kusangati se hone wale hania batate Hue chote bhai aap bahan ko Patra likhiye Patra lekhan
Answers
Answer:
Radius of copper ball, \sf{r=2\times10^{-3}\ m.}r=2×10 −3m.
Terminal velocity of copper ball, \sf{v=6.5\times10^{-2}\ m\,s^{-1}.}v=6.5×10
−2 ms −1 .
Density of copper, \sf{\rho=8.9\times10^3\ kg\,m^{-3}.}ρ=8.9×10
3 kgm −3
Density of oil, \sf{\sigma=1.5\times10^3\ kg\,m^{-3}.}σ=1.5×10
3kgm −3 .
Let \sf{g=10\ m\,s^{-2}.}g=10 ms
−2 .
The expression for terminal velocity,
\longrightarrow\sf{v=\dfrac{2gr^2(\rho-\sigma)}{9\eta}}⟶v=⟶v=
9η
2gr
2
(ρ−σ)
⟶v=
9η2gr 2 (ρ−σ)
Then, coefficient of viscosity,
\longrightarrow\sf{\eta=\dfrac{2gr^2(\rho-\sigma)}{9v}}⟶η=⟶η=
9v
2gr
2
(ρ−σ)
⟶η=
9v2gr 2
(ρ−σ)
\longrightarrow\sf{\eta=\dfrac{2\times10\times(2\times10^{-3})^2\times10^3(8.9-1.5)}{9\times6.5\times10^{-2}}}⟶η=⟶η=
9×6.5×10
−2
2×10×(2×10
−3
)
2
×10
3
(8.9−1.5)
⟶η=
9×6.5×10 −2
2×10×(2×10 −3 ) 2×10 3 (8.9−1.5)
\longrightarrow\sf{\eta=\dfrac{2\times10\times4\times10^{-6}\times10^3\times7.4}{9\times6.5\times10^{-2}}}⟶η=⟶η=
9×6.5×10
−2
2×10×4×10
−6
×10
3
×7.4
⟶η=
9×6.5×10 −22×10×4×10 −6×10 3 ×7.4
\longrightarrow\sf{\underline{\underline{\eta=1.01\ kg\,m^{-1}\,s^{-1}}}}⟶⟶
η=1.01 kgm
−1
s
−1
⟶
η=1.01 kgm−1s−1
Answer:
u r in which class?
I also want G-f
but i am In 8th
and you?