Paper Questions (Raw)
Q1. The graph of resistance of a conductor against temperature (in ($^{\circ} \mathrm{C}$) ) is
1) https://cdn.mathpix.com/cropped/1a3383cb-9929-4f02-8130-6e7122a42112-1.jpg?height=275&width=327&top_left_y=678&top_left_x=187
2) https://cdn.mathpix.com/cropped/1a3383cb-9929-4f02-8130-6e7122a42112-1.jpg?height=280&width=348&top_left_y=662&top_left_x=632
3) https://cdn.mathpix.com/cropped/1a3383cb-9929-4f02-8130-6e7122a42112-1.jpg?height=273&width=335&top_left_y=662&top_left_x=1107
4) https://cdn.mathpix.com/cropped/1a3383cb-9929-4f02-8130-6e7122a42112-1.jpg?height=266&width=333&top_left_y=662&top_left_x=1525
Hint: $$\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})=\mathrm{R}_{0}+\mathrm{R}_{0} \alpha \Delta \mathrm{T}$$
It is a graph of the type,
($\mathrm{y}=\mathrm{mx}+\mathrm{c}$) where
($\mathrm{c}=\mathrm{R}_{0}$) is positive intercept.
Q2. Which of the following will have negative slope of resistance temperature graph?
1) Conductors
2) Electrolytes
3) Semiconductors
4) Superconductors
Q3. A metal wire has a resistance of ($2.00 \Omega$) at ($50^{\circ} \mathrm{C}$) and ($2.04 \Omega$) at ($100^{\circ} \mathrm{C}$). The temperature coefficient of resistance of the metal is
1) ($\frac{10}{49} \times 10^{-3} \mathrm{~K}^{-1}$)
2) ($\frac{20}{49} \times 10^{-3} \mathrm{~K}^{-1}$)
3) ($\frac{30}{49} \times 10^{-3} \mathrm{~K}^{-1}$)
4) ($\frac{40}{49} \times 10^{-3} \mathrm{~K}^{-1}$)
Hint: ($\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{0}(1+\alpha \mathrm{t})$) ($\mathrm{R}_{50}=\mathrm{R}_{0}(1+50 \alpha) \quad \mathrm{R}_{100}=\mathrm{R}_{0}(1+100 \alpha)$) $$\therefore \frac{\mathrm{R}_{100}}{\mathrm{R}_{50}}=\frac{1+100 \alpha}{1+50 \alpha}=\frac{2.04}{2.00}$$
($2+200 \alpha=2.04+102 \alpha$) ($98 \alpha=0.04$) ($\alpha=\frac{40 \times 10^{-3}}{98}$) $$=\frac{20}{49} \times 10^{-3} \mathrm{~K}^{-1}$$
Q4. V-I graph for a conductor at temperature ($\mathrm{T}_{1}$) and ($\mathrm{T}_{2}$) are as shown in figure below. The resistance of the conductor is ($\mathrm{R}_{0}$) at ($0{ }^{\circ} \mathrm{C}$) and its temperature coefficient is ($\alpha$). Then ($\mathrm{T}_{2}-\mathrm{T}_{1}$) is
1) ($\frac{\sqrt{3}}{\mathrm{R}_{0} \alpha}$)
2) ($\frac{1+\sqrt{3}}{\sqrt{3} R_{0} \alpha}$)
3) ($\frac{\sqrt{3}-1}{\mathrm{R}_{0} \alpha}$)
4) ($\frac{\sqrt{3}-1}{\sqrt{3} R_{0} \alpha}$)
Hint: ($\Delta \mathrm{R}=\mathrm{R}_{0} \alpha(\Delta \mathrm{~T})$) ($\Delta \mathrm{T}=\frac{\Delta \mathrm{R}}{\mathrm{R}_{0} \alpha}$) At ($\mathrm{T}_{1}$) : ($\mathrm{R}_{1}=\mathrm{m}_{1}=\tan 30=\frac{1}{\sqrt{3}}$) At ($\mathrm{T}_{2}$) : ($\mathrm{R}_{2}=\mathrm{m}_{2}=\tan (30+15)=1$) ($\Delta \mathrm{R}=\mathrm{R}_{2}-\mathrm{R}_{1}=1-\frac{1}{\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{3}}$) ($\therefore \Delta \mathrm{T}=\frac{\sqrt{3}-1}{\sqrt{3} \mathrm{R}_{0} \alpha}$)
