Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • An air bubble of a radius 1cm in water has an upward acceleration 9.8 cm s-2. The density of water is 1gm cm-3 and water offers negligible drag force on the bubble. The mass of the bubble is (g=980 cm/s2)</sup

An air bubble of a radius 1cm in water has an upward acceleration 9.8 cm s-2. The density of water is 1gm cm-3 and water offers negligible drag force on the bubble. The mass of the bubble is (g=980 cm/s2)
Option: 1 4.51 gm
Option: 2 3.15 gm
Option: 3 4.15 gm
Option: 4 1.52 gm

Answers (1)

best_answer

According to Newtons 2nd law

B-mg=ma

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

Given

\begin{array}{l} \text { Volume } =\mathrm{V}=\frac{4 \pi}{3} \mathrm{r}^{3}=\frac{4 \pi}{3} \times(1)^{3}=4.19 \mathrm{~cm}^{3} \\ \mathrm{a}=9.8 \mathrm{~cm} / \mathrm{s}^{2} \end{array}

And

B=\mathrm{V} \rho_{w} \mathrm{g}

So

\begin{array}{l} \mathrm{m}=\frac{\left(\mathrm{V} \rho_{w} \mathrm{g}\right)}{\mathrm{g}+\mathrm{a}}=\frac{\mathrm{V} \rho_{\mathrm{w}}}{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} \end{array}

 

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question