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This sentence was said by Govinda to Abu Miyan.

This was said by Govinda to Abu Miyan.

This sentence was said by Abbu Miyan to Govinda.

When the clock struck twelve, Bennett, the director of the Earth Herald left the hall and sat down in a rolling armchair. In a few minutes he reached his dining room half a mile away, at the far end of the office where he had arranged to have lunch at the same time with Edith.

1.(iv) I' Mr. Herriot, the narrator of the story. He is a veterinary doctor.

2. (i) He is Tricki, Mrs. Pumphrey's dog. In two weeks, Tricki had transformed from a bloated sausage with legs into a lithe and hard-muscled animal who could keep up well with the pack.

3. (iii) Tricki's coat was golden.

4. (ii) The dogs were playing in the garden at the back of the house.

5. (iv) two weeks

1. (iv) Before admitting to the hospital, Tricki was in a bad shape. But after recovery, he was strong, healthy and active. So, she thinks the dog's recovery is 'a triumph of surgery'.

2. (i) Mrs. Pumphrey was very happy and sentimental to see Tricki recovered, looking strong, healthy and acitive. So, there were tears in her eyes.

3. (iv) She is feeling very grateful.

4. (iii) Tricki's recovery was not an impressive feat as he was simply ill due to overfeeding. However, since Mrs. Pumphrey loved her dog very much, it seemed like a triumph of surgery to her.

5. (ii) her lips were trembling

(b) Adherence to rules and regulations

Physical health is the purity of the body. Unsullied truth is the purity of speech. Unselfish love is the purity of heart. The righteous reason is the purity of thought and wholesome peace is the purity of mind. Unselfish service is the purity of action.

शारीरिक स्वास्थ्य शरीर की पवित्रता है। निष्कलंक सत्य वाणी की पवित्रता है और नि:स्वार्थ प्रेम हृदय की पवित्रता है। न्यायसंगत तर्क विचार की पवित्रता है। पूर्ण शांति मन की पवित्रता है। नि:स्वार्थ सेवा कर्म की पवित्रता है।

According to the author, the person whose mind is pure is pure. The purity of mind makes our vision, words and deeds pure. Such a person has the power to transform a bad man into a good man. 

लेखक के अनुसार वह व्यक्ति पवित्र है जिसका मन पवित्र है। मन की पवित्रता हमारे विचार, वाणी और कर्मों को पवित्र कर देती है। ऐसा व्यक्ति किसी बुरे व्यक्ति को अच्छे व्यक्ति में परिवर्तित करने की शक्ति (क्षमता) रखता है।

Gautam’s ego was purified by a pious lady and a pious man. Gautam was surprised to meet these two persons. They knew all that he did and thought in his mind. So his ego was purified.

एक पवित्र स्त्री और एक पवित्र पुरुष ने गौतम के अहंकार को पवित्र कर दिया। गौतम इन दो व्यक्तियों से मिलकर बहुत प्रसन्न हुए। वे वह सब कुछ जानते थे जो गौतम करते थे और अपने मन में सोचते थे। इस प्रकार गौतम का अहंकार पवित्र हो गया।

This statement suggests that the heart and mind of Pavaharibaba were completely pure. He did not see a thief in the thief. He saw a hungry man in the thief who needed food. He knew that it was his duty to feed a hungry man.

यह कथन यह प्रकट करता है कि पावाहरि बाबा के हृदय और मन पूर्णत: पवित्र थे। उन्होंने चोर में एक चोर (के हृदय) को नहीं देखा। उन्होंने उस चोर में एक भूखा आदमी देखा जिसे भोजन की आवश्यकता थी। वह जानते थे कि भूखे आदमी को भोजन कराना उनका कर्तव्य है।

Gautam Rishi observed rigorous austerities for several years. He did not eat and drink. He stood on his one foot. He controlled his senses and meditated on the chosen deity.

गौतम ऋषि ने कई वर्ष तक कठोर तपस्या की। उन्होंने न खाया, न पिया। वह अपनी एक टांग पर खड़े रहे। उन्होंने अपनी इन्द्रियों पर नियंत्रण किया और इष्ट देवता (अपने इष्ट देवता) का ध्यान किया।

When Pavaharibaba offered more food to the thief, he was wonder-struck at the affectionate words of Baba. There was a change in him. He prostrated himself before the saint and begged his pardon.

जब पावाहरि बाबा ने उस चोर को और अधिक भोजन प्रदान किया तो वह बाबा के स्नेहपूर्ण शब्दों को सुनकर आश्चर्यचकित रह गया। उस चोर में परिवर्तन आ गया। उसने उस सन्त के समक्ष दण्डवत् प्रणाम किया और उनसे क्षमा माँगी।

In order to liquefy a gas, it must be cooled to a certain temperature. This temperature is called critical temperature.

