letter to chacha nehru in 300 words
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Answer:
Jawaharlal Nehru’s Early Life
Jawaharlal Nehru was born on 14th November 1889 in Allahabad (now Prayagraj). His father’s name was Motilal Nehru who was a good lawyer. His father was very rich because of which Nehru got the best education.
At an early age, he was sent abroad for studies. He studied in two universities of England namely Harrow and Cambridge. He completed his degree in the year 1910.
Since Nehru was an average guy in his studies he was not much interested in law. He had an interest in politics. Though he later became a lawyer and practiced law in Allahabad High Court. At the age of 24, he got married to Smt. Kamla Devi. They gave birth to a daughter who was named Indira.
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Jawaharlal Nehru as a Leader
Most Noteworthy, Jawaharlal Nehru was the first Prime Minister of India. He was a man of great vision. He was a leader, politician, and writer too. Since he always India to become a successful country he always worked day and night for the betterment of the country. Jawaharlal Nehru was a man of great vision. Most importantly he gave the slogan ‘Araam Haram Hai’.
Jawaharlal Nehru was a man of peace but he saw how Britishers treated Indians. Due to which he decided to join the freedom movement. He had a love for his country because of which he shook hands with Mahatma Gandhi (Bapu). As a result, he joined the Non-Cooperation movement of Mahatma Gandhi.
In his freedom struggle, he had to face many challenges. He even went to jail many times. However, his love for the country did not get any
Explanation:
Answer:
since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.
=====================
final temperature of the mixture =
= [ m1 * T1 + m2 * T2 ] / (m1 + m2)
= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)
= 3,000 / 100 = 30⁰C
====================
another way using specific heats :
let the final temperature be = T ⁰C
Amount of heat given out by the hot water = m * s * (40⁰C - T)
= 50 gms * s* (40 -T)
Amount of heat taken in by the cold water = m * s * (T - 20⁰C)
= 50 gms * s * (T - 20 )
As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,
50 * s * (40 -T) = 50 gm * s * (T-20)
40 - T = T - 20
2 T = 60 => T = 40 °C
So Your Answer is Option (D) Which is 40 °C
since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.
=====================
final temperature of the mixture =
= [ m1 * T1 + m2 * T2 ] / (m1 + m2)
= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)
= 3,000 / 100 = 30⁰C
====================
another way using specific heats :
let the final temperature be = T ⁰C
Amount of heat given out by the hot water = m * s * (40⁰C - T)
= 50 gms * s* (40 -T)
Amount of heat taken in by the cold water = m * s * (T - 20⁰C)
= 50 gms * s * (T - 20 )
As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,
50 * s * (40 -T) = 50 gm * s * (T-20)
40 - T = T - 20
2 T = 60 => T = 40 °C
So Your Answer is Option (D) Which is 40 °C
since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.
=====================
final temperature of the mixture =
= [ m1 * T1 + m2 * T2 ] / (m1 + m2)
= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)
= 3,000 / 100 = 30⁰C
====================
another way using specific heats :
let the final temperature be = T ⁰C
Amount of heat given out by the hot water = m * s * (40⁰C - T)
= 50 gms * s* (40 -T)
Amount of heat taken in by the cold water = m * s * (T - 20⁰C)
= 50 gms * s * (T - 20 )
As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,
50 * s * (40 -T) = 50 gm * s * (T-20)
40 - T = T - 20
2 T = 60 => T = 40 °C
So Your Answer is Option (D) Which is 40 °C
since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.
=====================
final temperature of the mixture =
= [ m1 * T1 + m2 * T2 ] / (m1 + m2)
= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)
= 3,000 / 100 = 30⁰C
====================
another way using specific heats :
let the final temperature be = T ⁰C
Amount of heat given out by the hot water = m * s * (40⁰C - T)
= 50 gms * s* (40 -T)
Amount of heat taken in by the cold water = m * s * (T - 20⁰C)
= 50 gms * s * (T - 20 )
As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,
50 * s * (40 -T) = 50 gm * s * (T-20)
40 - T = T - 20
2 T = 60 => T = 40 °C
So Your Answer is Option (D) Which is 40 °C