$\Delta L = \frac{F}{Y}\frac{L}{\pi r^{2}}$
$\implies \Delta L \propto \frac{L}{r^{2}}$
$\frac{L}{r^{2}}$ is maximum for alternative 1.
$\Delta L = L$
$A = (2 \times 10^{-3})^{2} = 4 \times 10^{-6} \text{ m}^{2}$
$L = \frac{T}{10^{7}} \times \frac{L}{4 \times 10^{-6}}$
$T = 40\text{ N}$
Detailed solution not required.
Detailed solution not required.
$l_{a} = \frac{mg}{Y}\frac{L}{A} \implies l_{w} = \frac{mg(1 - \frac{d_{w}}{d_{m}})L}{YA}$
Now $l_{w} = l_{a}\left(1 - \frac{d_{w}}{d_{m}}\right)$
$\frac{d_{w}}{d_{m}} = 1 - \frac{l_{w}}{l_{a}}$
Relative density $\frac{d_{m}}{d_{w}} = \frac{l_{a}}{l_{a} - l_{w}}$
Since material and cross section are the same, hence breaking force will remain the same.
$1.0 = \frac{W}{Y}\frac{L}{A}$
With pulley $T = \frac{2W \cdot W}{\frac{W}{g} + \frac{W}{g}}g = W$
Since Tension remains same, elongation will be the same.
Stress will be developed only when rod is subjected to condition-3.
Stress $= \frac{F \sin \theta}{AB} = \frac{F \sin \theta}{A / \sin \theta} = \frac{F}{A} \sin^{2} \theta$
$Y = \frac{100}{10^{-6}} \times \frac{100}{0.1} = 10^{11} \text{ N/m}^{2}$
$\Delta L = \frac{F}{Y}\frac{L}{\pi r^{2}}$
Detailed solution not required.
$\frac{\Delta V}{V} = \frac{\Delta P}{B}$
$\frac{3\Delta l}{l} = \frac{1 \times 10^{5}}{1.33 \times 10^{11}}$
So $\frac{\Delta l}{l} = 0.25 \times 10^{-6}$
Let original length $= l_{0}$ and length at 9N is $l$.
$\Delta L = \frac{Fl_{0}}{AY} = kF$
$a - l_{0} = k(4)$
$b - l_{0} = k(5)$
$l - l_{0} = k(9)$
Now solving these equations gives $l = 5b - 4a$
Hooke's Law: within elastic limit, stress $\propto$ strain
$B = \frac{100 \times 10^{5}}{\frac{0.01}{100}} = 10^{11} \text{ N/m}^{2} = 10^{12} \text{ dyne/cm}^{2}$
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
$\frac{1}{2}\rho v^{2} = \rho gh$
$h = \frac{r^{2}\omega^{2}}{2g} = \frac{0.05 \times 0.05 \times 4 \times 4}{2 \times 10}$
$\implies h = 0.002 \text{ m}$
$r^{2}\omega^{2} = 2gh$
$= \frac{(0.05)^{2}(4)^{2}}{2 \times 10} = 200 \times 10^{-5} \text{ m}$
Terminal velocity $v = \frac{2}{9}\frac{r^{2}(d_{1} - d_{2})}{\eta}g$
Viscous force $F = 6\pi\eta rv$
$F = 6\pi\eta r \times \frac{2}{9}\frac{r^{2}(d_{1} - d_{2})}{\eta}g = \frac{4}{3}\pi r^{3}(d_{1} - d_{2})g = \frac{m}{d_{1}}(d_{1} - d_{2})g$
$\implies F = mg\left(1 - \frac{d_{2}}{d_{1}}\right)$
Pressure on the strip $= (P_{0} + \rho gx)$
Force on the strip $= (P_{0} + \rho gx) 2 R dx$
Total force on one half cylinder by another half cylinder is:
$F = \int_{0}^{h}(P_{0} + \rho gx)2R dx - 2RT$
$= 2P_{0}Rh + 2R\rho g\frac{h^{2}}{2} - 2RT = 2P_{0}Rh + R\rho gh^{2} - 2RT$
$R = \sqrt{2gh} \times \sqrt{\frac{2(H - h)}{g}} = 2\sqrt{h(H - h)}$
