A streamlined body falls through air from a height h on the surface of a liquid. If d and dd
d.Represents the densities of the material of the body and liquid respectively then the time after which the body will be instantaneously at rest, is
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Answer: 4
Explanation:
(d) Upthrust – weight of body = apparent weight
VDg-Vdg = Vda,
Where a = retardation of body
∴
a
=
(
D
−
d
d
)
g
∴a=D-ddg
The velocity gained after fall from h height in air,
v
=
√
2
gh
v=2gh
Hence, time to come in rest,
t
=
v
a
=
√
2
gh
×
d
(
D
−
d
)
g
=
√
2
h
g
×
d
(
D
−
d
)
t=va=2gh×d(D-d)g=2hg×d(D-d)
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