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Catalogs Discussion Forums -> Optics -> why the power of glass is zero? -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
3 replies   
thanks rtiit it helped me 2
Catalogs Discussion Forums -> Trignometry -> interesting ques. -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
4 replies   
if          2sinA
    ----------------------------=y    
     1 + cosA + sinA

then         1 - cosA + sinA
             --------------------------------=?
                   1 + sinA
Catalogs Discussion Forums -> Differential Calculus -> Interesting ques. on differentiation rates assured -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
8 replies   
can u give the detail?
Catalogs Discussion Forums -> Mechanics -> CHOICE -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
1 replies   
is da ans (b)?
Catalogs Discussion Forums -> Differential Calculus -> Interesting ques. on differentiation rates assured -> Go to message
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8 replies   
If  f(x)= (x-a)(x-b)(x-c)..............................................(x-z)
then find the value of
f'(x)=?
Catalogs Discussion Forums -> Mechanics -> HCV: Laws of motion -> Go to message
This Post 10 points    (Olaaa!! Perrrfect answer.   in 2 votes )   [?]
3 replies   
for 28
let T' be tension in the string att to m1 & T be for other
from observation, we get T'=2T
let a1 be acc of m1 a2 for m2 a3 for m3
let a1 dwnwards
hence, a2=a-a1 & a3=a+a1 .............a=common acc. for m2 n m3
for m1,
T'-m1g=m1a1
2T-m1g=m1a1 .......................1

now,
m3g-T=m3(a+a1) .........2
T-m2g=m2(a-a1) .................3

2m3g-m1g=2m3a+a1(m1+2m3)
putting values,
5g=6a+7a1 .................4
2m2g-m1g=-2m2a+a1(m1+2m2)
3g=-4a+5a1 ..............5

solving these 2 equations, we get,
a=2/29g m/s.s

a1=19/29g m/s.s

a2=17/29g m/s.s

a3=21/29g m/s.s

s=ut+1/2at2
0.2=0+ 1/2* 19/29g*t2

t=0.25sec
rate if u r convinced
Catalogs Discussion Forums -> Mechanics -> friction -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
4 replies   
Wt of the body(downwards)=0.2*10=2N
applying the force equation, we get
upward force=downward force ........since it's in equm
hence, frictional force=wt of the body ........f.f being the self adjusting force
f.f=2N is the ans.
Catalogs Discussion Forums -> Mechanics -> laws of motion -> Go to message
This Post 12 points    (Olaaa!! Perrrfect answer.   in 3 votes )   [?]
4 replies   
hey, it's simple
acc. of elevator is more than acc. due to gravity
so, the block present in the elevator will fall freely with acc.= g
as it starts frm rest, u=0
s=ut + 1/2at.t
s=0 + 1/2g.t.t
s=1/2*10*0.2*0.2
s=0.2m
Catalogs Discussion Forums -> Mechanics -> Relatitvity -> Go to message
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Space and Time in Classical Mechanics
It is not clear what is to be understood here by "position" and "space." I stand at the window of a
railway carriage which is travelling uniformly, and drop a stone on the embankment, without
throwing it. Then, disregarding the influence of the air resistance, I see the stone descend in a
straight line. A pedestrian who observes the misdeed from the footpath notices that the stone falls
to earth in a parabolic curve. I now ask: Do the "positions" traversed by the stone lie "in reality" on a
straight line or on a parabola? Moreover, what is meant here by motion "in space" ? From the
considerations of the previous section the answer is self?evident. In the first place we entirely shun
the vague word "space," of which, we must honestly acknowledge, we cannot form the slightest
conception, and we replace it by "motion relative to a practically rigid body of reference." The
positions relative to the body of reference (railway carriage or embankment) have already been
defined in detail in the preceding section. If instead of " body of reference " we insert " system of
co?ordinates," which is a useful idea for mathematical description, we are in a position to say : The
stone traverses a straight line relative to a system of co?ordinates rigidly attached to the carriage,
but relative to a system of co?ordinates rigidly attached to the ground (embankment) it describes a
parabola. With the aid of this example it is clearly seen that there is no such thing as an
independently existing trajectory (lit. "path?curve" 1)), but only a trajectory relative to a particular
body of reference.
In order to have a complete description of the motion, we must specify how the body alters its
position with time ; i.e. for every point on the trajectory it must be stated at what time the body is
situated there. These data must be supplemented by such a definition of time that, in virtue of this
definition, these time?values can be regarded essentially as magnitudes (results of measurements)
capable of observation. If we take our stand on the ground of classical mechanics, we can satisfy
this requirement for our illustration in the following manner. We imagine two clocks of identical
construction ; the man at the railway?carriage window is holding one of them, and the man on the
footpath the other. Each of the observers determines the position on his own reference?body
occupied by the stone at each tick of the clock he is holding in his hand. In this connection we have
not taken account of the inaccuracy involved by the finiteness of the velocity of propagation of light.
With this and with a second difficulty prevailing here we shall have to deal in detail later.
 
Catalogs Discussion Forums -> Mechanics -> laws of motion -> Go to message
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1 replies   
A smooth right circular cone of semi vertical angle =tan-15/12 is at rest on a horizontal plane. A rubber ring of mass 2.5 kg which requires force of 15N for an extension of 10cm is placed on cone. Find increase in the radius of ring in equilibrium.
 
 
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