Answer Type

Question 1.
Were the teachers interested in understanding Albert and bringing out his potential?
Answer:
The given extract mentions only three of the teachers—the history teacher, Mr Braun, the mathematics . teacher, Mr Koch, and the head teacher. The history teacher was not at all interested in bringing out Albert’s potential. He followed the traditional method of teaching history and laid more emphasis on the acquisition of knowledge, by rote learning, rather than the understanding of ideas. He also had a sarcastic attitude and mocked Albert for his views. He did not help the talented boy achieve his potential, rather he complained to the head teacher and got him expelled.

The mathematics teacher, Mr Koch, on the other hand, was very encouraging. He confessed that he could not teach Albert anymore; rather Albert would soon be able to teach him. The head teacher humiliated Albert and expelled him from the institution for being rebellious and for not allowing the class work to go on. Neither the history teacher nor the head teacher acknowledged Albert’s mathematical genius.

Question 2.
Why was Albert miserable in school? How did he plan to move out of it?
Answer:
Albert Einstein was bom in a modest family and was sent to Munich to study at school for his diploma. He was an intelligent student but not good at rote learning dates and facts in history. Albert was miserable in school, where his teacher detested him for not adopting the conventional method of rote learning, and at the place where he lived. To escape the torture at school, he had a plan. He wanted to get a doctor’s certificate that declared he had had a nervous breakdown and was unfit to go to school.

Question 3.
Why was Albert miserable in Munich?
Answer:
Albert Einstein was studying for his diploma in school, in Munich, which had a very conservative approach to education. Albert had an analytical and rational mind. He hated rote learning. This outraged Mr Braun, his history teacher. The teacher taunted him by calling his views the ‘Einstein theory of education’. The authorities believed that Albert had no desire to leam and was wasting his father’s money. The teacher punished Albert by detaining him for an extra period in school.

Albert was miserable as he returned to his lodging. He had got a room in an area that was ugly. Albert had no comfort, nor did he like the food there. The atmosphere was bad as his landlady kept beating her children and her husband came drunk and beat her. She even forbade him to play the violin. All this made him miserable. Expulsion from his school was a welcome relief to him.

Question 4.
Comment on the role of Yuri as described in the extract.
Answer:
During that traumatic period in the school at Germany, Einstein’s only saviour was his friend Yuri. He lived in a hostel with some other students and often encouraged Albert when he was upset. Yuri was Albert’s friend, philosopher, and guide. Yuri helped Albert obtain a certificate to say that he had had a nervous breakdown by referring him to his friend, a medical student, Ernst Weil.

Yuri advised Albert into taking a certificate of recommendation from the mathematics teacher before seeing the head teacher. Albert got a recommendation from him stating that he was fit to join an institute for higher ‘ education in mathematics. It was this certificate that helped him join a college in Italy.
Yuri understood his friend and admired his honesty, calling him, ‘the world’s worst liar’. Albert met with Yuri before leaving Munich. Yuri bade him a good bye and wished him the best for his future.

Question 5.
Keeping the whole passage in mind, briefly discuss Einstein’s character as it is revealed here.
Answer:
Albert Einstein was one of the greatest scientists of all times. As a young student, Einstein showed the sparks of his genius. His mathematics teacher had great respect for his ability and went so far as to say, “I can’t teach you more, and probably you’ll soon be able to teach me.”

Einstein was not interested in knowing factual details of historical events but their causes and effects, which infuriated his history teacher. The exchange between Einstein and the teacher show him as someone who is unwilling to compromise on his views, risking punishment. He is noted for his straightforwardness. He comes across as one who will chart his own course rather than follow the beaten track.

Yuri paid Albert’s virtue of honesty a compliment by calling him ‘the world’s worst liar’. Albert Einstein also loved music. He played the violin for his joy and comfort.
Albert abhorred violence and was unhappy in his lodgings because of his abhorrence for domestic violence.

Question 6.
Express your views on the educational system in Germany.
Answer:
The educational systerp in Germany was organized along traditional lines. Albert Einstein’s miserable five years in the school at Munich are a telling comment on the system. It had no room for individual brilliance, aptitude and aspirations. Students were required to study a regular number of subjects. Stress was laid on the learning of facts rather than its cause and effects.

Elsa assured Einstein, it was easy to pass an examination by learning by rote, like a parrot. The system discouraged genius and creativity. No effort was made to understand the students their problems or emotions. Mr Braun, the history teacher, was completely insensitive to Albert’s feelings and taunted him about his ideas by calling them the ‘Einstein theory of education’.

Teachers and authorities insisted on discipline and conformity. The head teacher expelled Albert Einstein from school for arguing his point of view with the teacher.
The educational system in Germany was rigid and laid more emphasis on mindless cramming of facts than on understanding or creativity.

Question 7.
Who were Yuri and Elsa? What role did they play in Einstein’s life?
Answer:
Albert Einstein was studying for his diploma in a school in Munich, where he was unhappy. He hated learning dates and facts by heart and candidly confessed his apathy of learning dates and dry facts; he believed in ideas. He was equally miserable when he went to his lodging because of the violence around him there.

The only people he could depend on were Elsa and Yuri. Elsa was his cousin who lived in Berlin. She encouraged him in his studies and tried to assure him that it was not difficult to pass his examination. All he had to do was learn like a parrot like other stupid boys who did that and passed. Yuri, his friend, too gave him a lot of support, and introduced him to Ernst Weil, helping him get out of his miserable existence in Munich.

Question 8.
Why did Albert feel the ‘certificate burning a hole in his pocket’?
Answer:
The doctor, Ernst Weil, referred by Yuri, falsely certified that Albert had had a nervous breakdown and should be kept away from school for six months. However, before he could go to the head teacher, Albert was summoned and expelled from school. The reason given was that he neither wanted to study nor was his presence conducive for others eager to learn. He knew that was not true. He wanted to show him the certificate to prove that he was equally miserable and wanted to get rid of the school as he was not in agreement with the methods of teaching there.

Here we are providing Online Education for Surface Areas and Volumes Class 9 Extra Questions Maths Chapter 13 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Online Education for Extra Questions for Class 9 Maths Surface Areas and Volumes with Answers Solutions

Extra Questions for Class 9 Maths Chapter 13 Surface Areas and Volumes with Solutions Answers

Surface Areas and Volumes Class 9 Extra Questions Very Short Answer Type

Surface Area And Volume Class 9 Extra Questions Question 1.
How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm?
Solution:
Here, radius (r) = 3.5 cm and height (h) = 12 cm
∴ Amount of ice-cream = \(\frac{1}{3}\) πr²
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3.5 × 3.5 × 12
= 154 cm³

Class 9 Surface Area And Volume Extra Questions Question 2.
Calculate the edge of the cube if its volume is 1331 cm³.
Solution:
Volume of cube = 1331 cm³
(Side)³ = 1331
Side = (11 × 11 × 11)^{\(\frac{1}{3}\)} = 11 cm

Class 9 Maths Chapter 13 Extra Questions Question 3.
The curved surface area of a cone is 12320 sq. cm, if the radius of its base is 56 cm, find its
height.
Solution:
Here, radius of base of a cone (r) = 56 cm
And, curved surface area = 12320 cm²
πrl = 12320
l = \(\frac{12320}{\pi r}\)
= \(\frac{12320 \times 7}{22 \times 56}\) = 70 cm
Again, we have
r² + h² = l²
h² = l² – r² = 70² – 56²
= 4900 – 3136 = 1764
h = √1764 = 42 cm
Hence, the height of the cone is 42 cm.

