Priyapatel

विजयनगर की जल-आवश्यकताओं क&#
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The given relation is aRb : a + 3b = 12
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Here, distance is same.∴  (Find detailed answer | clicking on this link)

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R = {(a, b) : 2 divides a – b}
where R is in the set Z of integers.
(i) a – a = 0 = 2 .0
∴ 2 divides a – a ⇒ (a, a) ∈ R ⇒ R is reflexive.
(ii) Let (a, a) ∈ R ∴ 2 divides a – b ⇒ a – b = 2 n for some n ∈ Z ⇒ b – a = 2 (–n)
⇒ 2 divides b – a ⇒ (b. a) ∈ R
(a, ft) G R ⇒ (b, a) ∈ R ∴ R is symmetric.
(iii) Let (a, b) and (b, c) ∈ R
2 divides a – b and b – c both ∴ a – b = 2 n1 and b – c = 2 n2 for some n1, n2 ∈ Z ∴ (a – b) + (b – c)= 2 n1 + 2 n2 ⇒ a – c = 2 (n1 + n2 )
⇒ 2 divides a – c
⇒ (a, c) ∈ R
∴ (a,b), (b,c) ∈ R ⇒ (a, c) ∈ R
∴ R is transitive
From (i), (ii), (iii) it follows th

Let energy E = Fx Ay T3Writing the dimensions on both sidesM L2 T -2 =M  L T -2 x L T -2 y&

We have given,
Mass percent of iron (Fe) =69.9%
Mass percent of oxygen (O)=30.1%
Number of moles of iron present in the oxide =69.90/55.85 =1.25
Number of moles of oxygen present in the oxide=30.1/16.0
=1.88
Ratio of iron to oxygen in the oxide,
1.25:1.88
=1:1.5
=2:3
Therefore the empirical formula of the oxide is Fe2O3.
Empirical formula mass of Fe2O3= [2(55.85) +3(16)]g
Molar mass of Fe2O3=159.69g
Thus, n= Molar mass/Empirical Formula mass =159.69/159.7g
 =0.999 = 1(approx.)
Molecular formula of a compound is obtained by multiplying the empirical formula with n. Thus, the emp

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यहाँ सरकार दो या अधिक स्तरों वाली होती है।
अलग - अलग स्तर की सरकारें एक ही नागरिक समूह पर शासन करती है ।
विभिन्न स्तरों की सरकारों के अधिकार - क्षेत्र संविधान में स्पष्ट रूप से वर्णित होते है।
संविधान के मौलिक प्रावधानों का बदलाव दोनों स्तर की सरकारों की सहमति से ही हो सकता है ।
अदालतों को संविधान और विभिन्न स्तर की सरकारों के अधिकारों की व्याख्या करने का अधिकार है ।
वित्तीय स्वायत्तता निश्चित करने के लिए विभिन्न स्तर की सरकारों के लिए राजस्व के अलग - अलग स्त्रोत निर्धारित है ।
इस प्रकार संघीय शासन व्यवस्था के दोहरे उद्देश्य है देश की एकता की सुरक्षा करना इसके साथ ही क्षेत्रीय विविधताओं

पुरातत्वविदों को खुदाई के दौरान अनेक प्रकार के शिल्प तथ्य प्राप्त होते हैं। वे इन शिल्प तथ्यों को अन्य वैज्ञानिकों की मदद से अन्वेषण व विश्लेषण करके व्याख्या करते हैं। इस कार्य में वर्तमान में प्रचलित प्रक्रियाओं, विश्वासों आदि का भी सहारा लेते हैं। हड़प्पा सभ्यता के संबंध में पुरातत्वविदों के अग्रलिखित निष्कर्षों से यह बात और भी स्पष्ट हो जाती है कि वे किस प्रकार अतीत का पुनर्निर्माण करते हैं:
जीवन निर्वाह का आधार कृषि: पुरावशेषों तथा पुरावनस्पतिज्ञों की मदद से पुरातत्वविद् हड़प्पा सभ्यता की जीविका निर्वाह प्रणाली के संबंध में निष्कर्ष निकालते हैं। उनके अनुसार नगरों में अन्नागारों का पाया ज

∠FOB + ∠BOE = 180° ...(1)| Linear Pair Axiom∠FOA + ∠AOE = 180° ...(2)| Linear Pair AxiomFrom (1) and (2),∠FOB + ∠BOE = ∠FOA + ∠AOE ...(3)But    ∠BOE = ∠AOE| ∵ Ray OE bisects ∠AOB∴ From (3),⇒    ∠FOB = ∠FOA.

