2020 A/L Physics Past Paper · Question 17 (English Medium)

Answer: (3) $35\ \text{cm}$

What the question asks

We must find: the distance the sliding key $J$ must move to regain balance after $R$ is changed from $70\ \Omega$ to $80\ \Omega$.

Relevant theory

The Potentiometer is an electrical arrangement constructed to determine potential

differences more accurately using a balancing method.

the potential difference per unit length or the potential gradient along the wire has to be determined.

Since no current passes through the cell in the balanced position, its internal resistance has no effect on its e.m.f..

V =IR, where R is a constant.

Required equations

$I = \frac{E_0}{R + 10}$

$k = \frac{I \times 10}{600}$

$E = kl$

Reasoning and calculation

What changes and what stays the same: the cell $E$, $E_0$, the wire's length ($600$ cm) and resistance ($10\ \Omega$) stay fixed. Only $R$ changes; because of it the current $I$, the gradient $k$, and so the balance length $l$ change. At balance, no current flows from the cell $E$ — it only acts as the e.m.f. being "measured".

$k_1 l_1 = k_2 l_2$

$l \propto \frac{1}{k} \propto (R + 10)$

$\frac{90}{80}$

$\frac{l_2}{l_1} = \frac{R_2 + 10}{R_1 + 10} = \frac{80 + 10}{70 + 10}$

$315$

$l_2 = 280 \times \frac{90}{80}$

$35$

$d = 315 - 280$

Reading the options

Comparing with the options: $J$ must move $35$ cm (toward $B$) — option (3).

Checks

$l1 * (R2 + Rw) / (R1 + Rw) - l1$

Dimension check: $\dfrac{l_2}{l_1} = \dfrac{R_2 + 10}{R_1 + 10}$ is dimensionless; so $l_2$ has the unit of $l_1$ (cm) ✓. Direction check: raising $R$ lowers the current, so every cm of wire has a smaller potential drop — the same $E$ needs a longer length; $315 > 280$ ✓. Boundary: $315 < 600$ cm — the balance point is still on the wire ✓.

Conclusion

$d = 280\left(\frac{90}{80} - 1\right) = 35\ \text{cm}$

$l \propto (R + 10)$: $l_2 = 280 \times \dfrac{90}{80} = 315$ cm; $J$ must move $35$ cm — the answer is (3).

Computing the current using only the resistance box (forgetting the wire's own resistance).

The wire and $R$ are in the same series circuit: $I = \dfrac{E_0}{R + 10}$.