Extra Questions On Surface Area And Volume Class 9 Question 4.
Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid.
Solution:
When two cubes are joined end to end, then
Length of the cuboid = 6 + 6 = 12 cm
Breadth of the cuboid = 6 cm
Height of the cuboid = 6 cm
Total surface area of the cuboid = 2 (lb + bh + hl)
= 2(12 × 6 + 6 × 6 + 6 × 12)
= 2(72 + 36 + 72) = 2(180)
= 360 cm²

Ch 13 Maths Class 9 Extra Questions Question 5.
A metallic sphere is of radius 4.9 cm. If the density of the metal is 7.8 g/cm², find the mass of the sphere (π = \(\frac{22}{7}\)).
Solution:
Here, radius of metallic sphere (r) = 4.9 cm

Extra Questions Of Surface Area And Volume Class 9 Question 6.
The volume of a solid hemisphere is 1152 π cm³. Find its curved surface area.
Solution:
Here, volume of hemisphere = 1152 π cm³
∴ \(\frac{2}{3}\)πr³ = 1152
⇒ r³ = (12)³ π
⇒ r\(\frac{1152 \times 3}{2}\) = 1728
⇒ r³ = (12)³
Now, curved surface area = 2πr²
= 2 × π × (12)² = 288π cm²

Class 9 Maths Ch 13 Extra Questions Question 7.
Find the diameter of a cylinder whose height is 5 cm and numerical value of volume is equal to

numerical value of curved surface area.
Solution:
Here, height of cylinder (h) = 5 cm
According to the statement of the question, we have
πr²h = 2πrh
r = 2 cm
Thus, diameter of the base of the cylinder is 2 × 2 i.e., 4 cm.

Extra Questions Of Chapter 13 Maths Class 9 Question 8.
In a cylinder, if radius is halved and height is doubled, then find the volume with respect to original volume.
Solution:
Here, r = \(\frac{r}{2}\), h = 2h

Surface Areas and Volumes Class 9 Extra Questions Short Answer Type 1

Class 9 Maths Surface Area And Volume Extra Questions Question 1.
A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm, find the volume of the spherical ball. [use π = 3.14]
Solution:
Since curved surface of half of the spherical ball = 56.57 cm²
2πr² = 56.57

= 113.04 cm³

Class 9 Maths Chapter 13 Extra Questions With Solutions Question 2.
Find the capacity in litres of a conical vessel having height 8 cm and slant height 10 cm.
Solution:
Height of conical vessel (h) = 8 cm
Slant height of conical vessel (l) = 10 cm
∴ r² + h² = l²
⇒ r² + 8² = 10²
⇒ r² = 100 – 64 = 36
⇒ r = 6 cm
Now, volume of conical vessel = \(\frac{1}{3}\)πr²h = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 8 = 301.71 cm³ = 0.30171 litre

Surface Area And Volume Extra Questions Class 9 Question 3.
Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside.
Solution:
Here, radius of hemispherical dome (r) = 14 m
Surface area of dome = 2πr²
= 2 × \(\frac{22}{7}\) × 14 × 14 = 1232 m²
Hence, total surface area to be whitewashed from outside is 1232 m².

Surface Areas And Volumes Class 9 Extra Questions Question 4.
A rectangular piece of paper is 22 cm long and 10 cm wide. A cylinder is formed by rolling the paper along its length. Find the volume of the cylinder.
Solution:
Since rectangular piece of paper is rolled along its length.
∴ 2πr = 22
r = \(\frac{22 \times 7}{2 \times 22}\) = 3.5 cm
Height of cylinder (h) = 10 cm
∴ Volume of cylinder = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 10 = 385 cm³

Surface Area And Volume Class 9 Questions Answers Question 5.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find it volume. If 1m³ wheat cost is ₹10, then find total cost.
Solution:
Diameter of cone = 10.5 m
Radius of cone (r) = 5.25 m
Height of cone (h) = 3 m
Volume of cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5.25 × 5.25 × 3
= 86.625 m³
Cost of 1m³ of wheat = ₹10
Cost of 86.625 m³ of wheat = ₹10 × 86.625
= ₹866.25

Question 6.
A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and1cm³ of water weighs 1 g, find the depth of water.
Solution:
Since 1 cm³ of water weighs 1 g.
∴ Volume of cylindrical vessel = 154 cm³
πr²h = 154
\(\frac{22}{7}\) × 3.5 × 3.5 × h = 154

h = \(\frac{154 \times 7}{22 \times 3.5 \times 3.5}\)
h = 4
cm Hence, the depth of water is 4 cm.

Surface Areas and Volumes Class 9 Extra Questions Short Answer Type 2

Question 1.
A wall of length 10 m is to be built across an open ground. The height of the wall is 5 m and thickness of the wall is 42 cm. If this wall is to be built with brick of dimensions 42 cm × 12 cm × 10 cm, then how many bricks would be required?
Solution:
Here, length of the wall (L) = 10 m = 1000 cm
Breadth of the wall (B) = 42 cm
Height of the wall (H) = 5 m = 500 cm
∴ Volume of the wall = L × B × H
= 1000 × 42 × 500 cm³
Volume of each brick = 42 × 12 × 10 cm³

= 4167
Hence, the required number of bricks is 4167.

Question 2.
The volume of cylindrical pipe is 748 cm. Its length is 0.14 m and its internal radius is 0.09 m. Find thickness of pipe.
Solution:
Internal radius (r) of cylindrical pipe = 0.09 m = 9 cm
Length (height) of cylindrical pipe (h) = 0.14 m = 14 cm
Let external radius of the cylindrical pipe be R cm.
Volume of cylindrical pipe = 748 cm³
⇒ π(R² – r²)h = 748
⇒ \(\frac{22}{7}\) (R² – 9²)14 = 748
⇒ R² – 81 = \(\frac{748 \times 7}{22 \times 14}\) = 17
⇒ R² = 81 + 17 = 98
⇒ R = √98 = 7√2 cm = 9.9 cm
Thus, thickness of the pipe = 9.9 -9 = 0.9 cm

Question 3.
The curved surface area of a cylinder is 154 cm. The total surface area of the cylinder is three times its curved surface area. Find the volume of the cylinder.
Solution:
Since curved surface area of cylinder = 154 cm² (given]
Total surface area of cylinder = 3 × curved surface area
2πrh + 2πr² = 3 × 154 3 154 + 2πr² = 462
2πr² = 462 – 154 = 308

= 539 cm³

Question 4.
A right-angled ∆ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. Find the volume of the solid generated. Also, find the total surface area of the solid.
Solution:
When rt. ∠ed ∆ABC is revolved about AB = 4 cm, it forms a right circular cone of radius 3 cm and height 4 cm. Slant height of the cone is 5 cm.

Question 5.
A semicircular sheet of metal of radius 14 cm is bent to form an open conical cup. Find the capacity of the cup.
Solution:
Radius of semicircular sheet (r) = 14 cm
∴ Slant height (1) = 14 cm
Circumference of base = Circumference of semicircular sheet

Surface Areas and Volumes Class 9 Extra Questions Long Answer Type

Question 1.
It costs ₹3300 to paint the inner curved surface of a 10 m deep well. If the rate cost of
painting is of ₹30 per m², find :
(a) inner curved surface area
(b) diameter of the well
(c) capacity of the well.
Solution:
Depth of well (h) = 10 m
Cost of painting inner curved surface is ₹30 per m² and total cost is ₹3300

Hence, inner curved surface area is 110 m², diameter of the well is 2 × 1.75 i.e., 3.5 m and capacity of the well is 96.25 m³.

Question 2.
Using clay, Anant made a right circular cone of height 48 cm and base radius 12 cm. Versha reshapes it in the form of a sphere. Find the radius and curved surface area of the sphere so formed.
Solution:
Height of cone (h) = 48 cm
Radius of the base of cone = 12 cm
Let R be the radius of sphere so formed
∴ Volume of sphere = Volume of cone
\(\frac{4}{3}\)πR³ = \(\frac{1}{3}\)πr²h
4R³ = 12 × 12 × 48
R³ = 12 × 12 × 12
R = 12 cm
Now, curved surface area of sphere = 4πR²
= 4 × \(\frac{22}{7}\) × 12 × 12
= 1810.29 cm

Question 3.
A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of ₹498.96. If the rate of whitewashing is ₹4 per square metre, find the :
(i) Inside surface area of the dome
(ii) Volume of the air inside the dome.
Solution:
Here, dome of building is a hemisphere.
Total cost of whitewashing inside the dome = ₹498.96
Rate of whitewashing = ₹4 per m²

Question 4.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm. Find the volume of the solid so obtained. If it is now revolved about the side 12 cm, then what would be the ratio of the volumes of the two solids obtained in two cases ?
Solution:
Here, right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm.
∴ Radius of the base of cone = 12 cm
Height of the cone = 5 cm

= 12 : 5

Question 5.
A right triangle of hypotenuse 13 cm and one of its sides 12 cm is made to revolve taking side 12 cm as its axis. Find the volume and curved surface area of the solid so formed.
Solution:
Here, hypotenuse and one side of a right triangle are 13 cm and 12 cm respectively.

