$FSR = MSR + n \times L.C$
$= 0.5 + 25 \times 0.01 = 0.525$
-0.004 is -ve error so add in FSH
$FSR = 0.525 + 0.004 = 0.529$
After 3 sec velocity & height of parachutist:
$v = -u + gt = -10 + 10 \times 3 = 20 \text{ m/s}$
$h^{\prime} = -ut + 1/2~gt^{2} = 15$
so height from ground $= 30 \text{ m}$
$h = ut - 1/2~at^{2}$
$30 = 20~t - 1/2 \times 5~t^{2}$
$5t^{2} - 40t - 60 = 0$
$t^{2} - 8t + 12 = 0 \implies (t-2)(t-6) = 0 \implies t = 2 \text{ sec}$
Total time $= 3 + 2 = 5 \text{ sec}$
$U = \frac{nRTf}{2}$
for hydrogen $f = \text{degrees of freedom} = 5$
$U_{1} = \frac{5n_{1}RT_{1}}{2} = \frac{5n_{1}RT}{2}$
for helium, $f = 3$
$U_{2} = \frac{3n_{2}RT_{2}}{2} = \frac{3n_{2}R(3T)}{2}$
$U_{1} = U_{2} \text{ (given)}$
$\frac{5n_{1}RT}{2} = \frac{9n_{2}RT}{2} \implies n_{1}/n_{2} = 9/5.$
Detailed solution not required.
$a = \frac{\text{Applied force}}{\text{Total mass}}$
$a_{1} = \frac{mg}{m}$, $a_{2} = \frac{mg}{3m}$, $a_{3} = \frac{mg}{2m}$
$\implies a_{1} > a_{3} > a_{2}$
$U = \frac{-Gm_{1}m_{2}}{r}$
$= -4\left(\frac{Gm^{2}}{l}\right) + \left(\frac{-2Gm^{2}}{l\sqrt{2}}\right)$
$= -5.41~G\frac{m^{2}}{l}$
Here $T = 2\pi\sqrt{\frac{M}{k}}$ and $\frac{5}{4}T = 2\pi\sqrt{\frac{M+m}{k}}$
$\left(\frac{5}{4}\right)^{2} = \frac{M+m}{M}$
or $\frac{25}{16} - 1 = \frac{m}{M}$ or $\frac{m}{M} = \frac{9}{16}$
$\sin~\beta = \frac{C}{B} = \frac{B}{2B} = \frac{1}{2}$
$\beta = 30^{\circ} = \frac{\pi}{6}$
$a = A + \frac{B}{S^{2}}$
$V\frac{dv}{ds} = A + \frac{B}{S^{2}}$
$\int V~dv = \int \left(A + \frac{B}{S^{2}}\right)ds$
$\left[\frac{V^{2}}{2}\right]_{0}^{V} = \left[AS - \frac{B}{S}\right]_{1}^{10}$
$\frac{V^{2}}{2} = 9A + \frac{9B}{10} \implies V = \sqrt{18\left(A + \frac{B}{10}\right)}$
$a = v\frac{dv}{dx}$
$a = 10\left(-\frac{2}{3}\right)$
$a = -\frac{20}{3} \text{ m/s}^{2}$
$a = -\alpha x^{2} \implies v\frac{dv}{dx} = -\alpha x^{2}$
$\int_{v_{0}}^{0} v~dv = -\int_{0}^{x} \alpha x^{2} dx$
$\int_{0}^{v_{0}} v~dv = \alpha \int_{0}^{x} x^{2} dx$
$\frac{v_{0}^{2}}{2} = \frac{\alpha x^{3}}{3} \implies x = \left(\frac{3v_{0}^{2}}{2\alpha}\right)^{\frac{1}{3}}$
$F = \frac{mv^{2}}{r} \implies v = \sqrt{\frac{rF}{m}}$
$2200 - 825 - 1100 + W_{4} = 3645$
$W_{4} = 3370 \text{ J}$
$-\frac{GMm}{10R_{e}} + \frac{1}{2}mv_{i}^{2} = -\frac{GMm}{R_{e}} + \frac{1}{2}mV_{t}^{2}$
$V_{t}^{2} = V_{i}^{2} + \frac{2GM}{R_{e}}\left[1 - \frac{1}{10}\right]$
Detailed solution not required.
