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The force experienced by a unit positive charge placed at a point is termed as the electric field intensity at that point. It is vector quantity it’s directi
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solution:The elevation in boiling point depend on the dissociation of particle in solution.0.1 M NaCl will dissociate into Na+ and CI– whereas 0
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Find CBSE, ICSE State Boards Textbook Solutions of Class 8 to 12 of all subjects. Read NCERT Textbooks Questions and Answers & Find Previous year solved papers.
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R= {(L1, L2) : L1 is perpendicular to L2} Since no line can be perpendicular to itself ∴ R is not reflexive.Let (L1, L2) ∈ R∴ L1 is perpendicular to L2
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Using ohm s law we have,                V = IR ⇒ vI = R Drift velocity is given by, vd =&
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ड़े भाई साहब के अनुसार जीवन की समझ केवल किताबी ज्ञान से नहीं आती बल्कि अनुभव से आती है। बड़े भाई के अनुसार जीवन की समझ ज्ञान के साथ अनुभव और व्यावहारिकता से आती है। इसके लिए उन्होंने अम्माँ, दादा व हैडमास्टर की माँ के उदाहरण भी दिए हैं कि वे पढ़े लिखे न होने पर भी हर समस्याओं का

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Given solution
Red litmus solution
Blue litmus solution
Phenolphthalein solution
Methyl orange solution
HCl
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The density of 3M solution of NaCl is 1.25g ml-1.  Calculate mo clicking on this link
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Width of central diffraction band = 2D λd.where, d is the width of the slit.So, on doubling the width of the slit, the size of the central d
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Let PQ be the tower and Y is a point vertically above X such that XY = 40 m. The angle of elevation of the top Q from a point X on the ground is 60° and from
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Given: In ∆ ABC, AD is the perpendicular bisector of BC.To Prove: A ABC is an isosceles triangle in which AB = AC.Proof: In ∆ ADB and ∆ADC,∠ADB = ∠AD
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⇒ Sn – 1 = 5(n – 1)2 + 3(n – 1)∴ Sn – Sn – 1 = (5n2 + 3n) – [5(n – 1)2 + 3(n – 1)}⇒ an = (5n2 + 3n) – {5 (n2 – 2n + 1) + 3n
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The given equation is x2+k(2x+k−1)+2=0.
⇒x2+2kx+k(k−1)+2=0
So, a = 1, b = 2k, c = k(k − 1) + 2
We know D=b2−4ac
⇒D=(2k)2−4×1×[k(k−1)+2]
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