Q5. Two conductors, have same resistance at all temperatures. If their temperature coefficients of resistivity are in the ratio ($2: 3$), then their resistances at ($0^{\circ} \mathrm{C}$) are in the ratio
1) ($2: 3$)
2) ($3: 2$)
3) ($1: 1$)
4) ($9: 4$)
Hint: ($\Delta \mathrm{R}_{1}=\Delta \mathrm{R}_{2}$) ($\alpha_{1}\left(\mathrm{R}_{01}\right) \Delta \mathrm{T}=\alpha_{2}\left(\mathrm{R}_{02}\right) \Delta \mathrm{T}$) ($\frac{\left(\mathrm{R}_{01}\right)}{\left(\mathrm{R}_{02}\right)}=\frac{\alpha_{2}}{\alpha_{1}}=\frac{3}{2}$)
Q6. For a rise of ($20^{\circ} \mathrm{C}$), the resistance of a conductor increases by ($5 \Omega$). When the rise in temperature is ($80^{\circ} \mathrm{C}$), the increase in resistance of the conductor will be
1) ($10 \Omega$)
2) ($40 \Omega$)
3) ($2.5 \Omega$)
4) ($20 \Omega$)
Hint: ($\Delta \mathrm{R}=\mathrm{R}_{0}(\alpha \Delta \mathrm{T}) \Rightarrow \Delta \mathrm{R} \propto \Delta \mathrm{T}$) ($\frac{(\Delta \mathrm{R})_{2}}{(\Delta \mathrm{R})_{1}}=\frac{(\Delta \mathrm{T})_{2}}{(\Delta \mathrm{T})_{1}}=\frac{80}{20}=4$) ($(\Delta \mathrm{R})_{2}=4 \times(\Delta \mathrm{R})_{1}=4 \times 5=20 \Omega$)
Q7. The resistance of a wire is ($10 \Omega$) at ($0^{\circ} \mathrm{C}$) and ($20 \Omega$) at ($273^{\circ} \mathrm{C}$). The temperature coefficient of resistance is
1) ($3.6 \times 10^{-4}$)
2) ($4.5 \times 10^{-3}$)
3) ($7.8 \times 10^{-4}$)
4) ($3.7 \times 10^{-3}$)
Hint: ($\alpha=\frac{\mathrm{R}-\mathrm{R}_{0}}{\mathrm{R}_{0} \mathrm{t}}=\frac{20-10}{10 \times 273}=\frac{1}{273}=0.00366 /{ }^{\circ} \mathrm{C}$)
Q8. Temperature coefficient of resistivity, of the material of a conductor, is ($2 \times 10^{-4} /{ }^{\circ} \mathrm{C}$). If it is cooled by ($50^{\circ} \mathrm{C}$), then percentage change in its resistance will be
1) - 1%
2) ($1 \%$)
3) - ($2 \%$)
4) ($2 \%$)
Hint: ($\Delta \mathrm{R}=\alpha \mathrm{R}_{0} \Delta \mathrm{~T}$) ($\frac{\Delta \mathrm{R}}{\mathrm{R}_{0}}=\alpha(\Delta \mathrm{T})$) ($\% \Delta \mathrm{R}=-2 \times 10^{-4} \times 50 \times 100=1 \%$)
Q9. A heating element made of nichrome is connected to a 240 V supply. Initially it draws a current of 2.40 A which settles down after a few seconds to steady value of 2.0 A . If the room temperature is ($40^{\circ} \mathrm{C}$), what is the temperature of the heating element when the current has become steady? The temperature coefficient of resistance of nichrome is ($2.0 \times 10^{-4}$) per ($^{\circ} \mathrm{C}$).
