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1 kilometer = 1000 meters 

2500 m = 2500/1000 = 2.5 km

Correct option is C) Arthshastra

I select a tape to which is a suitable instrument to measure the length.

Measuring cylinders.

• Each country had its own scale which differs from others, so it lead to commerce a lot of problems in trade and commerce.
• So each country in the world began to make their own sales. 
• Finally in France, it was decided that a certain length of rod made of a special material (Platinum – Iridium) would be called a meter. 
• This was finally accepted as an instrument of measuring lengths.

Correct option is A) Ml

The Area of Rectangular plot 

A = L1 × B1 Here 

L1 = 60ft, 

B1 = 50ft 

A1 = L1 × B1 = 60ft × 50ft = 3000 square feet. 

The Area of house 

A2 = L2 × B2 

Here L2 = 40ft, 

B2 = 40ft 

A2 = L2 × B2 = 40ft × 40ft = 1600 square feet 

The remaining area 

A3 = A1 – A2 = 3000 – 1600 = 1400s 

feet A3 = 14 × B2 = 40ft × 40ft = 1600 square feet 

The remaining area 

A3 = A1 – A2 = 3000 – 1600 = 1400s feet A3 = 1400 Square feet. 

So the area planned for garden = 1400 square feet.

The length of hall (L) = 20 m 

The breadth of hall (B) = 15 m 

The area of hall = l × b 

= 20 m × 15 m = 300 m² = 300 sq. m.

Measuring cylinder.

The answer is (B) Mercurus Nitrate

The answer is (A) Meghnad Saha

When a gas of Bose particles is cooled down to temperatures very close to absolute zero, then the kinetic energy of the particles decreases to a negligible amount and they condense to the lowest energy level state. This state is called as Bose-Einstein Condensation.

Brackett series and P-fund series.

(i) Infrared rays: It is used for fog or haze photography.

(ii) Gamma rays: If is used in medical treatment.

(iii) Micro-waves: If is used in radio and tele communication.

(iv) Ultraviolet radiations: If is used in sterilization of water.

Drawbacks of Rutherford’s model of atom:

• According to Rutherford‘s model of atom, electrons which are negatively charged particles revolve around the nucleus in fixed orbits. Thus, the electrons undergo acceleration. According to electromagnetic theory of Maxwell, a charged particle undergoing acceleration should emitelectromagnetic radiation. Thus, an electron in an orbit should emitradiation. Thus, the orbit should shrink. But this does not happen. 
• The model does not give any information about how electrons are distributed around nucleus and what are energies of these electrons 

James Chadwick.

Number of electrons = Number of protons = 35

Number of neutrons = 80 – 35 = 45

(i)  Rutherford‟s model cannot explain the stability of an atom. 

(ii)  It cannot explain the distribution of electrons around the nucleus or their energies.

The conclusions drawn by Rutherford from alpha particle scattering
experiment are: -

i. There is a positively charged centre in an atom called the nucleus. Nearly all the mass of an atom resides inside the nucleus.

ii. The electrons revolve around the nucleus in well-defined orbits.

iii. The size of the nucleus is very small as compared to the size of an atom.

Hence, statements ii and iv are correct and the correct option is B.

Correct option is c. nucleus

i. True 

ii. False,

In Rutherford’s experiment of scattering of α-particles by thin gold foil, very few α-particles bounced back.

iii. False, 

Cathode rays are a stream of very small, negatively charged particles.

1. The triple point of water is that point where water in a solid, liquid and gas state co-exists in equilibrium and this occurs only at a unique temperature and a pressure.

2. To know the triple point, one has to see that three phases coexist in equilibrium and no one phase in dominating. This occurs for each substance at a single unique combination of temperature and pressure.

3. Thus, if three phases of water solid ice, liquid water and water vapour coexist, the pressure and temperature are automatically fixed.

4. Internationally, triple point of water has been assigned as 273.16 K at pressure equal to 6.11 × 10² Pa or 6.11 × 10^-3 atmosphere, as the standard fixed point for calibration of thermometers.

5. The physical significance of triple point of water is that, it represents unique condition and it is used to define the absolute temperature.

1. The absolute scale of temperature, is so termed since ills based on the properties of an ideal gas and does not depend on the property of any particular substance.