$4~u = 4~v_{1}\cos 30^{\circ} + 1~v_{2}\cos 60^{\circ}$
$0 = 4~v_{1}\sin 30^{\circ} + v_{2}\sin 60^{\circ}$
$\frac{v_{1}}{v_{2}} = -\frac{\sqrt{3}}{2\times 2} = \frac{\sqrt{3}}{4}$
Also:
Write $P = \frac{a}{T}$ as $\frac{P^{2}V}{nR} = a \implies PV^{1/2} = \text{constant}$ ($x = 1/2$)
$W = \frac{nR(T - 4T)}{\frac{1}{2} - 1} = 6nRT$
$mc~d\theta = Q~dt \implies \frac{d\theta}{dt} = \frac{Q}{mc}$
$\text{Slope} \propto \frac{1}{\text{Heat capacity}}$
$\Delta P = mg\Delta t$
$= 0.5 \times 9.8 \times 10 = 49$
$\sqrt{\frac{l}{g}} = \sqrt{\frac{L \times 6}{g}} \implies L = \frac{l}{6}\text{ m}$
$\bar{x} = \frac{[1.34 + 1.38 + 1.32 + 1.36]}{4} = 1.35$
$\% \text{error} = \frac{\sum |x - \bar{x}|}{\bar{x}n} \times 100$
$= \frac{0.01 + 0.03 + 0.03 + 0.01}{4 \times 1.35} \times 100$
$= \frac{8}{4 \times 1.35} = \frac{200}{135} \approx 1.48\%$
Also:
$\frac{ML^{2}T^{-2}T^{3}}{Ma} = 1 \implies a = L^{2}T$
$b = ML^{-1}T^{-2}L^{2}T = MLT^{-1}$
$\frac{dx}{dt} = y$, $\frac{dy}{dt} = x$
$\frac{dy}{dx} = \frac{x}{y}$
$x~dx = y~dy \implies y^{2} = x^{2} + c$
$N = \frac{FM}{m+M} = \frac{F \times 50}{10+50} = \frac{5F}{6}$
$\mu_{s}N = mg$
$0.5 \times \frac{5F}{6} = 10 \times 10 \implies F = \frac{600}{5 \times 0.5} = 240\text{ N}$
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
From conservation of linear momentum in horizontal direction:
$mu = (M+m)v \implies v = \frac{mu}{M+m}$
From conservation of energy:
$\frac{1}{2}mu^{2} = \frac{1}{2}(M+m)v^{2} + mgh$
$\frac{1}{2}mu^{2} - \frac{1}{2}\frac{m^{2}u^{2}}{M+m} = mgh$
$Mu^{2} = 2gh(M+m) \implies u^{2} = 2gh\left(1 + \frac{m}{M}\right)$
Detailed solution not required.
Detailed solution not required.
$-G\frac{Mm}{R} + E = -\frac{GMm}{2r}$
$E = GMm\left[\frac{1}{3200} - \frac{1}{2[3200+6400]}\right]$
$= \frac{GMm}{3200}\left[1 - \frac{1}{6}\right] = \frac{5GMm}{6 \times 3200} = \frac{GMm}{38400}$
Detailed solution not required.
$2kA\frac{\Delta\theta}{L} = kA\frac{50}{L}$
$\Delta\theta = 25\text{ K}$
$Q = \epsilon\sigma(T^{4} - T_{0}^{4})$
$= 0.6 \times 5.67 \times 10^{-8} \times (400^{4} - 300^{4})$
$\approx 595 \text{ Jm}^{-2}\text{s}^{-1}$
$Q = 1500 \times 0.12 \times (400 - 25) \text{ Calorie} = 283.5 \text{ kJ}$
$W = P\Delta V = 10^{5} \times (5 \times 10^{-2})^{3} \times 3.5 \times 10^{-5} \times (400 - 25)$
$= 0.16 \text{ J} \approx 0.16 \times 10^{-3} \text{ kJ}$
$\Delta U = Q - W = 283.5 - 0.16 \times 10^{-3} \approx 283.5 \text{ kJ}$
$\lambda = \frac{kT}{\sqrt{2}\pi d^{2}P}$
$T = (273 - 50) \times 16 = 223 \times 16 = 3568\text{ K} = 3295^{\circ}\text{C}$
$v_{rms} = \sqrt{\frac{3P}{d}} = \sqrt{\frac{3 \times 10^{5}}{1.2}} = \sqrt{\frac{300}{12} \times 10^{4}} = 100\sqrt{\frac{100}{4}} = 500 \text{ m/s}$
$\omega = \frac{\pi}{6 \times 0.1} = \frac{5\pi}{3}$
Restoring force per unit mass = acceleration
$= R\omega^{2} = (0.36)\left(\frac{5\pi}{3}\right)^{2} = 9.87 \text{ m/s}^{2}$
$v_{p} = \omega\sqrt{A^{2} - x^{2}}$
$= 2\pi\frac{10}{0.5 \times 100}\sqrt{10^{2} - 5^{2}} \text{ cm/s}$
$= \frac{2\pi\sqrt{75}}{100} \text{ m/s} = \frac{\sqrt{3}\pi}{50} \text{ m/s}$
$v = 2N(l_{2} - l_{1})$
$= 2(320)(73 - 20) \text{ cm/s}$
$= 320 \times 1.06 \text{ m/s} \approx 339 \text{ m/s}$
Detailed solution not required.
Mean value $= \frac{90 + 91 + 95 + 92}{4} = 92$
$\text{Error} = \frac{|-2| + |-1| + |3| + |0|}{4} = \frac{6}{4} = 1.5$
Hence reported mean time is $92 \pm 1.5 \text{ s}$.
Detailed solution not required.