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A balloon having mass ‘m’ is filled with gas and is held in hands of a boy. Then suddenly it get released and gas start coming out of it with a constant rate. The velocities of the ejected gases is also constant 2 m/s with respect to the balloon. Find out the velocity of the balloon when the mass of gas is reduced to half.    Plz explain it briefly and step by step
Asked by vishakhachandan026 | 05 Aug, 2019, 08:44: AM
answered-by-expert Expert Answer
Upward thrust Ft acting on balloon due to ejected gas is given by ,    Ft = - v (dm/dt)  ........................ (1)
 
where v is velocity of ejected gas,  dm is decreass of mass of baloon in time dt.
Negative sign appears in eqn.(1) because (dm/dt) is negative.
 
Accelecration  dv/dt  = Ft / m  = - (v/m) dm/dt   ...................(2)
 
since we have constant ejection velocity 2 m/s, eqn.(2) is written as,     dv = -2 (dm/m)  ......................(3)
 
Let m0 be initial mass of balloon and V is velocity of balloon when mass is reduced to half
 
By integaration of eqn.(3),  we get,  begin mathsize 14px style integral subscript 0 superscript V d v space equals negative space 2 integral subscript m subscript 0 end subscript superscript bevelled m subscript 0 over 2 end superscript fraction numerator d m over denominator m end fraction end style  ..........................(4)
Hence V = -2 ln(1/2) = 1.386 m/s
Answered by Thiyagarajan K | 05 Aug, 2019, 10:32: AM

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