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Length of the fence 

= \(\frac{Total\,cost}{Rate}\)

= \(\frac{Rs.5280}{Rs.24/metre}\) = 220 m 

So, length of fence = Circumference of the field 

∴ 220m = 2πr = 2 x \(\frac{22}{7}\) x r 

So, r = \(\frac{220\times7}{2\times22}\) m = 35 m

EC = AC – AE = (7 - 3.5)cm = 3.5 cm 

\(\frac{AD}{BD}\) = \(\frac{2}{3}\) and \(\frac{AE}{EC}\) = \(\frac{3.5}{3.5}\) = \(\frac{1}{1}\)

So, \(\frac{AD}{BD}\) ≠ \(\frac{AE}{EC}\)

Hence, By converse of Thale’s Theorem, DE is not Parallel to BC.

By converse of Thale’s theorem DE II BC 

∠ADE = ∠ABC = 70° 

Given ∠BAC = 50° 

∠ABC + ∠BAC + ∠BCA = 180° (Angle sum prop of triangles) 

70° + 50° + ∠BCA = 180° 

∠BCA = 180° - 120° = 60°

Horse can graze in the field which is a circle of radius 28 cm. 

So, required perimeter = 2πr = 2.π(28) cm 

=2 x \(\frac{22}{7}\) x (28)cm = 176 cm

Reduction is a phase metabolism I.

Given:

Time taken by 5 persons to complete 1/48^th work = 1/2 days

Formula used:

(M₁ × T₁)/W₁ = (M₂ × T₂)/W₂

Calculation:

[5 × (1/2)]/(1/48) = (6 × T₂)/(1/40)

⇒ 5 × (1/2) × (1/40) = 6 × T₂ × (1/48)

⇒ 1/16 = T₂/8

⇒ T₂ = 1/2 days

∴ The work will be completed in 1/2 days

Answer is (1) (p ∧ (p → q)) → q

(2) Actinoids and Group 4

Element with atomic no. 101 is an Actinoid element. 12. 

(i) p ↔ q

= p ↔ q and q ↔ p

= (p ↔ q) ^ (q ↔ p)

Hence, p ↔ q = (p ↔ q) ^ (q ↔ p)

It means p ↔ q is logically equivalent to (p ↔ q) ^ (q ↔ p)

(ii)  p ↔ q

= q ↔ p and p ↔ q

= ~ p ↔ ~q and ~q ↔ ~p

= ~(~p) V ~ q and ~(~q) V ~p

= (p v ~q) ^ (q v ~p)

= (~pvq) ^ (~qvp)

Hence, (A) is also equivalent to p ↔ q

(iii) p ↔ q = p ↔ q and q ↔ p

= (~pvq) ^ (~qvp)

= ((~pvq) ^ ~q) V ((~pvq) ^ p)

= ((~ p ^ ~q)) v (q ^ ~q)) v ((~P^P)) V (q^p))

= ((~p ^ ~q)) V F) V (FV(P ^q))

(∵ ~p ^ P = F)

= (~p ^ ~q) V (P ^ q)

Hence, (B) is equivalent to p ↔ q.

i.e, (p ^ ~q) V (q ^ ~p) is not logically equivalent to p ↔ q.

2x-4–3x+9 = 5x-25

2x-3x-5x = -25+4–9

-6x = -30

x = -30/-6

x = 5

L.H.S. = (p ∧ q) ∨ (∼ p ∧ q) ∨ (p ∧ ∼ q)

≡ [(p ∨ ∼ p) ∧ q] ∨ (p ∧ ∼ q) ..........(Distributive Law)

≡ (T ∧ q) ∨ (p ∧ ∼ q) .........(Complement Law)

≡ q ∨ (p ∧ ∼ q) ...........(Identity Law)

≡ (q ∨ p) ∧ (q ∨ ∼ q) .........(Distributive Law)

≡ (q ∨ p) ∧ T ..........(Complement Law)

≡ q ∨ p ..........(Identity Law)

≡ p ∨ q ......(Commutative Law)

= R.H.S.

5.4/108 × NA = number of atoms of silver

Molecular weight of silver atom  = 108 g/m

Weight of silver wire = 5.4 g

\(\therefore\) Number of moles of silver atoms = \(\frac{5.4g}{108g/mol}\) = 0.05 mol

\(\because\) We know that

one mole atom contain = 6.023 x 10²³ atoms.

\(\therefore\) 0.05 mol silver atom contain = 6.023 x 10²³ x 0.05 atom

 = 0.30115 x 10²³ silver atom

 = 3.0 x 10²² silver atom.

Hence, 5.4 g silver wire contain 3.0 x 10²² silver atoms.

Concept:

Let \(\rm a_1 + a_2 + a_3 + .... + a_n \) be geometric progression series then s_n =  \(\rm \sum a_n = \dfrac {a_1(1 - r^n)}{1-r}\), Where r is common ratio

Calculations:

Consider 

\(1+i^2 + i^4 + i^6 + ...+ i^{100} \)

Here,  \(i =\sqrt{1} \)

⇒ \(\rm i^2 = -1\)

Series becomes,

⇒ 1 - 1 + 1 - 1 + .... + 1

which is GP with common ratio r = -1 and first term a = 1.

