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Given two circle radii are 5cm and 12cm
Then,
Area of circle of radius 5 cm =πr²=π(5)²
 =25π sq cm
Area of circle of radius 12 cm =πr²=π(12)² =144πsq cm
So area of circle whose area equal to sum of rheas two circle=25π+144π=169π sq cm
Let the radius of the resultant =R cm
∴πR²=169π
⇒R²
 =169
⇒R=13cm

i) a₃ = a + 2d = 4 …… (i) and 

a₉ = a + 8d = –8 ….. (ii) 

Subtracting (i) from (ii), we get 

6d = –12 ⇒ d = –2 

d = –2 and a₃ = a + 2d = 4 

⇒ a = 4 – 2(–2) = 4 + 4 = 8 

So, the AP is 8, 6, 4, 2, 0, -2,…. 

ii) Here, a = 8 and d = –2 and a_n = –160 

Also, a_n = a + (n-1)d 

⇒ –160 = 8 + (n-1)(–2) 

⇒ –160 – 8 = –2n+2 

⇒ n = 85

\(\frac{3x +4}{2x-5} = \frac{9}{4}\) 

4(3x+4) = 9(2x-5)

12x + 16 = 18x - 45

12x - 18x = -45 - 16

-6x = -61

x = 61/6

Given:-

In an AP, if the third term and seventh term are 4 and 8 respectively.

To find:-

Find which term value is zero ?

Third term of the given AP = 4

t₃ = 4

a + 2d = 4 ------(1)

Seventh term of the given AP = 8

t₇ = 8

a + 6d = 8 ------(2)

On solving (1) & (2)

a + 2d = 4

a + 6d = 8

(-)

_________

0 - 4d = -4

_________

=> -4d = -4

=> 4d = 4

=> d = 4/4

=> d = 1

Common difference of the AP = 1

On Substituting the value of d in (1)

=> a+2(1) = 4

=> a+2 = 4

=>a = 4-2

=> a = 2

First term of the AP = 2

Then the AP :2,3,4,5,6,7,8....

Let nth term of the AP will be zero

=> t_n = 0

=> a+(n-1)d = 0 

=> 2+(n-1)(1) = 0

=> 2+(n-1) = 0

=> 2+n-1 = 0

=> n+1 = 0

=> n = -1

Number of terms = -1

n = -1 , but n can not be negative

So there is no such term in the given AP can not be equal to zero.

And also It is an increasing series (AP)

15/ 1600

Factorizing the denominator, we get,

1600 = 2⁶5²

Since, the denominator is in the form of 2^m × 5ⁿ

Thus 15/1600 has a terminating decimal expansion.

We know that the length of an arc of a sector of angle θ = \(\dfrac{2\pi r \theta }{360} \) 

Using this formula, we get 

\(\dfrac{2\pi r \theta }{360} \) = 2π 

⇒ 6θ/ 360 = 1

∴  6θ = 360 ∴ θ = 60°

Hence, θ/30 = 60°/30° = 2°

Molar mass of Sodium sulphate (Na₂​​​​​​SO₄)-

= 2(23)+32+4(16)

= 46+32+64 

= 142 g/mol

Number of moles of Na​​​​​​₂SO₄ in 120 grams of it:

No. of moles = Molecular mass/ Molar mass

No. of moles = 120/142 = 0.845 moles

Now, in each mole of sodium sulphate, there are 2 moles of Sodium (Na). Therefore, 0.845×2 = 1.69 moles of sodium are present in 120 grams of sodium sulphate.

Na₂SO₄(sodium sulphate)

Molecular weight - 142 g/mol

Number of moles of sodium sulphate = \(\frac{120g}{142g/mol} = 0.84 mole\) 

\(\because\) 1 molecule of Na₂SO₄ contains  = 2 atom of sodium

\(\therefore\) 0.84 mole of Na₂SO₄ contain = 1.68 moles sodium atom.

Hence, 120 g Na₂SO₄ contain 1.68 moles sodium atom

The acceleration due to gravity does not vary with the change in the mass of the object as it is a dimensionless quantity representing the amount of matter in a particle or object.

Zinc is an essential trace element commonly found in red meat, poultry, and fish. It is necessary in small amounts for human health, growth, and sense of taste.

India is the 7^th largest country in area and the second-most populous country in the world. 

