Reasoning: Glue does not have a regular, long-range arrangement of particles. Its molecules are arranged irregularly, so it is classified as an amorphous solid.

Reasoning: Amorphous solids do not have a sharp melting point. They gradually soften over a range of temperatures and pass through a glass-transition temperature.

Reasoning: In solid silicon, each silicon atom is covalently bonded to neighbouring silicon atoms, forming a continuous three-dimensional covalent network structure.

Reasoning: The selected point is $\frac{\mathbf{3L}}{\mathbf{4}}$ above the lower end.
Therefore, it supports: $\mathbf{Suspended\ weight\ }\mathbf{W}_{\mathbf{1}}\mathbf{\ }$ Weight of $\frac{\mathbf{3L}}{\mathbf{4}}$ of the wire $\mathbf{=}$ $\frac{\mathbf{3W}}{\mathbf{4}}$
$\mathbf{Stress =}$ $\frac{\mathbf{Total\ supported\ force}}{\mathbf{Area}}$ $\mathbf{\Longrightarrow}$ $\mathbf{Stress =}$ $\frac{\mathbf{W}_{\mathbf{1}}\mathbf{\ + \ (3W\ /\ 4)\ }}{\mathbf{S}}$

Reasoning: Assuming the volume of the wire remains constant: $\mathbf{A}_{\mathbf{1\ }}\mathbf{L}_{\mathbf{1}}\mathbf{=}\mathbf{A}_{\mathbf{2\ }}\mathbf{L}_{\mathbf{2}}$
Since the final length is double: $\mathbf{L}_{\mathbf{2}}\mathbf{= 2\ }\mathbf{L}_{\mathbf{1}}$ Therefore, $\mathbf{A}_{\mathbf{2}}\mathbf{=}$ $\frac{\mathbf{A}}{\mathbf{2}}$ Instantaneous stress is calculated using the new area:
$\mathbf{Instantaneous\ Stress =}$ $\frac{\mathbf{mg}}{\mathbf{A\ /\ 2}}$
$\mathbf{Stress =}$ $\frac{\mathbf{2mg}}{\mathbf{A}}$

Reasoning: $\mathbf{d = 4}\mathbf{\ }\mathbf{mm}$ $\mathbf{\Longrightarrow}$ $\mathbf{r = 2}\mathbf{\ }\mathbf{mm = 2 \times}\mathbf{10}^{\mathbf{- 3}}\mathbf{\ }\mathbf{m}$
Area: $\mathbf{A = \pi}\mathbf{r}^{\mathbf{2}}$ $\mathbf{\Longrightarrow}$ $\mathbf{A = \pi\ }\left( \mathbf{2 \times}\mathbf{10}^{\mathbf{- 3}} \right)^{\mathbf{2}}\mathbf{= 4\pi \times}\mathbf{10}^{\mathbf{- 6}}\mathbf{\ }\mathbf{m}^{\mathbf{2}}$ Strain: $\mathbf{0.1\% =}$ $\frac{\mathbf{0.1}}{\mathbf{100}}\mathbf{=}\mathbf{10}^{\mathbf{- 3}}$ Therefore, $\mathbf{F = Y \times Strain}\mathbf{\times}\mathbf{A}$
$\mathbf{F =}\mathbf{(}\mathbf{10 \times}\mathbf{10}^{\mathbf{10}}\mathbf{)(}\mathbf{10}^{\mathbf{- 3}}\mathbf{)(}\mathbf{4\pi \times}\mathbf{10}^{\mathbf{- 6}}\mathbf{)}$ ; $\left\lbrack \mathbf{\because}\mathbf{\ }\mathbf{Y =}\frac{\mathbf{Stress}}{\mathbf{Strain}}\mathbf{=}\frac{\mathbf{F\ /\ A}}{\mathbf{Strain}} \right\rbrack$
$\mathbf{F = 400\ \pi N}$

Reasoning: When a load is applied at the free end of a cantilever, the beam bends. The deformation produced by this bending causes elastic potential energy, called strain energy, to be stored in the beam.

Reasoning: Bulk modulus is: $\mathbf{K =}$ $\frac{\mathbf{Applied\ Pressure}}{\mathbf{Fractional\ change\ in\ volume}}$
Pressure: $\mathbf{P = 100}\mathbf{\ }\mathbf{atm}$ Since: $\mathbf{1\ atm \approx 1.013 \times}\mathbf{10}^{\mathbf{6}}\mathbf{\ dyne\ }\mathbf{cm}^{\mathbf{- 2}}$
$\mathbf{P \approx 100 \times 1.013 \times}\mathbf{10}^{\mathbf{6}}\mathbf{= 1.013 \times}\mathbf{10}^{\mathbf{8}}\mathbf{\ dyne\ }\mathbf{cm}^{\mathbf{- 2}}$
Volume change $\mathbf{(\mathrm{\Delta}V)}$: $\mathbf{0.01\% =}$ $\frac{\mathbf{0.01}}{\mathbf{100}}\mathbf{=}\mathbf{10}^{\mathbf{- 4}}$ Therefore, $\mathbf{K =}$ $\frac{\mathbf{1.013}\mathbf{\ }\mathbf{\times}\mathbf{\ }\mathbf{10}^{\mathbf{8}}}{\mathbf{10}^{\mathbf{- 4}}}$ $\mathbf{\Longrightarrow}$ $\mathbf{K = 1.013 \times}\mathbf{10}^{\mathbf{12}}\mathbf{\ dyne\ }\mathbf{cm}^{\mathbf{- 2}}$ Approximately: $\mathbf{K = 1 \times}\mathbf{10}^{\mathbf{12}}\mathbf{\ dyne\ }\mathbf{cm}^{\mathbf{- 2}}$

Reasoning: Resilience is the ability of a material to absorb and store energy when it is elastically deformed and release that energy when the force is removed.

Reasoning: According to Hooke’s law, within the elastic limit: $\mathbf{F = k}\mathbf{x}$
Therefore, $\mathbf{F}\mathbf{\ }\mathbf{/}\mathbf{\ }\mathbf{x}\mathbf{= k}$ The gradient of a force-extension graph is the spring constant $\mathbf{k}$, which remains constant within the elastic limit.