Youngs’ modulus is given by $$Y = \frac{{F \times l}}{{A \times \Delta l}}\,.....\left( {\text{i}} \right)$$
As, $$V = A \times l = {\text{constant}}\,.....\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get
$$\eqalign{ & Y = \frac{{F \times {l^2}}}{{V \times \Delta l}} \Rightarrow \Delta l = \frac{F}{{V \times Y}} \times {l^2} \cr & \Rightarrow \Delta l \propto {l^2} \cr} $$

$$\eqalign{ & {F_e} = T\sin {15^ \circ }\,\,\,\,\,\,;\,\,\,\,\,mg = T\cos {15^ \circ } \cr & \Rightarrow \tan \,{15^ \circ } = \frac{{{F_e}}}{{mg}}\,.....\,(1) \cr & {\text{In liquid, }}{F_e}' = T'\sin {15^ \circ }\,.....\,({\text{A}}) \cr & mg = {F_B} + T'\cos {15^ \circ } \cr} $$

$$\eqalign{ & {F_B} = V\left( {d - \rho } \right)g = V\left( {1.6 - 0.8} \right)g = 0.8\,Vg \cr & = 0.8\frac{m}{d}g = \frac{{0.8mg}}{{1.6}} = \frac{{mg}}{2} \cr & \therefore \,mg = \frac{{mg}}{2} + T'\cos {15^ \circ } \cr & \Rightarrow \frac{{mg}}{2} = T'\cos {15^ \circ }\,.....\,(B) \cr} $$
From (A) and (B), $$\tan {15^ \circ } = \frac{{2{F_e}'}}{{mg}}\,.....(2)$$
From (1) and (2),
$$\frac{{{F_e}}}{{mg}} = \frac{{2{F_e}'}}{{mg}}\,\,\,\,\, \Rightarrow {F_e} = 2{F_e}'\,\,\,\,\,\, \Rightarrow {F_c}'\, = \frac{{{F_c}}}{2}$$

Let the density of water be $$\rho ,$$ then the force by escaping liquid on container $$ = \rho S{\left( {\sqrt {2gh} } \right)^2}$$
$$\therefore $$ Acceleration of container
$$\eqalign{ & a = \frac{{2\rho Sgh - \mu \rho Vg}}{{\rho V}} = \left( {\frac{{2Sh}}{V} - \mu } \right)g \cr & {\text{Now,}}\,\mu = \frac{{Sh}}{V} \cr & \therefore a = \frac{{Sh}}{V}g \cr} $$

Let $${\ell _0}$$ be the unstretched length and $${\ell _3}$$ be the length under a tension of $$9N.$$  Then
$$Y = \frac{{4{\ell _0}}}{{A\left( {{\ell _1} - {\ell _0}} \right)}} = \frac{{5{\ell _0}}}{{A\left( {{\ell _2} - {\ell _0}} \right)}} = \frac{{9{\ell _0}}}{{A\left( {{\ell _3} - {\ell _0}} \right)}}$$
These give
$$\eqalign{ & \frac{4}{{{\ell _1} - {\ell _0}}} = \frac{5}{{{\ell _2} - {\ell _0}}} \Rightarrow {\ell _0} = 5{\ell _1} - 4{\ell _2} \cr & {\text{Further,}}\,\,\frac{4}{{{\ell _1} - {\ell _0}}} = \frac{9}{{{\ell _2} - {\ell _0}}} \cr} $$
Substituting the value of $${{\ell _0}}$$ and solving, we get $${\ell _3} = 5{\ell _2} - 4{\ell _1}$$

Initial length (circumference) of the ring $$ = 2\pi r$$
Final length (circumference) of the ring $$ = 2\pi R$$
Change in length $$ = 2\pi R - 2\pi r$$
strain $$ = \frac{{2\pi \left( {R - r} \right)}}{{2\pi }} = \frac{{R - r}}{r}$$
Young's modulus $$E = \frac{{\frac{F}{A}}}{{\frac{l}{L}}} = \frac{{\frac{F}{A}}}{{\frac{{\left( {R - r} \right)}}{r}}}$$
$$\therefore F = AE\left( {\frac{{R - r}}{r}} \right)$$

$$\gamma = \frac{F}{{\pi {r^2}}} \times \frac{\ell }{{\Delta \ell }} \Rightarrow \gamma \propto \frac{1}{{{r^2}}}$$

During the streamline flow of viscous and incompressible fluid through a pipe varying cross-section, the product of area of cross-section and normal fluid velocity $$\left( {Av} \right)$$  remains constant throughout the flow.
Consider a cylindrical tube of a spray pump has radius $$R,$$ one end having $$n$$ fine holes, each of radius $$r$$ and speed of liquid in the tube is $$v$$ as shown in figure.

According to equation of continuity, $$Av = {\text{constant}}$$
where, $$A$$ is a cylindrical tube and $$v$$ is velocity of liquid in a tube.
Volume in flow rate = volume in out flow rate
$$\pi {R^2}v = n\pi {r^2}v' \Rightarrow v' = \frac{{{R^2}v}}{{n{r^2}}}$$
Thus, speed of the ejection of the liquid through the holes is $$\frac{{{R^2}v}}{{n{r^2}}}.$$

The elasticity of a material depends upon the temperature of the material. Hammering & annealing reduces elastic property of a substance.

Pressure difference $${P_2} - {P_1} = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right)$$
$$ = \frac{1}{2} \times 1.2\left( {{{\left( {150} \right)}^2} - {{\left( {100} \right)}^2}} \right) = 7500\,N{m^{ - 2}}$$

$$B = \frac{{\Delta p}}{{\frac{{\Delta V}}{V}}} = \frac{{\left( {1.165 \times {{10}^5} - 1.01 \times {{10}^5}} \right)}}{{0.1}} = 1.55 \times {10^5}\,Pa$$