The special features of Indian history are:

a. Continuity of civilization and culture: India has one of the earliest histories in the world. The physical features of our country, full of variety, richness and contrasts tend to divide India into different local zones. However, it has 4000 years of continuous history and continuity of civilization and culture, like China.

b. Evolution in phases: Its has developed in various stages with necessary improvements. We find a connecting link of events from the Indus to the Vedic period, Vedic to Islamic and Christian influences.

c. Foreign invasions: The natural barriers on the frontiers of India provided security from foreign invasions. However, foreigners like Greeks, Persians, Huns, Shakas, Arabs, Turks, Kushans, Afghans, and others entered India from the Khyber and Bolan passes. All these invaders contributed to the Indian culture.

The historical monuments and other structures like forts built by these invaders are attracting tourists even today. South India had immunity from such invasions and developed a distinct culture of its own.

d. Religious tolerance (dominant and tolerant Hindu faith): India is home for Hindus, Jains, Buddhists, Sikhs, Muslims, Parsis, Christians, and several tribal faiths and practices. Indians believe in the concept of ‘Vasudhaivaka kutumbakam’ and ‘Sarve janaha sukhino bhavantu’, which means that the whole world is one family and let all the people be happy.

e. Indian contributions to the world: India has contributed immensely in the fields of literature, philosophy, science, art, culture, architecture, mathematics, medicine, astronomy, etc. UNO has recognized more than 30 Indian historical sites as centers of world heritage, such as the hill forts of Rajasthan, Khujaraho, Konark, Taj Mahal, Bodh Gaya, Sanchi, Ajanta, Ellora, Hampi, Aihole, Pattadakallu, Madurai, Kanchi, Churches of Goa, etc.

Yoga, Ayurveda, and other artistic specimens are the special contributions of Indians to the world. The great contributions of Indian mathematicians have enriched the world with the concept of zero and the decimal system.

The ancient universities of Nalanda, Takshashila, Ujjain, Prayag, Vikramshila, Kashi and Kanchi attracted students from different countries of the world. India was at the height of its intellectual and spiritual glory.

f. Unity in diversity: India possesses diverse physical and geographical features and also shows diversity racially, linguistically, socially, economically, religiously and almost in every sphere of human activities. In spite of all these diversities, there are many unifying forces that have kept India united.

1. Social conditions: The early vedic people developed a highly organised society, that was based on the principle of monogamy. Polygamy was practiced-only among the royal families. The eldest male member was the head of the family and was called ‘Kulapathi’ or ‘Grihapathi’.

There was no system of child marriage but widow remarriage prevailed. Marriage was considered a sacred bond and after marriage the bride lived in the house of the bridegroom. Usually a joint family system prevailed among the Aryans.

2. Social divisions: The social divisions, chati vamas were based on professions. They were Brahmana, Kshatriya, Vaishya, and Sudra. People could change professions and hence change their vamas. Thus, there was mobility among the vamas.

3. Position of the women : The status of women in the family and in the society was high and they had equal rights with men. Women were educated and highly civilized for e.g.: Gargi, Maithreyi, Apala, Ghosha, Vishwavara, and others. Girls had considerable freedom in selecting their life partners. Women freely moved out of their houses and attended public functions. A high standard of morality was maintained.

4. Food and entertainment: People consumed wheat, barley, rice, fruit, vegetables, fish and meat and intoxicating drinks like soma and sura. Aryans wore clothes made of cotton and wool. Ornaments were used by both men and women, made of gold, silver, and flowers. Gambling, chariot and horse racing, hunting and dance were the popular entertainments. Education on the whole was oral. It aimed at the development of character and was religions in nature.

During the later vedic period, polygamy and polyandry came into practice. Patriarchal system still continued, and the joint family system was quite common. Women were still allowed to get higher education and participate in the religious rites. But the women were now under the protection of father or husband or a son. On the whole, position of the women had considerably come down.

Varnas turned into many castes. Caste system became hereditary and very rigid. Brahmanas and Kshatriyas enjoyed a higher status compared to Vaishyas and Shudras. Life of an individual was divided into four stages called ashramas.

They were Brahmacharya, Grihastha, Vanaprastha, and Sanyasa. Education was imparted by learned teachers to the students. The aim of education was to develop knowledge, character, truthfulness and devotion. Gurus enjoyed great respect.

Living standard of the people was usually the same as it was in the early vedic civilization. People still lived in villages and small towns. Agriculture was the main profession of the people.

Multiple fission is the reproductive cycle in which several individuals are formed or created out of the parent cell. The nucleus repeatedly divides in this method to create a huge number of nuclei. Every nucleus absorbs a little amount of cytoplasm and forms a membrane around each structure. All the daughter cells are similar and identical in dimension. The organism responsible for the multiple fission is Plasmodium.

