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सही विकल्प है (a) ओडिशा।  

परमाणु ऊर्जा आयोग की स्थापना अगस्त 1948 में की गई।

खनिज तेल (पेट्रोलियम) की उत्पत्ति – खनिज तेल टर्शियरी युग की बालू और चूने की अवसादी शैलों में उसी तरह विद्यमान रहता है जैसे स्पंज में जल। करोड़ों वर्षों तक बड़ी मात्रा में कीचड़, मिट्टी और बालू आदि में वनस्पति एवं जीवों के दबे रहने, उन पर गर्मी, दबाव, रसायन, जीवाणु और रेडियो-सक्रियता आदि क्रियाओं के प्रभाव के फलस्वरूप खनिज तेल की उत्पत्ति होती है।

सही विकल्प है (a) झारखण्ड।

सही विकल्प है (a) सन् 1954

In 1893, Gandhi went to South Africa as the legal advisor of a Gujarati Business Firm. He spent two decades there. During that time he became the undisputed leader of the Indians there. He led the fight against the racial discrimination policy of the South African government.

Historian Devanesan says that Gandhiji was made in South Africa because of the following :

(a) It was here that Gandhi formulated his non-violent method of protest was known as satyagraha.

(b) It was here that Gandhi tried to encourage religious tolerance.

(c) It was here that he demanded to end the discrimination against women and lower classes

Answer June 3 Plan

(a) सन् 1956

By 1922, Gandhi was able to make the Indian Nationalism a highly popular Movement. Until then it was a Movement of intellectuals and professionals. But soon, with the efforts of Gandhi, thousands of farmers, artisans and workers joined the Movement. Many of them started calling Gandhi ‘Mahatma’ showing him their respect. He was not like other leaders who preferred to keep some distance from people. He sympathized with them and became one of them. He lived like them, and dressed like them. He also spoke in their language.

Khan Abdul Ghaar Khan

Gandhi began his speech by severely criticizing the upper classes of India. He accused them of neglecting the poor working-class people. He praised the glorious inaugural function of the Banaras Hindu University. He then talked about the disparity between the people fortunate enough to attend the function and the millions of people who had no chance of attending such functions. He told the rich and the specially invited guests to contribute their ornaments for the welfare of the poor people in India.

He then told them that India won’t be really free when there is such a huge disparity between the rich and the poor. He added that by exploiting the work of the peasants or helping in such exploitation, autonomous government or freedom will be meaningless. Only through farmers, India can be free. Advocates, doctors, the rich and landowners won’t bring us freedom. The inauguration of the Banaras Hindu University was a time for celebration.

The University was established by using the money and efforts of Indians. Gandhi, in his speech, was trying to show the absence of farmers and workers, who formed the vast majority of Indians, on such an important occasion. His Banaras speech was to show that the Indian National Movement had become a movement of the upper-class people.

Gandhiji’s first Satyagraha was that of Champaran in Bihar. The indigo farmers there were exploited by the European estate owners. The farmers were forced to cultivate indigo in place of other crops. Then they were forced to sell the indigo at the prices determined by the estate owners. In 1917, Gandhi went to Champaran and studied in detail about the pathetic situation of the farmers there. Authorities asked Gandhi to go away from the place but he did not obey. He continued his research. He wanted to ensure that the farmers could cultivate what they wanted and not what others wanted. Ultimately, the government appointed a Commission. The Commission approved most of the demands of the farmers. Gandhi succeeded in his first attempts at Satyagraha.

In 1918, Gandhiji led two protest movements. One was in Ahmedabad and the other was in Kheda. Gandhi interfered in a dispute between the workers and the owners of a cloth mill in Ahmedabad. In 1919, the workers embarked on a strike demanding increased wages. Gandhi took up this issue.

He started a fast unto death demanding increased wages to the workers and better working conditions for them. The mill owners were ready to negotiate. The wages of the workers were increased by 35%.