2. The zero of this scale is ideally the lowest temperature possible although it has not been achieved in practice.

3. It is termed as Kelvin scale after Lord Kelvin with its zero at -273.15 °C and temperature intervals same as that on the Celsius scale. It is written as K (without °)

1. The two fixed point scale, Celsius scale and Fahrenheit scale had a practical shortcoming for calibrating the scale.

2. It was difficult to precisely control the pressure and identify the fixed points, especially for the boiling point as the boiling temperature is very sensitive to changes in pressure.

3. Hence, a one fixed point scale was adopted to define a temperature scale.

4. This scale is called the absolute scale or thermodynamic scale.

सहोपकारिता (Mutualism) – यह दो भिन्न प्रजाति के जीवों के बीच पाया जाने वाला सहजीवी (symbiotic) सम्बन्ध होता है जिसमें दोनों जीवों को परस्पर लाभ (benefit) होता है। कवक और प्रकाश-संश्लेषी शैवाल या सायनोबैक्टीरिया के बीच घनिष्ठ सहोपकारी सम्बन्ध का उदाहरण लाइकेन में देखा जा सकता है। इसी प्रकार कवकों और उच्चकोटि पादपों की जड़ों के बीच कवकमूल (माइकोराइजी) साहचर्य है। कवक मृदा से अत्यावश्यक पोषक तत्वों के अवशोषण में पादपों की सहायता करते हैं, जबकि बदले में पादप कवकों को ऊर्जा-उत्पादी कार्बोहाइड्रेट देते हैं। इसी प्रकार लेग्यूम-राइजोबियम सह-सम्बन्ध है। सहोपकारिता के सबसे शानदार और विकास की दृष्टि से लुभावने उदाहरण पादप-प्राणी सम्बन्ध में पाए जाते हैं। पादपों को अपने पुष्प परागित करने और बीजों के प्रकीर्णन के लिए प्राणियों की सहायता चाहिए। स्पष्ट है कि पादप को जिन सेवाओं की अपेक्षा प्राणियों से है उसके लिए शुल्क तो देना होगा। पुरस्कार अथवा शुल्क के रूप में ये परागणकारियों को पराग और मकरंद तथा प्रकीर्णकों को रसीले और पोषक फल देते हैं।

लेकिन परस्पर लाभकारी तंत्र की ‘धोखेबाजी’ से रक्षा होनी चाहिए, उदाहरण के लिए, ऐसे प्राणी जो परागण में सहायता किए बिना ही मकरंद चुराते हैं। अब आप देख सकते हैं कि पादप-प्राणी पारस्परिक क्रिया में सहोपकारियों के लिए प्राय: सह-विकास’ क्यों शामिल है, अर्थात् पुष्प और इसके परागणकारी जातियों के विकास एक-दूसरे से मजबूती से जुड़े हुए हैं।

अंजीर के पेड़ों के अनेक जातियों में बर्र की परागणकारी जातियों के बीच मजबूत सम्बन्ध है। इसका अर्थ यह है कि कोई दी गई अंजीर जाति केवल इसके साथी’ बर्र की जाति से ही परागित हो सकती है, बर्र की दूसरी जाति से नहीं। मादा बर्र फल को न केवल अंडनिक्षेपण के लिए काम में लेती है; बल्कि फल के भीतर ही वृद्धि कर रहे बीजों को डिंबकों के पोषण के लिए प्रयोग करती है।

अंडे देने के लिए उपयुक्त स्थल की तलाश करते हुए बर्र अंजीर पुष्पक्रम को परागित करती है। इसके बदले में अंजीर अपने कुछ परिवर्धनशील बीज, परिवर्धनशील बर्र के डिंबंकों को, आहार के रूप में देता है।

1. परजीविता 

ये विषमपोषी (heterotrophic) जीव हैं जो अपने परपोषी (host) के शरीर से आहार प्राप्त करते हैं। परजीवी विकल्पी (facultative) तथा अविकल्पी (obligate) दो प्रकार के हो सकते हैं। विकल्पी परजीवी मुख्यत: मृतोपजीवी (saprophytic) होते हैं, जो विशेष परिस्थितियों में ही परजीवी बनते हैं; जैसे-अधिकांश कवक या जीवाणु। अविकल्पी परजीवी, परपोषी (host) के परभक्षी (predator) होते हैं। कुछ आवृतबीजी (angiosperms) भी परजीवी स्वभाव को व्यक्त करते हैं, जैसे- कुस्कुटा (Cuscuta) – पूर्ण तना परजीवी; विस्कम (Viscum) तथा लोरेन्थस (Loranthus) – आंशिक तना परजीवी, रेफ्लेसिया (Rufflesia) और औरोबैकी (Orobanchae) – पूर्ण जड़ परजीवी; सैन्टेलम एलबम (Suntalum album) – आंशिक जड़ परजीवी है।