Buddacharita was written by Ashwagosha.

Anglo-Mysore wars (1767-1799) :

1. The first Anglo-Mysore war (1767-1769): The British after establishing supremacy in Bengal, waged war against Mysore to expand their Empire. Tippu had participated in his father’s campaigns and had gained sufficient military experience. In 1766, he fought against the Paliagars of Balam. In 1767-1769, in the first Anglo-Mysore war, he took his army towards Madras. Later, he helped his father capture the forts of Tirupattur and Vaniyambadi.

2. The second Anglo-Mysore war (1780-1784): Hyder Ali died in 1782. His son Tippu Sultan continued the war. Tippu defeated the British at Wandiwash in 1783, and marched against Mangalore and besieged the fort. Negotiations for peace started between Tippu and British through signing the treaty of Mangalore in 1784. and the second Anglo-Mysore war ended with that.

3. Third Anglo-Mysore war (1790-1792) : The third Anglo-Mysore war was again fought between Tippu Sultan and the British. Tippu’s rise caused fear and jealousy among the Britishers. Tippu was trying to get the help of the French to expel the British from India. War broke out with Tippu’s unprovoked attack on Travancore in 1789, whose ruler was an ally of the British. British Governor-General, Lord Cornwallis was waiting for a pretext to wage a war against Tippu.

He formed a coalition consisting of the British, the Nizam and the Maratbas against Tippu, and attacked Sirangapattana. Tippu could not fight this combined army arid he began to lose ground. They besieged his capital Srirangapattana in 1792. Forced by circumstances, Tippu signed the most humiliating treaty of Srirangapattana in March 1792.

4. Treaty of Srirangapattana in 1792 :

The terms of the treaty were: 

• Tippu had to surrender half of his Kingdom to the British and their allies. 
• Tippu agreed to pay a war indemnity of 3.5 crores (30 lakh pounds) to the British. As he did not have enough money, he had to send two of his sons to the British as hostages.

5. Fourth Anglo-Mysore war (1798-1799): Tippu could not reconcile to the defeat and humiliation in the third Anglo-Mysore war and was determined to drive out the British from India. He again started negotiations with France, Turkey, Kabul, Afghanistan, etc. by sending his delegations but he could not get any help. Lord Wellesley forced him to sign the subsidiary Alliance, which he refused, As a result war became inevitable.

Lord Wellesley sent a powerful army along with the Marathas and Nizam. Tippu was defeated in the battle of Siddeshwara and Malavalli. On fourth May 1799, the British besieged the fort of Srirangapattana. The fort was bombarded and the enemy entered the fort. Tippu died fighting in the battle and the British captured Srirangapattana.

After the death of Tippu, his territories Were divided among the British, the Marathas and the Nizam. A portion of his Kingdom was given to the Wodeyars of Mysore. Krishnaraja Wodeyar – III became the King of Mysore.

1²⁰¹¹ + 2²⁰¹¹ + 3²⁰¹¹ +...... +2010²⁰¹¹

= (1²⁰¹¹ + 2010²⁰¹¹) + (2²⁰¹¹ + 2009²⁰¹¹) + (3²⁰¹¹ + 2008²⁰¹¹) ...... (1011²⁰¹¹ + 1000²⁰¹¹)

These terms are of the form aⁿ + bⁿ, so they will be divided by a + b, as n = 2011 which is an odd number. 

Clearly, in the series a + b = 2011. Hence, the remainder is 0. 

Answer:  (A)  straight line 

The electronegativity of Nitrogen is much more than phosphorous. Thus the N-H bonds in NH3 are polar and the Hydrogen atoms of NH3 acquire a partially positive charge while the H atoms of PH3 remain neutral.
The partially charged H atoms of NH3 repels each other and thus Bond Angle increases. While PH3 lacks this effect. Thus bond angle of NH3 is greater than PH3.

The given AP is 6,13,20,........,216.

First term, a = 6

Common difference, d = 13 - 6 = 7

Suppose these are n terms in the given AP. Then,

a_n = 216

⇒ 6 + (n - 1) × 7 = 216 [ a_n = a + (n-1) d]

⇒ 7(n - 1) = 216 - 6 = 210 

⇒ n - 1 = 210/7 = 30

⇒ n = 30 + 1 = 31 

Thus, the given AP contains 31 terms,

∴ Middle term of the given AP

= (31 + 1/2)th term

= 16th term

= 6 + (16 - 1) × 7 

= 6 + 105

= 111

Hence, the middle term of the given AP is 111.