Now, given triangle is revolved, taking 12 cm as its axis
∴ Radius of the cone (r) = 5 cm
Height of the cone (h) = 12 cm
Slant height of the cone (1) = 13 cm
∴ Curved surface area = πrl = π(5)(13) = 65π cm²
Volume of the cone = \(\frac{1}{2}\)πr²h = \(\frac{1}{2}\)π × 5 × 5 × 12 = 100π cm³
Hence, the volume and curved surface area of the solid so formed are 100 π cm³ and 65 π cm² respectively.

Surface Areas and Volumes Class 9 Extra Questions HOTS

Question 1.
Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.
Solution:
Let each edge of the cube be a cm.

Question 2.
A rectangular tank is 225 m × 162 m at base. With what speed should water flow into it through an aperture 60 cm × 45 cm so that the level of water is raised by 20 cm in 2.5 hours?
Solution:
Volume of water to be flown in 2.5 hour
= 225 m × 162 m × 20 cm

Hence, the speed of flow of water = 10.8 km/hour

Surface Areas and Volumes Class 9 Extra Questions Value Based (VBQs)

Question 1.
To maintain beauty of a monument, the students of the school cleaned and painted the dome of the monument. The monument is in the form of a hemisphere. From inside, it was white washed by the students whose area is 249.48 m².
(a) Find the volume of the air inside the dome. If white washing costs ₹2 per m², how much does it costs ?
(b) Which value is depicted by the students? (π = \(\frac{22}{7}\))
Solution:
(a) Here, dome of the monument is hemispherical in shape, which was whitewashed by the students.
Now, total area to be white washed = 249.48 m²
Cost of white washing = ₹2 per m²
∴ Total cost of white washing = ₹2 × 249.48
= ₹498.96
Also, 2πr² = 249.48

= 4191.264 m³
(b) Value: Cleanliness, beautification as well as preserving the heritage along with social values.

Question 2.
Salim provides water to a village, having a population of 4000 which requires 150 litres of water per head per day. He has storage tank measuring 20 m × 15 m × 6 m. For how many days will the water of his tank last? He increased the rate for providing water as the dependence of villagers increased on him. Which value is depicted by Salim?
Solution:
(i) Here, the population of the village = 4000
Requirement of water per head per day = 150 litres
∴ Total requirement of water per day = 4000 × 150 litres
= 600000 litres
Volume of water tank = 20 × 15 × 6
= 1800 m³
= 1800 × 1000 litres
Now, number of days for which water of the tank will last = \(\frac{1800 \times 1000}{600000}\) = 3 days
Hence, water tank can serve for 3 days.
(ii) Helping the needy.

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Online Education for Extra Questions for Class 10 Maths Some Applications of Trigonometry with Answers Solutions

Extra Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry with Solutions Answers

Some Applications of Trigonometry Class 10 Extra Questions Short Answer Type 1

Applications Of Trigonometry Class 10 Extra Questions With Solutions Question 1.
If a man standing on a platform, 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.

Solution:
False, θ₁ ≠ θ₁ (Fig. 11.15)

Applications Of Trigonometry Hot Questions Question 2.
Find the angle of elevation of the sun when the shadow of a pole h m high is √3 h m long.
Solution:

In ∆ABC

θ = 30°

Application To Trigonometry Class 10 Extra Questions With Answers Question 3.
The height of a tower is 12 m. What is the length of its shadow when 10 Sun’s altitude is 45°?
Solution:

Let AB be the tower [Fig. 11.17].
Then, ∠C = 45°, AB = 12 m

∴ The length of the shadow is 12 m.

Some Applications Of Trigonometry Class 10 Extra Questions With Solutions Pdf Question 4.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° [Fig. 11.18].
Solution:
Let AB be the vertical pole and AC be the long rope tied to point C.
In right ∆ABC, we have

Therefore, height of the pole is 10 m.

Some Applications of Trigonometry Class 10 Extra Questions Short Answer Type 2

Applications Of Trigonometry Extra Questions Question 1.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution:
Let BC be the tower whose height is h metres and A be the point at a distance of 30 m from the
foot of the tower. The angle of elevation of the top of the tower from point A is given to be 30°.
Now, in right angle ∆CBA we have,

Hence, the height of the tower is 10 √3 m.

Some Applications Of Trigonometry Extra Questions Question 2.
A tree breaks due to storm and the broken part bends, so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution:
In right angle ∆ABC, AC is the broken part of the tree (Fig. 11.20).
So, the total height of tree = (AB + AC)
Now in right angle ∆ABC,

Class 10 Maths Chapter 9 Extra Questions With Solutions Question 3.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:

Let OA be the tower of height h metre and P, l be the two points at distance of 9 m and 4 m respectively from the base of the tower.
Now, we have OP = 9 m, OQ = 4 m,
Let ∠APO = θ, ∠AQO = (90° – θ)
and OA = h metre (Fig. 11.21)
Now, in ∆POA, we have


Height cannot be negative.
Hence, the height of the tower is 6 metre.

Applications Of Trigonometry Questions Question 4.
Determine the height of a mountain if the elevation of its top at an unknown distance from the base is 30° and at a distance 10 km further off from the mountain, along the same line, the angle of elevation is 15o. (Use tan 15° = 0.27)
Solution:
Let AB be the mountain of height h kilometres. Let C be a point at a distance of x km, from the base of the mountain such that the angle of elevation of the top at C is 30°. Let D be a point at a distance of 10 km from C such that angle of elevation at D is of 15°.
In MBC (Fig. 11.22), we have

Substituting x = √3h in equation (i), we get
⇒ 0.27 ( √3h + 10) = h
= 0.27 × √3h + 0.27 × 10 = h
⇒ 2.7 = h – 0.27 × √3h
⇒ 27 = h (1 – 0.27 × √3)
⇒ 27 = h (1 – 0.46)
⇒ h = \(\frac{2.7}{0.54}\) = 5
Hence, the height of the mountain is 5 km.

Applications Of Trigonometry Class 10 Extra Questions Question 5.
The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.
Solution:
In Fig. 11.23, AB is the tower and BC is the length of the shadow when the Sun’s altitude is 60°, i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is 30°.
Now, let AB be h m and BC be x m.
According to the question, DB is 40 m longer than BC.
So, BD = (40 + x) m
Now, we have two right triangles ABC and ABD.

Using (i) in (ii), we get (x √3 ) √3 = x + 40, i.e., 3x = x + 40
i.e., x = 20
So, h = 20 √3 [From (i)]
Therefore, the height of the tower is 20 √3 m.

Applications Of Trigonometry Class 10 Questions With Solutions Question 6.
From a point P on the ground, the angle of elevation of the top of a 10m tall building is 30°. A flag is hosted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45o. Find the length of the flagstaff and the distance of the building from the point P. (You may take √3 = 1.732).
Solution:

In Fig. 11.24, AB denotes the height of the building, BD the flagstaff and P the given point. Note that there are two right triangles PAB and PAD. We are required to find the length of the flagstaff, i.e., BD and the distance of the building from the point P, i.e., PA.
Since, we know the height of the building AB, we will first consider the right ∆PAB.

i.e., x = 100(√3 – 1) = 7.32
So, the length of the flagstaff is 7.32 m.

