March 31, 2019 – Physicspupil

March 31, 2019January 19, 2023

201903310628 Solution to 1980-AL-PHY-I-23

Faraday’s law:

$\varepsilon = -N \displaystyle{\frac{\mathrm{d}\Phi}{\mathrm{d}t}}$

where $\varepsilon$ is the e.m.f. induced, $N$ the number of turns in the coil, $\Phi$ the magnetic flux, also ( $\Phi =BA$ ) the product of magnetic flux density $B$ and area $A$ , $t$ the time, and the negative sign due to Lenz’s law.

The magnitude of the e.m.f. induced in the coil is therefore

$\varepsilon = \bigg| -N\displaystyle{\frac{\Delta \Phi}{\Delta t}} \bigg| = \bigg| -N \displaystyle{\frac{\Phi -0}{t}} \bigg| = N\Phi /t$ .

And the answer is C.

March 31, 2019January 19, 2023

201903310609 Solution to 1980-AL-PHY-I-14

By Snell’s law,

$\begin{aligned} n_1\sin\theta_1 & = n_2\sin\theta_2\\ n_1 \sin 45^\circ & = 1 \times \sin 90^\circ \\ n_1 & = \displaystyle{\frac{1}{\sin 45^\circ}} \\ & = \sqrt{2} \\ \end{aligned}$

And the answer is C.

March 31, 2019January 19, 2023

201903310557 Solution to 2004-AL-PHY-II-16

The original electric potential $V_B$ at point $B$ is due to charge $+Q$ at point $A$ :

$V_B =V_{BA} = k \displaystyle{\frac{(+Q)}{(2r)}}=k \displaystyle{\frac{Q}{2r}}$ .

The final electric potential $V_B$ is due to both charges $+Q$ and $-2Q$ at points $A$ and $C$ :

$V_B' = V_{BA} + V_{BC} = k \displaystyle{\frac{Q}{2r}} + k \displaystyle{\frac{(-2Q)}{(r)}}=-3 \bigg( k \displaystyle{\frac{Q}{2r}} \bigg) =-3V_B$ .

And the answer is B.

March 31, 2019January 19, 2023

201903310540 Solution to 1980-AL-PHY-I-2

The original gravitational force $F$ is

$F=G \displaystyle{\frac{(m)(m)}{(2r)^2}}=G \displaystyle{\frac{m^2}{4r^2}}$ .

The final gravitational force $F'$ is

$F'=G \displaystyle{\frac{(m)(m)}{(2r+4r)^2}}=G \displaystyle{\frac{m^2}{36r^2}}=\displaystyle{\frac{1}{9}}\bigg( G \displaystyle{\frac{m^2}{4r^2}} \bigg) = \displaystyle{\frac{1}{9}}F$ .

Thus $F:F'=9:1$ .

And the answer is B.

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