28-02-09 What will be the mass?

answer to bhaiyyas prev. question :-

"@everyone.. is the mass of air measured when we put a box on a weight?

I mean an open box of mass M has a small mass of air dM

thenis the weight given by

Mg or (M+dm)g ?
"

no not in open box ,
not even in a thin closed box(thick box mein toh it will be lesser than the box's weight)
but if the closed box is weighed in vacuum , then the mass of air will be measured too........

but bhaiya plz explain that last case(in original question) a bit more elaborately.....

un able to understand situation clearly in (3)....[2] ....

i think the bird will feel an upthrust because of the air in block when it is flying .............

let that be B N ................and the bird prodices flight similar to a rocket in this case[3] ...

so an equal and opposite rxn acts on the box ................ now if F is the thrust for flight created by the bird then and if the bird is flying upward (but same time not touchin the roof) then F>mg - B ....................or F+B>mg ...

LOOK AT THIS CAREFULLY ULL SEE THE FORCE ON THE BASE DUE TO THE BIRD IS F+B...

hence the readin will be greater that M+m !!!!!!!!!!!!!

if the bird is steady (ie not goin up nor comin down) then F+B =mg so readin = M+m ....

and if bird is decending then F+B<mg so reading will be less than M+m............

now if its standing on the floor then its surely M+m .......

For (3)
For closed cage
If the cage and bird move upward with acceleration 'a' then weight recorded wud be (Mg+mg+Ma+ma)/g
If slowly lifted then M+m

For open cage....
i dont know..

For (2)
For a closed cage..... it will be (Mg+mg)/g if the bird is flying horizontally or very slowly upwards or downwards... if it moves with some acceleration then obviously answer will be different.. if it moves with acceleration 'a',
then reading will be (Mg + mg + ma)/g if acceleration is upwards or (Mg + mg - ma)/g if acceleration is downwards.

For an open cage.... nothing can be said as it depends on movement of air ... however if air only moves horizontally. then answer will be (Mg)/g. as the air exerts no downward force.. however if the air exerts downward force (say F) then reading = (Mg+F)/g

sry posting my answer again i hadnt read the question properly...

For 1) if we neglect air pressure, then
for both open and closed cages reading will be (Mg+mg)/g i.e. M+m

in first case it is mg+M

second case it cant be said bcoz the bird is flying so their is change in mass at every instance

third case (M+m)g

let nishant point some mistakes in our ans and Nishant plz see the pirates question

OPTION 2

and in both the cases u r mentioning my ans will be the same too

there is a case which is eating at my mind rite now..

will the mass be same (oops the reading i mean ) even when the bird is accelerating in air and when she (or it if you like) is stationary..

anyways i still answer M+m each time..(for closed box)

reason matching to quite extent to gaurav bhaiya

i think i have given the reason too

by the pressure of the air bird is doveloping by its wings to remain in air

how would that matter? in the 2nd case u mean?

i will back phillip with the ans

1) M + m
2) M
3) M - m
???????????

Looks very easy but i've learnt that such questions are major traps..........

M+m each time i think....