Ch 9 Maths Class 10 Extra Questions Question 7.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
Let AC be a steep slide for elder children and DE be a slide for younger children. Then AB = 3 m and DB = 1.5 m (Fig. 11.25).
Now, in right angle ∆DBE, we have

So, the length of slide for elder children is 2 √3 m.

Applications Of Trigonometry Class 10 Important Questions Question 8.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution:
Let AB be the horizontal ground and K be the position of the kite and its height from the ground is 60 m and let length of string AK be x m. (Fig. 11.26)
∠KAB = 60°
Now, in right angle ∆ABK we have

So, the length of string is 40 √3 m.

Some Applications Of Trigonometry Class 10 Extra Questions Question 9.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution:
Let AB be the building and PQ be the initial position of the boy (Fig. 11.27) such that
∠APR = 30°
and AB = 30 m
Now, let the new position of the boy be P’Q’ at a distance QQ’.
Here, ∠AP’R = 60°
Now, in ∆ARP, we have

Therefore, required distance, QQ = PP’ = PR – P’R
= 28.5 √3 – 9.5 √3 = 19√3
Hence, distance walked by the boy is 19√3 m.

Chapter 9 Maths Class 10 Extra Questions Question 10.
From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 3 m from the banks, find
the width of the river.
Solution:
In Fig. 11.28, A and B represent points on the bank on opposite sides of the river, so that AB is the width of the river. P is a point on the bridge at a height of 3 m, i.e., DP = 3m. We are interested to determine the width of the river, which is the length of the side AB of the ∆APB.
In right ∆ADP, ∠A = 30°

∴ DB = 3m
Now, AB = BD + AD = 3 + 3 √3 = 3 (1 + √3) m
Therefore, the width of the river is 3(√3 + 1) m.

Class 10 Chapter 9 Maths Extra Questions Question 11.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:

Let AB be the lighthouse of height 75 m and P, Q be the position of the two ships whose angles of depression are 45° and 30°, respectively (Fig. 11.29). Let BP = x m and PQ = y m, we have
∠APB = 45° and ∠AQB = 30°
Now, in ∆ABP we have
tan 45° = \(\frac{AB}{BP}\)
⇒ 1 = \(\frac{70}{x}\)
⇒ x = 75 m …..(i)
Again, in ∆ABQ we have
tan 30° = \(\frac{AB}{BQ}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{75}{x+y}\)
⇒ x + y = 75 √3 ……(ii)
From (i) and (ii), we have
75 + y = 75 √3
y = 75 √3 – 75
⇒ y = 75(√3 – 1)
Hence, the distance between two ships is 75(√3 – 1) metres.

Class 10 Applications Of Trigonometry Important Questions Question 12.
Two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. The angles of depression of two ships as observed from the top of the light house are 60° and 45°. If the height of the light house is 200 m, find the distance between the two ships. (Use √3 = 1.73]
Solution:
Let the distance between the two ships be d.
Let the distance of one ship from the light house be x metres. Then, the distance of the other ship from the light house will be (d – x) metres.

Class 10 Maths Ch 9 Extra Questions Question 13.
The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of 3000 √3 m, find the speed of the aeroplane.
Solution:
Let A be point of observation and P and Q be positions of the plane. Let ABC be the line through A and it is given that angles of elevation from point A to two positions P and Q are 60° and 30°.
∠PAB = 60°, ∠QAB = 30°
Height = 3000 √3 m
So, in ∆ABP, we have

Some Applications Of Trigonometry Class 10 Important Questions Question 14.
From the top of a 60 m high building, the angles of depression of the top and the bottom of a tower are 45° and 60° respectively. Find the height of the tower. [Take √3 = 1.73]
Solution:
Let the height of the building be AE = 60 m, the height of the tower is ‘h’. The distance between the base of the building and the tower be ‘d’.
In ∆ADE,

⇒ AC = 34.60 m
Now, height of tower = AE – AC = 60 – 34.60 = 25.4 m

Question 15.
Two ships are approaching a light-house from opposite directions. The angles of depression of the two ships from the top of the light-house are 30° and 45°. If the distance between the two ships is 100 m, find the height of the light-house. (Use √3 = 1.732]
Solution:
Let AD be the light house and its height be h. The distance of one ship from the light house is x and that of other ship is 100 – x.
In ∆ADB,

= 50 × (1.732 – 1)
h = 36.6 m

Question 16.
Two men on either side of a 75 m high building and in line with base of building observe the angle of elevation of the top of the building as 30° and 60°. Find the distance between the two men. (Use √3 = 1.73)
Solution:

(i) In ∆ABM₁
\(\frac{A B}{B M_{1}}\) = tan 30° = \(\frac{1}{\sqrt{3}}\)
⇒ BM₁ = 75 √3 m

(ii) In ∆ABM₂
\(\frac{A B}{B M_{2}}\) = tan 60° = √3
⇒ BM2 = \(\frac{75}{\sqrt{3}}\) = 25 √3 m
∴ M₁ M₂ = M₁B + BM₂
= 75√3 + 25 √3 = 100 √3 m = 173 m
∴ Distance between two men = 173 m.

Question 17.
A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h.
Solution:

Let the speed of boat be x m/min
∴ CD = 2x
In ∆ABC
\(\frac{150}{y}\) = tan 60°
⇒ y =\(\frac{150}{\sqrt{3}}\) = 50√3 m
In ∆ABD
\(\frac{150}{y+2 x}\) = tan 45°
⇒ 150 = 50√3 + 2x
⇒ x = 25(3 – √3)
Speed = 25(3 – √3 ) m/min
∴ = 25 × 60 (3 – √3) m/h = 1500 (3 – √3) m/h

Some Applications of Trigonometry Class 10 Extra Questions Long Answer Type

Question 1.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60°, respectively. Find the height of the tower.
Solution:
Let AB be a building of height 20 m and BC be the transmission tower of height x m and D be any point on the ground (Fig. 11.36).
Here, ∠BDA = 45° and ∠ADC = 60°
Now, in ∆ADC, we have

⇒ x = 20√3 – 20 = 20 (√3 – 1) = 20 (1.732 – 1) = 20 × 0.732 = 14.64 m
Hence, the height of tower is 14.64 m.

Question 2.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the bottom of the pedestal is 45°. Find the height of the pedestal.
Solution:
Let AB be the pedestal of height h metres and BC be the statue of height 1.6 m (Fig. 11.37).
Let D be any point on the ground such that,
∠BDA = 45° and ∠CDA = 60°
Now, in ∆BDA, we have


Hence, height of the pedestal is 0.8 (√3 + 1) m.

Question 3.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
OR
From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower. (Use √3 = 1.732]
Solution:

Let PQ be the building of height 7 metres and AB be the cable tower. Now it is given that the angle of elevation of the top A of the tower observed from the top P of building is 60° and the angle of depression of the base B of the tower observed from P is 45° (Fig. 11.38).
So, ∠APR = 60° and ∠QBP = 45°
Let QB = x m, AR = h m then, PR = x m
Now, in ∆APR, we have
tan 60° = \(\frac{AR}{PR}\)
⇒ √3 = \(\frac{h}{x}\)
⇒ √3x = h
⇒ h = √3x ……(i)
Again, in ∆PBQ we have
tan 45o = \(\frac{PQ}{QB}\)
⇒ 1 = \(\frac{7}{x}\)
⇒ x = 7 ……(ii)
Putting the value of x in equation (i), we have
h = √3 × 7 = 7 √3
i.e., AR = 7 √3 metres
So, the height of tower = AB = AR + RB = 7 √3 + 7 = 7(√3 + 1) m.

Question 4.
At a point, the angle of elevation of a tower is such that its tangent is \(\frac{5}{12}\) On walking 240 m to the tower, the tangent of the angle of elevation becomes \(\frac{3}{4}\). Find the height of the tower.
Solution:
In the Fig. 11.39, let AB be the tower, C and D be the positions of observation from where given that


Hence, the height of the tower is 225 metres.

