Solve this following
Question:

A air bubble of radius $1 \mathrm{~cm}$ in water has an upward acceleration $9.8 \mathrm{~cm} \mathrm{~s}^{-2}$. The density of water is $1 \mathrm{gm} \mathrm{cm}^{-3}$ and water offers negligible drag force on the bubble. The mass of the bubble is $\left(\mathrm{g}=980 \mathrm{~cm} / \mathrm{s}^{2}\right)$

 

  1. $3.15 \mathrm{gm}$

  2. $4.51 \mathrm{gm}$

  3. $4.15 \mathrm{gm}$

  4. $1.52 \mathrm{gm}$


Correct Option: , 3

Solution:

Volume $\mathrm{V}=\frac{4 \pi}{3} \mathrm{r}^{3}=\frac{4 \pi}{3} \times(1)^{3}=4.19 \mathrm{~cm}^{3}$

$a=9.8 \mathrm{~cm} / \mathrm{s}^{2}$

$\mathrm{B}-\mathrm{mg}=\mathrm{ma}$

$\mathrm{m}=\frac{\mathrm{B}}{\mathrm{g}+\mathrm{a}}$

$\mathrm{m}=\frac{\left(\mathrm{V} \rho_{\omega} \mathrm{g}\right)}{\mathrm{g}+\mathrm{a}}=\frac{\mathrm{V} \rho_{\omega}}{1+\frac{\mathrm{a}}{\mathrm{g}}}$

$=\frac{(4.19) \times 1}{1+\frac{9.8}{980}}=\frac{4.19}{1.01}=4.15 \mathrm{gm}$

 

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