In Kheda Gandhi fought for the farmers. Because of a serious draught, farmers in Kheda had a serious problem as their crops were damaged. Gandhi told the farmers not to give taxes until some reductions in the taxes were made. Finally, the government approved the demand of the farmers and Gandhi ended his satyagraha.

• The Champaran, Kheda and Ahmedabad satyagaha of Gandhi were a preparation for his entry into Indian National Movement. These activities helped him in understanding the problems of the people.
• They helped Gandhi to have close contacts with people. The sympathy he showed to the poor people helped him to get a place in their hearts.

All the early struggles of Gandhiji were local ones. Soon the British themselves gave him a chance to have a bigger platform for his working. Some incidents in 1919 led Gandhi to be in the forefront of the Indian National Movement.

Lahore Meeting did the Congress declare Poorna Swaraj

Gandhiji did not take part in the Independence Day Celebrations that took place on August 15, 1947. He was then in Calcutta. He did not take part in any celebration or raise the Indian National Flag. Instead, he fasted for 24 hours. His was a life or death struggle for independence. But when freedom came, a big price had to be paid. The country was divided into two. The Hindus and Muslims murdered one another. Gandhiji did not want such freedom. B.G. Tendulkar, the biographer of Gandhiji, tells us how he was working during the Partition period. In September and October Gandhi was visiting hospitals and refugee camps comforting the suffering people there. He exhorted the Hindus, Sikhs and Muslims to let bygones be bygones, to forget and forgive, and live in peace and harmony.

The Annual Meeting of the Congress was held in Lahore in December 1929. It was a historic meeting.

Here are the reasons for its importance. 

1. Jawaharlal Nehru was elected as the President of the Congress. It was an indication that the leadership of the Congress was going into young hands.

2. It passed a resolution saying that the ultimate aim of the Congress was ‘Poorna Swaraj’.

3. It decided to celebrate 26 January 1930 all over India as the Day of Independence.

4. It decided to embark on Civil Disobedience Movement under Gandhiji.

The Dandi March opened the eyes of the British. They realized the need to give more representation to the Indians in the government. With this aim, the British government called for Round Table Conferences in London. The first Round Table Conference (RTC) was summoned when the Civil Disobedience Movement was going on. It was in 1930. But it was boycotted by all important political leaders of India. Congress also boycotted it. Thus the first RTC was without any use.

In September 1931, the 2nd RTC was summoned in London, Gandhi said Congress would represent India. But this was objected by 3 groups-Muslim League, the local kings and Ambedkar.

The Muslim League claimed that it represented the interests of the Muslims.

The local kings said that in their kingdoms, Congress, had no support.

B.R. Ambedkar, who was a lawyer and thinker, said that Congress did not represent the lower class people.

In November 1932, the British Government summoned the 3rd RTC. Congress representatives did not attend this Conference. This Conference formulated certain principles regarding the future Indian Constitution. It was on the basis of these that the India Act of 1935 was passed.

(a) The India Act of 1935 marked the beginning of the Representative government in India. There were basic changes in the Provinces. It ended dual administration and brought in provincial autonomy. 

(b) This Act allowed the right to vote in a limited manner. In 1937, there were elections to the Provinces. Congress got big victories.

In 8 of the 11 Provinces, Congress was able to form ministries. Even then Congress Ministers were to work under the British Governor’s supervision and control.

Gandhi was very critical of machines saying that they made people their slaves and they took away employment opportunities of people. He objected to the extensive use of machines and technology. He saw the spinning wheel as a symbol of humanity. He though that the spinning wheel brought extra income to people and made them self sufficient and self- reliant.

Autobiographies are a great source for formulating history. They give us a description of the past. When we read and interpret autobiographies, we ought to be careful. Autobiographies talk of past things. They are written from memories. Writers of autobiography would want readers to evaluate their lives in a particular way.

