Physics

(c) constant velocity and a varying speed

Constant velocity means both magnitude and direction are constant, so speed must be constant. The other situations are possible (for example, a body at the top of its path has zero speed with non-zero acceleration; uniform circular motion has constant speed with varying velocity).

(b) Yes

The bullet and the monkey both fall under gravity by the same amount $\frac12gt^2$ below the straight line of aim, so the bullet hits the monkey (provided it reaches that point before hitting the ground).

(b) bend with copper on convex side

Copper expands more than steel on heating, so it must lie on the outer (longer) side of the curve, which is the convex side.

(c) Oscillatory

The particle first falls freely to the tunnel mouth, then moves inside under a restoring force proportional to the distance from the centre, and goes to the other side and back. The motion is periodic, but because the free-fall part is not SHM, the whole motion is oscillatory and not simple harmonic.

(a) zero

In a conductor, free charges rearrange until the net field inside the conductor is zero (electrostatic shielding).

(c) directly proportional to their squares of marked voltages

For the same power $P$, $R=\frac{V^2}{P}\propto V^2$. In series the current is the same, so the heat produced $I^2R\propto V^2$.

(b) W - N direction

The field of the coil at its centre is perpendicular to the plane of the coil, i.e. along the E-W line. Since the current is large, the coil's field dominates over the horizontal field of the earth. For a clockwise current (seen from the east side), the field at the centre points towards the west, so the north pole of the needle points W-N.

(d) have neither dispersion nor average deviation

A combination of two prisms of different materials can give dispersion without average deviation (direct-vision combination) or deviation without dispersion (achromatic combination), and can give both. It cannot be made to give neither while still being the prism combination, because the materials have different dispersive powers.

(d) is the same for all the metals and independent of the intensity

$K_{max}=h\nu-\phi$, so the graph of $K_{max}$ against $\nu$ is a straight line of slope $h$ (Planck's constant), the same for all metals.

(d) $n^2$

From Bohr's model, $r_n=\frac{n^2h^2\varepsilon_0}{\pi mZe^2}\propto n^2$.

(b) 7 min

By Torricelli, emptying time $t\propto\sqrt H$. So $t=\frac{10}{\sqrt2}\approx7.07\text{ min}\approx7\text{ min}.$

(a) $\frac{5r}{2}$

At the top of the loop, just completing the circle needs $mg=\frac{mv^2}{r}$, so $v^2=gr$. Energy conservation from release height $h$ (measured from the bottom of the loop) to the top at height $2r$: $mg h = mg(2r)+\tfrac12 m v^2 = 2mgr+\tfrac12 mgr\Rightarrow h=\frac{5r}{2}.$

(a) $\frac{a}{2a+g}$

Upthrust $U$ is constant: $U-Mg=Ma\Rightarrow U=M(g+a)$. After detaching mass $m$: $U-(M-m)g=(M-m)(2a)$, so $M(g+a)=(M-m)(g+2a)$.

$\frac{m}{M}=1-\frac{g+a}{g+2a}=\frac{a}{2a+g}$

(c) 11

Each plank gives the same retardation over the same thickness $s$. After one plank the speed is $\frac{19}{20}v$, so $v^2-\left(\tfrac{19}{20}v\right)^2=\tfrac{39}{400}v^2=2as.$ Planks needed to stop the bullet: $\frac{v^2}{(39/400)v^2}=\frac{400}{39}\approx10.26$, so the least whole number is 11.

(b) $2I\alpha\Delta\theta$

$I\propto L^2$ (for a rod, $I=\frac{ML^2}{12}$ or similar), so $\frac{\Delta I}{I}=2\frac{\Delta L}{L}=2\alpha\Delta\theta$. Hence $\Delta I=2I\alpha\Delta\theta$.

(a) 1:1

Reading each wave as $y=A\sin2\pi(ft-kx)$: $A_1=0.06,\ f_1=0.04$ and $A_2=0.03,\ f_2=0.08$. Intensity $\propto A^2f^2$, so $\frac{I_1}{I_2}=\frac{0.06^2\times0.04^2}{0.03^2\times0.08^2}=4\times\frac14=1.$

(c) 33 m/s

Listener moving away from a stationary source: $f_{app}=f\left(\frac{v-v_L}{v}\right)=0.9f$. So $v_L=0.1v=0.1\times330=33$ m/s.

(a) 5

With a slab of thickness $t$ and dielectric constant $K$, the effective separation is $d-t+\frac tK$. To keep the same capacitance (same p.d. for the same charge), the plate separation must be increased by $1.6$ mm to compensate: $t\left(1-\frac1K\right)=1.6\Rightarrow 2\left(1-\frac1K\right)=1.6\Rightarrow\frac1K=0.2\Rightarrow K=5.$

(b) $-\frac Q4$

Let the two charges $Q$ be $2r$ apart with $q$ at the midpoint. For one charge $Q$ to be in equilibrium, the repulsion from the other $Q$ must be balanced by attraction to $q$: $\frac{kQ^2}{(2r)^2}+\frac{kQq}{r^2}=0\Rightarrow q=-\frac Q4.$ Then $q$ is in equilibrium by symmetry.

(b) 3 $\Omega$

Series: $I=\frac{2E}{3+2r}$. Parallel: $I=\frac{E}{3+r/2}$. Equating: $2\left(3+\frac r2\right)=3+2r\Rightarrow 6+r=3+2r\Rightarrow r=3\ \Omega.$

(b) 100 W bulb

In parallel both bulbs get the rated 220 V, so each consumes its rated power. The 100 W bulb dissipates more power and glows brighter.

(b) $60^\circ$

For the first minimum, $a\sin\theta=\lambda$, so $\sin\theta=\frac{6\times10^{-7}}{12\times10^{-7}}=0.5\Rightarrow\theta=30^\circ$. The angular width of the central maximum is $2\theta=60^\circ$.

(b) 4.5 m/s

A bird at actual height $h$ above the water appears at $\mu h$ to a fish below, so its apparent speed is $\frac43 v$. The fish rises at 3 m/s, so the relative apparent speed is $3+\frac43v=9\Rightarrow v=\frac{6\times3}{4}=4.5\text{ m/s}.$

(c) $2.67\times10^{20}$

Taking a 100 W lamp, the light power is $0.9\times100=90$ W. Energy per photon: $E=\frac{hc}{\lambda}=\frac{6.63\times10^{-34}\times3\times10^8}{590\times10^{-9}}=3.37\times10^{-19}\text{ J}.$ $n=\frac{90}{3.37\times10^{-19}}\approx2.67\times10^{20}\text{ photons/s}.$

(c) 3

The fourth excited state is $n=5$ and the second excited state is $n=3$. The levels involved are $n=5,4,3$, giving the emission transitions $5\to4,\ 4\to3,\ 5\to3$, i.e. 3 transitions.