Physics

(a) $[ML^2T^{-2}A^{-1}]$

Magnetic flux $\Phi=\dfrac{\text{emf}\times t}{1}$ (volt-second). Volt $=ML^2T^{-3}A^{-1}$, so $\Phi=ML^2T^{-2}A^{-1}$.

(a) Speed

For a circular orbit $v=\sqrt{GM/r}$, which is independent of the satellite's mass. Energy, momentum and angular momentum all depend on the satellite's mass.

(c) 1:3:5

Starting from rest, distances in successive equal intervals are in the ratio $1:3:5$ (Galileo's law of odd numbers), since $s_n\propto(2n-1)$.

(c) High average thermal speed

Free electrons move with very high thermal speeds and quickly carry kinetic energy from the hot end to the cold end by collisions, so metals conduct heat well.

(b) Y

Young's modulus is a property of the material, not of the dimensions of the wire. The wire's length does not change it.

(b) $qEy$

Force $=qE$ is constant, so by the work-energy theorem $KE=\text{work done}=qE\,y$.

(b) Hypermetropia

Power $+2$ D means a converging (convex) lens, which corrects far-sightedness (hypermetropia).

(a) 2:1

$P=\frac13\rho v_{rms}^2$, so $\dfrac{P_1}{P_2}=\dfrac{\rho_1}{\rho_2}\cdot\dfrac{v_1^2}{v_2^2}=\dfrac12\times\dfrac41=\dfrac21$.

(c) maximum

$a=-\omega^2x$, so the magnitude of acceleration is greatest at maximum displacement (the extreme position), where it is $\omega^2A$.

(b) 3 times

$f=\frac1{2l}\sqrt{T/\mu}\propto\sqrt T$. With $T\to9T$, $f\to3f$.

(b) 9m $s=\int_0^3 t^2\,dt=\left[\frac{t^3}{3}\right]_0^3=9\text{ m}.$

(a) $\frac{2L}{3V}$ Each particle moves directly toward the next one, so its velocity makes $60^\circ$ with the line joining it to the particle ahead. The rate at which the side shrinks is $v+v\cos60^\circ=\frac{3v}{2}.$ They meet when the side reaches zero: $t=\dfrac{L}{3v/2}=\dfrac{2L}{3v}$.

(b) $45^\circ$ Taking the weight $mg=10$ N (m = 1 kg, g = 10 m/s²) and the pendulum held at angular displacement $\theta$ (at rest), the wire tension balances the weight and the horizontal holding force, so $T=\dfrac{mg}{\cos\theta}$. The wire breaks when $T=10\sqrt2$ N: $\cos\theta=\frac{10}{10\sqrt2}=\frac1{\sqrt2}\Rightarrow\theta=45^\circ.$

(d) 11 Let the initial speed be $u$. After one plank the speed is $\frac{19}{20}u$, so the retardation $a$ and plank thickness $s$ satisfy $u^2-\left(\tfrac{19}{20}u\right)^2=2as\Rightarrow 2as=\frac{39}{400}u^2.$ To stop the bullet completely it must travel a distance $n s$ with $u^2=2a\,ns$, so $n=\frac{400}{39}\approx10.26.$ The least whole number of planks is 11.

(b) 0, v For a head-on elastic collision between identical masses, momentum and kinetic energy conservation give $v_A=0$ and $v_B=v$: the bodies exchange velocities.

(a) 1:32

At terminal velocity $v \propto r^2$ (Stokes balance of weight and drag), and mass $m \propto r^3$. So momentum $p = mv \propto r^5$, giving $p_1:p_2 = 1:2^5 = 1:32$.

(b) 15 cc

Volume of metal $= 40/8 = 5$ cc. Loss of weight in water $= 40-20 = 20$ g, so the total volume of the sphere is 20 cc. Cavity volume $= 20-5 = 15$ cc.

(a) $42.85^\circ C$

Newton's law: $\dfrac{\theta_1-\theta_2}{t} = k\left(\dfrac{\theta_1+\theta_2}{2}-\theta_0\right)$.

First interval: $\dfrac{10}{10} = k(55-25)$, so $k = \dfrac{1}{30}$ per minute.

Second interval: $\dfrac{50-T}{10} = \dfrac{1}{30}\left(\dfrac{50+T}{2}-25\right) = \dfrac{T}{60}$, so $300 - 6T = T$ and $T = 300/7 \approx 42.85^\circ C$.

(d) 13/9

$C_V = 1\cdot\tfrac32 R + 3\cdot\tfrac52 R = 9R$ and $C_P = 1\cdot\tfrac52 R + 3\cdot\tfrac72 R = 13R$ (for the 4 moles in total).

$\gamma = \frac{C_P}{C_V} = \frac{13}{9}$

(b) 1:100

$\Delta L = 60-20 = 40$ dB, so $10\log(I_2/I_1)=40$ and $I_2/I_1 = 10^4$. Since $I \propto A^2$, $A_2/A_1 = 10^2$, so the ratio of amplitudes is $1:100$.

(b) $6\times10^{-4}C$

$C_{eq} = \dfrac{3\times6}{3+6} = 2\,\mu F$. In series every capacitor carries the same charge: $Q = C_{eq}V = 2\times10^{-6}\times300 = 6\times10^{-4}$ C.

(b) $T\alpha r^0$

$T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}$, which does not depend on $r$ (as $v = qBr/m$ grows with $r$). So $T \propto r^0$.

(b) $\sqrt{\dfrac{I_1^2+I_2^2}{2}}$

$i = I_1\sin\omega t + I_2\cos\omega t = I_0\sin(\omega t+\phi)$ with $I_0=\sqrt{I_1^2+I_2^2}$. Hence $I_{rms} = I_0/\sqrt2 = \sqrt{\dfrac{I_1^2+I_2^2}{2}}$.

(c) -20 cm, 15 cm

Achromatic condition: $\dfrac{\omega_1}{f_1}+\dfrac{\omega_2}{f_2}=0$, so $f_1 = -\dfrac{4}{3}f_2$. Also $\dfrac1{f_1}+\dfrac1{f_2}=\dfrac1{60}$.

Substituting: $-\dfrac{3}{4f_2}+\dfrac1{f_2}=\dfrac{1}{4f_2}=\dfrac1{60}$, so $f_2=15$ cm and $f_1=-20$ cm.

(c) $8.4\times10^7ms^{-1}$

$\tfrac12 mv^2 = eV \Rightarrow v=\sqrt{\frac{2eV}{m}}=\sqrt{\frac{2\times1.6\times10^{-19}\times2\times10^4}{9.1\times10^{-31}}}\approx 8.4\times10^7\ m/s$