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I have given you the free body diagrams of the sphere and

the plank in the figure below:

Let "a1" be the acceleration of the plank And

Let "v1" be the --Velocity--of the plank.

Then we have:

F = ma1 = -umg

a1 = -ug

And

v1 = (-ug)T

v1 = -ugT

For the sphere,

acm = ug

and

vcm = ugT

The torque acting on the sphere about the centre of mass

is given by:

= umgr

I

= -umgr

I

Here,

I = (2/5)mr2

Therefore we will get:

= -5ug

2r

Now,

wt = (wo - 5ugt)

2r

For rolling,

V1 = Vcm - wt*r

V1 = Vcm - (wo - 5ugt)r

2r

After solving the above expression we will get:

t = 2wor

9ug

Now the distance travelled by the plank is given by:

s1 = (1/2)(-ug)(2wor/9ug)2

s1 = -2wo2r2/81ug

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Answer:Option(D)

Because the speed of sound is not affected by the motion of

source or observer.It depends only on the medium and on its

state of motion with respect to the reference frame.

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people please check this in for "IITJEE-07" paper solution

You will find it there.

Chloroform reacts with strong base to form the chloroform

carbanion (2),which will quickly

-eliminate to give dichloro

carbene (3).

Dichlorocarbene will react in the ortho- and para- position

of the phenate (5) to give the dichloromethyl substituted

phenol (7).After basic hydrolysis,the desired product (9)

is formed.

Hope you find it useful.

This answer has been taken from the link provided by

@sandeepramesh.

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The Cut-offs for the year 2007 in IITJEE for PCM are:

Maths :1

Physics :4

Chemistry:3

So cut off overall was : 1 + 4 + 3 = 8/480+

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Here is the solution:
Assuming mass of pulley as "M"
mass of block given as "m = 1 kg"

First we will find the mass of pulley:

I = MR²/2
Given:
I = 0.2 kg-m²
R = 0.2m.

(0.2) = M(0.2)²/2
M = 0.4/0.04
M = 10 kg.

Now writing equations of motion for block:
T₁ = ma .........(1)
m = 1
T₁ = a ...........(1)

Now supposing that the block accelerates with "a" then:
From (b) FBD pulley will accelerate with "a/2".

Now writing equations of motion for "pulley":

Mg - T₁ - T₂ = M(a/2)

(10*9.8) - (T₁+T₂) = 5a

98 - (T₁+T₂) = 5a ............(2)

Now for linear motion of the block the pulley is going down
and hence the pulley is rotating:

= I@
T₂R - T₁R = I@
T₂(0.2) - T₁(0.2) = (0.2)@
T₂ - T₁ = @

We know that @ = a/R
Therefore we get,
@ = (a/2)0.2
as we know pulley is accelerating down with "a/2"
@ = 5a/2

T₂ - T₁ = 5a/2...........(3)

Now from eq.(1) w know T₁ = a.
So we have,
T₂ - a = 5a/2
T₂ = 7a/2

Substituting the values of "T₂" and "T₁" in equation (2) :
98 - (T₁+T₂) = 5a
5a + 7a/2 + a = 98
6a + 7a/2 = 98
19a/2 = 98
19a = 196
a = 196/19
a

10 m/s².

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1)Optical Isomerism:

Compounds having Similar Physical and Chemical properties but
differing in the behaviour toward polarised light are called as
"Optical Isomers" and This phenomenon is called as "Optical-
Isomerism"

For A compound to be Optically active it must be:
(a)Chiral
(b)non-Super imposable on its mirror image

The molecules which are not superimposable on their mirror
images are called as "Chiral Molecules"

Eg:

(a)     CH₃               (b)        CH₃
   l                              l
   H - C - OH                HO - C - H
     l                              l
     CH₂CH₃                       CH₂CH₃

Here above (a) and (b) are mirror images of each other.
And it is clear that (a) 2-Butanol is "non-superimposable"

Non-Super Imposable mirror images are also called as
"Enantiomers"

Hope it helps you.