Question 5.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (Fig. 11.40). Find the distance travelled by the balloon during the interval.
Solution:
Let A and B be two positions of the balloon and G be the point of observation. (eyes of the girl)
Now, we have
AC = BD = BQ – DQ = 88.2 m – 1.2 m = 87 m .
∠AGC = 60°, ∠BGD = 30°
Now, in ∆AGC, we have

Hence, the balloon travels 58 √3 metres.

Question 6.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution:
Let OA be the tower of height h, and P be the initial position of the car when the angle of depression is 30°.
After 6 seconds, the car reaches to such that the angle of depression at Q is 60°. Let the speed of the car be v metre per second. Then,
PQ = 6υ (∵ Distance = speed × time)
and let the car take t seconds to reach the tower OA from Q (Fig. 11.41). Then, OQ = υt metres.
Now, in ∆AQO, we have

Hence, the car will reach the tower from Q in 3 seconds.

Question 7.
In Fig. 11.42, ABDC is a trapezium in which AB || CD. Line segments RN and LM are drawn parallel to AB such that AJ = JK = KP. If AB = 0.5 m and AP = BQ = 1.8 m, find the lengths of AC, BD, RN and LM.
Solution:
We have,
AP = 1.8 m
AJ = JK = KP = 0.6 m
AK = 2AJ = 1.2 m
In ∆ARJ and ∆BNJ’ we have
AJ = BJ, ∠ARJ = ∠BNJ = 60°
and ∠AJR = ∠BJ’N = 90°
∴ ∆ARJ ≅ ∆BNJ
⇒ RJ = NJ (By AAS congruence criterion)
Similarly, ∆ALK ≅ ∆BMK”
⇒ LK = MK”
In ∆ARJ,


Since ∆ACP ≅ ∆BDQ
So, BD = AC = 2.0784 m
Now, RN = RJ + JJ + J’N
= 2RJ + AB [∵ RJ = J’ N and JJ = AB]
= 2 × 0.3464 +0.5 = 1.1928 m
Length of step LM = LK + KK + KM
= 2LK + AB [∵ LK = K M and KK = AB]
= 2 × 0.6928 + 0.5 = 1.8856 m
Thus, length of each leg = 2.0784 m = 2.1 m
Length of step RN = 1.1928 m = 1.2 m
and, length of step LM = 1.8856 m = 1.9 m

Question 8.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Solution:

Let AB and CD be two poles of equal height h metre and let P be any point between the poles, such that
∠APB = 60° and ∠DPC = 30°
The distance between two poles is 80m.(Given)
Let AP = x m, then PC = (80 – x) m.
h’m Now, in ∆APB, we have

Now, putting the value of x in equation (i), we have
h = √3 × 20 = 20 √3
Hence, the height of the pole is 20 √3 m and the distance of the point from first pole is 20 m and that of the second pole is 60 m.

Question 9.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (Fig. 11.44). Find the height of the tower and the width of the canal.
Solution:
Let height of the tower be h metres and width of the canal be x metres, so AB = h m and BC = x m
Now in ∆ABC, we have

Now, putting the value of x in equation (i), we have
h = √3 × 10 = 10√3
⇒ h = 10 73 m
Hence, height of the tower is 10 √3 m and width of the canal is 10 m.

Question 10.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 metres away from the bank, he finds the angle of elevation to be 30°. Find the height of the tree and the width of the river.
Solution:
Let AB be the tree of height metres standing on the bank of a river. Let C be the position of man standing on the opposite bank of the river such that BC = x m. Let D be the new position of the man. It is given that CD = 40 m and the angles of elevation of the top of the tree at C and D are 60° and 30°, respectively, i.e.,
∠ACB = 60° and ∠ADB = 30°.
In ∆ACB, we have


Hence, the height of the tree is 34.64 m and width of the river is 20 m.

Question 11.
The angles of elevation and depression of the top and bottom of a lighthouse from the top of a building, 60 m high, are 30° and 60° respectively. Find
(i) the difference between the heights of the lighthouse and the building.
(ii) distance between the lighthouse and the building.
Solution:
Let AB be the building and CE be the lighthouse (Fig. 11.46).
In right-angled ∆ABC,

= √3 DE = 20 √3
⇒ DE = 20
∴ (i) Difference between the heights of lighthouse and building = EC – DC = DE = 20 m
and (ii) Distance between the lighthouse and the building = BC = 34.64 m.

Question 12.
In Fig. 11.47, from the top of a building AB, 60 metres high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60°, respectively. Find
(i) the horizontal distance between AB and CD.
(ii) the height of the lamp post.
Solution:
Given AB is the building and CD is the vertical lamp post. Then, DE is the horizontal distance between AB and CD. Let CD = h metres.
Then
∠EDB = 30° and ∠ACB = 60°, AE = CD = h metres and EB = (60 -h) m
Now, in ∆ABC

⇒ 60 – h = 20, i.e., h = 40 m
∴ (i) Horizontal distance between AB and CD = 20 √3 m = 20 × 1.732 = 34.64 m
(ii) The height of lamp post = 40 m

Question 13.
A boy standing on a horizontal plane finds a bird flying at a distance an elevation of 30°. A girl standing on the roof of 20 metre high building, finds the angle of elevation of the same bird to be 45°. Both the boy and the girl are on opposite sides of the bird. Find the distance of bird from the girl.
Solution:
Let B be the position of bird. O and P be the positions of boy and girl respectively and PQ be the building
We have, ∠AOB = 30° and ∠BPM = 45°
Now, in ∆AOB we have

Hence, distance of bird from girl is 30 √2 m.

Question 14.
The angles of depression of the top and the bottom of a 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi-storeyed building and the distance between the two buildings.
Solution:
In Fig. 11.49, PC denotes the multi-storeyed building and AB denotes the 8 m tall building. We are interested to determine the height of the multi-storeyed building, i.e., PC and the distance between the two buildings, i.e., AC.
Let PD = x m
In right ∆PDB, we have

From (i), BD = [4(√3 + 1)] √3 = 4(3 + √3 )]
So, the height of the multi-storeyed building is {4( √3 + 1) +8} m = 4(3 + √3) m
and the distance between the two buildings is also 4(3 + √3) m.

Question 15.
The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is 60°, then find the height of the flagstaff. [Use √3 = 1.73]
Solution:

Let CD be the tower which is at a distance of 120m from A.
BD = x be the length of flagstaff.

120 √3 = x + 120
= 120 × 1.73 – 120
x = 120 (1.73 – 1) = 120 × 0.73
x = 87.6 m

Question 16.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:
Let AB be the building of height h m and CD be the tower of height 50 m. We have,
∠ACB = 30° and ∠DAC = 60°
Now, in ∆ACD, we have

Hence, the height of the building is 16\(\frac{2}{3}\)m.

Question 17.
A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 5 m. From a point on the ground the angles of elevation of the top and bottom of the flagstaff are 60° and 30° respectively. Find the height of the tower and the distance of the point from the tower. (Take √3 = 1.732)
Solution:
Let height of tower be x m and distance of
point from tower be y m.


