Question

A solid sphere of radius r made of a material of bulk modulus B is surrounded by a liquid in a
cylindrical container. A massless piston of area a floats on the surface of the liquid. When a mass
mis placed on the piston to compress the liquid, the fractional change in the radius of the sphere
(dr/r) is :
(A) Balmg (B) B a/3 mg
(C) mg/3 B a (D) mg/B a​

Answer 1

Answer:

• Fractional Change Will be Mg/3 AB.

Given:

1. Bulk Modulus = b,
2. Weight = mg.

Explanation:

$\rule{300}{1.5}$

From Volume of Sphere,

$\large\bigstar \; {\boxed{\tt V = \dfrac{4}{3} \pi R^3}}$

$\bold{Here}\begin{cases}\text{V Denotes Volume} \\ \text{R Denotes Radius}\end{cases}$

$\large{\boxed{\tt V = \dfrac{4}{3} \pi R^3}}$

For Fractional Change

$\large{\tt \longmapsto \dfrac{\Delta V}{V} = 3\Bigg[\dfrac{\Delta R}{R}\Bigg]}$

• As 4/3 & π are Constants.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Bulk Modulus Formula.

$\large\bigstar \; {\boxed{\tt B = \dfrac{P.V}{\Delta V}}}$

$\bold{Here}\begin{cases}\text{V Denotes Volume} \\ \text{V Denotes volume} \\ \text{B Denotes Bulk Modulus}\end{cases}$

Now,

$\large{\boxed{\tt B = \dfrac{P.V}{\Delta V}}}$

$\large{\tt \longmapsto B = \dfrac{P.V}{\Delta V}}$

We Know,

Pressure = Force/area

⇒ P = F/A

⇒ P = Mg/A

∵ [Here Force is Caused due to Weight of the body]

Substituting,

$\large{\tt \longmapsto B = \dfrac{Mg}{A} \times \dfrac{V}{\Delta V}}$

Now,Interchanging.

$\large{\tt \longmapsto B \times \dfrac{\Delta V}{V} = \dfrac{Mg}{A}}$

But we Know,

∵ [ΔV/V = 3ΔR/R]

Substituting,

$\large{\tt \longmapsto B \times \dfrac{3 \Delta R}{R} = \dfrac{Mg}{A}}$

$\large{\tt \longmapsto B \times \dfrac{3 \Delta R}{R} = \dfrac{Mg}{A}}$

$\large{\tt \longmapsto 3B \times \dfrac{\Delta R}{R} = \dfrac{Mg}{A}}$

$\large{\tt \longmapsto \dfrac{\Delta R}{R} = \dfrac{Mg}{A \times 3B}}$

$\large\longmapsto{\underline{\boxed{\red{\tt \dfrac{\Delta R}{R} = \dfrac{Mg}{3AB}}}}}$

∴ Fractional Change Will be Mg/3 AB.

$\rule{300}{1.5}$

Answer 2

Answer:

Given:

Radius of sphere is R

Bulk modulus of sphere = B

Mass of "m" is placed on Piston to compress the fluid.

Surface area of Piston = a

To find:

Fractional change in radius of sphere.

Calculation:

We know that,

Volume of sphere = 4/3 πr³

For small changes in radius, the corresponding change in volume can be represented as follows:

∆V/V = 3(∆r/r)

Now force applied on Piston = mg

∴ pressure = F/A = mg/a

Now, bulk modulus (B):

B = volume stress/vol. strain

B = P(V/∆V)

=> B = P { ⅓(r/∆r)}

=> B = {mg/a}{⅓(r/∆r)}

=> ∆r/r = ⅓mg/aB

So the final answer is

$frac. \: change \: = \frac{mg}{3aB}$