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Hello mate,

I think if you are not understanding whole of morrison and

boyd then please instead of wasting further time just do

not refer this book anymore.

I already have said this several times to many goiitians on

this site that "morrison and boyd" is really very complex &

congested + Huge to understand.

I would suggest you to go for a better,easy to understand

book like :

"MTG - Interactive Organic chemistry".

I use this book and i feel that this book is really very easy

to understand as well as grasp the topics that can be very

tough to know.

Hope you find it useful.

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And no need to go back to 10th class books.

"The Law of conservation of angular momentum states that

if no external torque acts on a body rotating about a fixed

point, the angular momentum of the body remains constant"

when external forces act on the particles,they exert torque

on them. Now the angular momentum of the particles change

with time. The rate of change of angular momentum is given

by:

dL =

d (r x p)

dt dt

On solving we will get

dL = (Torque)ext

dt

Now if in a system no external torque acts then we have :

dL = 0

dt

means here L = constant

Hence,This shows that in a rotatory motion, if no external

torque acts on the system of particles, the angular momen

tum remians constant. This is the law of " conservation of

momentum"

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Physics Analysis:

Class XI :- 53.7%
Class XII :-46.3%

Waves & Sound = 8%
Mechanics & General Physics = 33%
Heat = 13%
Modern Physics = 11%
Optics = 13%
Electricity & Magnetism = 22%

Hope you find it useful.
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Here is the link to the homepage of the University.

You can check out for the different queries directly from
this site itself.

http://www.annauniv.edu/

Hope you find it useful.
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yes mate
Even I think there is no need for HC Verma solutions coz
goiitians are always there to help you out when it comes
to solving your doubts.

So feel free to ask anyone if u have any doubts.You got
some great goiitians to help you out over here.

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Congrats dude
I really think that you deserve these 3K points that you
have got but mate "Life does not end even after achieving
what we wanted,It goes on and on till the very fate"

So Your fate is across 10K atleast as in 2 months if 3K
then I think you may easily cross an altitude of 10K in a
time of 5-6 months that means you are not far away.

Best of Luck for your rest of the time at GoIIT and do
what you are here for........"Rocking"

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Here are the contact details if u need them for Fiitjee:

Delhi(North-West)
31 , 32 , 33 Central Market
West Punjabi Bagh
New Delhi - 26
011 - 25222373,25222456

Delhi(East)
5th Floor,Roots tower,District centre
Laxmi nagar,Delhi-92
011 - 42448484,42448485

Delhi(South)
ICES House,29-A,Kalu Sarai
Sarvapriya Vihar
New Delhi-92
011 - 26515949,26569493

Delhi(NCR)(Gurgaon)
Classes conducted at D.P.S,
Sector - 45
For inquiry contact at Delhi
South.

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29)

The Force which may cause the tendency of motion or the
motion in the body is its own Weight & the applied horizontal
force of 15N.

We have Resultant of the forces as:
F_res =

(20)²+(15)²
F_res =

625
F_res = 25N.

In a Direction of:
@ = Tan^-1(15/20) = Tan^-1(3/4) = 37⁰ with the Vertical.

The Friction will always oppose the tendency of Relative
motion and will act in a direction opposite of Resultant F.
F = Force.

Now for the acceleration to be minimum:
Minimum Force required: (N=40=F applied against the wall)
F_min = F - uN
F_min = 25 - (0.5)40
F_min = 5N.

So the minimum Acceleration will be:
F_min=ma_min
5 = 2(a_min)
a_min = 2.5m/s².

Since @ = 37⁰ with the vertical so @ = 53⁰ with the
Horizontal.

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Rate if useful.

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Well

U got a great opportunity of taking Fiitjee in "Delhi" then

why miss it.

Fiitjee - Delhi is the best fitjee institute in India and is

heard of having the best Ex-iitians as professors that r

there to teach you.Moreover Fiitjee "Delhi" has got the

best passing rate as compared to all of its other branch

es in India.

I hope it helps.

Cheers !!!!!!!!!!!!!