33Q. 36
P = dw/dt,
dw = F.dS
P=d(F.S)/dt
P = F.dS/dt ............ so here haven't we assumed F to be constant..
P = F. V
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94 Answers
331while i was giving tha paper , i knew that the answer will be d to question 38... :P :).thats y , i went staight to chk its answer and pointed that out immediately
1Q 8 is correct,
yeah, a little ambiguous..f(x) and g(x) are differentiable..assume this.
becoz when using d/dx(f.g)..we say f and g are differntiable
1q 51. in phy, there we have given dK/dt & we have to find dp/dt.but solution given is totally reversed...........
33Q 32....
??
What about m1=m2..... nothing was mentioned so if m1=m2 then it will not tilt..
62so what is the answer of 12 it is a power 4 only!!
let me check.. i may have been wrong!
38 ur point is taken... And u are absolutely correct
62in 32 you and abhishek have posted for the case when m1=m2
The answer comes by finding torque not by equating masses. radius!
The tension is not same on both the sides :)
The torque here is tension times distance. Not mass times distance!
138.. both wrong suppose... but din read the second statement wen i saw th efirst one is wrong...
62checking the question 55
btw in that question srinath... do remember that the net tension acting on the right side will be 2 times that value!
(that wont change the answer though)
62vishal what do u mean by net force.. I am not sure if that is the right expression! :O
In fact when M1=M2.... the balance will not tilt!
1Q.55
this method gives approx 25% .
x(46) + 92(1-x) = 57.5 where x is mol fraction
thus . solving for (1-x) we get 5 of N2O4 is 25 % approx . around 23% I think.
1Q32 cheak this too
1.net force lft=(M1+M2)/g
2. net force rt=|M1-M2|/g
62check the explantion for 49 there in the solution
the y square terms get cancelled out
1Q38
statement 2 is ambiguous-->>
not always true, we cant really say that
on the velocity of that point is zero-->> the horizontal plane on which it is moving may not be stationary..so we must take all possibilities
both statements are wrong in this way :))
1Q.49.
let the 3 projectiles be at 3 opsitions at a time t.
it's given ttha θi is diff as also vi
then, having their co-ordinates as (x i , y i) i=1,2,3
solving for the area of triangle with co-ordinates.
we get the area is proportional to t3
1yup. there , first velocity is maximum , then decreases , then attains zero at max. height? so acc is constant.
62my wealth can remain continuous for sometime..
Like say i have 1000 rupees now and it can remain 1000 for some bit of time!
Well try to derive mathematically.. u will get a slope exactly negative to the slope of the line first equation (did u notice that the first graph was velocity vs displacemnt!)
1and Q.4 .how's that possible ? it's not continuos . it changes everytime no ? then its not continous.
1also Q.30 . why can't it be the one with constant negative acceleration? eg. projectile motion.