2. He would (6)
to kiss the trees lovingly.
3. He would whisper kind words to (c)
4. This (d)
odd but beautiful.
5. It was indeed (e)
noble act in this wicked world.
(ii) on
1. (a) (i) to
2. (6) bend
3. (c) (i) these
4. (d) (i) had
5. (e) (i) an
(ii) bent
(ii) them
(iii) at
(iii) bound
(iii) their
(iii) is
(iii) a
(iv) by
(iv) band
(iv) those
(iv) were
(iv) of
(ii) was
(ii) the
Answers
Answer:
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.