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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: HC Verma Page no. 80 ch:5 Newton's Laws of motion Exercise Q no. 19
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coolboy910 (5)

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HC Verma Page no. 80 ch:5 Newton's Laws of motion Exercise Q no. 19
    
chetan_tg (8)

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buoyancy = B


resistance = R


i. balloon moving downwards


B + R = Mg


ii. balloon moving upwards


B = (M-x)g + R


solve for x.

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varun.tinkle (1130)

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I WAS SOLVING THE SUM TODAY ITSELF
BUYOANCY=B (CONSTANT)
LET THE CONSTANT OF AIR RESISTANCE BE K
THEREFORE IN BOTH THE CASES ITS VALUE BE KV
IN THE FIRST CASE SINCE THE BALLOON IS NOT ACC THE MAG OF OPP FORCES WILL BE EQUAL
ACCORDING TO QUES THE AIR RESISTANCE WILL BE OPP TO VEL
SO
B+KV=MG
IN THE 2ND PART
LET THE MASS REMOVED BE m
THEREFORE THE NEW MASS = (M-m)
THE BAOON IS RISING UP SO THE AIR RESISTANCE WILL BE IN THE OPP DIRECTION
B=(M-m)G+KV
SOVING THE EQUATIONS WE GET
m =2KV/G
IN THE 2ND PART
KV=MG-B
SO 2KV.G
=2(MG-B)/G=
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