⇒ 1 - 1 + 1 - 1 + .... + 1 = \(\rm \dfrac {a (1 - r^n)}{1- r}\)

Here n = 51.

⇒ 1 - 1 + 1 - 1 + .... + 1 = \(\rm \dfrac {1 [1 -(-1)^{51}]}{1- (-1)}\)

⇒ 1 - 1 + 1 - 1 + .... + 1 = 1

Hence,  \(1+i^2 + i^4 + i^6 + ...+ i^{100}\) = 1

The correct answer is 1 only.

• The Arthashastra is a primaeval Indian discourse on statecraft, economic policy, and military strategy. Hence, Statement 1 is correct.
• It is written in Sanskrit. Hence, Statement 2 is incorrect.
• This treatise is divided into sixteen covering topics concerned with the running of a state, taxation, law, diplomacy, military strategy, economics, bureaucracy, statecraft, politics, strategy, selection, and training of employees, leadership skills, legal systems, accounting systems, taxation, fiscal policies, civil rules, internal and foreign trade, etc.
• The emphasis is on the uniformity of law throughout the empire.
• Kautilya's Arthasashtra was composed between the 3^rd to 2^nd Century B.C.
• It is mainly a work on the art of government.
  • In his view, for the smooth functioning of administration and for the welfare of the people, the king had to be acquainted with the four Vedas and four sciences of government (Anvikashaki, Trai, Varta, and Dandniti). Hence, Statement 3 is incorrect.
  • Kautilya's administrative and judicial structure was hierarchical in nature.
  • He emphasized the principle of equity.
  • He believed that law was an imperial command enforced by sanctions.
• He was both a thinker and statesman.
• He contributed to the formulation of some general principles capable of universal application and effective in all times and ages.
• Arthsashtra sets the conceptual groundwork for making India the first welfare state.

• In Indian theory, State was usually thought to cover seven elements:

1. The king
2. The ministers
3. The populace
4. The fort
5. The treasury
6. The army
7. The ally

CONCEPT : 

• If a, b, c are in AP, then by the rule of arithmetic mean we can write as \(b = \rm\frac{a + c}{2}\).

CALCULATION : 

As given p – r = q.

⇒ r = p - q . . . . . . . . . . . . .  (1)

And we know that \(q = \rm\frac{p + r}{2}\)

Putting the value of r from equation (1) to the above equation we get

\(\Rightarrow q = \rm\frac{p + (p -q)}{2}\)

On solving we get 3q = 2p

\(\Rightarrow p = \rm\frac{3q}{2}\)

Therefore, option (1) is the correct answer.

Concept:

If series is in AP then nth term of AP = a_n = S_n - S_{(n - 1)}

Calculations:

Given series is in AP and the sum of first n terms is 2n2 + 3n

⇒ S_n =2n2 + 3n.

Then, the first term will be when n=1 in the above expression

⇒ S1 = 2 + 3 = 5

Sum of first two terms,

n = 2 , S₂ = 2 × 2² + 3 × 2 = 14

So, the second term will be 

a₂ = S₂ - S₁ = 14 - 5 = 9

Similarly, 

n = 3, S3 = 2 × 3² + 3 × 3 = 27

So, the third term will be 

a3 = S3 - S₂ = 27 - 14 = 13

Similarly, 

n = 4, S4 = 2 × 4² + 3 × 4 = 44

So, the fourth term will be 

a4 = S4 - S3 = 44 - 27 = 17

Hence, the 4th term of an AP is 17.

The first Prime Minister of India was Jawaharlal Nehru.

Jaipur is known as pink city

(1) R is reflexive, as (a, a) ∈ R 

⇒ a is divisible by a 

∴ R is reflexive. 

(2) Let (1, 2) ∈ R ⇒ 2 is divisible by 1 

But (2,1) ∈ R ⇒ 1 cannot divide by 2 

∴ R is not symmetric. 

(3) Let (1, 2)∈ R ⇒ 1(2, 4)ER 

⇒ 2 is divisible by I and 4 is divisible by 2 

Now, (1,4) ∈ R ⇒ 4 is divisible by 1 

∴ (1, 2)∈ R I(2,4) ∈ R ⇒ (1, 4)∈ R 

∴ R is transitive.

When we suck the straw, pressure inside straw becomes lower than atmospheric pressure. This pressure difference cause the soft drink to rise through the straw.

The capillaries of old earthen pots will get blocked with passage of time. For new earthen pots, water oozes out through capillaries, gets evaporated at the surface and makes it cool

Correct option is an air conditioner

Correct option is Australia

The correct answer is option (2) i.e. 918.

• National average of Sex Ratio (Females/Thousand Males) is 940.

The correct answer is option (2) i.e. House Sparrow.

• Passer domesticus is the binomial name of House Sparrow.
• Mere Bharat Ke Kanth Haar is the state song and 22nd March is celebrated as Bihar's formation day.