Answer is : D. (0, 3/2) ∪ (3, ∞)

⇒ f(x) = [x(x - 3)]²

⇒ f'(x) = 2[x(x - 3)] = 0

⇒ x = 3 and x = 3/2

When x > 3/2 the function is increasing

x < 3 function is increasing.

⇒ (0, 3/2) ∪ (3, ∞) Function is increasing.

Answer is : A. k > 3

Given f(x) = kx³ - 9x²+ 9x+3

⇒ f’(x) = 3kx² - 18x+9

⇒ f’(x) = 3(kx² – 6x + 3) > 0

⇒ kx² – 6x + 3 > 0

For quadratic equation to be greater than 0. a > 0 and D < 0

⇒ k > 0 and (-6)²- 4(k)(3) < 0

⇒ 36 – 12k < 0

⇒ 12k > 36

⇒ k > 3

∴ k > 3.

Answer is : A. -2

f(x) = x²+kx+1

For increasing

f’(x) = 2x+k

k ≥ -2x

thus,

k ≥ -2.

Least value of -2

Given, a, b, c are in G.P.

∴ \(\frac{b}{a}\) = \(\frac{c}{b}\)

\(\frac{a_{k+1}}{a_k}\) = constant

⇒ b² = ac

Hence proved

Answer is : B. (0, 2)

Given : f(x) = x².e^-x

⇒ f’(x) = 2x. e^-x – x² e^-x

⇒ Put f’(x) = 0

⇒ - (x²– 2x)e^-x = 0

⇒ x = 0 or x = 2.

Now as there is a -ve sign before f’(x)

When x > 2 the function is decreasing

x < 0 function is decreasing

But in the interval (0,2) the function is increasing.

Answer is : D. (1, ∞)

Given:

⇒ f(x) = (x+1)³.(x - 3)³

⇒ f’(x) = 3(x+1)²(x - 3)³ + 3(x - 3)³ (x+1)³

Put f’(x) = 0

⇒ 3(x+1)²(x - 3)³ = -3(x - 3)²(x+1)³

⇒ x - 3 = -(x+1)

⇒ 2x = 2

⇒ x = 1

When x>1 the function is increasing.

x<1 function is decreasing.

So, f(x) is increasing in (1, ∞).

(c) H.P.

a, b, c are in G.P ⇒ b² = ac

⇒ \(\frac{1}{\text{log}_a\,10}\) + \(\frac{1}{\text{log}_c\,10}\) = log₁₀ a + log₁₀ c = log₁₀ (ac)

= = log₁₀ b² = 2 log₁₀ b = \(\frac{2}{\text{log}_b\,10}\)

⇒ \(\frac{1}{\text{log}_a\,10}\), \(\frac{1}{\text{log}_b\,10}\), \(\frac{1}{\text{log}_c\,10}\) are in A.P.

⇒ log_a 10, log_b 10, log_c 10 are in H.P.

We have 5^{\(\frac{1}{2}\)}. 5^{\(\frac{1}{4}\)}. 5^{\(\frac{1}{8}\)} … ∞

= 5[\(\frac{1}{2}\)+ \(\frac{1}{4}\) + \(\frac{1}{8}\) +⋯∞]

= \(5\bigg[\frac{\frac{1}{2}}{1-\frac{1}{2}}\bigg]\)

\(\bigg[\) ∴ \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) + ⋯ ∞ = \(\frac{\frac{1}{2}}{1-\frac{1}{2}}=1\bigg]\)

= 5¹

= 5

Let x = 0.\(\bar{3}\)

x = 0.33333 … ….

x = 0.3 + 0.03 + 0.0003 + 0.0003 + ………

∴ a = 0.3, r = \(\frac{0.03}{0.3}\) = 01 n = ∞

∴ Sum of infinite terms in G.P,S = \(\frac{a}{1-r}\)

∴ x = \(\frac{0.3}{1−0.1}\) = \(\frac{0.3}{0.9}\) = \(\frac{1}{3}\)

∴  0.\(\bar{3}\) = \(\frac{1}{3}\)

(a) 2

H being the harmonic mean between P and Q.

H = \(\frac{2\,PQ}{P+Q}\)

∴ \(\frac{H}{P}+\frac{H}{Q}=\) \(\frac{2\,Q}{P+Q}\) + \(\frac{2\,Q}{P+Q}\) = \(\frac{2\,(P + Q)}{P+Q}\) = 2.

Given,

a = 2 and S_∞ = 6

∴ S_∞ = \(\frac{a}{1-r}\)

∴ 6 = \(\frac{2}{1-r}\)

⇒ 6 − 6r = 2

⇒ r = \(\frac{4}{6}\) = \(\frac{2}{3}\)

(d) S\(\big\{1-\big(1-\frac{1}{S}\big)^{n}\big\}\)

Let the first term of the G.P. be a and common ratio r. 

