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hi dude!!
dunno much bout the 2 colls u r talkin bout but as a branch cse is in more demand then IT at the moment........
so my advice is dat if dere is not much diff bw the 2 colls den u shud opt for cse
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hi lynette!!
in the prepration of jee or ne other entrance xams......... too much deep knowledge of the topic elasticity is not reqd............
jus read carefully the hc verma theory on the topic............. practice its questions n dats more than enof for the topic
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sry yaar did 1 mistake:
N shud be replaced by half a carbon atom
so i have edited my previous reply........
reason: in an org comp if we want to insert a N atom inside (ie the N shud not be at the end of the chain) its chain then no of H atoms reqd is 1 (ie half of 2) n dat for C is 2
no of S atom is equi to 0 carbon atom coz if we wanna insert it in the org comp chain then the no of H atom reqd is 0 .
do rate me if it was helpful
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hey dude..........solved it
connect the above system wid a 1 ohm resistor.......
solve the circuit using kirchoff rules....... get the amount of current flowing through the the 1 ohm resistor in ampere........dat is numerically equal to the emf of the above sys.
m gettin the emf as 2V..........
and the left side is on higher potential
do rate me if it was helpful
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dude the figure is not clear.......
it is not clear dat which battery has positive term on right n which one has the same on left
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hi nithy!!
to find double bond eq:
1) replace Br as H (halides r equi to H in terms of bonds to C reqd)
2)N by .5 C
3)n jus remove S or O as it is (as it does not effect no of double bonds)
the resultant comp will be: C13.5H11
when dere r no double bonds in a hypothetical compound wid 13.5 carbon atoms then no of H atoms r=29
diff of H atoms=18
no. of double bonds=18/2=9
henc the ans is 9
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do note dat the rod wud be moving linearly and the wedge wud be moving horizontally.................... so just consider the linear forces in the case of rod and horizontal forces for wedge
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hi kritika!!
dat was a tough one......... still solved it for u ..... dunno if i am right or not......do tell me bout that
let the acc of the rod be a1 n dat of wedge be a2
let the angle be t instead of theta:
using constraint eqn we get the realtionship bw the accen of the 2 bodies
a2tant=a1 ...................1
now we solve acc to laws of motion:...................... takin normal reaction as N bw them.
ma1=mg-Ncost..........................2
nma2=Nsint...............................3
now solve the 3 eqs n get the ans, m gettin:
a1=(tant)^2*g/[(tant)^2+n]
a2=tant*g/[(tant)^2+n]
do rate me if it was helpful
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the other fuse wire will blow if the power produced in it is the same
so
i1^2*R1=i2^R2
i1^2/a1=i2^2/a2
(cutting resistivity n length)
i1/r1=i2/r2 (put i1=5 n value of r1 n r2 as radius)
i2=7.5 A
hence the ans is 7.5 A
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hi hsg2189
the iits dont have aeronautical engg at present......it is rather known as aerospace engg dere
kanpur,kharagpur,madras and bombay have this course
kanpur has the best facilities for it
cr:
k(1470)
b(1278)
m(1669)
kgp(1836)
dude rate me if it was helpgul
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the ans to this q is b)
as the electric field in the volume of dat blanket wud become less......derefore the overall potential on the shell wud lessen
dont forget to rate me!!
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hi rini
solved the first 1:
first replace 11 by a variable f ........ we will put the value 11 at the end
and let x=ans
therefore x=(f-a)(f-a^2)(f-a^3)(f-a^4).........(f-a^n-1)
=> (f-1)x=(f-1)(f-a)(f-a^2)...........(f-a^n-1)
=> (f-1)x=f^n - 1
(as 1, a,a^2 etc r the nth roots of unity)
putting f=11
u will get x=(11^n - 1)/10
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hi miss striving4perfection
practice practice and more practice is the only way out of confusion
regarding this q:
at first: both of them have equal capacitance c so the circuit capacitance wud be c/2 hence charge on each capacitor wud be cv/2
potential diff across each capacitor wud be v/2
when u insert a slab then one of the capacitor will have capacitance kc
and the circuit capacitance will be kc/(k+1)
hence charge on each capacitor wud be kcv/(k+1)
potential diff on slab capacitor=charge/capacitance=v/(k+1)
and on the other 1=kv/(k+1)
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the force bw the plates is attractive ,
so for the plates to be moved closer so as to not inc their kinetic
energy from 0, we will need a force dat is opp to the direction of the
displacement of the plates as they r moved closer
hence the work done wud be negative
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hi, fused bulb
considered one of the coolest persons on this site
wat have u got at roorkee
i got chem btech roorkee
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chemical btech
iit roorkee
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hey nayan
the answer to qs number 2 is C) ellipse
rate me if i am correct
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hey, vinay
the answer most probably is 3)HNO2 +HNO3
rate me if i am correct
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hey tina (anothe emma watson fan)
is the answer cyclopentane and the chloride is cyclopentenyl chloride
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in this question
all three capacitors are connected in parrallel
and capacitance of each=C
so equivalect capacitance=C/3
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