2. सहभोजिता 

यह दो जीवों के बीच परस्पर सम्बन्ध है जिसमें एक जीव को लाभ होता है और दूसरे जीव को न हानि न लाभ होता है। आम की शाखा पर अधिपादप के रूप में उगने वाला ऑर्किड और हेल की पीठ को आवास बनाने वाले बार्नेकल को लाभ होता है जबकि आम के पेड़ और ह्वेल को उनसे कोई लाभ नहीं होता। पक्षी बगुला और चारण पशु निकट साहचर्य में रहते हैं। सहभोजिता का यह उत्कृष्ट उदाहरण है। जहाँ पशु चरते हैं उसके पास ही बगुले भोजन प्राप्ति के लिए रहते हैं क्योंकि जब पशु चलते हैं तो वनस्पति को हिलाते हैं और उसमें से कीट बाहर निकलते हैं। बगुले उन कीटों को खाते हैं अन्यथा वानस्पतिक कीटों को ढूंढ़ना और पकड़ना बगुले के लिए कठिन होता। सहभोजिता का दूसरा उदाहरण समुद्री ऐनिमोन दंशन स्पर्शक हैं, जिसमें उनके बीच क्लाउन मछली रहती है। मछलियों को परभक्षियों से सुरक्षा मिलती है जो दंशन स्पर्शकों से दूर रहते हैं। क्लाउन मछली से ऐनिमोन को कोई लाभ मिलता हो ऐसा नहीं लगती।

3. सहजीवन 

यह दो भिन्न प्रजाति के जीवों के बीच पाया जाने वाला सहजीवी (Symbiotic) सम्बन्ध होता है। जिसमें दोनों जीवों को परस्पर लाभ (benefit) होता है। कवक और प्रकाश संश्लेषी शैवाल या सायनोबैक्टीरिया के बीच घनिष्ठ सहोपकारी सम्बन्ध का उदाहरण लाइकेन में देखा जा सकता है। इसी प्रकार कवकों और उच्चकोटि पादपों की जड़ों के बीच कवकमूल (माइकोराइजी) साहचर्य है। कवक, मृदा से अत्यावश्यक पोषक तत्त्वों के अवशोषण में पादपों की सहायता करते हैं जबकि बदले में पादप, कवकों को ऊर्जा-उत्पादी काबोहाइड्रेट देते हैं।

अगर समुद्री मछली को अलवणजल (freshwater) की जल-जीवशाला में रखा जाए तो वह परासरणीय समस्याओं के कारण जीवित नहीं रह पाएगी तथा मर जाएगी। तेज परासरण होने के कारण रक्त दाब तथा रक्त आयतन बढ़ जाता है जिससे मछली की मृत्यु हो जाती है।

Diseases are spreading faster due to globalisation and increased movement of people because it is responsible for rapid spreading of infectious diseases.

Four important components of poultry farm management are:

i. Selection of disease-free and suitable breeds

ii. Proper and safe farm conditions

iii. Proper feed and water

iv. Hygiene and health care

Rapid increase in the fishery industry in recent years is called blue revolution. For example, through aquaculture and pisciculture we have been able to increase the production of aquatic plants and animals, both fresh-water and marine. This has led to the development and flourishing of the fishery industry, and it has brought a lot of income to the farmers in particular and the country in general. We now talk about the ‘Blue Revolution’ which is being implemented along the same lines as ‘Green Revolution’.

Bees are the pollinators of many of our crop species such as sunflower, Brassica, apple and pear. Keeping beehives in crop fields during flowering period increases pollination efficiency and improves the yield beneficial both from the point of view of crop yield and honey yield.

25 Primary spermatocytes and 100 primary oocytes will be required for the formation of 100 spermatozoa and ova respectively.

a) Gemmule

b) Bulbil

Phagocytase, monocytes are motile and phagocytic leucocytes. They destroy microbes engulfs and destroys the antigen/microbes and constitute cellular barriers of innate immunity.

The girl is collecting seashells

“The given poem ‘Kindness to animals’ is penned by an anonymous poet. In the poem, the poet is requesting little children to never give pain to all things that feel and live. He pleads with them to let the gentle robin (a small brown bird marked with red on its breast) to come and eat the crumbs of food that we have saved in our home.

