Author: mrchuakh

24P1Q20

20 B

$\displaystyle R=\rho \frac{L}{A}=\rho \frac{x}{yz}$

For minimum resistance, the current should flow between the top and bottom surfaces, since this will incur the shortest L and largest A.

24P1Q14

14 C

At $\displaystyle t=0$ , oscillation is at the lowest extreme position and $\displaystyle v=0$ . So the E_K graph must start from 0.

Each oscillation contains two complete energy cycles. So the E_K graph must show two complete cycles during one T. (See below how the velocity-time graph is overlayed onto the E_K graph)

24P1Q19

19 C

$\displaystyle \mathrm{I}=nAvq$

The number density n must be the same since both wires are made of the material.

The drift velocity v must be higher in order to support the same current.

24P1Q18

18 D

Two images may become indistinguishable due to the spreading of light when it passes through the aperture of optical instruments. Rayleigh criterion defines the condition when this becomes.

24P1Q17

17 D

$\displaystyle A={{A}_{0}}\cos 15{}^\circ =0.966{{A}_{0}}$

24P1Q16

16 C

$\displaystyle \begin{aligned}(\mathrm{I}\propto \frac{1}{{{r}^{2}}}) \\ \frac{{{\mathrm{I}}_{M}}}{{{\mathrm{I}}_{E}}}&=\frac{{{r}_{E}}^{2}}{{{r}_{M}}^{2}}&={{(\frac{1.5}{2.3})}^{2}}&=0.4253 \\ \end{aligned}$

$\displaystyle \begin{aligned}(P&=\mathrm{I}A) \\ {{\mathrm{I}}_{E}}{{A}_{E}}&={{\mathrm{I}}_{M}}{{A}_{M}} \\ {{A}_{M}}&=\frac{{{\mathrm{I}}_{E}}}{{{\mathrm{I}}_{M}}}{{A}_{E}} \\ &=\frac{1}{0.4253}(1.0) \\ &=2.35\text{ }{{\text{m}}^{2}} \end{aligned}$

24P1Q15

15 C

Option A: Undamped oscillation. (Amplitude is constant)

Option B: Damped oscillation. (Amplitude decreases over time)

Option C: Critical damping. (System returns to equilibrium position in the fastest time, without ever crossing the equilibrium position)

Option D: Nonsense. (System mysteriously remains forever displaced from equilibrium position)

24P1Q13

13 D

Use m to denote the mass of volume V of water.

Method 1

$\displaystyle \frac{4m\times 60+m\times 20}{5m}=52\,{}^\circ \text{C}$

Method 2

$\displaystyle \begin{aligned} \text{heat gained by }m&=\text{heat lost by 4}m \\ mc(T-20)&=4mc(60-T) \\ T-20&=240-4T \\ T&=52\text{ }\!\!{}^\circ\!\!\text{ C} \end{aligned}$

24P1Q11

11 A

$\displaystyle \phi =-\frac{GM}{r}$ . So the “magnitude of gravitational potential” refers to the $\displaystyle \frac{GM}{r}$ term. As r increases, this magnitude decreases.
From the curvature of the graph, it is clear that the difference between the gravitational potentials of P and Q also decreases as they move towards infinity.

24P1Q12

12 A
Time interval between collisions, $\displaystyle \Delta t=2x\div v\text{ }$
Change in momentum per collision, $\displaystyle \Delta p=2mv$
Average force, $\displaystyle \left\langle F \right\rangle =\frac{\Delta p}{\Delta t}=\frac{2mv}{2x\div v}=\frac{m{{v}^{2}}}{x}$