A stone of relative density K is released from rest on the surface of a lake. If various effects are ignored, the stone sinks in water with an acceleration of

g(1 – K)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

g(1 + K)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C

The net force acting on the stone in a downward direction would be equal to F = Vσg – Vρg, where m is the mass of the stone. The force acting downwards would be equal to Vσg due to its weight and the force acting upwards would be equal to Vρg due to the buoyant force. Hence we get

$F = V σ g - V ρ g = V σ g (1 - \frac{ρ}{σ}) = m g (1 - \frac{ρ}{σ}) = m g (1 - \frac{1}{K}) T h u s, a = g (1 - \frac{1}{K})$

Suggest Corrections

Similar questions

Q. A stone of relative density k is released from rest on the surface of a lake. If viscous effects are ignored, the stone sinks in water with an acceleration of

Two planets have the same average density but their radii are $R_{1}$ and $R_{2}$ . If acceleration due to gravity on these planets be $g_{1}$ and $g_{2}$ respectively, then

The radii of two planets are respectively $R_{1}$ and $R_{2}$ and their densities are respectively $ρ_{1} a n d ρ_{2}$ . The ratio of the accelerations due to gravity at their surface is

Q. If

a, a_{1}, a_{2}, . . . . . a_{2 n}, b

are in A.P.,

a, g_{1}, g_{2}, . . . . . g_{2 n}

are in G.P
and

a, h, b

are in H.P then:

\frac{a_{1} + a_{2 n}}{g_{1} g_{2 n}} + \frac{a_{1} + a_{2 n - 1}}{g_{1} g_{2 n - 1}} + . . . . . + \frac{a_{n} + a_{n + 1}}{g_{n} g_{n + 1}} =

Q. If n geometric means between a and b be

G_{1}, G_{2}, . . . . . . . G_{n}

and a geometric mean be G, then the true relation is

Join BYJU'S Learning Program