TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Given, diameter of air bubble, d=1mm
=1×10−3m
Viscosity, η=0.85N−s∕m2
Density, ρ=900kg∕m3
Air density, σ=1.293kg∕m3
Let terminal velocity be vT.
∵‌‌vT=‌

2

9

rg(‌

ρ−σ

η

)
where, r is radius of bubble,
ρ is density of bubble
and σ is density of air.
∴‌‌vT‌‌=‌

2

9

×(‌

d

2

)2×10(‌

900−1.293

0.85

)
‌‌=‌

2

9

×(‌

1×10−3

2

)2×10(‌

900−1.293

0.85

)
‌‌=587.4×10−6=0.587×10−3
‌‌=0.587mms−1
‌‌=0.5mms−1