1. Allauddin Khilji (1296-1316 C.E.): Allauddin Khilji’s early name was Aligurshap. He lost his father in his boyhood and was brought up in the care of Jalaluddin. Allauddin married Jalaluddin’s daughter and was appointed as the Governor of Khara province. He was highly ambitious and aspired to become the ruler of Delhi. In 1294, he set his eyes on Devagiri. It’s ruler Ramachandradeva was defeated by him.

Allauddin returned to Khara with a heavy amount of booty. Jalaluddin was unaware cf the evil intentions of Allauddin. He went to receive Allauddin with only a few unarmed guards and was murdered by the supporters of Allauddin. Thus, Allauddin became the Sultan of Delhi in 1296 C.E.

Military Achievements of Allauddin:

a. The conquests of North India:

i. Conquest of Gujarat in 1297 C.E.: Allauddin sent Ulugh Ki an and Nazarath Khan, his generals to conquer Gujarat in 1297 C.E. Raja Karnadeva – II was defeated and he fled to Devagiri along with his daughter Devaladevi. The generals captured Kamaladevi (Queen of Karnadeva) and she was taken to Delhi and Allauddin married her. The Delhi troops plundered the rich ports of Gujarat.

ii. Conquest of Ranathambore in 1301 C.E.: Allauddin turned his attention towards Ranathambore. Hamira Deva, the ruler of Ranathambore, had given shelter to a few muslims (Neo muslims) who were enemies of Allauddin. So, Allauddin invaded and took over Ranathambore.

iii. Expedition on Mewar (Chittor) in 1303 C.E.: Allauddin led an expedition against Rana Ratan Singh of Chittoor (Mewar). He desired to possess Rani Padmini of Mewar, Queen of Ratan Singh, renowned for her beauty and talent. The fort of Chittor was captured with great hardship. Padmini and other rajput women committed ‘Jauhar’. Chittor was captured and Khizer Khan (son of Allauddin) was made the Governor of Chittor.

v. Other conquests: Allauddin took an expedition to Malwa in 1305 C.E. Mahakaladeva, the ruler of Malwa was defeated by him. The territories of Ujjain, Mandu, Dhara, Chanderi and Jolur were subjugated to Allauddin. He became the master of the whole of north India.

b. The Mongol Invasion (Raids): In 1299 C.E., Mongols attacked Delhi under Qutlugh Khwaja Frequent raids by the Mongols were a constant threat to the Empire. Allauddin and his general Malik Kafur successfully drove back the Mongols. He defeated them and imprisoned many of them.

c. South Indian compaign: Allauddin turned his attention towards south India. He sent an expedition under his eminent general, Malik Kafur to conquer the south. He coveted the enormous wealth of south India and its temples. The four main southern rulers were defeated.

i. Expedition to Devagiri (1306-1307 CE): Ramachandradeva, the ruler of Devagiri, had not paid tribute for nearly three years and he had given shelter to Karnadeva-II of Gujarat. For that reason, Malik Kafur raided Devagiri and defeated Ramachandradeva and collected a lot of booty.

ii. Conquest of Warangal (1309 CE): The Delhi forces marched via Devagiri and attacked Telangana. Pratapa Rudradeva, the ruler of Warangal, put up a stiff resistance. However, he was defeated and he had to surrender a lot of wealth which was carried away to Delhi by Malik Kafur.

iii. Expedition to Hoysalas in 1310 C.E.: Malik Kafur attacked Dwarasamudra, when Veera Ballala-III was busy interfering in the Chola politics. Malik Kafur occupied Dwarasamudra and Ballala-III was forced to plead for peace and he also accepted the sovereignty of Allauddin.

iv. Conquests of Madhurai in 1311 C.E.: A civil war was raging between Sundrapandya and Veerapandya, when Malik Kafur attacked the capital of Pandyas (Madhurai) and plundered the city. The wealth looted in south India was transported to Delhi on a herd of elephants.