Given S = \(\frac{a}{1-r}\) ⇒ S = \(\frac{1}{1-r}\)

⇒ 1 - r = \(\frac{1}{S}\) ⇒ r = 1 - \(\frac{1}{S}\)

Let S_n be the sum of first n terms of the series. Then,

S_n = \(\frac{1(1-r^n)}{1-r}\)               (∵ | r | < 1)

= \(\frac{1\big(1-\big(1-\frac{1}{S}\big)^n\big)}{\big(1-\big(1-\frac{1}{S}\big)\big)}\) =  S\(\big\{1-\big(1-\frac{1}{S}\big)^{n}\big\}\)

(c) \(\frac{n(1-x)-x(1-x^n)}{(1-x)^2}\)

S_n = 1 + (1 + x) + (1 + x + x²) + .... n terms

⇒ Sn = \(\frac{1}{1-x}\) [(1 – x) + (1 – x) (1 + x) + (1 – x) (1 + x + x²) + .... to n terms]

= \(\frac{1}{1-x}\) [(1 – x) + (1 – x²) + (1 – x³) + .... to n terms]

= \(\frac{1}{1-x}\) [n – (x + x² + x⁴ + .... to n terms]

= \(\frac{1}{1-x}\) \(\bigg[n-\frac{x(1-x^n)}{1-x}\bigg]\)                    \(\bigg(\because\,S_n = \frac{a(1-r^n)}{1-r}\bigg)\)

= \(\frac{n(1-x)-x(1-x^n)}{(1-x)^2}\)

(c) \(\frac{2}{b}\)

a, b, c are in G.P ⇒ b² = ac                     ...(i)

\(x\) is the A.M. of a, b ⇒ \(x\) = \(\frac{a+b}{2}\)                 ....(ii)

y is the A.M. of b, c ⇒ y = \(\frac{b+c}{2}\)                  ....(iii)

Now \(\frac{1}{x}+\frac{1}{y}\) = \(\frac{2}{a+b}\) + \(\frac{2}{b+c}\) = \(\frac{2(b+c)+2(a+b)}{(a+b)(b+c)}\)

= \(\frac{2(a+2b+c)}{ab+b^2+ac+bc}\) = \(\frac{2(a+2b+c)}{ab+b^2+b^2+bc}\) = \(\frac{2(a+2b+c)}{b(a+2b+c)}\) = \(\frac{2}{b}\).

Work done by the net force during its displacement from x = 0 to x = 2 m is - (b) 50 J

Volume density of charge is charge per unit volume. Its SI unit is Cm^-3.

In a non-uniform electric field an electric dipole experiences both the torque and a net force. 

The magnetic field lines prefer to pass through iron than air because the permeability of iron is much larger than air.

The work done by the cycle on the road is - (d) – 20,000J

(b) the magnetic forces do no work on each particle. 

Iron is a ferromagnetic substance while and copper is diamagnetic, the susceptibility of iron is much larger.

Yes, According to Curie law,x ∝\(\frac{1}{T}\) . X^T= constant is true for paramagnetic substances.

Ceramics are used for coating magnetic tapes in a cassette player or for building stores in a modern computer. Ceramics are specially treated barium-iron oxides and are also called ferrites.

For an electromagnet, the substance should have high retentivity but low coercivity. Soft iron is such a substance.

The sage named Grutsmad discovered the process of writing zero after figures.

Zero was discovered by the great Indian mathematician named Aryabhatta.

Bhaskaracharya discovered signs of addition and subtraction.

Correct option is : (b) 1215

(5 × 3), (15 × 3), (45 × 3), (135 × 3), 

so, next number is (405 × 3) = 1215

Once, during a hunt, Harlhara saw a big rabbit and sent his hunting dog after it. However, the rabbits chased the dog and escaped. While returning from the hunt, Harihara saw a holyman, and narrated the strange incident to him. The holyman was Vidyaranya. The two men went to the place where the rabbit chased the dog. Vidyaranya told him that the place was sacred, and advised him to establish the capital of his new kingdom there.

The battle of Rakkasi Tangadi took place in the year 1565 CE.

The another name is battle of Tallikota.

The twinkling of a star is due to atmospheric refraction of starlight. The atmospheric refraction occurs in a medium of the graduallv changing refractive index. Since the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position. This apparent position of the star is not stationary but keeps on changing slightly, as the physical conditions of the earth's atmosphere are not stationary. Since the stars are very distant, they approximate point-sizes sources of light. As the path of rays of light corning from the star goes on varying slightly, the apparent position of the star fluctuates and the amount of starlight entering the eye flickers-the star sometimes appear brighter, and at some other time, fainter, which is the twinkling effect.

The distance between eve lens and retina is the image distance inside the eye. The image distance is fixed. It cannot be changed at all. Therefore, when we increase the distance of an object from the eye, there is no change in the image distance, inside the eye.

The pupil of our eye contracts.