Suppose that R1 and R2 are two equivalence relations on a non-empty set X.First we prove that R1 ∩ R2 in an equivalence relation on X.(i) R2 ∩ R2 is reflexive :Let a ∈ X arbitrarily.Then (a, a) ∈ R1 and (a, a) ∈ R2 , since R1, R2 both being equivalence relations are reflexive.So. (a, a) ∈ R1 ∩ R2⇒ R1 ∩ R2 is reflexive.(ii) R1 ∩ R2 is symmetric :Let a, b ∈ X such that (a, b) ∈ R1 ∩ R2 ∴ (a, b) ∈ R1 and (a, b) ∈ R2 ⇒ (b, a) ∈ R1and (b, a) ∈ R2, since R1 and R2 being equivalence relations are also symmetric.(b, a) ∈ R1∩ R2(a, b) ∈ R1 ∩ R2 implies that (b, a) ∈ R1 ∩ R2.∴ R1 ∩ R2 is a symmetric re

Answer :- in 1952

Given, Radius of the circular track, r = 30 m
Speed of the train, v = 54 km/h = 15 m/sMass of the train, m = 106 kg The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail. The angle of banking θ, is related to the radius (r) and speed (v) by the relation:             tan θ = v2 / rg                       = 152 / (30 × 10)                       = 225 / 300                    θ = tan-1 (0.75) = 36.870 Ther

Mukesh brings the writer to his own hut. It is a half built shack. Its one part is thatched with dry grass. Then she sees a firewood stove having a vessel containing spinach leaves for boiling. In a large aluminium platter, chopped vegetables can be seen. A frail young woman cooks food for the whole family.

एकदिष्ट अधिमान: एकदिष्ट अधिमान का अर्थ यह हैं कि एक उपभोक्ता किन्हीं दो बंडलोंमें से उस बंडल को अधिमान देता है जिसे इन वस्तुओं में से कम-से-कम एक वस्तु की अधिक मात्रा हो और दूसरे बंडल की तुलना में दूसरी वस्तु की भी कम मात्रा न हो।

(i) First order(ii) s-1 or min-1(iii)           t1/2 = 0.693K(iv)           k=2.303tlogR0RWhere               R0 = Initial conc. of the reactant at t = 0R = Concentration of the reactant at time it.              OrLet the law be r = kAxBywhere              x = order with respect to A                        y = order with respect to BTaking data from experiment (i) to experiment (iii), we get0.096 = k(0.30)x (0.30)y ...(ii)0.384 = k(0.60)x (0.30)y ...(iii)0.192 = k(0.30)x (0.60)y ...(iv)0.768 = k(0.60)x (0.60)y ...(v)Divide eq. (v) by eq. (iv)0.7080.192 = 2x = 4 = 22               x=2Divide eq

Characteristics of the periodical markets:
1. These markets are held weekly or fortnightly at a particular place.
2. In villages, the periodical markets are called by different names such as penth etc.  In these periodical markets the local items are sold and purchased by the local people.
3. Mostly daily used articles are sold in these markets. Different articles are sold on cheap rates as the purchasers belong to the village people.

Putting x = 0 in 9x - 12y = 108,         0 - 12y = 108,    y = -9Putting y = 0 in 9x - 12y = 108          9x - 0 = 108⟹   x = 12∴     OA = 12,   OB = 9

Statement: For the streamline flow of non-viscous and incompressible liquid, the sum of potential energy, kinetic energy and pressure energy is constant.Proof: Let us consider the ideal liquid of density ρ flowing through the pipe LM of varying cross-section. Let P1 and P2 be the pressures at ends L and M and A1 and A2 be the areas of cross-sections at ends L and M respectively. Let the liquid enter with velocity V1 and leave with velocity v2. Let A1 > A2. By equation of continuity,

Let r cm be the radius and h cm be the height of a pipe (cylinder), then                           r = 7 cm = And, h = 5 km = 5000 m
[distance covered in 1 hr. = height of pipe]Now, Volume of water that flows out of the circular pipe in 1 hourLet l, b and h are respectively the length, breath and height of the tank (cuboid), then                         l = 50 m                        b = 44 mand                   h =  7 cm =           Now, Volume of water that flows in the rectangular tank                 = l x b x h           time taken to flow 77 m3 of water in to the tank K = 1 hr time t