Messages posted by ganesha1991

consider a small part of the semicircular wire dl

now dE at the centre due t o this part is = Gdm/R^2 sin@

teh cos componenets get cancelled

dm =M/L *dl

dl= Rd@

dm = M Rd@ / L

dE = G M Rd@ / L *R^2sin@

= GM/RL sin@ d@

inetgrate both sideslimits to righ t hadn side wil eb 0 to pie

E = GM/RL *[cos pie - cos 0]

= 2GM/RL

I f u want it in terms of pie

pieR = L

so E = 2GM/pieR^2 eliminating L

or

E = 2GMpie /L^2 eliminating R

2x^logx +1/ 3x^logx = 5x^logx= y2y+3/y = 52y^2 -5y + 3 =0solving y=1 or y=3/2 case 1-x^logx=1 log x= 1for this x=1this is the required soltuion x=1

x / |x |or | x | / x is signum function

a small block of mass m is kept at the left end of a larger block of mass M and length L. the syste

2^lo^g₃x + 3.x^log₃2 = 2

3.x^log₃2can be written as 3 2^log₃x

now let 2^log₃x=y

so y+3y=2

y=1/2

2^log₃x=2^-1

log₃x= -1

x = 1/3

electric field at the cirlce due to Q = kQ/l^2

electric filed at teh circle due to -Q = kQ/l^2

both in same directionso they add up total electric filed = 2kQ/l^2

now flux acroos the circle = E.s

= 2kQ./l^2 * pie R^2

= 2kQ pie R^2 / l^2

u r right

but here its about height

i dont think its necessary to add teh signs

it just confuses ppl

so didnt use them

anyways urs is absolutely correct

consider a small portion of the arc

now let the small portion be dl

so charge dq= λdl

but dl = a d@ where @ is teh angle subtended by it at the cnetre

thus dq=λ a d@

dE=K dq/a^2 sin@ bcos the cos@ components gets cancelled due to symmetry

dE = K λa d@ / a^2 sin@

integrating both sides with limits 0 to E for E and 0 to pie for right hand side

E =K λ/a *[cos pie - cos 0]

= 2K λ / a

let the intial velocityof th eball b4 the last 6m be =U

a=10 s=6 t=0.2

s=ut+1/2at^2

6=0.2 U+1/2*10*0.2*0.2

0.2U=5.8 U=29

now consider the journey from rest to this height

V=29m/s U=0 a=10

v^2=u^2+2as

29*29=2*10*s

therefore s= 29*29/20 = 42.05

thereforethe total height = 42.05+6 = 48.05m

for example u can consider two bodies of mass 4kg and 3 kg conencted to each other by a spring

so now if u want to solve any problem based on this and if u follow teh regular method it gets complicated

instead u can replace the system by a single body connected to the same spring of mass = 4*3/4+3 =12/7

this is called reduced mass

well relative speed = 4m/srelative distance=10mtiem taken by the boy to move from one side of trolley to the other =distance/speed = 10/4 = 2.5s initial speed of trolley=36km/h=10m/s time taken = 2.5 sdistance moved by the trolley = 2.5*10= 25 m

when u heat then the mettalic sphere expand there by volume increasesupthrust= mg - V*ro* g, here V increasesso upthrust decreases

thats because the electric field lines indicate the direction of the electric field so if it does form a loop then at a point it wil have 2 direction for the sam,e electri field- which isnt correctso electric field lines donot form loops

well as far as i know the forces are internal only seconds before th ebomb explodes buut when it does explode the internal force acts as the external force for the pieces of the bombso its no longer internal froce as the bomb on the hwole does not existso he law isnt violatedexternal force is applied

yup always welcome :D

ok ......if for any problem like the one given in the question if two masses are attached to eahc other by a srping the system is equivalent to one single mass attached to same spring whose mass is given as, mu = (m*M) / m+M where m and M are the masses attached to the spring this is called reduced mass.....hope u understud!!!

work done by spring is given as = -1/2kx^2

so its negative when the final extension is more than the initial extension

but if initial extensionm is m ore than final extension then its positive

so it may be negative or positive

the answer is c)

reduced mass = mM/m+M = 3*6/3+6 = 2

so the given system is equivalent to a block of ass 2 kg attached to same spring

so f= rootk/m /2pie = root 300/2 / 2 pie

= 1.95 = 2

answer is 2) 2Hz

1/2mr^2 = mk^2 K = r/root2 = 2.5 / root 2 = 1.76 so the answer is c

let the length of train be L

at this moment the train has velocity V1

after some time the rear part crosses teh same place with velocity V2

so initial velocity = V1

final velocity = V2

distance covered = L

V^2=u^2+2as

thus a =v2^2-v1^2/ 2L

now consider the middle part inital velocity = v1

distance covered = L/2

a=v2^2-v1^2/ 2L

V^2 = V1^2 + 2* (v2^2-v1^2/ 2L)*L/2

v^2 = V1^2 +V2^2 /2

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