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Catalogs Discussion Forums -> Mechanics -> Cylinderical can........... -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]


 



Wt of the cylinder=Wt of the liquid displaced= 2(pie)rtps (r+l)g


Volume of the liquid displaced =Volume of cylinder submerged= 2(pie)rtps (r+l)/ pl


Total volume of cylinder= (pie) r2


Hence

 f=Volume of cylinder submerged / total volume of cylinder


f= [2(pie)rtps (r+l)/ pl ]/ (pie) r2 l




 


f=2tps (r+l)/pl rl




 


 

Catalogs Discussion Forums -> Mechanics -> A uniform rod of length l is suspended from one end such that it is free to rotate about a -> Go to message
This Post 7 points    (Olaaa!! Perrrfect answer.   in 2 votes )   [?]

Applying conservation of momentum:


Hence velocity of combined body V= u/2


where u is velocity of sphere before collision:


Total energy required to be imparted to the system to make full rotation:


2mgl=1/2 (2m)(u/2)2


u=2 root(2gl)

Catalogs Discussion Forums -> Mechanics -> laws of motion........... -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]

All have solved it correctly:


Now let us explain it in my own way


The man feel his wt. because of normal reaction on his feet. Let us consider downward direction +tive


There only two forces acting on the man


1. Normal reaction "N" upward


2. Wt i.e. mg downward


Hence  Summation of all forces= ma


mg-N=mg


N=m(g-a)


Hence result


 


Now in second case:


only direction of accn has changed:


Hence a in above equation must be replaced by  (-a )


N=m(a+g)   = W/g (a+g)=W(a/g +1)

Catalogs Discussion Forums -> Mechanics -> Cylinderical can........... -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

Wt of the cylinder=Wt of the liquid displaced= 2(pie)rtps (r+l)g


 


Volume of the liquid= 2(pie)rtps (r+l)/ pl


 


Total volume of cylinder= (pie) r2 l


 


Hence


 


f= [2(pie)rtps (r+l)/ pl ]/ (pie) r2 l


f=2tps (r+l)/pl rl


 


 

Catalogs Discussion Forums -> Mechanics -> Please solve it -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

m


From above fig:


Let m=6kg, M=15kg


=300


 


N1=m(g-a1)


But a1=a2


Hence N1=m(g-a2 ) ..............(1)


For M:


Ma2=(N1+Mg)..........................(2)


From (1), (2)


a2=g(m+M)/{M+m2 }


a2=70/11 m/sec2


a1=35/11 m/sec2


 

Catalogs Discussion Forums -> General Physics -> IS MOTION OF PULLEYS & CONSTRAINT A PART OF COURSE FOR JEE IN 2010? -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

Hi


I am solving Erodov prob u have asked me


Let at time t the position of the particle is P(x1, y1) and at t+dt is Q(x1+dx, y1+dy)


ds=dx i + dy j where ( x1=R, y1=R )


ds/dt= -Rd i + Rd j


v= -Rd/dt i + Rd /dt j


Vav where limit for Theta is "0" to 1800


=-2R/T i + 0. j


|Vav | = 2R/T   This is average of velocity vector.


 


Mean velocity is easy to find:


Distance= Pie R Time =T


V=(Pie R) / T


 


,

Catalogs Discussion Forums -> Mechanics -> Mechanics -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

Initial height h is gained with uniform velocity v


Hence t1 = h/v .....................(1)


Now engine fails , hence its speed will be zero after further going h1 height in next t2 sec.


h1 = v2 /2g..........................(2)


t2 = v/g....................................(3)


Total height H= h+h1 =1/2 g t3 2  where t3 is time to descend


v2 /2g +h =1/2 g(t -t1 -t2)2  ...........(4)


Now substitute the value of t1 and t2 in equation (4) and solve for h


You will get the answer.


 

Catalogs Discussion Forums -> Mechanics -> NLM -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

Conside the figure sent by Sivam:


The triangle shows the corners that will contain Centre of mass of base spheres. Normal reaction(R)due to top sphere will make an angle = 1/


Tension T will make 300 with each of the bisectors of the base c.g. angles of triangle.