= x + 5 = 3x
⇒ x = \(\frac{5}{2}\) = 2.5
Height of tower = 2.5 m
Distance of point from tower = y = √3x
= (2.5 x 1.732) or 4.33 m

Question 18.
From the top of a tower, 100 m high, a man observes two cars on the opposite sides of the tower and in same straight line with its base, with angles of depression 30° and 45o. Find the distance between the cars.[Take √3 = 1.732]
Solution:
Let AQ be the tower of height 100m. Car B and Car Care in opposite direction and at distance
of x m and y m respectively.
In ∆ABO,

Distance between the cars = x + y
= 100 + 100√3 [From equation (i) and (ii)]
= 100 (1 + √3 )
= 100 (1 + 1.732) = 273.2 m

Some Applications of Trigonometry Class 10 Extra Questions HOTS

Question 1.
A man standing on the deck of a ship, which is 10 m above the water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill.
Solution:
In Fig. 11.54, let C represents the position of the man on the deck of the ship, A represents the top of hill and D its base.
Now in right-angled triangle CWD,

Now, AD = AB + BD = 30 m + 10 m = 40 m.
Therefore, the distance of the hill from the ship = 17.3 m and the height of the hill = 40 m

Question 2.
A spherical balloon of radius r subtends an angle θ at the eye of an observer. If the angle of elevation of its centre is Ø, find the height of the centre of the balloon.
Solution:
In Fig. 11.55, O is the centre of balloon, whose radius OP = rand ∠PAQ = 0. Also, ∠OAB = Ø.
Let the height of the centre of the balloon be h. Thus OB = h.
Now, from ∆OAP,

Question 3.
From a window (h metres high above the ground) of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are θ and Ø respectively. Show that the height of the opposite house is h (1 + tan θ cot Ø).
Solution:
Let W be the window and AB be the house on the opposite side.
Then, WP is the width of the street (Fig. 11.56).
Let AP = h’ m

⇒ h’ = WP tan θ
⇒ h’ = h cot Ø tan θ
∴ Height of house = h’ + h
= h cot Ø tan θ + h = h (1 + tan θ cot Ø)

Question 4.
The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of
1500√3 m find the speed of the jet plane.
Solution:
Let P and Q be the two positions of the plane and let A be the point of observation. Let ABC be the horizontal line through A. It is given that angles of elevation of the plane in two positions P and Q from a point A are 60° and 30°, respectively,
Then, ∠PAB = 60°, ∠QAB = 30°
It is also given that PB = 1500√3 metres
In ∆ABP, we have

Question 5.
If the angle of elevation of a cloud from a point h metres above a lake is a and the angle of depression of its reflection in the lake is B, prove that the height of the cloud is \(\frac { h(tanß-tanα) }{ tan ß – tan α }\)

Solution:
Let AB be the surface of the lake and let P be a point of observation (Fig. 11.58) such that AP = h metres. Let C be the position of the cloud and C’ be its reflection in the lake. Then, CB = C’ B. Let PM be perpendicular from P on CB. Then, ∠CPM = a and ∠MPC’ = B.
Let CM = x.
Then, CB = CM + MB = CM + PA = x + h
In ∆CPM, we have

Question 6.
The shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the sun is 60°. Find the angle of elevation of the sun at the time of the longer shadow.
Solution:
Let AB be the flagstaff and BC be the length of its shadow when the Sun rays meet the ground at an angle of 60°. Let o be the angle between the Sun rays and the ground when the length of the shadow of the flagstaff is BD. Let h be the height of the flagstaff (Fig. 11.59).
Let BC = x
∴ BD = 3x and CD = 2x
In ∆ABC, we have

Question 7.
From an aeroplane vertically above a straight horizontal plane, the angles of depression of two consecutive kilometre stones on the opposite sides of the aeroplane are found to be α and β. Show that the height of the aeroplane is = \(\frac { tanα.tanβ }{ tanα+tanβ }\)
Solution:
In Fig. 11.60, let P be the position of plane, A and B be the positions of two stones one kilometre apart. Angles of depression of stones A and B are a and ß respectively.
Let PC = h.
In right-angled triangle ACP, we have

Question 8.
The angle of elevation of a cliff from a fixed point is θ. After going up a distance of k metres towards the top of the cliff at an angle of Ø, it is found that the angle of elevation is α. Show that the height of the cliff is \(\frac {k(cos\phi-sin\phi cot\alpha)}{cot\theta-cot\alpha}\) metres.
Solution:
Let AB be the cliff and O be the fixed point such that ∠AOB = θ. Let ∠AOC = Ø and OC = k m.
From C, draw CD and CE perpendiculars on AB and OA, respectively. Now, ∠DCB = a. Let h be the height of the cliff AB.
In ∆COE, we have

Hence Proved.

Question 9.
The angle of elevation of the top of a tower from a point A due south of the tower is a and from B due east of the tower is β. if AB = d, show that the height of the tower is \(\frac {d}{\sqrt{{cot}^{2}}\alpha+{cot}^{2}\beta}\).
Solution:
Let OC be the tower. Let height of tower OC be h. A and B be two points due south and east of tower at O.
In ∠AOC, ∠O = 90°

Question 10.
From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be 45° and 30° respectively. Find the height of the hill.
Solution:
Let the height of the hill be hm, C and D are two consecutive stones having distance 1000 m between them and AC = x m.
In ∆ABC,
tan 45° = \(\frac{h}{x}\)
⇒ x = h ….(i)
In ∆ABD,

Hence, the height of the hill = 500(√3 + 1)m.

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Online Education for The Ailing Planet: the Green Movement’s Role Important Extra Questions and Answers Class 11 English Hornbill

The Ailing Planet: the Green Movement’s Role Extra Questions and Answers Short Answer Type’

The Ailing Planet Extra Questions Question 1.
What awareness according to Nani Palkhivala is growing worldwide? Why?
Answer:
The movement, which has gripped the imagination of the entire human race, is the worldwide consciousness that the earth itself is a living organism of which we are parts. It has its own metabolic needs and vital processes that need attention because the earth’s vital signs reveal its declining health.

The Ailing Planet Class 11 Extra Questions Question 2.
What is propagated by the concept of sustainable expansion?
Answer:
The World Commission on Environment and Development popularized the concept of sustainable development in 1987. It stressed the idea of development that meets the needs of the present, without compromising the ability of future generations to meet their needs, i.e., without stripping the natural world of resources future generations would need.

Ailing Planet Class 11 Extra Questions Question 3.
What is the global concern raised by Mr Lester R. Brown that threatens the very existence of man?
Answer:
Mr Lester R. Brown has noted the earth’s principal biological systems as fisheries, forests, grasslands, and croplands the foundation of the global economic system. They provide us with our food and virtually all the raw materials for industries. However, human claims are exhausting these resources leading to the collapse and disappearance of fisheries and grasslands.

Ailing Planet Extra Questions Question 4.
What are the reasons that are leading to depletion of our natural resources?
Answer:
In a protein-conscious and protein-hungry world, over-fishing is common. In poor countries, local forests are being destroyed in order to procure firewood for cooking. As a consequence, in some places, firewood has become so expensive that fuel costs more than the food.

The Ailing Planet Question And Answer Question 5.
What steps has the Indian government taken to ensure the protection of the environment? What is the impact?
Answer:
The Indian government through Article 48A of the Constitution of India provides that the State shall try to protect and improve the environment and safeguard the forests and wildlife of the country. But the law has not had the due impact as laws are neither valued nor enforced in India.

Ailing Planet Class 11 Ques Ans Question 6.
The population in the world is growing at an alarming rate. Comment.
Answer:
The growing world population is deforming the future of human society. Human population reached its first billion around the year 1800. By the year 1900, a second billion was added, and another 3.7 billion in the twentieth century. Presently, the world population is estimated at 5.7 billion. Every day the world population increases by one million.

The Ailing Planet Extra Questions And Answers Question 7.
Why is the growing population detrimental to the world’s progress?
Answer:
Development will not be possible if the present increase in numbers continues. The rich get richer and the poor produce more children, hampering their economic growth. More children do not mean more workers, merely more people without work. Excessive population perpetuates poverty. People would die of hunger unless population growth is controlled.

Class 11 English Chapter 5 Extra Question Answer Question 8.
Why is our age the ‘Era of Responsibility’?
Answer:
A growing anxiety about the survival of our planet has surfaced for the first time in human history. The emerging new world vision a holistic view of the source of our survival has steered in the Era of Responsibility; an ecological view of the world as a complete whole and not a disconnected collection of parts.

The Ailing Planet: the Green Movement’s Role Extra Questions and Answers Long Answer Type

The Ailing Planet Class 11 Important Questions Question 1.
“The earth’s vital signs reveal a patient in declining health.” What are these ‘vital signs’?
Answer:
A three-year study using satellites and aerial photography conducted by the United Nations, warns that the environment has worsened so badly that it is ‘critical’ in many of the eighty-eight countries investigated.

