Messages posted by 123098

20 divides every number from 5! giving remainder 0. So 1!+2!+3!+4!=25 when divide by 20 gives 5 as remainder.

my rank is 916 n my bitsat score is 394........................wat shud i do

my bits score 394.maths n physics was quite easy. cheemistry n english i felt little bit tougher. I hav attempted all the questions n took the extra questions which has a lot of advantage

how did u count without the paper? n where's the key dude?

sabaharish nikhil is from goiit-guy with 8 rank.

a guy with 235 got 8199 and with 228 not qualified

explain with reactions dude

more replies required

Alcl3 is right. others i dont know some of them. u should solve it

Alcoholic KOH is a very strong base compared to aq. KOH. when both react with alkyl halides,alcohols etc , alcoholic KOH favours elimination reaction by removing beta hydrogen and forming unsaturated compund due to its strong basic natur,whereas aq. KOH cannot remove beta hydrogen instead it favours nucleophilic substitution reactions.

hope u understood, let me cross 500 dude....................

major part is from 3-A

q) Substance A IS AYELLOWISH WHITE DELIQUISCENT SOLID WHICH SUBLIMES AND HAS A VAPOUR DENSITY OF 133. (A) REACTED VIOLENTLY WITH WATER, FORMING SOLUTION (B). A SAMPLE OF (B) GAVE A WHITE CURDY PPT (C) ON ADDITION OF DILUTE HNO3 AND AGNO3, BUT THIS READILY DISSOLVED ON ADDITION OF DIL. NH4OH, THOUGH A GELATINOUS WHITE PPT WAS FORMED IN ITS PLACE. (D) WAS FILTERD OFF AND DISSOLVED IN EXCESS NAOH, FORMING A CLEAR SOLUTION (E). WHEN CO2 WAS PASSES INTO (E) COMPOUND (D) WAS REPRECIPITATED.

SUBSTANCE (A) DISSOLVED UNCHANGED IN DRY ETHER AND WHEN THIS SOLUTION WAS REACTED WITH LIH ONE OF THE TWO PRODUCTS (F) OR (G) WAS FORMED, DEPENDING ON WHETHER THE LIH WAS IN EXCESS OR NOT. QUALITATIVE ANALYSIS OF SOLUTION (B) GAVE A WHITE GELATINOUS PPT IN GROUP 3

IDENTIFY THE COMPOUNDS (A) TO (G)

OH is very poor leaving group than bromine. Benzene with a +M group takes part in electrophilic substitution reactions where leaving group is also important.

CCl4 is a non polar solvent. so Br2 exists in elemental form. It cannaot replace OH in phenol. As it is an o-p director u get mono bromination in oand p positions

for the first oneangle AOB=pi/4angle CED=pi/2put area =1/(a)(b)sinqhere a,b =r and q=pi/2 and pi/4 respectively for CED and AOB

65sol: 5^3-3c15^2+3c25=65

Yup, i too heard that. U can look up 2006 paper

1) A long horizontal rod has a bead which can slide along its lenght, and initially placed at a distance L from one end of A of rod. The rod is set in angular motion about A with constant angular acceleration a. If coefficient of friction between the rod and the bead is u and gravity is neglected then the time after which the bead starts slipping is

a) root u/a b) u/roota c)_ 1/rootua d) infinitesimal

2)A spring of force constant k is cut into two pieces such that one piece is double the lenght of other. Them the long piece will hav a force constant of

a)2/3k b)3/2k c)3k d)6k

can u give proof for the second one , i mean the derivation

since it is a circular wire and current leaves at the diametrically opposite end, current divides into equal parts in both semicircles. The magnetic field due to one is balanced by other, hence force on charge=0

@sujit, but the rod is not fixed

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