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Chapter 25: Problem 2804

A solid sphere of radius \(\mathrm{R}\) made of a material of bulk modulus \(\mathrm{K}\) 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 \(\mathrm{M}\) is placed on the pisten to compress the liquid, the fractional change in the radius of the sphere \((\delta \mathrm{R} / \mathrm{R})\) is.......... (A) \(\\{(\mathrm{Mg}) /(3 \mathrm{KA})\\}\) (B) \(\\{(\mathrm{Mg}) /(\mathrm{KA})\\}\) (C) \(\\{(\mathrm{Mg}) /(4 \mathrm{KA})\\}\) (D) \(\\{(\mathrm{Mg}) /(2 \mathrm{KA})\\}\)

Short Answer

Expert verified
The short answer is: \(\frac{δR}{R} = \frac{Mg}{3KA}\).

Step by step solution

01

Write the formula for bulk modulus

The bulk modulus (K) is the ratio of the stress (force per unit area) acting on the object to the relative change in volume (ΔV/V). Mathematically, it can be written as: \[K = -\frac{P}{ΔV/V} \]
02

Analyze the forces exerted on the piston

When the mass M is placed on the piston, it exerts a force equal to \(Mg\) (where \(g\) is acceleration due to gravity). This force compresses the liquid and thus affects the sphere as well. The pressure P exerted by the piston on the liquid can be found by dividing the force on the piston by the area A where the piston contacts the liquid: \[P = \frac{Mg}{A}\]
03

Combine equations from Step 1 and Step 2

Insert the expression for the pressure P from Step 2 into the formula for bulk modulus K from Step 1: \[K = -\frac{(Mg)/A}{ΔV/V} \] Now let's solve for the fractional change in volume: \[\frac{ΔV}{V} = -\frac{Mg}{KA}\]
04

Relate the fractional change in volume to the fractional change in radius

The volume V of the sphere is given by the formula: \[V = \frac{4}{3}πR^3\] Since the sphere's volume changes uniformly in all directions due to compression, we can write the fractional change in the radius as: \[\frac{ΔV}{V} = 3 \times \frac{δR}{R}\]
05

Solve for the fractional change in radius

Now, replacing \(\frac{ΔV}{V}\) in the equation from Step 3 with the expression found in Step 4: \[3 \times \frac{δR}{R} = -\frac{Mg}{KA}\] Finally, we solve for the fractional change in the radius (δR/R): \[\frac{δR}{R} = -\frac{Mg}{3KA}\] The negative sign indicates that the radius decreases due to compression. Therefore, the correct answer is (A): \[\frac{δR}{R} = \frac{Mg}{3KA}\]

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Most popular questions from this chapter

A uniform rod of length \(L\) and density \(\rho\) is being pulled along A smooth floor with a horizontal acceleration \(\alpha\) what is the magnitude of the stress at the transverse cross-section through the midpoint of the rod? (A) \((1 / 3) \operatorname{L} \rho \alpha\) (B) \(\operatorname{L} \rho \alpha\) (C) \((2 / 3) \mathrm{L} \rho \rho\) (D) \(\\{(\operatorname{L} \rho \alpha) / 2\\}\)

Two rods of different materials having coefficients of thermal expansions \(\alpha_{1}, \alpha_{2}\) and Young's module \(\mathrm{y}_{1}, \mathrm{y}_{2}\) respectively are fixed between two rigid walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rod. if \(\alpha_{1}: \alpha_{2}=2: 3\) the thermal stresses developed in the rod are equal provided \(\mathrm{y}_{1}: \mathrm{y}_{2}\) equals. (A) \(3: 2\) (B) \(2: 3\) (C) \(4: 9\) (D) \(1: 1\)

Arrange rubber, steel and glass in the order of decreasing elasticity. (A) glass, steel, rubber (B) rubber, glass, steel (C) glass, rubber, steel (D) steel, glass, rubber

The compressibility of a substance equals......... (A) \((\Delta \mathrm{V} / \mathrm{PV})\) (B) \(\\{(\mathrm{P} \Delta \mathrm{V}) / \mathrm{V}\\}\) (C) \(\\{\mathrm{V} /(\mathrm{P} \Delta \mathrm{V})\\}\) (D) \((\mathrm{PV} / \Delta \mathrm{V})\)

A solid cylindrical steel column is \(4 \mathrm{~m}\) long and \(9 \mathrm{~cm}\) in diameter $\cdot\left(\mathrm{Y}_{\text {steel }}=1.9 \times 10^{11} \mathrm{Nm}^{-2}\right)$. The decrease in length of the column, while carrying a load of \(80000 \mathrm{~kg}\) is........ (A) \(3.2 \mathrm{~mm}\) (B) \(1.8 \mathrm{~mm}\) (C) \(4.4 \mathrm{~mm}\) (D) \(2.6 \mathrm{~mm}\)
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