1) ($848^{\circ} \mathrm{C}$)
2) ($948^{\circ} \mathrm{C}$)
3) ($1048^{\circ} \mathrm{C}$)
4) ($1148^{\circ} \mathrm{C}$)
Hint: ($\mathrm{V}=240 \mathrm{~V}$) ($\mathrm{I}_{1}=2.4 \mathrm{~A} \quad \mathrm{I}_{2}=2.0 \mathrm{~A}$) The resistance at room temperature
($\mathrm{t}_{1}$) is
($\mathrm{R}_{1}=\frac{\mathrm{V}}{\mathrm{I}_{1}}=\frac{240}{2.4}=100 \Omega$) The resistance at final (steady) temperature
($\mathrm{t}_{2}$) is
($\mathrm{R}_{2}=\frac{\mathrm{V}}{\mathrm{I}_{2}}=\frac{240}{2}=120 \Omega$) ($\mathrm{R}_{1}=\mathrm{R}_{0}\left(1+\alpha \mathrm{t}_{1}\right) \quad \mathrm{R}_{2}=\mathrm{R}_{0}\left(1+\alpha \mathrm{t}_{2}\right)$) $$\therefore \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{1+\alpha \mathrm{t}_{2}}{1+\alpha \mathrm{t}_{1}}=\frac{120}{100}=\frac{6}{5}$$
($5+\left[5 \times 2 \times 10^{-4} \times \mathrm{t}_{2}\right]=6+\left[6 \times 2 \times 10^{-4} \times 40\right]$) ($10^{-3} \times \mathrm{t}_{2}=6+0.048-5=1.048$) ($t_{2}=\frac{1.048}{10^{-3}}=1048^{\circ} \mathrm{C}$)
Q10. A conductor is connected to a voltage source. The graph of current passing through it against time is
1) https://cdn.mathpix.com/cropped/1a3383cb-9929-4f02-8130-6e7122a42112-6.jpg?height=240&width=262&top_left_y=571&top_left_x=173
2) https://cdn.mathpix.com/cropped/1a3383cb-9929-4f02-8130-6e7122a42112-6.jpg?height=240&width=259&top_left_y=571&top_left_x=676
3) https://cdn.mathpix.com/cropped/1a3383cb-9929-4f02-8130-6e7122a42112-6.jpg?height=245&width=255&top_left_y=566&top_left_x=1073
4) https://cdn.mathpix.com/cropped/1a3383cb-9929-4f02-8130-6e7122a42112-6.jpg?height=240&width=256&top_left_y=571&top_left_x=1471
Hint: As time increases, heat dissipated in the conductor increases. Its temperature increases. Therefore, resistance increases. Now, ($\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}$) ($\mathrm{R} \uparrow$ es ($\Rightarrow \mathrm{i} \downarrow$) es. Hence option is (4).
Q11. If the resistance of the conductor is 6 ohm at ($100^{\circ} \mathrm{C}$), then its resistance at ($0^{\circ} \mathrm{C}$) is ( ($\alpha=1.75 \times 10^{-3} /{ }^{\circ} \mathrm{C}$) )
1) ($5.10 \Omega$)
2) ($3.25 \Omega$)
3) ($6.2 \Omega$)
4) ($1.75 \Omega$)
Hint: ($\mathrm{R}_{1}=\mathrm{R}_{0}(1+\alpha \mathrm{t})$) $$\mathrm{R}_{0}=\frac{\mathrm{R}_{\mathrm{t}}}{1+\alpha \mathrm{t}}=\frac{6}{1+\left(1.75 \times 10^{-3} \times 100\right)}$$
$$=\frac{6}{1+0.175}=\frac{6}{1.175}=5.10 \Omega$$
Q12. Two resistances ($\mathrm{R}_{1}$) and ($\mathrm{R}_{2}$) are made of different materials. The temperature coefficient of the material of ($\mathrm{R}_{1}$) is ($\alpha$) and of the material of ($R_{2}$) is ($-\beta$). The resistance of the series combination of ($R_{1}$) and ($R_{2}$) will not change with temperature, if ($R_{1} / R_{2}$) equals
1) ($\frac{\alpha}{\beta}$)
2) ($\frac{\alpha+\beta}{\alpha-\beta}$)
3) ($\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}$)
4) ($\frac{\beta}{\alpha}$)
Hint: ($\mathrm{R}_{1}+\mathrm{R}_{2}=\mathrm{R}_{1}(1+\alpha \mathrm{t})+\mathrm{R}_{2}(1-\beta \mathrm{t})$) ($\mathrm{R}_{1}+\mathrm{R}_{2}=\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{1} \alpha \mathrm{t}-\mathrm{R}_{2} \beta \mathrm{t}$) ($\mathrm{R}_{1} \alpha \mathrm{t}=\mathrm{R}_{2} \beta \mathrm{t}$) ($\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\beta}{\alpha}$)
Q13. The resistance of a bulb filament is ($100 \Omega$) at a temperature of ($100^{\circ} \mathrm{C}$). If its temperature coefficient of resistance be 0.005 per ($^{\circ} \mathrm{C}$), its resistance will become ($200 \Omega$) at a temperature of
1) ($200^{\circ} \mathrm{C}$)
2) ($300^{\circ} \mathrm{C}$)
3) ($400^{\circ} \mathrm{C}$)
4) ($500^{\circ} \mathrm{C}$)
Hint: ($100=\mathrm{R}_{0}[1+0.5]$) ($\ldots$ (i)) ($\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{0}[1+0.005 \mathrm{t}]$) ($200=\mathrm{R}_{0}[1+0.005 \mathrm{t}] \ldots$. (ii)) Divide (i) by (ii), we get ($\frac{100}{200}=\frac{1.5}{1+0.005 \mathrm{t}}$) ($1+0.005 \mathrm{t}=3$) ($\therefore \mathrm{t}=\frac{2}{0.005}=400^{\circ} \mathrm{C}$)