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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 May 2007 22:55:17 IST
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1)a pipe ABCD is bent into three sections horizontal AB vertical BC (C below B) and horizontal CD . In AB liquid flowing velocity is v1 and pressure is p1 similarly it is v2,p2 in CD then ... v1=v2 dn p1=p2 TRUE OR FALSE!! 2)the kinetic energy of a gas molecule is 3PV/2 for monoatomic gas only- T/F 3)a stick is thrown in air then the CM will travell in parabolic path in all cases - T/F
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27 May 2007 22:57:30 IST
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F,T,T
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27 May 2007 23:02:43 IST
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WHAT WOULD BE THE ANSWER TO THE FIRST ONE!
VI=V2 IS CORRECT FOR SURE SINCE AREA IS UNIFORM. PI AND P2 MEIN KYA RELATION HAI?
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27 May 2007 23:19:02 IST
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From Bernoulli's equation: P 1 + 1/2  v 12 + gh 1 = P 2 + 1/2 v 22 + gh 2 As v1 = v2 So , the equation reduces to : P 1 + gh 1 = P 2 + gh 2 But since h1 is not equal to h2 So, P1 not equal to P2 Hence false. II) No single gas molecule can have 3/2.PV. The energy of one gas molecule (ideal gas irespective of whether it is monoatomic or diatomic) is 3PV/2NA. Hence false.
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27 May 2007 23:20:43 IST
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It is not given in the question that the c.s.area of the pipe is uniform.
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Let us build a new world with love, peace, happiness and engineering! (DON'T CHOOSE THE ODD ONE OUT)
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27 May 2007 23:29:03 IST
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how last one is true
consider the case of stick going directy upwards
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28 May 2007 08:34:09 IST
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for first:: p2 should be greater that p1, & v1 should be equal to v2..
for 2nd:: 1/3mnu^2 = pv -->according to kinetic theory
so, kinetic energy should be 3pv/2 irrespective of no. of atoms in a molecule of the gas...
for 3rd:: the center of mass is constant, so if we just consider the center of mass, then it will be a point following a parabolic path...
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28 May 2007 09:07:59 IST
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The second one is false because in the question it is asked about the kinetic energy of one single molecule. Moreover there is no requirement of being monoatomic, the gas just has to be ideal. The third one is correct.
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28 May 2007 09:25:25 IST
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the third one has to be true...
it is because there is no EXTERNAL force to "disturb" the COM and hene it will continue to move in the original path....
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28 May 2007 10:38:56 IST
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how do you know it has been thrown at a some angle
plz calrify
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28 May 2007 11:02:01 IST
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hey arnnie.
Kinetic energy for monoatomic and diatomic gases are different.
KE=fRT/2 or fPV/2
monoatomic has f=3 while diatomic has f=5
so A,C options both seem correct .. what is wrong??...
options are
A)3pv/2 for monoat. gas only
B)3pv/2 for diat. gas only.
C)>3PV/2 for diat.
D)3PV/2 in all cases.
@prakriteesh
the kinetic energy of a single molecule is fiven by fR/2 and kinetic energy per mole is fKt/w(K being the botlzman const) @arsenal..(liverpool rules btw) the stick being thrown exactly upwards is noot considered as an ideal condition, it is not possible to throw it directly upwards. moreover the options were.. a)in all cases b)only if stick is uniform c)only if stick is not rotating d)only if com is INSIDE the body and not at any point outside it. You are right in saying that if we are able to throw is directly up the com wouldtravell in a straight line. But seeing the options that case is rejected!!
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30 May 2007 20:27:14 IST
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joy, according to your statement i can draw a conclusion of just option A) coz' diatomic gases cannot hav KE=4pv/2 which is included in option C) so, A) must be correct
P.S. i think i have done this question wrong in the exam..
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