Question:

A Spherical ball is subjected to a pressure of 100 atmosphere. If the bulk modulus of the ball is 1011 N/m2,then change in the volume is:

Updated On: Apr 7, 2025
  •  10-1%

  •  10-2%

  •  10-3%

  •  10-4%

  •  10-5%

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The Correct Option is B

Approach Solution - 1

Step 1: Understand the Given Parameters

We are given:

  • Applied pressure (P) = 100 atm
  • Bulk modulus (B) = 1011 N/m2

Step 2: Convert Pressure to SI Units

1 atm = 1.01325 × 105 N/m2

\[ P = 100 \text{ atm} = 100 × 1.01325 × 10^5 \text{ N/m}^2 = 1.01325 × 10^7 \text{ N/m}^2 \]

Step 3: Apply Bulk Modulus Formula

The bulk modulus formula relates pressure change to volume change:

\[ B = -V \frac{\Delta P}{\Delta V} \]

For small changes, the fractional volume change is:

\[ \frac{\Delta V}{V} = -\frac{P}{B} \]

Step 4: Calculate Fractional Volume Change

\[ \frac{\Delta V}{V} = -\frac{1.01325 × 10^7}{10^{11}} = -1.01325 × 10^{-4} \]

The negative sign indicates volume decreases under pressure.

Step 5: Convert to Percentage

\[ \% \text{ change} = \left| \frac{\Delta V}{V} \right| × 100 = 1.01325 × 10^{-4} × 100 ≈ 10^{-2} \% \]

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Approach Solution -2

1. Recall the definition of bulk modulus:

Bulk modulus (B) is defined as the ratio of the change in pressure (ΔP) to the fractional change in volume (ΔV/V):

\[B = -V\frac{\Delta P}{\Delta V}\]

The negative sign indicates that the volume decreases as pressure increases.

2. Convert pressure to SI units:

1 atmosphere = \(1.013 \times 10^5 \, Pa\) (Pascals, which are equivalent to N/m²). Therefore:

\[\Delta P = 100 \, atm = 100 \times 1.013 \times 10^5 \, Pa = 1.013 \times 10^7 \, Pa\]

3. Solve for the fractional change in volume:

\[\frac{\Delta V}{V} = -\frac{\Delta P}{B} = -\frac{1.013 \times 10^7 \, Pa}{10^{11} \, Pa} = -1.013 \times 10^{-4}\]

4. Calculate the percentage change in volume:

Percentage change = \(\frac{\Delta V}{V} \times 100\%\)

Percentage change = \(-1.013 \times 10^{-4} \times 100\% = -1.013 \times 10^{-2} \% \approx -10^{-2}\%\)

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Concepts Used:

Mechanical Properties of Solids

Mechanical properties of solids intricate the characteristics such as the resistance to deformation and their strength. Strength is the ability of an object to resist the applied stress, to what extent can it bear the stress.

Therefore, some of the mechanical properties of solids involve:

  • Elasticity: When an object is stretched, it changes its shape and when we leave, it retrieves its shape. Or we can say it is the property of retrieving the original shape once the external force is removed. For example Spring
  • Plasticity: When an object changes its shape and never attains its original shape even when an external force is removed. It is the permanent deformation property. For example Plastic materials.
  • Ductility: When an object is been pulled in thin sheets, wires or plates, it will be assumed that it has ductile properties. It is the property of drawing into thin wires/sheets/plates. For example Gold or Silver
  • Strength: The ability to hold out applied stress without failure. Many types of objects have higher strength than others.