Class 12th
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New answer posted
4 months agoContributor-Level 9
Covalent solids have high melting point due to strong covalent bonds, also they are insulator in both solid as well as in molten state.
New answer posted
4 months agoContributor-Level 9
In basic medium H? O? can behaves as both oxidizing and reducing agent, so it can oxidize Mn²? to Mn? and reduce I? to I?
New answer posted
4 months agoContributor-Level 10
F = (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + .
⇒ F = (qQ)/ (4πε? ) [1/1² + 1/2² + 1/4² + 1/8² + .]
⇒ F = (qQ)/ (4πε? ) [1/ (1 - 1/4)]
⇒ F = (qQ)/ (4πε? ) [4/3] = 10? * 1 * 9 * 10? * (4/3) = 12 * 10³ N
New answer posted
4 months agoContributor-Level 10
Δλ = (v/c)λ = (286/3000000) * 630 * 10? = 6.006 * 10? ¹? ≈ 6 * 10? ¹? m
New answer posted
4 months agoContributor-Level 9
Both C? H? OH and AgCN can generate nucleophile.
KCN generate nitrile as nucleophile while AgCN generate isonitrile as nucleophile in nucleophilic substitution reaction.
New answer posted
4 months agoContributor-Level 10
R_eq = (R? )/ (R? +R? ) ⇒ p (l)/ (2A) = [ (p? l/A) (p? l/A)] / [ (p? l/A) + (p? l/A)]
⇒ ρ/2 = (p? )/ (p? +p? ) ⇒ ρ = (2p? p? )/ (p? +p? ) = (2 * 6 * 3)/9 = 4 Ωcm
New answer posted
4 months agoContributor-Level 10
d_lm = Distance covered by transmitting antenna + Distance covered by receiving antenna
⇒ d_lm = √ (2Rh_transmitting) + √ (2Rh_receiving)
⇒ d_lm = √ (2 * 64 * 10? * 20) + √ (2 * 64 * 10? * 5) = 16000 + 8000 = 24000m
When h_receiving = 0 then
d_2m = √ (2 * 64 * 10? * 20) = 16000m
% increment = (8000/16000) * 100 = 50
New answer posted
4 months agoContributor-Level 10
As we know that direction of propagation of electromagnetic wave is perpendicular to plane containing mutually perpendicular electric field and magnetic field, so option D will be correct answer.
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