आपके क्षेत्र में स्थाई स्थिति स्थानीय संस्था ग्राम पंचायत ग्राम ग्राम पंचायत नगर निगम नगर पालिका में जाकर पता कीजिए उस में महिलाओं की संख्या अनुसूचित जनजाति अनुसूचित जाति के कितने कितने लोग हैं उनके कार्यों के बारे में पता कर ली बद्दल लिखिए क्लास 10th
Answers
Explanation:
Answer:
Rate at which volume of the bubble is increasing = 72 π cm³/s
Step-by-step explanation:
Given:
Radius of an air bubble is increasing at the rate of 2 cm/s
To Find:
The rate at which the volume of the bubble is increasing when the radius of the air bubble is 3 cm.
Solution:
Let the radius of the air bubble be r cm and volume be V.
Here the air bubble is in the shape of a sphere.
Volume of a sphere = 4/3 × π × r³
Now the rate of volume change with respect to time is given by,
\sf \dfrac{dV}{dt} = \dfrac{d}{dt} (\dfrac{4}{3}\: \pi r^{3})
dt
dV
=
dt
d
(
3
4
πr
3
)
Using chain rule,
\sf \dfrac{dV}{dt} =\dfrac{d}{dr} (\dfrac{4}{3}\: \pi r^{3} ).\dfrac{dr}{dt}
dt
dV
=
dr
d
(
3
4
πr
3
).
dt
dr
Differentiating,
\sf \dfrac{dV}{dt} =\dfrac{4}{3}\: \pi \times 3r^{2} .\dfrac{dr}{dt}
dt
dV
=
3
4
π×3r
2
.
dt
dr
Now the rate of increase of radius of the air bubble is given as,
\sf \dfrac{dr}{dt} =2\:cm/s
dt
dr
=2cm/s
Also by given, the radius of the air bubble is 3 cm.
Substitute the data,
\sf \dfrac{dV}{dt} =\dfrac{4}{3}\: \pi \times 3\times 3^{2} \times 2
dt
dV
=
3
4
π×3×3
2
×2
Simplifying we get,
\sf \dfrac{dV}{dt} =4\times \pi \times 9\times 2
dt
dV
=4×π×9×2
⇒ 72 π cm³/s
Therefore the rate at which the volume of the air bubble is increasing is 72 π cm³/s.