If you allow him to feed on the ‘Meat’ (crumbs), he will be greateful to you, and in return for that favour, he will aing a sweet song to cheer you.

Again the poet pleads the children not to hurt the timid hare, peeping franctically from her green grass lair. He requests the children to allow it come and play on the lawn during evening and you can watch it playing and entertain yourself.

The poet gives a vivid imagery of a little lark that goes soaring high into the bright sky. It sings cheerfully as if it was spring everyday. The little lark is so active that he does not get tired of fluttering and singing all day long.

Hence the poet pleads with the children to let him sing his happy song .and to never ‘do wrong’ i.e, never hurt or kill these gentle creatures.

I am a  dog.

12 men can complete a work in 30 days by working 9 hours a day.

\(\therefore\) 1 man can complete a work in 1 day by working (12 × 30 × 9) hours.

Let the number of men required be x.

Then x men can complete 10 times the work in 24 days by working 5 hours a day.

\(\therefore\) 1 man can complete 1 work in 1 day by working \(\frac{X\times24\times5}{10}\) hours

\(\therefore\) 12 × 30 × 9 = \(\frac{X\times24\times5}{10}\)  \(\Rightarrow\) x = 270 men.

(c) 1350 kg

12 cows in 7 days eat 756 kg of grass

12 cows in 1 day eat \(\frac{756}{7}\) kg of grass

(Less days \(\Rightarrow\) Less grass )

1 cow in 1 day eats \(\big(\frac{756}{12\times7}\big)\)kg of grass

 (More days \(\Rightarrow\) More grass)

15 cows in 10 days eat \(\big(\frac{756\times10\times15}{7\times12}\big)\)kg of grass = 1350 kg

(More cows \(\Rightarrow\) More grass)

A’s 1 days’ work = \(\frac{1}{80}\)

A’s 10 days’ work = \(\frac{10}{80}\) = \(\frac{1}{8}\)

Remaining work = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)

\(\frac{7}{8}\) of the work is completed by B in 42 days.

\(\therefore\) The whole work is completed by B in \(\frac{42\times8}{7}\) days = 48 days

\(\therefore\) B’s 1 days’ work = \(\frac{1}{48}\)

(A + B)’s 1 days’ work = \(\frac{1}{80}\) + \(\frac{1}{48}\) = \(\frac{3+5}{240}\) = \(\frac{8}{240}\) = \(\frac{1}{30}\)

\(\therefore\) A and B can complete the whole work together in 30 days.

(b) \(11\frac{2}{3}\) days

20 men working 7 hours a day can complete the work in 10 days.

20 men working 1 hour a day can complete the work in (10×7) days

(Less hours \(\Rightarrow\) More days)

1 man working 1 hour a day can complete the work in (10×7×20) days (Less men \(\Rightarrow\) More days)

1 man working 8 hours a day can complete the work in \(\frac{(10\times7\times20)}{8}\) days

(More hours \(\Rightarrow\) Less days)

15 men working 8 hours a day can complete the work in \(\big(\frac{10\times7\times20}{8\times15}\big)\) days

(More men \(\Rightarrow\) Less days)

= \(\frac{35}{3}\) days = \(11\frac{2}{3}\) days

First man’s 1 days’ work = \(\frac{1}{7}\)

Other man’s 1 days’ work = \(\frac{1}{8}\)

Let the boy complete the whole work in x days. Then

Boys’ 1 days’ work = \(\frac{1}{X}\)

Given, 3\(\big(\frac{1}{7}+\frac{1}{8}+\frac{1}{X}\big)\) = 1

\(\Rightarrow\) \(\frac{1}{7}+\frac{1}{8}+\frac{1}{X}\) = \(\frac{1}{3}\)

\(\Rightarrow\) \(\frac{1}{X}\) = \(\frac{1}{3}-\big(\frac{1}{7}+\frac{1}{8}\big)\) = \(\frac{56-(24+21)}{168}\) = \(\frac{56-45}{168}\) = \(\frac{11}{168}\)

\(\therefore\) The boy can complete the work in \(\frac{168}{11}\) days.

\(\therefore\) Ratio of their shares = Ratio of their one days’ work = \(\frac{1}{7}\): \(\frac{1}{8}\): \(\frac{11}{168}\) = 24 : 21 : 11

\(\therefore\) The boy’s share = \(\frac{11}{56}\) x Rs 1400 = Rs 275

(b) Rs 400

Let C does the same work in x days. Then,

\(3\big(\frac{1}{6}+\frac{1}{8}+\frac{1}{X}\big)\) = 1 \(\Rightarrow\) \(\frac{7}{24}\) + \(\frac{1}{X}\) = \(\frac{1}{3}\)

\(\Rightarrow\) \(\frac{1}{X}\)= \(\frac{1}{3}\)- \(\frac{7}{24}\) = \(\frac{1}{24}\)

\(\therefore\) C will do the same work in 24 days.