• Autobiography means that a writer is presenting a picture of himself for the readers to evaluate him. Most often writers of autobiographies don’t write those things that will harm their good image. They might also exaggerate the good things.
• Therefore we have to look for things the writers have not said in their autobiography. We should also find the reason for their silence on certain issues. We should know if the silence was deliberate or accidental.
• However, the “The Story of My Experiments with Truth”, the autobiography of Gandhi, overcomes some of these limitations of common autobiographies.

(A) Acharya Vinoba Bhave

B) were not benefitted much

Herbert Macaulay founded the first Nigerian political party won consecutive elections.

The army of Bengal was the best army of the East India Company. A majority of the soldiers in the army belonged to Avadh. Lord Dalhousie annexed Avadh into the British Empire. Avadh soldiers did not like it and turned against the British. The English disbanded the Avadh army as a result of which thousands of soldiers became unemployed. They decided to rise in revolt in protest.

Maharashtra .

25th April, 1809 A.D.

Sanyasis did pilgrimages regularly along agriculture. They become annoyed due to ban on visiting pilgrimage places.

Given:

 \(\frac{2{\text{x}}-3}{3{\text{x}}-7}\)< 0, x ∈ R

Signs of 2x – 3:

2x – 3 = 0 → x = \(\frac{3}{2}\)

(Adding 3 on both the sides and then dividing both sides by 2)

2x – 3 < 0 → x > \(\frac{3}{2}\)

(Adding 3 on both the sides and then dividing both sides by 2)

Signs of 3x – 7:

3x – 7 = 0 → x = \(\frac{7}{3}\)

(Adding 7 on both the sides and then dividing both sides by 3)

3x – 7 < 0 → \(x< \frac{7}{3}\)

(Adding 7 on both the sides and then dividing both sides by 3)

3x – 7 > 0 → \(x> \frac{7}{3}\)

(Adding 7 on both the sides and then dividing both sides by 3) 

Zeroes of denominator: 

3x – 7 = 0

x = \(\frac{7}{3}\)

(Adding 7 on both the sides and then dividing both sides by 3) 

Interval that satisfies the required condition: < 0

\(\frac{3}{2}< x < \frac{7}{3}\)

Given :

\(\frac{3}{{\text{x}}-2}<2\), x ∈ R

Subtracting 2 from both the sides in the above equation,

\(\frac{3}{{\text{x}}-2}-2\)<2-2

\(\frac{3-2({\text{x}}-2)}{{\text{x}}-2}<0\)

\(\frac{3-2{\text{x}}+4}{{\text{x}}-2}\) <0

\(\frac{7-2{\text{x}}}{{\text{x}}-2} <0\)

Signs of 7 – 2x:

7 - 2x = 0 → x = \(\frac{7}{2}\)

(Subtracting by 7 on both the sides, then multiplying by -1 on both the sides and then dividing both the sides by 2)

 7 - 2x < 0 → x > \(\frac{7}{2}\)

(Subtracting by 7 on both the sides, then multiplying by -1 on both the sides and then dividing both the sides by 2)

7 – 2x > 0 → x < \(\frac{7}{2}\)

(Subtracting by 7 on both the sides, then multiplying by -1 on both the sides and then dividing both the sides by 2)

Signs of x – 2: 

x – 2 = 0 → x = 2 (Adding 2 on both the sides) 

x – 2 < 0 → x < 2 (Adding 2 on both the sides) 

x – 2 > 0 → x > 2 (Adding 2 on both the sides)

Zeroes of denominator:

x – 2 = 0 → x = 2

At x = 2, \(\frac{7-2{\text{x}}}{{\text{x}}-2} \) is not defined

intervals satisfying the condition: <0

x<2 and x > \(\frac{7}{2}\)

Therefore,

x ∈ (-∞,2) U \(\big(\frac{7}{2}, ∞\big)\)

Given:

