Harshal Patil
@Harshal Patil
Best posts made by Harshal Patil

RE: Sound
@yashingle in the formula of frequency u hav to replace L by L+e for closed pipe and if it is open on both sides we replace L by L+2e.
In case of resonance experiment we use one end closed pipe so use L+e.

RE: O c
@yashingle in a, hydride shift wont take place so after elimination of OH the molecule will hav chiral centre at topmost carbon. So since elimination can occur at either left or right side, racemic mixture will be formed
In b, double bond will be formed b/w middle 2 carbons so number of alpha H will be 8
In c, carbocation cannot be formed at bridgehead posn so no dehydration
In d, there is no H for double bond to be formed at middle so elimination will occur at sideways posn amd number of alpha H will be 4.

RE: Theory of equations
@saumilparsodkar
$f(0)=C$ so $C$ is integer
$f(1)=A+B+C$ so since $C$ is integer, $A+B$ is also integer
$f(−1)=A−B+C$ so since $C$ is integer, $A−B$ is also integerSince $A−B$ and $A+B$ are both integers, $A$ and $B$ are also integers. So all $A,B$ and $C$ are integers
Now try proving converse by urself
Latest posts made by Harshal Patil

RE: Polynomial
$a−2=2_{1/3}+2_{2/3}$
now cube both sides
$a_{3}−8−6a(a−2)=$
$2+4+3.2_{1/3}.2_{2/3}(2_{1/3}+2_{2/3})$replace $2_{1/3}+2_{2/3}$ in RHS by $a−2$ and simplify

RE: Kinematics
@TechWithUs 50v1=20v2 (conserving momentum) and v1+v2=0.7 which gives v2=0.2

RE: ELECTROCHEMISTRY
@KanishkKumar0 10mmoles of NaBr will be formed...
so conductivity will be 10×10^(3)×(8+4)×10^(3)÷(10^(4))=12×10^(1)
100ml corresponds to 10^(4)m^3 so i have divided it with that volume

RE: Functions
@HarshalGajendra $3_{y}=3−3_{x}$ and $3_{y}$ should be positive so x should be less than 1

RE: Goc
@SiddharthaGhosh ring becomes aromatic if electrophile attacks at 1... because the carbon marked at 2 will have positive charge

RE: What will be the equation below?
@Indranil it is the law of indices... x^a/y^a=(x/y)^a

RE: Complex no
putting values of $ω$ and $ω_{2}$ we get
$S=(a+2b+c )_{2}+43(b−c)_{2} $
we have to minimise this
now $(a+2b+c )_{2}$ cannot be zero as all $a,b,c$ cannot be zero at the same time
but $43(b−c)_{2} $ can be zero so let $b=c$
now we have to minimise $(a+2b+c )_{2}$
as two numbers can be equal, let $b=c=0$
and since $a$ cannot be zero and it is also an integer, $a=1$

RE: Complex no
@nishantthakre z=2 represents circle at origin and zi=z+5i represents the line y=2 as it is collection of points where the distance of every point from i and 5i is equal.
so common soln is (0,2)