Mechanical Properties of Solids

$\begin{array}{l}Y=\frac{F\times L}{A\times \Delta l}\\ F=50\times 3\pi \\ A=\pi \times {\left({3\times 10}^{-3}\right)}^2\\ =5\times {10}^{11}\\ P=5\end{array}$

Given Values

Mass (\(M\)): $500\text{ g}$

Extension (\(l\)): $2\text{ cm}$

Least Count of \(M\) ($\Delta M$): $5\text{ g}$ (smallest scale division along load axis)

Least Count of \(l\) ($\Delta l$): $0.02\text{ cm}$ (smallest scale division along extension axis)

Calculation of Percentage Error

Correct Option: D (2%)

Given Values

Mass (\(M\)): $500\text{ g}$

Extension (\(l\)): $2\text{ cm}$

Least Count of \(M\) ($\Delta M$): $5\text{ g}$ (smallest scale division along load axis)

Least Count of \(l\) ($\Delta l$): $0.02\text{ cm}$ (smallest scale division along extension axis)

Calculation of Percentage Error

Correct Option: D (2%)

For string A :

$L=l,Y=Y$

For string B :

$L=2l,Y=2Y$

total extension is

$\hat{L}l=\hat{L}{l}_A+\hat{L}{l}_g$

$=\frac{\mathrm{m}g}{\mathrm{AV}}+\frac{\mathrm{m}\left(2n\right)}{\mathrm{A}(2 \mathrm{V})}=\frac{2\mathrm{mg}}{\mathrm{AV}}$

$=\frac{2\times 0.8\times 10\times \left(0314\right)}{3.14\times {\left(2\times {10}^{-4}\right)}^2\times 2\times {10}^{10}}$

$\Delta l=2\times {10}^{-3} \mathrm{m}=2 \mathrm{mm}$

Given ratios
$\frac{L_A}{L_B}=\frac{1}{3}$
$\frac{d_A}{d_B}=2$
Elongation formula
$\mathrm{\Delta L}=\frac{FL}{AY}$
Same force and material imply
$\frac{{\mathrm{\Delta L}}_A}{{\mathrm{\Delta L}}_B}=\frac{L_A{A}_B}{L_B{A}_A}$
Area–diameter relation
$A=\frac{\pi }{4}{d}^2$
Substitute areas
$\frac{{\mathrm{\Delta L}}_A}{{\mathrm{\Delta L}}_B}=\frac{L_A}{L_B}{\left(\frac{d_B}{d_A}\right)}^2$
Apply given ratios
$\frac{{\mathrm{\Delta L}}_A}{{\mathrm{\Delta L}}_B}=\frac{1}{3}\times \frac{1}{4}$
Final ratio
$\frac{{\mathrm{\Delta L}}_A}{{\mathrm{\Delta L}}_B}=\frac{1}{12}$

• 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:

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

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

$$x \approx 1.23\%$$

$$\boxed{1.23}$$​

Bulk modulus is given by

$B=-\frac{\mathrm{\Delta }P}{\mathrm{\Delta }V/V}$

Hence $\frac{∆V}{V}=\frac{∆P}{B}$

Substituting the given values,

$\frac{\mathrm{\Delta }V}{V}=-\frac{6.3\times {10}^7}{2.1\times {10}^9}=-3\times {10}^{-2}$

Therefore, the percentage decrease in volume is 3%

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

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

$\begin{array}{l}Y=\frac{F\times L}{A\times \Delta l}\\ F=50\times 3\pi \\ A=\pi \times {\left({3\times 10}^{-3}\right)}^2\\ =5\times {10}^{11}\\ P=5\end{array}$

(c) From $Y=\frac{FL}{A\Delta L}$, we can see that $L\propto Y$ when $F,A$, and $\Delta L$ are constant. As per the equation, $F,A$ and are identical for both wires.

$\frac{L_A}{L_B}=\frac{Y_A}{Y_B}$
$\frac{2.2}{L_B}=\frac{20}{11}$
$L_B=\frac{24.2}{20}=1.21m$

(3)

• 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:

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

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

$$x \approx 1.23\%$$

$$\boxed{1.23}$$​

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

Given ratios
$\frac{L_A}{L_B}=\frac{1}{3}$
$\frac{d_A}{d_B}=2$
Elongation formula
$\mathrm{\Delta L}=\frac{FL}{AY}$
Same force and material imply
$\frac{{\mathrm{\Delta L}}_A}{{\mathrm{\Delta L}}_B}=\frac{L_A{A}_B}{L_B{A}_A}$
Area–diameter relation
$A=\frac{\pi }{4}{d}^2$
Substitute areas
$\frac{{\mathrm{\Delta L}}_A}{{\mathrm{\Delta L}}_B}=\frac{L_A}{L_B}{\left(\frac{d_B}{d_A}\right)}^2$
Apply given ratios
$\frac{{\mathrm{\Delta L}}_A}{{\mathrm{\Delta L}}_B}=\frac{1}{3}\times \frac{1}{4}$
Final ratio
$\frac{{\mathrm{\Delta L}}_A}{{\mathrm{\Delta L}}_B}=\frac{1}{12}$

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

(c) Young’s modulus is a material property given by

$Y=\frac{\text{ stress }}{\text{ strain }}=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$

It depends only on the nature of the material and is independent of the dimensions (r) and (L).

Hence, even if radius and length are doubled, the value of (Y) remains unchanged.

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

Assertion (A) is a definition of elasticity (a macroscopic property).
Reason (R) describes the microscopic origin of restoring force, not the definition of elasticity itself.
Elasticity refers to the ability to regain original shape, whereas the restoring force is a consequence of elasticity, not its explanation.
So, although both statements are true, R does not directly explain A.

Assertion (A) is a definition of elasticity (a macroscopic property).
Reason (R) describes the microscopic origin of restoring force, not the definition of elasticity itself.
Elasticity refers to the ability to regain original shape, whereas the restoring force is a consequence of elasticity, not its explanation.
So, although both statements are true, R does not directly explain A.

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

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

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