 \(\cfrac{{\text{x}}-7}{{\text{x}}-2}\)\(\ge\) 0, X ∈ R

 \(\cfrac{{\text{x}}-7}{{\text{x}}-2}\)\(\ge\) 0

Signs of x – 7: 

x – 7 = 0 → x = 7(Adding 7 on both the sides) 

x – 7 > 0 → x > 7 (Adding 7 on both the sides) 

x – 7 < 0 → x < 7 (Adding 7 on both the sides) 

Signs of x – 2:

x – 2 = 0 → x = 2 (Adding 2 on both the sides) 

x – 2 > 0 → x > 2 (Adding 2 on both the sides) 

x – 2 < 0 → x < 2 (Adding 2 on both the sides) 

Zeroes of denominator:

x – 2 = 0 → at x = 2 \(\cfrac{{\text{x}}-7}{{\text{x}}-2}\) will be undefined

intervals that satisfy the required condition: \(\ge\) 0

x < 2 or x = 7 or x >7

Therefore,

x є (-∞, -2) υ [7, ∞)

(C) x + y = 7

(a) Take x + 2y, on putting x = 5 and y = 2, we get
5 + 2(2) = 5 + 4 = 9 ≠ 7.

So, (5, 2) is not a solution of x + 2y = 7

(b) Take 5x + 2y, on putting x = 5 and y = 2, we get
5 x 5 + 2 x 2 = 25 + 4 = 29 ≠ 7

So, (5, 2) is not a solution of 5x + 2y = 7.

(c) Take x + y, on putting x = 5 and y = 2, we get 5 + 2 = 7

So, (5,2) is a solution of x + y = 7.

(d) Take 5x + y, on putting x = 5 and y = 2, we get
5 × 5 + 2 = 25 + 2 = 27 ≠ 7

So, (5, 2) is not a solution of 5x + y = 7.

\(\frac{1}{{\text{x}} - 2}<0\)

We have to find values of x for which \(\frac{1}{{\text{x}} - 2}\) is less than  zero that is negative

Now for \(\frac{1}{x - 2}\) to be negative x - 2 should be negative that is x - 2 < 0

⇒ x – 2 < 0 

⇒ x < 2

Hence x should be less than 2 for \(\frac{1}{{\text{x}} - 2}<0\)

x < 2 means x can take values from -∞ to 2 hence x ∈ (-∞, 2) 

Hence the solution set for \(\frac{1}{{\text{x}} - 2}<0\) is (-∞, 2)

\(\frac{|{\text{x}}-2|}{({\text{x}}-2)}\) < 0 means we have to find values of x for which \(\frac{|{\text{x}}-2|}{({\text{x}}-2)}\) is negative Observe that the numerator |x – 2| is always positive because of mod, hence for \(\frac{|{\text{x}}-2|}{({\text{x}}-2)}\) to be a negative quantity the denominator (x - 2) has to be negative

 That is x - 2 should be less than 0

⇒ x – 2 < 0

⇒ x < 2

Hence x should be less than 2 for\(\frac{|{\text{x}}-2|}{({\text{x}}-2)}\) < 0

x < 2 means x can take values from -∞ to 2 hence x ∈ (-∞, 2)

Hence the solution set for  \(\frac{|{\text{x}}-2|}{({\text{x}}-2)}\) < 0 is (-∞, 2)

\(\frac{|{\text{x}} -3|}{{\text{x}} -3}\) < 0 , x ϵ R.

|x - 3| < 0

The above condition can’t be true because the absolute value cannot be less than 0 

Therefore, 

There is no solution for x є R.

