Question

A small spherical object having density 2p is placed inside the liquid of density p and connected to an ideal string of length / whose other end is kept fixed so that the object can move in a vertical circle. Find minimum speed (in m/s) which must the given at the bottom to just complete the vertical circle. [Ignore viscosity or any such other drag forces] (g = 9.8 m/s2, 1 = 2m)​

Answer 1

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>>Physics

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>>Variation of Pressure With Depth

>>A small spherical ball of density p is g

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A small spherical ball of density p is gently released in a liquid of density o(p> o) Find the initial acceleration of the ball

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Solution

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Given that,

Density of water = o

Density of spherical ball = p

We know that,

A small spherical ball of density p is gently released in a liquid of density o(p> o) Find the initial acceleration of the ball

Medium

Solution

verified

Verified by Toppr

Given that,

Density of water = o

Density of spherical ball = p

We know that,

Net force = Weight – buoyancy force

W−Fbuoyancy

=Fmg−oVg=ma....(I)

Now, mass = density× Volume

Put the value of mass in equation (I)

pVg−oVg=pVa

Now, volume = mass/density

So, the initial acceleration is

p( pm )g−o( pm)g=p(pm (p−o)g=ma

p(p−o)g =a

Hence, the initial acceleration is

p(p−o)g m/s 2

Answer 2

Let $g_{eff}$ be the effective acceleration due to gravity.

Let $F_{b}$ the force acting on the sphere in the upward direction.

Then,

$mg_{eff} = mg-F_{b}$

$\rho_{s} .V.g_{eff} = \rho_{s}.V.g - \rho_{L} .V.g\\\\2.\rho .V.g_{eff} =2. \rho.V.g - \rho .V.g\\\\g_{eff} = \frac{\rho Vg}{2\rho V} \\\\g_{eff} = \frac{g}{2}$

The minimum speed required to complete the vertical circle =

$V_{min} = \sqrt{5g_{eff}l } \\\\V_{min} = \sqrt{\frac{5}{2} gl }$