Administrative achievements of Allauddin:

a. Sultanship: Allauddin followed an independent policy towards political matters. He set up a strong central administration. He did not permit the interference by religious leaders in administrative matters. He believed in the divine rights of Kingship (Shadow of God).

b. Espionage: He established an elaborate spy network, to get the information regarding all the activates of his nobles. He also tried to prevent the outbreak of rebellions within the Empire. He deprived his nobles of all pensions and endowments. He forbade social parties and secret meetings of the nobles, even in their houses.

c. Prohibition of drinking: He banned the sale and the use of intoxicating drinks and drugs at Delhi. He knew that, gambling dens and drinking bouts were the breeding grounds. of sedition.

d. Military reforms : The Standing army: Allauddin maintained a large standing army for maintaining internal law and order and to prevent the invasions of the Mongols. Ariz-i- Mumalik was the in charge for the appointment of soldiers. The state maintained a record of the Huliya or register of each soldier and his mount in the royal service. He also introduced the branding of horses or Dagh system.

e. Revenue reforms : Allauddin introduced scientific methods of measurement of land for the assessment of land revenue. He appointed a special officer called ‘Mustakhraj’ to collect land revenue from the peasants. To check bribery and corruption among the revenue officials, their salaries were increased. Steps were taken to safeguard the peasants from the demands of corrupt revenue officials.

f. Market regulation : The most remarkable of all these was an attempt to control the market prices by determining the cost of most of the essential commodities. Prices of all the articles of common use were fixed. Separate officers were appointed to regulate the market prices on a daily basis.

g. Personality of Allauddin : He is renowned not only for his conquests but also for his administrative and economic reforms. He was vigorous, efficient, bold and original as a reformer. He established an absolute state, free from the control of religion. H:s resourcefulness, energy, capacity for work, his unbounded courage tempered with calculation and penetrating common sense stand out.

S_n = 3n² + n 

S₂₂ = 3(22)² + 22

S₂₂ = 1452 + 22 ∴ S₂₂ = 1474

S₂₁ = 3(21)² + 21

S₂₁ = 1344

a₂₂ = S₂₂ - S₂₁ 

∴ a₂₂ = 1474 - 1344 ∴ a₂₂ = 130

Hence, the 22nd term of A.P is 130

a₂₂ = S₂₂ - S_22-1

a₂₂ = 3(22)²+22 - [3(21)²+21]

a₂₂ = 1474 - 1344

a₂₂ = 130

Hence the 22nd term of the AP is 130.

Option (B) is correct.

Reason- When acid reacts with base (NaOH) CO₂​​​ is not released while the other two are basic salts (NaHCO₃ and Na₂CO₃) which release CO₂ on reaction with acid.

Given tangents are-

5x + 12y = 3 ...(i)

5x + 12y = 29...(ii)

The tangents are parallel in this case.

Now, the chord formed by two parallel tangents of a circle is the diameter of the circle.

\(d = |\frac{c_{2}-{c_{1}}}{\sqrt{a²+b²}}|\\d=|\frac{-29+3}{5²+12²}|\\d=|\frac{-26}{13}| \\d=2\)

Now, d =2r 

r = 1 unit

In the given figure, 5Ω, 4Ω and 10Ω resistors are in parallel. In parallel combination, voltage across each resistor is same.

It is given that,

R = 5Ω

I = 1A

V = IR = 5V

Current across 4Ω resistor I = 5/4 ​A = 1.25 A

Current across 10Ω resistor I = 5/10 ​= 0.5 A

Given,

Current flowing through 5 ohms resistor = 1 Ampere

R₁= 5 ohms, R₂= 4 ohms and R₃= 10 ohms

Now, Potential Difference across 5 ohms resistor will be:

V = I × R (By Ohm's Law)

V = 1 × 5

V = 5 Volts

Let the current flowing through the other two resistors be I₂ and I_3.

Since all the resistors are connected in parallel, the potential difference across the three resistances will be same.

Therefore, in R₂ :

V = I₂ × R₂

5 = I₂ × 4

I₂ = 5/4 = 1.25 Amperes

Similarly in R₃ :

V = I₃ × R₃

5 = I₃ × 10

I₃ = 5/10 = 0.5 Amperes

= 2x²-2√6x+3=0

= 2x²-√6x-√6x+3=0

= √2x(√2x-√3)-√3(√2x-√3)=0

= (√2x-√3)(√2x-√3)

= x₁ = √3/√2 and

   x₂ = √3/√2

(x - 3)(2x + 1) = x(x + 5)

⇒ 2x² - 5x - 3 = x² + 5x

⇒  x² - 10x - 3 = 0

It is of the form ax² + bx + c = 0.

Hence, the given equation is quadratic equation.

The correct option is (B) 4 : 1.