Given:

 \(\frac{|{\text{x}} |-1}{|{\text{x}} |-2}\) ≥0 x ϵ R. –{–2, 2}

Intervals of |x|:

x ≥ 0, |x| = x and x < 0, |x| = -x

Domain of \(\frac{|{\text{x}} |-1}{|{\text{x}} |-2}\) ≥ 0

\(\frac{|{\text{x}} |-1}{|{\text{x}} |-2}\) is not defined for x = -2 and x = 2

Therefore, Domain: x < -2 or -2 < x < 2 or x > 2 

Combining intervals with domain: 

x < 2, -2<x<0, 0≤x<2, x≤2

For x < -2:

\(\frac{|{\text{x}} |-1}{|{\text{x}} |-2}\) = \(\frac{-{\text{x}} -1}{-{\text{x}} -2}\)

\(\frac{-{\text{x}} -1}{-{\text{x}} -2}\) \(\ge\)0

Signs of – x – 1: 

-x -1 = 0 → x = -1 

(Adding 1 to both the sides and then dividing by -1 on both the sides) 

-x – 1> 0 → x < -1 

(Adding 1 to both the sides and then multiplying by -1 on both the sides) 

-x – 1 < 0 → x > -1 

(Adding 1 to both the sides and then multiplying by -1 on both the sides) 

Signs of – x – 2: 

-x -2 = 0 → x = -2 

(Adding 2 to both the sides and then dividing by -1 on both the sides) 

-x – 2> 0 → x < -2 

(Adding 2 to both the sides and then multiplying by -1 on both the sides) 

-x – 2 < 0 → x > -2 

(Adding 2 to both the sides and then multiplying by -1 on both the sides) 

Intervals satisfying the required condition: ≥ 0 

x < - 2 or x = -1 or x > -1 

Merging overlapping intervals: 

x < -2 or x ≥ -1 

Combining the intervals: 

x < -2 or x ≥ -1 and x < -2 

Merging overlapping intervals: 

x < -2 

Similarly, for -2 < x < 0:

 \(\frac{|{\text{x}} |-1}{|{\text{x}} |-2}\) = \(\frac{-{\text{x}} -1}{-{\text{x}} -2}\)

 \(\frac{-{\text{x}} -1}{-{\text{x}} -2}\) \(\ge\)0

Therefore, 

Intervals satisfying the required condition: ≥ 0 

x < - 2 or x = -1 or x > -1 

Merging overlapping intervals: 

x < -2 or x ≥ -1 

Combining the intervals: 

x < -2 or x ≥ -1 and -2 < x < 0 

Merging overlapping intervals: 

-1 ≤ x < 0 

For 0 ≤ x < 2,

 \(\frac{|{\text{x}} |-1}{|{\text{x}} |-2}\) = \(\frac{-{\text{x}} -1}{-{\text{x}} -2}\)

\(\frac{-{\text{x}} -1}{-{\text{x}} -2}\) \(\ge\)0

Signs of x – 1: 

x – 1 = 0 → x = 1(Adding 1 to both the sides) 

x – 1 > 0 → x > 1(Adding 1 to both the sides) 

x – 1 < 0 → x < 1(Adding 1 to both the sides) 

Signs of x – 2: 

x – 2 = 0 → x = 2(Adding 2 to both the sides) 

x – 2 < 0 → x < 2(Adding 2 to both the sides) 

x – 2 > 0 → x > 2(Adding 2 to both the sides)

At x = 2, ^{\(\cfrac{x-1}{x-2}\)} is not defined

Intervals satisfying the required condition: ≥ 0 

x < 1 or x = 1 or x > 2 

Merging overlapping intervals: 

x ≤ 1 or x > 2 

Combining the intervals: 

x ≤ 1 or x > 2 and 0 ≤ x < 2 

Merging overlapping intervals: 

0 ≤ x ≤ 1 

Similarly, 

for x > 2:

 \(\frac{|{\text{x}} |-1}{|{\text{x}} |-2}\) = \(\frac{-{\text{x}} -1}{-{\text{x}} -2}\)