A recent report of our Parliament’s Estimates Committee has highlighted the near disastrous exhaustion of India’s forests over the last four decades. India, according to reliable data, is losing its forests at the rate of 3.7 million acres a year. Large areas, officially designated as forestland, “are already virtually treeless”.

The actual loss of forests is estimated to be about eight times the rate indicated by government statistics.
The world’s ancient patrimony of tropical forests is now eroding at the rate of forty to fifty million acres a year, and the growing use of dung for burning deprives the soil of an important natural fertilizer. The World Bank estimates that a five-fold increase in the rate of forest planting is needed to cope with the expected fuel wood demand.

Extra Questions Of Chapter Ailing Planet Question 2.
Various visionaries and academicians have warned against hazardous consequences if we do not wake up to our responsibilities towards the environment. Elaborate.
Answer:
Mr L.K. Jha, a member of an international commission which dealt with the question of ecology and environment, raised the question of whether we would leave our successors a parched planet with increasing deserts, impoverished landscapes and ailing environment. Mr Lester R. Brown in his thought-provoking book, The Global Economic Prospect, points out that the earth’s principal biological systems, i.e., fisheries, forests, grasslands, and croplands form the foundation of the global economic system.

In large areas of the world, human claims on these systems are reaching an unsustainable level, a point where their productivity is being impaired. Dr Myers warns against the depletion of forests as a result of which several species face extinction. James Speth, the President of the World Resources Institute, said that we are losing the forests at an acre-and-a-half to a second.

The Ailing Planet Summary Questions And Answers Question 3.
What are the four systems that sustain life on earth? What threats are they facing?
Answer:
Mr Lester R. Brown in his thought-provoking book, The Global Economic Prospect, points out that the earth’s principal biological systems, i.e., fisheries, forests, grasslands, and croplands form the foundation of the global economic system. In addition to supplying our food, these four systems provide almost all the raw materials for industry except minerals and petroleum-derived synthetics. It is because of this that fisheries collapse, forests disappear, grasslands are converted into barren wastelands and croplands deteriorate. In a protein-conscious and protein-hungry world, over-fishing is common. In poor countries, local forests are being demolished in order to procure firewood for cooking. Since tropical forests house various species of life, they face extinction as a result of its destruction.

Ailing Planet Important Questions Question 4.
“Fertility falls as incomes rise, education spreads, and health improves.” Justify.
Answer:
Undoubtedly, the growth of world population is one of the strongest factors disfiguring the future of human society. The present world population is estimated at 5.7 billion. Every four days the world population increases by one million. Development is not possible if the present increase in numbers continue. The rich get richer and the poor beget more children, which in turn makes them poorer. More children do not mean more workers, merely more people without work.

The choice is really between control of population and perpetuation of poverty. The population of India has crossed 1.3 billion today. This leaves little doubt that hordes of people would die in their hungry hutments unless population control is given topmost priority.Compulsory sterilization is not the solution. Masses should be educated so that they volunteer for family planning without introducing an element of compulsion. It ought to be understood that the choice is between control of population and continuation of poverty.

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Online Education for Mother’s Day Important Extra Questions and Answers Class 11 English Snapshots

Mother’s Day Extra Questions and Answers Short Answer Type

Mother’s Day Extra Questions Question 1.
Who is Mrs Fitzgerald? What does she advise Mrs Pearson?
Answer:
Mrs Fitzgerald is Mrs Pearson’s neighbour and friend. A fortune teller, who had learnt the art from the East, she tells Mrs Pearson that her fortune could turn either way. With effort and counsel, the situation would swing in her favour. She advised her to assert herself as the boss of the house.

Mother’s Day Class 11 Questions And Answers Question 2.
What was Mrs Pearson’s reaction to Mrs Fitzgerald’s advice?
Answer:
Mrs Pearson said that it would not be easy to put her family members in place as she was very fond of them. She knew that they were thoughtless and selfish but felt, perhaps, they did not mean to be so.

Mother’s Day Class 11 Extra Questions Question 3.
What was Mrs Fitzgerald’s opinion of Mrs Pearson’s attitude?
Answer:
Mrs Fitzgerald said that Mrs Pearson’s family was undoubtedly spoilt. She felt that it was Mrs Pearson’s attitude that did them no good, tending to their needs, taking their orders, and staying at home every night while they went out enjoying themselves.

Mothers Day Questions And Answers Question 4.
What does Mrs Fitzgerald offer to do for her?
Answer:
Mrs Fitzgerald sensed that Mrs Pearson was far too gentle, submissive and generous to tackle her family. Mrs Fitzgerald offered to make them realize the error of their ways not as Mrs Fitzgerald but as Mrs Pearson. She offered to change their bodies and change back again.”

Mother’s Day Class 11 Extra Questions And Answers Question 5.
How did the two women react after their bodies were changed?
Answer:
When Mrs Pearson looked down at herself in Mrs Fitzgerald’s body, she gave a scream of fright. On the other hand, Mrs Fitzgerald is rather pleased and feels that the transition was so neat that she did not even know that she had it in her.

Mother’s Day Extra Questions And Answers Question 6.
What is Doris’s first reaction on seeing her mother? Why?
Answer:
Doris was taken aback to see her mother smoking and playing cards. When Doris asks her what she was doing, she is startled to get her answer—‘whitewashing the ceiling.’ Moreover, her conduct was not nervous and apologetic but cool and incisive.

Mother’s Day Important Questions Question 7.
What did Doris want her mother to do? How did the mother react?
Answer:
Doris wanted her to iron her yellow silk dress that she ‘must wear’ that night. She also wanted her mother to make tea for her. She refused to get her tea and iron her dress, telling her that she put in twice the hours Doris did but got neither wages, nor thanks for it.

Mother’s Day Questions And Answers Question 8.
What does Mrs Pearson say to Doris that really bothered her?
Answer:
Mrs Pearson asked where Doris would wear her yellow silk dress. She said that she planned to go out with Charlie Spence. Mrs Pearson told her to find somebody better, and insulted Charlie Spence by calling her buck-toothed and was half-witted.

Extra Questions Of Mother’s Day Class 11 Question 9.
What does Mrs Pearson have to say to Cyril that shocks him?
Answer:
When Cyril walk in and insists on her getting the tea and his clothes ready, he is stunned to hear that she doesn’t ‘like mending’. She goes on to tell him that when he does not want to do something, he does not do it. She planned to do the same. Cyril could not believe his ears.

Mother’s Day Class 11 Important Questions And Answers Question 10.
What do Doris and Cyril feel about Mrs Pearson’s changed behaviour?
Answer:
Doris and Cyril discuss that there is something wrong with their mother as she is not behaving in character. They discuss how Mrs Pearson behaved oddly with each of them. They try to fathom if she had gone crazy or had a concussion.

Mother’s Day Class 11 Questions And Answers Pdf Question 11.
What is Mrs Pearson’s reaction to see her children giggling when she returns to the room?
Answer:
Mrs Pearson asks them the reason for their amusement. Doris answers that she had never understood their jokes. To which Mrs Pearson retorts, rudely, that she was bored at their jokes even before they were bom. Doris is tearful and Mrs Pearson blames them for being selfish about their needs.

Mother’s Day Question Answer Question 12.
What reason does she give Cyril for not making the tea?
Answer:
When Cyril again asks for tea as he had been working for an eight-hour day, Mrs Pearson replies that she had done her eight hours and henceforth she would work only for forty hours a week. She declared that she would have her two days off on the weekend.

Important Questions Of Mother’s Day Class 11 Question 13.
What, according to Mrs Pearson, were her plans for the weekends?
Answer:
Mrs Pearson tells her children that at the weekend she would have her two days off. She agreed to make beds and cook a little as a favour, conditional to how she was treated. Mrs Pearson tells her children that in case they did not like the arrangement, she would go elsewhere for the weekend.

Mothers Day Class 11 Question Answer Question 14.
Why was George Pearson surprised when he came home? What was the answer that he got?
Answer:
Mr George Pearson was surprised to see Doris crying and was shocked to see Mrs Pearson sipping beer. He said that it did not look right. Mrs Pearson replied that it was ‘a nice change’ and it had been quite some time since he was surprised at her.