The ratio of their one days’ work = \(\frac{1}{6}:\frac{1}{8}:\frac{1}{24}\) = 4 : 3 : 1

The amount to be paid to C

= \(\frac{1}{(4+3+1)}\)x Rs 3200 = Rs 400.

(a) Liverpool is the place where Fogg disembarks from his steamer, and from where he has to catch a train to London. Liverpool is also the place where he is arrested by Fix and thus misses his train. He arranges a special train in an attempt to reach London on time.

(b) Fogg lives in London. The Reform Club, which Fogg frequented on a regular basis, is also in London. Fogg’s accepts the wager in London; his journey starts from London and ends in London. Hence, London is the central place in the novel.

(c) Reform Club is the place which Fogg frequented on a regular basis. It is at the Reform Club that Fogg gets involved in an argument over an article, and where the wager with his fellow club members takes place. It is the place where Fogg was supposed to return before 8.45 p.m. on 21 December, 80 days later. It is the place where his antagonists are waiting anxiously for him, and which he reaches at practically the last second to win his wager.

(d) Phileas Fogg’s residence is in Saville Row. This is also the place where he takes Aouda. Fogg has always stayed quietly at this place. When he returned from his trip around the world and thought he had lost the wager, he remained there so quietly that no one even knew he had returned.

(d) Detective Fix arrested Phileas Fogg in Liverpool thinking that he was a bank robber. However, the real bank robber, James Strand, had been arrested on 17th December at Edinburgh.

Fogg’s residence was in London. The Reform Club, which Fogg frequented on a regular basis and where the allimportant wager took place, is also in London. Fogg’s journey starts from London and ends in London. Hence, London is the central place in the novel, and thus this setting is suited to the theme.

Fogg embarks on his journey to preserve his honour and prove his worth to the men at the Reform Club. He spends nearly all of his money along the way, showing that riches are not what he is truly out for. He is honourable – when he thinks he is penniless, he does not want Aouda to marry him. Aouda, by proposing to him, shows that she is not materialistic. He forgives Passepartout his mistakes. Passepartout shows his loyalty and love for his master at every step.

In the end, when he wins the bet, he divides whatever money is left between Passepartout and Detective Fix, showing that he had no grudges against him. The writer shows that with human effort and willpower enormous obstacles can be overcome. The writer also tells us in the end that Fogg had won something more important than money, by travelling around the world. He had won a charming woman, who made him the happiest of men. The moral at the end is that love and its attainment is more important than all the challenges and money in the world.

05 July The Eco Club of L. R. Senior Secondary School organized Van Mahotsav Programme on 04 July. On this very occasion, many saplings of Ashok, bottle palm, neem and some plants of fruits and flowers were planted. Chief guest Dr R. N. Raj expressed his views on the importance of trees in our life. Our principal said that trees sustain our lives. They reduce the level of pollution and provide many important things. All the members of the club and school staff, students and other invitees were present in the programme. Ankit Kumar, the president of the club delivered the vote of thanks.
Prakashdeep
Secretary, Eco Club

Human life can become perfect by achieving beauty and nobility of character. The beauty and nobility of character can be achieved even in a short span of life. A short spanned worth-while life is more desirable than a meaningless long life.

चरित्र की सुन्दरता और श्रेष्ठता की उपलब्धि से ही मानव-जीवन संपूर्ण हो सकता है। चरित्र की सुन्दरता और श्रेष्ठता कम समय में भी पायी जा सकती है। अर्थपूर्ण गतिविधियों से परिपूर्ण अल्प जीवन भी दीर्घ किन्तु अर्थहीन जीवन से अधिक आनंद देता है।

Respected teachers and dear students. You will be very glad to know that our school has won the National Soft Ball Trophy. The National Soft Ball Tournament was held on the court of Lotus Public School from 4th March 20– to 6th March 20–. Twenty teams of different states participated in the tournament. There were various students to see the matches. Despite several ups and downs, our team qualified for the final and won the National Soft Ball Trophy after defeating the rival team. I feel proud of my team and once again I congratulate all the players of my team.
I thank you all for encouraging us.
Thank you