\(\frac{-{\text{x}} -1}{-{\text{x}} -2}\) \(\ge\)0

Therefore, Intervals satisfying the required condition: ≥ 0 

x < 1 or x = 1 or x > 2 

Merging overlapping intervals: 

x ≤ 1 or x > 2

Combining the intervals: 

x ≤ 1 or x > 2 and x > 2 

Merging overlapping intervals: 

x > 2 

Combining all the intervals: 

x < -2 or -1 ≤ x < 0 or 0 ≤ x ≤ 1 or x >2 

Merging the overlapping intervals: 

x < -2 or -1 ≤ x ≤ 1 or x > 2 

Therefore

x ϵ (-∞, -2) Ս [-1,1] Ս (2, ∞)

Given: 

X-4 > 1, x ≠ 4. Adding 4 to both the sides in above equation 

x – 4 + 4 > 1 + 4 

x > 5 

Therefore, x є (5, ∞)

False

Given, x is an even number. Then, the next even number is (x + 2).

False.

Given, (15/8) – 7x = 9

Transposing 15/8 to RHS it becomes – (15/8)

– 7x = 9 – (15/8)

False.

Given, (x/3) + 1 = (7/15)

Transposing 1 to RHS it becomes – 1

(x/3) = (7/15) – 1

(x/3) = (7 – 15)/15

(x/3) = -8/15

False.

Given, x/11 = 15

Multiplying both LHS and RHS by 11, we get

(x/11) × 11 = 15 × 11

x = 165

True.

Given, 6x = 18

Multiplying both LHS and RHS by 3, we get

6x × 3 = 18 × 3

18x = 54

If (2/5)x – 2 = 5 – (3/5)x, then x = 7

Given,

(2/5)x – 2 = 5 – (3/5)x

Transposing -2 to RHS and it becomes 2 and (3/5)x to LHS it becomes –(3/5)x.

(2/5)x + (3/5)x = 5 + 2

(2x + 3x)/5 = 7

5x = 7 × 5

x = 35/5

x = 7

(x/5) + 30 = 18 has the solution as -60.

Given, (x/5) + 30 = 18

Transposing 30 to RHS and it becomes -30.

(x/5) = 18 – 30

(x/5) = -12

x = -12 × 5

x = -60

9 is subtracted from the product of p and 4, the result is 11. The value of p is 5.

From the question, it is given that,

9 is subtracted from the product of p and 4, the result is 11 = 4p – 9 = 11

4p – 9 = 11

Transposing -9 to RHS and it becomes 9.

4p = 11 + 9

4p = 20

P = 20/4

P = 5

When a number is divided by 8, the result is –3. The number is -24.

Let the number be x,

Then,

x/8 = -3

x = -3 × 8

x = -24

The share of A when Rs 25 are divided between A and B so that A gets Rs. 8 more than B is Rs 16.50.

Let us assume B share be x

As per the condition in the question A share be x + 8

Then,

x + (x + 8) = 25

x + x + 8 = 25

2x + 8 = 25

2x = 25 – 8

2x = 17

x = 17/2

x = 8.5

So, A gets x + 8

= 8.5 + 8

= Rs 16.5

आवेशितं कण अविचलित सरल रेखीय गति करता है, इसका यह अर्थ है कि कण पर चुम्बकीय क्षेत्र के कारण कोई बल नहीं लगा है। इससे प्रदर्शित होता है कि कण का प्रारम्भिक वेग या तो चुम्बकीय क्षेत्र की दिशा में है अथवा उसके विपरीत है।

LHS means left hand side expression in a equation.

(b) -13/19

Given, (5x/3) – 4 = (2x/5)

(5x/3) – (2x/5) = 4

LCM of 3 and 5 is 15

(25x – 6x)/15 = 4

19x = 4 × 15

19x = 60

x = 60/19

Then, Substitute the value of x in 2x -7

= (2 × (60/19)) – 7

= (120/19) – 7

= (120 – 133)/19

= – 13/19

A term of an equation can be transposed to the other side by changing its sign.

For example:- 2x + 3 = 0

Transposing 3 to RHS and it becomes -3

2x = -3

x = -3/2