Extra Questions Of Chapter Mother’s Day Class 11 Question 15.
What did Mrs Pearson say to her husband when he was angry with her for not making tea?
Answer:
Mr Pearson informed Mrs Pearson that he did not want tea but grew angry at being told that tea was not ready. She taunted him that if he went up to the bar at the club and refused a glass of beer and showed irritation because they had not served it earlier, he would invite ridicule.

Question 16.
What was the truth about Mr George Pearson that hurt him the most?
Answer:
Mrs Pearson told George that that he was one of the standing jokes in the club. He was called ‘Pompy-ompy Pearson’ because they thought that he was slow and pompous. She was surprised that he spent so much time at a place where people always ridiculed him, leaving his wife at home.

Question 17.
What was Mrs Fitzgerald’s reaction to Mrs Pearson shouting at Cyril? Why?
Answer:
Mrs Fitzgerald was in reality Mrs Pearson, so when she saw Mrs Pearson (the real Mrs Fitzgerald) shouting at Cyril, she protested as she was actually Cyril’s mother. But Mrs Pearson told her not to interfere.

Question 18.
What were the two slips that could have let out the real identity of Mrs Fitzgerald?
Answer:
Mrs Fitzgerald, in her nervousness, addresses Mr Pearson, as George. Mr Pearson is surprised to be called thus, but Mrs Pearson covers up for Mrs Fitzgerald. Later, when Mrs Fitzgerald attempts to slap George, following an argument, the real Mrs Pearson exclaims and calls out to her, ‘Mrs Fitzgerald’, which confuses George.

Question 19.
How was the experience for the two women after the change of bodies?
Answer:
The real Mrs Pearson (now Mrs Fitzgerald) had not enjoyed the experience as she had seen her family being treated roughly and rudely by Mrs Fitzgerald. On the other hand, Mrs Fitzgerald had enjoyed the experience, as she had been able to teach Doris,Cyril and George Pearson a lesson to value Mrs Pearson.

Question 20.
What was Mrs Fitzgerald’s advice to Mrs Pearson after she had put back the family members in their proper place?
Answer:
Mrs Fitzgerald advised Mrs Pearson not to be soft and waste all the effort she had put in to change the attitude of her family for the better. Mrs Pearson is unsure how she would explain her behaviour. But Mrs Fitzgerald warns her not to yield.

Question 21.
What was the change that came over the Pearson family in the end?
Answer:
When Mrs Fitzgerald left, Mrs Pearson’s family was relieved to see her smile. Mrs Pearson decides to stay home for a family game of rummy, and have the children prepare dinner. They readily agree and gather around Mrs Pearson as the play ends.

Mother’s Day Extra Questions and Answers Long Answer Type

Question 1.
Write in your words the conversation between Mrs Pearson and Mrs Fitzgerald in the beginning of the play. What is the outcome of the meeting?
Answer:
Mrs Fitzgerald predicts her friend Mrs Pearson’s fate and tells her that it was high time she asserted herself as the head of the family. Mrs Pearson says that it was not easy because she loves her family although they are very thoughtless and selfish. But Mrs Fitzgerald insists that they ought to learn to appreciate her and treat her appropriately. She tells her not to run after them and oblige. Mrs Pearson agrees with Mrs Fitzgerald, but wonders if anything would affect them.

She is afraid of creating unpleasantness in the family. As Mrs Pearson is about to rush off to prepare dinner for her family, Mrs Fitzgerald comes up with an idea. She tells . Mrs Pearson that they could exchange their bodies. She then holds her hand and asks her to keep quiet for a minute. They stare at each other and Mrs Fitzgerald mumbles ‘Arshtatta dum—arshtatta lam—arshtatta lamdumbona…’ and they assume each other’s personality.

Question 2.
What does Mrs Pearson have to say to Doris that disturbs her?
Answer:
When Mrs Pearson’s daughter Doris returns and tells Mrs Fitzgerald, in the body of Mrs Pearson, to iron her yellow silk dress as she would like to wear it that night, her mother continues playing patience. Doris asks her what she is doing and she answers her smugly that she was not whitewashing the ceiling. She also says that there is no law against smoking. She informs Doris that she had already had her tea but had not made tea for the others.

She had not cooked dinner either and would have her meal at the Clarendon. She tells Doris that she worked twice as hard as the others and got no wages or thanks for it. She then inquired from Doris where she wanted to wear her yellow dress. Doris tells her that she was going out with Charlie Spence. Mrs Pearson tells her to find someone better than the buck-toothed and half-witted man. Doris is offended and runs out.

Question 3.
Describe Mrs Pearson’s conversation with Cyril when he walks in and his reaction.
Answer:
Mrs Pearson’s son Cyril walks in and insists on Mrs Pearson getting the tea and his clothes ready. He reminds her of the promise she made the same morning, to mend his clothes. He is surprised to hear that she does not Tike mending’ and that she would not do anything that she did not want to do. Cyril could not believe his ears. Cyril again asks for tea, telling her that he had been working for an eight-hour day, to which Mrs Pearson says that she had done her eight hours and henceforth she would work for only forty hours a week. On weekends, she would have her two days off.

She might make a bed or two and do a bit of cooking as a favour but that would be conditional to the fact that they asked her very nicely and thanked her for everything and generally made a fuss of her. Cyril and Doris are surprised and wait for their father to arrive.

Question 4.
“Sometimes it does people good to have their feelings hurt.” Who says this and what does she say to hurt Mr Pearson?
Answer:
When Mr Pearson announces that he did not wish to drink tea, after he returned from work, the real Mrs Fitzgerald informs him that his tea was not ready. He is angry and she reminds him that he was annoyed because he did not get the tea that he did not want in the first place. She adds that if he did that at bar—went up to the bar at the club and told them he did not want a glass of beer but got irritated because they had not already poured it out for him, they would laugh at him even more than they did already.

George was indignant and she added that he was one of their standing jokes and was called ‘Pompy-ompy Pearson’ because they thought that he was slow and pompous.
George checks with Cyril on the truth of the matter, and Cyril accuses his mother of not being fair and sensitive. To which, she replies that sometimes it does people good to have their feelings hurt.

Question 5.
Pick out the instances that bring out the element of humour in the play.
Answer:
The play is a light-hearted comedy. A mother accomplishes changing the attitude of her family by exchanging her body with her friend, Mrs Fitzgerald, a bold and assertive woman, thus, introducing humour in the play. The moment the spirits change, Mrs Pearson notices the cigarette, snatches it off Mrs Fitzgerald, while she looks down at her changed body and screams out of fright.

Mrs Pearson answers to her daughter’s query, sarcastically, that she was not whitewashing the ceiling. Mrs Pearson insults Doris’ date, comically, calling him ‘buck-toothed and half-witted’. Cyril is told that she does not ‘like mending’ his clothes, stunning him to silence. When Mrs Pearson walks out, Doris and Cyril laugh at the idea of her having gone crazy and decide to wait till the father comes. She tells Mr Pearson how he was a standing joke in the club and was called ‘Pompy-ompy Pearson’ because they thought that he was slow and pompous. She also tells him that he was George, and not the Duke of Edinburgh.

Question 6.
Mrs Fitzgerald’s effort does not go in vain, as the family changes for the better. Justify.
Answer:
Mrs Pearson has a selfish and a thoughtless family, who is insensitive to the feelings of the mother. But after the staged trick, and when Mrs Fitzgerald leaves, the three—George, Doris, and Cyril look anxiously at Mrs Pearson, who smiles. They are much relieved, and smile back at her. Mrs Pearson tells them that since they have decided to stay at home, they would have a nice family game of rummy and then the children could get the supper ready while she talked with their father. All of them agree. Mrs Pearson wishes Mrs Fitzgerald goodbye and the family surrounds Mrs Pearson, implying a happy ending. It looks as if Mrs Fitzgerald’s effort is rewarded and they seem eager to appease the real Mrs Pearson.