$= 2\sqrt{\frac{H}{3}\left(H - \frac{H}{3}\right)} = 2\sqrt{\frac{H}{3} \times \frac{2H}{3}}$
$= \frac{2H\sqrt{2}}{3} = \frac{\sqrt{8}}{3}H$
$P_{P} = \frac{4T}{r}$
$P_{Q} = \frac{4T}{2r}$
So $\Delta P = P_{P} - P_{Q} = \frac{2T}{r}$
Suppose $R$ is radius of curvature at common interface then
$\frac{4T}{R} = \frac{4T}{r} - \frac{4T}{2r} \implies \frac{4T}{R} = \frac{2T}{r} \implies R = 2r$
Also, combining volumes:
$3 \times \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi R_{new}^{3} \implies R_{new} = 3^{1/3}r$
Since $v \propto r^{2}$
$\frac{v_{1}}{v_{2}} = \frac{r^{2}}{(3^{1/3}r)^{2}} = \frac{r^{2}}{3^{2/3}r^{2}} \implies v_{2} = (9)^{1/3}v_{1}$
Work done formatting calculation (from subsequent problem part):
$W = 2T(4\pi(2r)^{2} - 4\pi r^{2})$
$= 2 \times 30 \times 10^{-3} \left(12 \times \frac{22}{7} \times (2 \times 10^{-2})^{2}\right) \approx 9.051 \times 10^{-4} \text{ J}$
Work done = Surface tension $\times$ increase in area
Since soap bubble has two free surfaces:
Increase in area $= 2[10 \times 10 - 10 \times 6] = 80 \text{ cm}^{2}$
Work done $= 0.030 \times 80 \times 10^{-4} = 2.4 \times 10^{-4} \text{ J}$
Detailed solution not required.
From work energy theorem:
$V\rho g(H + h) - V\sigma gH = 0$
$H = \frac{\rho h}{\sigma - \rho}$
$\frac{T_{w}}{T_{m}} = \frac{h_{w}d_{w}}{\cos~\theta_{w}} \times \frac{\cos~\theta_{m}}{h_{m} \times d_{m}}$
$= \frac{10 \times 1 \times \cos~135^{\circ}}{1 \times 3.42 \times 13.6} = \frac{1}{6.57}$
Excess Pressure $\propto \frac{1}{r}$
Air goes from higher pressure to lower pressure.
Therefore, air flows from the smaller bubble to the bigger bubble.
$F = 6\pi\eta Rv$
$F \propto R$ and $F \propto v$
$(\rho h_{1}g)\pi(2r)^{2} + \pi(2r)^{2}H\frac{\rho}{3}g = \rho(h_{1} + H)g[\pi(2r)^{2} - \pi r^{2}]$
$h_{1} = \frac{5H}{3}$
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
$\left(P_{0} + \frac{4T}{r}\right)V = \left(P + \frac{4T}{r/2}\right)\frac{V}{8}$
$\implies P = 8\left(P_{0} + \frac{3T}{r}\right)$
Detailed solution not required.
Dynamic Lift
$= (\Delta P)A$
$= \left[\frac{1}{2}\rho(\sqrt{2}v)^{2} - \frac{1}{2}\rho v^{2}\right]A$
$= \frac{1}{2}\rho v^{2}A$
Detailed solution not required.
$F = \eta A\frac{dv}{dz}$
$2 = 10^{-2} \times 1 \times \frac{v - 0}{2 \times 10^{-2}}$
$v = 4 \text{ m/s}$
Given that $F/A = 4.8 \times 10^{6} \text{ Nm}^{-2}$
$\therefore F = 4.8 \times 10^{6} \times A$
$\frac{mv^{2}}{r} = 4.8 \times 10^{6} \times 10^{-6} = 4.8$
or $r\frac{mr^{2}\omega^{2}}{r} = 4.8 \implies \omega^{2} = \frac{4.8}{mr}$
$\omega = \sqrt{\frac{4.8}{10 \times 0.03}} = \sqrt{16} = 4 \text{ rad/sec.}$
Detailed solution not required.