Mechanical Properties of Solids

$$\begin{gathered} \mathrm{Bulk} \mathrm{Modulus} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{B} =\frac{\mathrm{P} \mathrm{V}}{∆\mathrm{V}} \\ \mathrm{so}, \\ \mathrm{P} = \frac{\mathrm{B} ∆\mathrm{V}}{\mathrm{V}} \\ \mathrm{P} = 25\times {10}^{11}\times \frac{0.2}{100}=5\times {10}^9 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

The relationship between force ($F$) and extension ($x$) in a spring follows Hooke’s law:

$$F = kx$$

Where:

• $k$ is the spring constant.
• $F$ is the applied force.
• $x$ is the extension.

1. From the given data:
   For $F = 5 \, \text{N}$:

$$5 = kx_1 \quad \Rightarrow \quad k = \frac{5}{x_1}$$​

For $F = 7 \, \text{N}$:

$$7 = kx_2 \quad \Rightarrow \quad k = \frac{7}{x_2}$$​

Equating $k$:

$$\frac{5}{x_1} = \frac{7}{x_2} \quad \Rightarrow \quad x_2 = \frac{7}{5}x_1$$

1. Find the tension for the extension $5x_1 - 2x_2$
   Substitute $x_2 = \frac{7}{5}x_1$

$$5x_1 - 2x_2 = 5x_1 - 2 \left(\frac{7}{5}x_1\right) = 5x_1 - \frac{14}{5}x_1 = \frac{25x_1 - 14x_1}{5} = \frac{11x_1}{5}$$​​

The corresponding force (tension) is:

$$F = kx = k \cdot \frac{11x_1}{5}$$​​

Substitute $k = \frac{5}{x_1}$:

$$F = \frac{5}{x_1} \cdot \frac{11x_1}{5} = 11 \, \text{N}$$

Correct answer: (a) 11N

The relationship between force ($F$) and extension ($x$) in a spring follows Hooke’s law:

$$F = kx$$

Where:

• $k$ is the spring constant.
• $F$ is the applied force.
• $x$ is the extension.

1. From the given data:
   For $F = 5 \, \text{N}$:

$$5 = kx_1 \quad \Rightarrow \quad k = \frac{5}{x_1}$$​

For $F = 7 \, \text{N}$:

$$7 = kx_2 \quad \Rightarrow \quad k = \frac{7}{x_2}$$​

Equating $k$:

$$\frac{5}{x_1} = \frac{7}{x_2} \quad \Rightarrow \quad x_2 = \frac{7}{5}x_1$$

1. Find the tension for the extension $5x_1 - 2x_2$
   Substitute $x_2 = \frac{7}{5}x_1$

$$5x_1 - 2x_2 = 5x_1 - 2 \left(\frac{7}{5}x_1\right) = 5x_1 - \frac{14}{5}x_1 = \frac{25x_1 - 14x_1}{5} = \frac{11x_1}{5}$$​​

The corresponding force (tension) is:

$$F = kx = k \cdot \frac{11x_1}{5}$$​​

Substitute $k = \frac{5}{x_1}$:

$$F = \frac{5}{x_1} \cdot \frac{11x_1}{5} = 11 \, \text{N}$$

Correct answer: (a) 11N

• Depth of the sea = $h = 2.5$km = $2500$ m
• The bulk modulus of the material of the ball = $B = 2 \times 10^9$Pa
• The volume decrease percentage ($x$) is related to the bulk modulus using the formula:

$$B = \frac{\text{Applied Pressure} (\Delta P)}{\text{Volumetric Strain} (\frac{\Delta V}{V})}$$​

Since the pressure at depth $h$ is given by:

$$\Delta P = \rho g h$$

where:

• $\rho$ = Density of seawater = $1000$ kg/m³
• $g$ = Acceleration due to gravity = $9.8$ m/s²
• $h = 2500$ m

$$\Delta P = (1000) (9.8) (2500)$$ $$= 2.45 \times 10^7 \text{ Pa}$$

Using the bulk modulus formula:

$$\frac{\Delta V}{V} = \frac{\Delta P}{B}$$​ $$= \frac{2.45 \times 10^7}{2 \times 10^9}$$​ $$= 1.225 \times 10^{-2}$$

$$x = 1.225 \times 10^{-2} \times 100$$

$$x \approx 1.23\%$$

$$\boxed{1.23}$$​

$$\begin{gathered} B=\frac{\rho gh}{\left(\frac{\Delta v}{v}\right)} \\ \frac{\Delta v}{v}\times 100=\frac{\rho gh}{B}\times 100 \\ =\frac{1000\times 10\times 2.5\times {10}^3}{2\times {10}^9}\times 100\% \\ =1.25\% \end{gathered}$$

$$\begin{gathered} B=\frac{\rho gh}{\left(\frac{\Delta v}{v}\right)} \\ \frac{\Delta v}{v}\times 100=\frac{\rho gh}{B}\times 100 \\ =\frac{1000\times 10\times 2.5\times {10}^3}{2\times {10}^9}\times 100\% \\ =1.25\% \end{gathered}$$

We use the formula for bulk modulus (B):

$$B = \frac{\text{Pressure Change} (\Delta P)}{\frac{\Delta V}{V}}$$​

where:

• B = 1.4 × 10¹¹ Pa (Bulk modulus)
• ΔP = 7 × 10⁶ Pa (Applied pressure)
• V = a³ = (10 cm)³ = (0.1 m)³ = 10⁻³ m³ (Initial volume of cube)

Rearranging for ΔV:

$$\Delta V = \frac{\Delta P}{B} \times V$$

Substituting the values:

$$\Delta V = \frac{(7 \times 10^6)}{(1.4 \times 10^{11})} \times (10^{-3})$$

$$\Delta V = \frac{7}{1.4} \times 10^{-8}$$

$$\Delta V = 5 \times 10^{-8} \text{ m³}$$

Converting to mL:

$$\Delta V = 0.05 \text{ mL}$$

Correct option: (C) 0.05 mL