A small ball of mass 2x10^-3 kg having a charge of 1 microC is suspended by a string of length 0.8 m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity, which shud b imparted to the lower ball so tat it can make complete revolution.

yupp...COEP is 1 of d best engineering coll in pune.....entrance is through MH-CET....

its der in Wellsley road..Shivaji Nagar.....

havnt heard of ane coll of engineering under BVP...

n as far as i know....BVP pune isnt gud at all... 

oleum = H2SO4 + SO3

if wt of H2S2O7 = 100g   

wt of H2SO4 = 118g

wt of H2O = 18g

moles of SO3 =1

wt of SO3 = 80

so % = 80

:)

see

in electrolysis of aq NaCl, Cl2 is evolvd at anode bcoz the discharging potential of OH- is greater thn tat of Cl-

so discharging tendency of Cl- is more

O2 will b evovld at anode wen the soln is vry dil.....coz thn the concentration of OH- will b more

 pyrolusite ~ ore of manganese dioxide ( MnO2)

Here, pyrolusite is d limiting reagent.....oxalic acid being takn in excess....

resulting mixture is thn titrated wid KMnO4......an eg of back titration

Givn tat 25 cc  soln req 32 cc of KMnO4

equivalence of any substance = Normality x Volume

no of equi of KMnO4 = 32 x 0.1/1000 = 0.0032

by law of equivalence,  no of equi of oxalic acid = 0.0032  

nw in 250 cc soln, no of equi of oxalic acid = 0.032

total no of equi of oxalic acid = 50 x 1/ 1000 = 0.05 ( initially )

no of equi of oxalic acid wic react wid MnO2 = 0.05 - 0.032 = 0.018

no of equi of MnO2 left  unreacted = 0.018

henc mass of MnO2 left = no of equi x equi wt = 0.018 x 43.5 = 0.783g        [(55+ 32)/2 = 43.5]

so %  = 0.783 x 100/ 1.6 = 48.9 %

available oxygen....????