Hence  3R=2   (weight of top sphere)


3R( ) =2  


R=2


2T=R=2/


T=2/3


Since safety of faxtor is 3


Tmax =2

Catalogs Discussion Forums -> Mechanics -> centre of mass -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]

Let at time t velocity of trolley is vt


mass of trolly=(m+ut)


momentum of the system=v(ut) + (m+ut)vt


At time t+dt, corresponding velocity of trolley vt + dvt , mass=m+u(t+dt)


momentum of the system=vu(t+dt) +(m+ut+udt)(vt + dvt )


Since there is no external forces acting on the system, hence momentum is conserved:


v(ut) + (m+ut)vt =vu(t+dt) +(m+ut+udt)(vt + dvt )


Dividing both sides by dt we get:


(m+ut)dvt /dt= -u(v+vt )


Solving above dirrerential equation for vt , we get:


(m+ut)(v+vt ) = K( a constant)


when t=o, vt =0


Hence k=mv


Hence vt = - uvt/(m+ut)


dvt / dt= accn = -muv/(m+ut)2


 

Catalogs Discussion Forums -> Mechanics -> NLM -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

Tmax=2

Catalogs Discussion Forums -> Mechanics -> Two blocks of mass 2 kg and M are at rest on an inclined plane and are separated by a..... -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

e=0.8415


M=16.475kg


I will explain if this answer is correct

Catalogs Discussion Forums -> Mechanics -> a -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

When its accn :


(50)2 =100+2as


2as=2400............................(1)  where a=accn,  s=distance


When its direction of  acce is reversed, its velocity will first become zero then goes in reverse direction with accn  a


0=(50)2 - 2as1


2as1=2500...............................(2)


Hence velocity at starting point: Total distance =s+s1


v2=0 + 2a(s+s1 )=4900


v=70m/s Ans

Catalogs Discussion Forums -> Mechanics -> centre of mass -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

Let vt is velocity of trolley after t sec.


Hence (v- vt) u=mdvt /dt


Solving above equation you get:


vt=v(1- e-u/m t )   Ans.


Assumptions:


1. No dust accumulate in the trolley.

Catalogs Discussion Forums -> Mechanics -> collision-impulse-momentum......... click image to enlarge -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

Hi Zeosthegod


 


Although you have cosen right options,


but you have solved incorrectly the question (ii)


because u/2i + u j= root of(5) X u/2


So use here energy equation to find velocity of the ball, then solve it

Catalogs Discussion Forums -> Mechanics -> Find the acceleration of the following:- -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

The resultant accn of block B is vector sum of  b and c. So add them vectorically and get the answer.

Catalogs Discussion Forums -> Mechanics -> a unifrm sq plate side "A",mass "M" a sq portion,side A/2 removed,calc MOI remainig al -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]

3/16 Ma2

Catalogs Discussion Forums -> Mechanics -> work power energy -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

I am trying to solve this:


Let distance travelled on the plane = d


Elongation in spring= dx=Square root of(l0 2 + d2)  - l0  


Tension in spring= T = k(dx) = (5mg/l0 )(dx)


Now let spring makes an angle  with horizontal plane


Remember just at the time of taking off the mass from the floor (mg) equals the vertical component of T


Hence T =mg


(5mg/l0 )(dx){l0/root(l02 + d2)}=mg


But dx=Square root of(l0 2 + d2)  - l0   Solving for (d/l0)


We get : d= (3/4)l0


Now writting energy equation for the system we get:


k(dx)2 + (2mv2)  = 2mgd


Now put the value of d, we get


v= square root of (19/32 gl0 ) = 1.7m/s


  


 


 


  

Catalogs Discussion Forums -> Mechanics -> work power energy -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

nobody can understand this problem unless u attach the diagram or indicate the exact location of spring , masses and points  O, A and P

Catalogs Discussion Forums -> Mechanics -> Find the acceleration of the following:- -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]

Hey


Correct answers are:


B, C and A

Catalogs Discussion Forums -> Mechanics -> work energy, centre of mass -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

Hi ABHI


In an elastic or plastic collision reduction in energy appears in the form of heat and some part is consumed in deforming the bodies in question.


So tell me where this energy is going? Is there any deformation ?

 
 
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