I₁ = 1

I₂ = 9

I_max = (√I₂ + √I₁)²

I_min = (√I₂ - √I₁)²

\(\frac{I_{max}}{I_{min}}=\frac{(\sqrt{I_2}+\sqrt{I_1})^2}{(\sqrt I_2-\sqrt I_1)^2}\)

\(\frac{I_{max}}{I_{min}}=\frac{(\sqrt{9}+\sqrt{1})^2}{(\sqrt 9-\sqrt 1)^2}\) 

\(\frac{I_{max}}{I_{min}}=\frac{(3+1)^2}{(3-1)^2}\) 

\(\frac{I_{max}}{I_{min}}=\frac{(4)^2}{(2)^2}\) 

^{\(\frac{I_{max}}{I_{min}}=\frac{16}{4}\)} 

\(\frac{I_{max}}{I_{min}}=\frac4{1}\) 

The correct option is (B) transistor and (c) Diode.

A non-ohmic resistance is a resistance which does not follow the Ohm’s law. Ohm’s law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

The main Characteristic of Ohmic resistance is the VI curve which is a straight line.

The diode does not have a straight line as a VI curve.

The copper wire has a straight line as a VI graph.

The filament lamp has a straight line as a VI graph.

The carbon resistor has a straight line as a VI graph.

So, the diode is a non-ohmic resistance.

When two coils are brought in proximity with each other the magnetic field in one of the coils tend to link with the other. This further leads to the generation of voltage in the second coil. This property of a coil which affects or changes the current and voltage in a secondary coil is called mutual inductance.

Equivalent resistance= 12ohm

Total current=v/r=12/3=4A

pd on 4ohm resistor=4/4=1volt

If ∫ sin^-1 \(\sqrt{x}\) dx = x sin^-1 \(\sqrt{x}\) + a\(\sqrt{x}\) \(\sqrt{1-x}\) + b sin^-1 \(\sqrt{x}\) + C

\(\frac{d}{dx}\) ∫ sin^-1 \(\sqrt{x}\) dx = \(\frac{x}{2\sqrt{x}\sqrt{1-x}}+sin^{-1}\sqrt{x}+a\sqrt{x}(\frac{-1}{2\sqrt{1-x}}+\frac{a}{2\sqrt{x}}\sqrt{1-x}+\frac{b}{\sqrt{1-x}}.\frac{1}{2\sqrt{x}}+0)\)

⇒ \(sin^{-1}\sqrt{x}=\frac{\sqrt{x}}{2\sqrt{1-x}}+sin^{-1}\sqrt{x}-\frac{a\sqrt{x}}{2\sqrt{1-x}}+\frac{a\sqrt{1-x}}{2\sqrt{x}}+\frac{b}{2\sqrt{x}\sqrt{1-x}}\)

⇒ \(\frac{x-ax+a(1-x)+b}{2\sqrt{x}\sqrt{1-x}}\)

⇒ x - ax + a - ax + b = 0

⇒ x(1 - 2a) + (a - b) = 0 = or + 0

∴ 1 - 2a = 0 and a - b = 0

⇒ a = 1/2 & b = a = 1/2

∴ 3(a - b) = 3 x 0 = 0

Answer: Both B and C

The answer is (3)

\(J^2\rho=(\frac{I}{A})^2\times\frac{RA}{L}\)

\(=\frac{I^2}{AL}.R\)

= Heat generated / unit volume

बडे-बडे व्यक्ति, महान योद्धा जानते हैं कि वह जिस ओर से लड़ रहे हैं, वह अन्यायी हैं। वे अन्यायी शासक वर्ग अपनी शक्ति और अधिकार रूपी ब्रह्मास्त्र से निरायुध व्यक्ति को कुयल देना चाहते हैं। ऐसी अवसर पर मैं रथ का टूटा हुआ पहिया मानव-मूल्य बनकर निरायुध के हाथ में आ जाता हूँ और ब्रह्मास्त्रों से लोहा ले सकता हूँ। यहाँ महारथी शोषक वर्ग का और ब्रह्मास्त्र शासक वर्ग के द्वारा शक्ति और अधिकार का दुरुपयोग का प्रतीक हैं। इस कविता में महाभारत के अभिमन्यु की कहानी को प्रतीकात्मक बनाकर वर्तमानयुग की जटिलता का चित्रण किया है।

सामना करना’ के अर्थ में कवितांश में प्रयुक्त मुहावरा लोहा लेना है।

निरायुध व्यक्ति की है। जिस केलिए संसार में कोई बड़ा स्थान या धन नहीं।