Section A: Assertion-Reasoning
- (a) Both A and R are true and R is the correct explanation. (Thermal energy breaks
bonds, freeing electrons).
- (c) A is true (high R for heat, low MP to break), but R is false (Used for both
small and large currents if rated correctly).
- (a) Both A and R are true. (Volume const $\Rightarrow L'=3L,$ $A'=A/3$
$\Rightarrow R
\propto L^2$).
- (a) Both A and R are true. Current has direction but adds algebraically.
- (a) Both A and R are true. Series: $P \propto R$. Lower wattage bulb (40W) has
higher R ($R=V^2/P$), so it glows brighter.
- (b) A is true, but R explains nothing relevant. Real reason: No potential
difference exists across the bird's body (both feet on same wire).
- (a) Both A and R are true.
- (b) Both are true but R is not explanation. R depends on dimensions, but bending
doesn't change L or A significantly. Resistivity is material property.
- (a) Both A and R are true.
- (a) Both A and R are true. $V \propto R$ in series.
Section B: Case-Study Based Questions
- Plot $V$ on y-axis, $I$ on x-axis. Slope = $R$. $R \approx 3.3-3.4 \Omega$. (Using
average $V/I$).
- Straight line passing through origin.
- Yes. The ratio $V/I$ is approximately constant ($\approx 3.3$).
- $R_{new} = 2R = 2 \times 3.3 = 6.6 \Omega$. (Since $R \propto L$).
- Refrigerator: $P=400W=0.4kW$. Time= $10 \times 30 = 300$ h. $E = 0.4 \times 300 =
120$ kWh.
- Fans: $2 \times 80 = 160$W. Tubes: $6 \times 18 = 108$W. Total P = 268W = 0.268kW.
Daily fan hours = 12, Tube hours = 6. Let's calculate separately.
Fans: $160 \times 12 =
1920$ Wh. Tubes: $108 \times 6 = 648$ Wh. Total Daily = $2568 \text{ Wh} = 2.568$ kWh.
- Total Monthly: Ref (120) + Others ($2.568 \times 30 = 77.04$) = 197.04 kWh. Bill =
$197.04 \times 3 = \text{Rs } 591.12$.
- Use LED bulbs instead of tubes; Switch off fans when not in use.
Section C: Advanced Competency & Numericals
- Circle leads to two semicircles in parallel. Each semicircle has R/2 = 10
$\Omega$. Parallel of 10 and 10 is $5 \Omega$.
- Triangle ABC. Resistance between A and B: Two sides (6+6=12) are in series, and
this combination is parallel to the third side (6). $R_{eq} = (12 \times 6)/(12+6)$ $= 72/18 = 4
\Omega$.
- Circuit needs careful checks. If one 60W is in series with 100W, total R of branch
1 = $R_{100} + R_{60}$. Branch 2 has just $R_{60}$. Branch 2 (parallel) gets full 250V. Current
in Branch 2 bulb = $P/V$ (rated) if rating matches source. If rating is 250V, $I = 60/250 =
0.24$ A.
- Max: $nR$ (Series). Min: $R/n$ (Parallel). Ratio: $n^2$.
- (i) Series: $I = 6/(1+2) = 2$A. $P = I^2 R = 4 \times 2 = 8$ W. (ii) Parallel:
$V=4$V across 2$\Omega$. $P = V^2/R = 16/2 = 8$ W. Ratio 1:1.
- $P_{total} = 160$W. $I = 160/220 = 0.727$ A.
- TV: 250 Wh. Toaster: 200 Wh. TV uses more.
- Rate = Power = $I^2 R = 225 \times 8 = 1800$ W.
- kWh (Kilowatt-hour). 1 kWh = $3.6 \times 10^6$ Joules.
- Ratio of heat $H_s : H_p$ for same time t. $H_s = V^2 t / (2R)$.
$H_p = V^2 t
/
(R/2)$ $= 2V^2 t / R$.
Ratio = $ (1/2) : 2 = 1:4$.
- Nichrome. High resistivity and high melting point, doesn't oxidize.
- Single line: Same potential, no PD, no current. Two lines: Large PD exists,
current flows through bird, fatal.
- $W = QV$ and $Q = It \Rightarrow W = VIt$. By Ohm's Law $V=IR$, so $W = (IR)It$ $=
I^2Rt$. This work is converted to Heat. $H=I^2Rt$.
- Cord (Cu) has very low R, negligible heat produced ($I^2R$). Element (Nichrome)
has high R, large heat produced, glows red hot.
- Resistivity ($\rho$) remains unchanged. It is a material property, independent of
dimensions.
