IMO Practice Test: Electricity | Class 10 Physics

Question 1 of 12

A wire of resistance $R$ is cut into 4 equal parts and these are joined in parallel. The new resistance is:

$4R$

$R$

$\frac{R}{4}$

$\frac{R}{16}$

Explanation: Each part is $\frac{R}{4}$; four in parallel give $\frac{R/4}{4}=\frac{R}{16}$.

Question 2 of 12

When a wire is stretched to double its length (volume constant), its resistance becomes:

$2R$

$4R$

$\frac{R}{2}$

$\frac{R}{4}$

Explanation: Length doubles, area halves, so $R'=\rho\frac{2l}{A/2}=4R$.

Question 3 of 12

Two bulbs rated 60 W and 100 W (same voltage) are connected in series. Which glows brighter?

the 100 W bulb

the 60 W bulb

both equally

neither glows

Explanation: In series, current is common; the 60 W bulb has higher resistance, so $P=I^2R$ is larger for it.

Question 4 of 12

If the current through a resistor is doubled, the heat produced in the same time becomes:

double

half

four times

unchanged

Explanation: $H=I^2Rt$; doubling $I$ makes heat $4$ times.

Question 5 of 12

Equivalent resistance between A and B for $2\ \Omega$ in series with ($3\ \Omega$ parallel $6\ \Omega$) is:

$2\ \Omega$

$4\ \Omega$

$5\ \Omega$

$11\ \Omega$

Explanation: Parallel $=2\ \Omega$; total $=2+2=4\ \Omega$.

Question 6 of 12

A 5 A current flows for 4 minutes. The charge transferred is:

$20\ \text{C}$

$1200\ \text{C}$

$240\ \text{C}$

$60\ \text{C}$

Explanation: $Q=It=5\times240=1200\ \text{C}$.

Question 7 of 12

A device of resistance $R$ at voltage $V$ dissipates power $P$. If voltage halves, the power becomes:

$2P$

$\frac{P}{2}$

$\frac{P}{4}$

$4P$

Explanation: $P=\frac{V^2}{R}$; halving $V$ makes power $\frac{1}{4}$.

Question 8 of 12

$n$ resistors each of $R$ in parallel give an equivalent resistance of:

$nR$

$\frac{R}{n}$

$\frac{n}{R}$

$R$

Explanation: $\frac{1}{R_p}=\frac{n}{R}$, so $R_p=\frac{R}{n}$.

Question 9 of 12

Two resistors $R_1$ and $R_2$ in series carry the same current because:

voltage is the same

charge cannot accumulate, so $I$ is common

resistance is equal

power is shared

Explanation: In a series path the same current flows everywhere (charge conservation).

Question 10 of 12

A 1 kW appliance runs 2 hours/day. Energy used in 30 days (in units) is:

$30$

$60$

$120$

$15$

Explanation: $1\ \text{kW}\times2\ \text{h}\times30=60\ \text{kWh}=60$ units.

Question 11 of 12

For a fixed power $P$ and voltage $V$, the resistance of an appliance is:

$\frac{V}{P}$

$\frac{V^2}{P}$

$\frac{P}{V^2}$

$PV$

Explanation: $P=\frac{V^2}{R}\Rightarrow R=\frac{V^2}{P}$.

Question 12 of 12

The V–I graph of a filament bulb is a curve (not a straight line) because:

the bulb is broken

resistance increases as it heats up

voltage is negative

current is zero

Explanation: As the filament heats, its resistance rises, so it is non-ohmic and the graph bends.

IMO Practice Test — Electricity