A streamlined body falls through air from a height h on the surface of a liquid. If d and D (D > 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
Text Solution
Verified by ExpertsD
Up thrust – weight of body = apparent weight
$\nabla Dg - \nabla dg \equiv \nabla da,$
Where a = retardation of body ∴ ∴ $a=\left(\frac{D-d}{d}\right)g$
The velocity gained after fall from h height in air, $v = \sqrt{2gh}$
Hence, time to come in rest,
$t = \frac{v}{a} = \frac{\sqrt{2gh} \times d}{(D - d)g} = \sqrt{\frac{2h}{g}} \times \frac{d}{(D - d)}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems