Current Electricity MCQs with Answers – Part 2 (NEET Physics)

Explanation: Resistance describes the opposition to current offered by a particular conducting specimen. Its SI unit is the ohm, written \(\Omega\), so P matches 2. Resistivity is a material property appearing in the relation \(R=\rho L/A\). Rearranging gives \(\rho=RA/L\), whose SI unit is \(\Omega\,\text{m}\). Therefore, Q matches 3. Electrical conductance is the reciprocal of resistance and measures how readily a particular specimen conducts current. Its SI unit is the siemens, written \(\text{S}\), where \(1\,\text{S}=1\,\Omega^{-1}\). Consequently, R matches 4. Electrical conductivity is the reciprocal of resistivity. Its SI unit is therefore \((\Omega\,\text{m})^{-1}\), or \(\text{S m}^{-1}\). Thus, S matches 1. The distinction is that resistance and conductance refer to a particular specimen and depend on its dimensions. Resistivity and conductivity describe material properties under specified physical conditions. Dividing a specimen's conductance by an arbitrary length does not generally give conductivity unless its geometry is accounted for correctly. The complete matching is P–2, Q–3, R–4 and S–1.

Explanation: \(\textbf{Determine the microscopic conductivity ratio:}\) For electron conduction, \[ \sigma=ne\mu \] The electronic charge magnitude \(e\) is the same for both materials. Therefore, \[ \frac{\sigma_Y}{\sigma_X} = \frac{n_Y\mu_Y}{n_X\mu_X} \] Substituting, \[ \frac{\sigma_Y}{\sigma_X} = \frac{(n_X/2)(4\mu_X)}{n_X\mu_X} \] \[ \frac{\sigma_Y}{\sigma_X}=2 \] Although Y has half the carrier number density, its fourfold greater mobility produces twice the conductivity. \(\textbf{Convert conductivity to resistivity:}\) The two material properties are reciprocals: \[ \rho=\frac1\sigma \] Hence, \[ \frac{\rho_Y}{\rho_X} = \frac{\sigma_X}{\sigma_Y} \] \[ \frac{\rho_Y}{\rho_X}=\frac12 \] Thus, Y has lower material resistivity. \(\textbf{Account for specimen dimensions:}\) For a uniform specimen, \[ R=\frac{\rho L}{A} \] Therefore, \[ \frac{R_Y}{R_X} = \frac{\rho_Y}{\rho_X} \frac{L_Y}{L_X} \frac{A_X}{A_Y} \] The given geometric ratios are \[ \frac{L_Y}{L_X}=3 \] \[ \frac{A_X}{A_Y}=2 \] Substituting, \[ \frac{R_Y}{R_X} = \frac12\times3\times2 \] \[ \frac{R_Y}{R_X}=3 \] \(\textbf{Physical conclusion:}\) Specimen Y has half the resistivity of X, so its material is the better conductor under the stated conditions. Nevertheless, Y has three times the total electrical resistance because it is longer and has a smaller cross-sectional area. This is not contradictory. Resistivity describes the material, whereas resistance depends on the material together with the dimensions of the specimen. The result requires both microscopic conductivity reasoning and macroscopic geometric reasoning. \(\textbf{Final answer:}\) \[ \boxed{\frac{\rho_Y}{\rho_X}=\frac12,\quad\frac{R_Y}{R_X}=3} \]

Explanation: \(\textbf{Apply conservation of wire volume:}\) The wire is drawn without changing the amount of material. Therefore, \[ AL=A'L' \] The new length is \[ L'=3L \] Hence, \[ AL=A'(3L) \] Cancelling \(L\), \[ A'=\frac A3 \] The increased length is accompanied by a threefold decrease in cross-sectional area. \(\textbf{Calculate the resistance change:}\) Initially, \[ R=\frac{\rho L}{A} \] After drawing, \[ R'=\frac{\rho L'}{A'} \] Substituting the new dimensions, \[ R'=\frac{\rho(3L)}{A/3} \] \[ R'=\frac{9\rho L}{A} \] Therefore, \[ R'=9R \] The resistance increases by a factor of nine because both length and cross-sectional area change. \(\textbf{Identify the electrical constraint:}\) The new experiment uses an ideal constant-current source. Therefore, the current remains \(I\), even though the resistance has increased. This does not mean the source voltage remains unchanged. \(\textbf{Compare power dissipation:}\) For a resistor carrying steady current, \[ P=I^2R \] Initially, \[ P=I^2R \] After drawing, \[ P'=I^2R' \] Substituting \(R'=9R\), \[ P'=I^2(9R) \] \[ P'=9I^2R \] \[ P'=9P \] The constant-current source must provide nine times the original potential difference across the wire. This increased voltage allows the higher-resistance wire to carry the same current. Under a constant-voltage source, the power change would be different, because current would decrease as resistance increased. Therefore, the phrase constant-current source is a decisive part of the problem. \(\textbf{Final answer:}\) \[ \boxed{R'=9R,\quad P'=9P} \]

Explanation: \(\textbf{Convert the power observation into resistance information:}\) Both wires are connected across the same ideal source voltage. For an ohmic wire under constant conditions, \[ P=\frac{V^2}{R} \] Therefore, at the same voltage, \[ \frac{P_Q}{P_P}=\frac{R_P}{R_Q} \] The measured power ratio is \[ \frac{P_Q}{P_P}=\frac18 \] Hence, \[ \frac{R_Q}{R_P}=8 \] Wire Q has eight times the resistance of P. \(\textbf{Determine the cross-sectional area ratio:}\) For cylindrical wires, \[ A=\pi r^2 \] Given, \[ r_Q=\frac{r_P}{\sqrt2} \] Therefore, \[ A_Q=\pi\left(\frac{r_P}{\sqrt2}\right)^2 \] \[ A_Q=\frac{\pi r_P^2}{2} \] Thus, \[ \frac{A_Q}{A_P}=\frac12 \] \(\textbf{Infer the resistivity ratio:}\) For a uniform wire, \[ R=\frac{\rho L}{A} \] Taking ratios, \[ \frac{R_Q}{R_P} = \frac{\rho_Q}{\rho_P} \frac{L_Q}{L_P} \frac{A_P}{A_Q} \] Substituting, \[ 8=\frac{\rho_Q}{\rho_P}\times2\times2 \] \[ 8=4\frac{\rho_Q}{\rho_P} \] Hence, \[ \frac{\rho_Q}{\rho_P}=2 \] \(\textbf{Determine the current-density ratio:}\) Since the source voltages are equal, \[ I=\frac VR \] Therefore, \[ \frac{I_Q}{I_P}=\frac{R_P}{R_Q}=\frac18 \] Current density is \[ j=\frac IA \] Hence, \[ \frac{j_Q}{j_P} = \frac{I_Q}{I_P}\frac{A_P}{A_Q} \] Substituting, \[ \frac{j_Q}{j_P} = \frac18\times2 \] \[ \frac{j_Q}{j_P}=\frac14 \] Although wire Q has half the area of P, it carries only one-eighth of the current. Its current density is therefore one-quarter of that in P. The observed power ratio encodes resistance information, which must be combined with geometry to infer the resistivity ratio. It would be incorrect to identify the resistance ratio directly with the resistivity ratio because the wires have different dimensions. \(\textbf{Final answer:}\) \[ \boxed{\frac{\rho_Q}{\rho_P}=2,\quad\frac{j_Q}{j_P}=\frac14} \]

Explanation: \(\textbf{Identify the effective current path:}\) The block is made from the same material in both experiments. Its resistivity therefore remains unchanged. However, changing the electrode positions changes the length of the current path and the area normal to current. For a uniform conductor, \[ R=\frac{\rho L}{A} \] \(\textbf{Experiment P — Current along the longest dimension:}\) The electrodes cover the \(2\,\text{cm}\times1\,\text{cm}\) end faces. Thus, the current path length is \[ L_P=6\,\text{cm} \] The cross-sectional area is \[ A_P=(2)(1)=2\,\text{cm}^2 \] Therefore, \[ R_P=\rho\frac{6\,\text{cm}}{2\,\text{cm}^2} \] \[ R_P=3\rho\,\text{cm}^{-1} \] Here, \(\rho\) is to be expressed in units consistent with centimetres. \(\textbf{Experiment Q — Current along the width:}\) The electrodes now cover the \(6\,\text{cm}\times1\,\text{cm}\) faces. Thus, \[ L_Q=2\,\text{cm} \] The cross-sectional area is \[ A_Q=(6)(1)=6\,\text{cm}^2 \] Therefore, \[ R_Q=\rho\frac{2\,\text{cm}}{6\,\text{cm}^2} \] \[ R_Q=\frac{\rho}{3}\,\text{cm}^{-1} \] \(\textbf{Compare the two resistances:}\) The resistivity and length units cancel in the ratio. \[ \frac{R_P}{R_Q} = \frac{L_P/A_P}{L_Q/A_Q} \] Substituting directly, \[ \frac{R_P}{R_Q} = \frac{6/2}{2/6} \] \[ \frac{R_P}{R_Q} = \frac{3}{1/3} \] \[ \frac{R_P}{R_Q}=9 \] The resistance is greater in experiment P because the current must travel a longer distance through a smaller cross-sectional area. The block has not undergone any physical modification. The change in resistance arises entirely from changing the direction of current flow relative to its dimensions. For an isotropic material, resistivity is the same in both directions, even though the measured resistances differ. \(\textbf{Final answer:}\) \[ \boxed{\frac{R_P}{R_Q}=9} \]

Explanation: \(\textbf{Resistance of the original wire:}\) For a uniform conductor, \[ R=\frac{\rho L}{A} \] Its conductance is \[ G=\frac1R \] The wire is cut into four equal pieces without changing the material or cross-sectional area. \(\textbf{Resistance of each piece:}\) Each piece has length \[ L'=\frac L4 \] Therefore, \[ R_{\text{piece}}=\frac{\rho(L/4)}A \] \[ R_{\text{piece}}=\frac R4 \] \(\textbf{Equivalent resistance in parallel:}\) Four identical resistances of \(R/4\) are connected in parallel. For parallel resistances, \[ \frac1{R_{\text{eq}}} = \frac4{R/4} \] \[ \frac1{R_{\text{eq}}}=\frac{16}{R} \] Therefore, \[ R_{\text{eq}}=\frac R{16} \] \(\textbf{Equivalent conductance:}\) Conductance is the reciprocal of resistance: \[ G_{\text{eq}}=\frac1{R_{\text{eq}}} \] \[ G_{\text{eq}}=\frac{16}{R} \] Since \(G=1/R\), \[ G_{\text{eq}}=16G \] \(\textbf{Material conductivity:}\) Electrical conductivity is \[ \sigma=\frac1\rho \] Cutting and reconnecting the wire changes its effective geometry and total conductance. It does not change the intrinsic conductivity of the material, provided the physical conditions remain unchanged. Thus, the four-piece arrangement has sixteen times the original conductance, but the material conductivity remains \(\sigma\). Confusing conductance with conductivity would incorrectly assign the factor sixteen to a material property. \(\textbf{Final answer:}\) \[ \boxed{G_{\text{eq}}=16G,\quad \sigma'=\sigma} \]

Explanation: \(\textbf{Infer resistivity from macroscopic measurements:}\) The resistance of a uniform metallic wire is \[ R=\frac{\rho L}{A} \] Therefore, \[ \rho=\frac{RA}{L} \] Convert the area to SI units: \[ A=1.0\,\text{mm}^2=1.0\times10^{-6}\,\text{m}^2 \] Substituting, \[ \rho=\frac{(0.050)(1.0\times10^{-6})}{2.0} \] \[ \rho=2.5\times10^{-8}\,\Omega\,\text{m} \] \(\textbf{Calculate conductivity:}\) Conductivity and resistivity are reciprocal material properties: \[ \sigma=\frac1\rho \] Thus, \[ \sigma=\frac1{2.5\times10^{-8}} \] \[ \sigma=4.0\times10^7\,\text{S m}^{-1} \] \(\textbf{Connect conductivity with relaxation time:}\) For electron conduction, \[ \sigma=\frac{ne^2\tau}{m} \] This can be written using the supplied charge-to-mass ratio: \[ \sigma=ne\left(\frac em\right)\tau \] Therefore, \[ \tau=\frac{\sigma}{ne(e/m)} \] \(\textbf{Evaluate the denominator:}\) \[ ne=(5.0\times10^{28})(1.6\times10^{-19}) \] \[ ne=8.0\times10^9\,\text{C m}^{-3} \] Hence, \[ ne\left(\frac em\right) =(8.0\times10^9)(2.0\times10^{11}) \] \[ ne\left(\frac em\right)=1.6\times10^{21} \] \(\textbf{Determine relaxation time:}\) \[ \tau=\frac{4.0\times10^7}{1.6\times10^{21}} \] \[ \tau=2.5\times10^{-14}\,\text{s} \] The measured resistance alone cannot directly determine the microscopic relaxation time. The wire dimensions are first needed to obtain resistivity. The reciprocal relation then gives conductivity, which can be connected to electron number density and relaxation time. This calculation illustrates how a macroscopic electrical measurement can be linked to a microscopic transport parameter. \(\textbf{Final answer:}\) \[ \boxed{\rho=2.5\times10^{-8}\,\Omega\,\text{m},\quad\tau=2.5\times10^{-14}\,\text{s}} \]

Explanation: \(\textbf{Temperature-resistance relation:}\) For the stated linear temperature range, \[ R_T=R_0[1+\alpha(T-T_0)] \] Here, \(R_0\) is the resistance at the reference temperature \(T_0\). The question chooses \[ T_0=0^\circ\text{C} \] and \[ R_0=10.0\,\Omega \] \(\textbf{Determine the temperature coefficient:}\) At \[ T=100^\circ\text{C}, \] the resistance is \(14.0\,\Omega\). Substituting into the relation, \[ 14.0=10.0[1+\alpha(100-0)] \] Dividing by \(10.0\), \[ 1.4=1+100\alpha \] Therefore, \[ 100\alpha=0.4 \] \[ \alpha=0.004\,\text{K}^{-1} \] \[ \alpha=4.0\times10^{-3}\,\text{K}^{-1} \] \(\textbf{Resistance at }75^\circ\textbf{C:}\) Using the same reference temperature, \[ R_{75}=10.0[1+\alpha(75-0)] \] Substituting, \[ R_{75}=10.0[1+(0.004)(75)] \] \[ R_{75}=10.0(1+0.30) \] \[ R_{75}=13.0\,\Omega \] \(\textbf{Physical interpretation:}\) The positive temperature coefficient indicates that the metallic wire's resistance increases as temperature rises over the stated range. The coefficient has units of inverse kelvin because it describes fractional resistance change per unit temperature difference. A temperature difference of \(75^\circ\text{C}\) is numerically equal to a difference of \(75\,\text{K}\). Therefore, the Celsius temperature scale can be used for differences in this calculation. The coefficient is defined relative to the specified reference resistance; changing the reference temperature generally changes the corresponding coefficient value. Extrapolating the linear relation far beyond its stated interval would require additional physical justification. \(\textbf{Final answer:}\) \[ \boxed{\alpha=4.0\times10^{-3}\,\text{K}^{-1},\quad R_{75}=13.0\,\Omega} \]

Explanation: \(\textbf{First change — Heating the wire:}\) The temperature dependence of resistance is \[ R_T=R_0[1+\alpha(T-T_0)] \] Initially, \[ R_0=4.0\,\Omega \] The reference temperature is \(0^\circ\text{C}\), and the final temperature is \(100^\circ\text{C}\). Therefore, \[ \Delta T=100\,\text{K} \] Substituting, \[ R_{100}=4.0[1+(5.0\times10^{-3})(100)] \] \[ R_{100}=4.0(1+0.50) \] \[ R_{100}=6.0\,\Omega \] Thus, heating alone increases the resistance to \(6.0\,\Omega\). \(\textbf{Second change — Stretching at constant temperature:}\) The wire is now uniformly stretched to twice its original length. Let its length and cross-sectional area before stretching be \(L\) and \(A\). The stretched dimensions satisfy \[ L'=2L \] Because volume is conserved, \[ AL=A'L' \] Therefore, \[ AL=A'(2L) \] \[ A'=\frac A2 \] \(\textbf{Determine the geometrical resistance factor:}\) At the maintained temperature, resistivity is unchanged by stretching. Using \[ R=\frac{\rho L}{A}, \] the resistance ratio is \[ \frac{R'}{R_{100}} = \frac{L'}L\frac A{A'} \] Substituting, \[ \frac{R'}{R_{100}} = 2\times2 \] \[ \frac{R'}{R_{100}}=4 \] Therefore, \[ R'=4R_{100} \] \[ R'=4(6.0) \] \[ R'=24\,\Omega \] \(\textbf{Final electrical condition:}\) The stretched wire is connected across an ideal \(12\,\text{V}\) source. Using Ohm's law, \[ I=\frac VR \] \[ I=\frac{12}{24} \] \[ I=0.50\,\text{A} \] The resistance increases through two separate physical mechanisms. Heating changes the material resistivity, while stretching changes the wire's length and cross-sectional area. The geometrical factor must be applied to the resistance at the final temperature, not incorrectly treated as a change in resistivity. The final current is determined only after both resistance changes have been included. \(\textbf{Final answer:}\) \[ \boxed{R'=24\,\Omega,\quad I=0.50\,\text{A}} \]

Explanation: \(\textbf{Identify the reference temperature:}\) The temperature coefficient of resistance is defined relative to a specified reference resistance. For a linear resistance-temperature relation, \[ R_T=R_{T_0}[1+\alpha_{T_0}(T-T_0)] \] The numerical value of the coefficient depends on the chosen reference temperature. The question specifically requests the coefficient referred to \(120^\circ\text{C}\), not \(20^\circ\text{C}\). \(\textbf{Determine the resistance-temperature slope:}\) The resistance changes from \(10.0\,\Omega\) to \(14.0\,\Omega\) over a temperature interval of \(100\,\text{K}\). Therefore, \[ \frac{\Delta R}{\Delta T} =\frac{14.0-10.0}{120-20} \] \[ \frac{\Delta R}{\Delta T} =\frac{4.0}{100} \] \[ \frac{\Delta R}{\Delta T}=0.040\,\Omega\,\text{K}^{-1} \] \(\textbf{Coefficient referred to }120^\circ\textbf{C:}\) The defining relation is \[ \alpha_{120} =\frac1{R_{120}}\frac{\Delta R}{\Delta T} \] Substituting, \[ \alpha_{120}=\frac{0.040}{14.0} \] \[ \alpha_{120}=\frac1{350}\,\text{K}^{-1} \] The corresponding coefficient referred to \(20^\circ\text{C}\) would instead be \[ \alpha_{20}=\frac{0.040}{10.0} \] \[ \alpha_{20}=4.0\times10^{-3}\,\text{K}^{-1} \] These coefficients differ even though they describe the same straight-line resistance-temperature graph. \(\textbf{Resistance at }70^\circ\textbf{C:}\) Using the reference at \(20^\circ\text{C}\), \[ R_{70}=R_{20}[1+\alpha_{20}(70-20)] \] Substituting, \[ R_{70}=10.0[1+(0.004)(50)] \] \[ R_{70}=10.0(1.20) \] \[ R_{70}=12.0\,\Omega \] As a check, \(70^\circ\text{C}\) lies halfway between the two measured temperatures, and its resistance is halfway between \(10.0\,\Omega\) and \(14.0\,\Omega\). A temperature difference of \(50^\circ\text{C}\) is numerically equal to \(50\,\text{K}\). The reference-temperature distinction is essential; using \(\alpha_{20}\) as though it were \(\alpha_{120}\) would be incorrect. \(\textbf{Final answer:}\) \[ \boxed{\alpha_{120}=\frac1{350}\,\text{K}^{-1},\quad R_{70}=12.0\,\Omega} \]

Explanation: \(\textbf{Translate equal current into a resistance condition:}\) Both specimens are connected separately across the same ideal source voltage. Their currents are \[ I_P=\frac V{R_P},\qquad I_Q=\frac V{R_Q} \] At the common temperature where their currents are equal, \[ I_P=I_Q \] Since the applied voltage is identical and non-zero, \[ R_P=R_Q \] Thus, the required temperature is where the two resistance-temperature lines intersect. \(\textbf{Resistance-temperature relation for P:}\) The slope of P's graph is \[ m_P=\frac{12-8}{100-0} \] \[ m_P=0.040\,\Omega\,\text{K}^{-1} \] Therefore, \[ R_P(T)=8+0.040T \] Here, \(T\) is the numerical temperature in degrees Celsius. \(\textbf{Resistance-temperature relation for Q:}\) The slope of Q's graph is \[ m_Q=\frac{8-16}{100-0} \] \[ m_Q=-0.080\,\Omega\,\text{K}^{-1} \] Hence, \[ R_Q(T)=16-0.080T \] \(\textbf{Determine the intersection temperature:}\) Equating the resistances, \[ 8+0.040T=16-0.080T \] Rearranging, \[ 0.120T=8 \] \[ T=\frac8{0.120} \] \[ T=\frac{200}{3}{}^\circ\text{C} \] This temperature lies within the stated experimental range. \(\textbf{Determine the common resistance:}\) Substituting into P's relation, \[ R=8+0.040\left(\frac{200}{3}\right) \] \[ R=8+\frac83 \] \[ R=\frac{32}{3}\,\Omega \] The same result follows from Q's relation. At temperatures below the intersection, P has the smaller resistance and therefore draws the larger current at the same voltage. Above the intersection, Q has the smaller resistance and draws the larger current. The equality of current occurs because the resistances become equal, not because the materials have identical temperature coefficients. \(\textbf{Final answer:}\) \[ \boxed{T=\frac{200}{3}{}^\circ\text{C},\quad R=\frac{32}{3}\,\Omega} \]

Explanation: \(\textbf{Use the constant-voltage condition:}\) The applied voltage remains unchanged throughout the experiment. For the resistor, \[ R=\frac VI \] Thus, resistance is inversely proportional to the measured current when voltage is fixed. \(\textbf{Infer the resistance ratio:}\) Let the resistances at \(20^\circ\text{C}\) and \(120^\circ\text{C}\) be \(R_{20}\) and \(R_{120}\). Then, \[ \frac{R_{120}}{R_{20}} =\frac{V/I_{120}}{V/I_{20}} \] Cancelling the common voltage, \[ \frac{R_{120}}{R_{20}}=\frac{I_{20}}{I_{120}} \] Substituting, \[ \frac{R_{120}}{R_{20}}=\frac{3.0}{2.0} \] \[ \frac{R_{120}}{R_{20}}=\frac32 \] \(\textbf{Determine the temperature coefficient:}\) Using the linear model, \[ R_{120}=R_{20}[1+\alpha_{20}(120-20)] \] Dividing by \(R_{20}\), \[ \frac32=1+100\alpha_{20} \] Therefore, \[ 100\alpha_{20}=\frac12 \] \[ \alpha_{20}=5.0\times10^{-3}\,\text{K}^{-1} \] \(\textbf{Resistance at }70^\circ\textbf{C:}\) The temperature increase above the reference is \[ \Delta T=70-20=50\,\text{K} \] Hence, \[ R_{70}=R_{20}[1+\alpha_{20}(50)] \] Substituting, \[ R_{70}=R_{20}[1+(0.005)(50)] \] \[ R_{70}=1.25R_{20} \] \(\textbf{Determine the new current:}\) The voltage is unchanged, so \[ \frac{I_{70}}{I_{20}}=\frac{R_{20}}{R_{70}} \] \[ \frac{I_{70}}{3.0}=\frac1{1.25} \] Therefore, \[ I_{70}=\frac{3.0}{1.25} \] \[ I_{70}=2.4\,\text{A} \] The resistance varies linearly with temperature under the stated model, but the current does not. At fixed voltage, current is inversely proportional to the temperature-dependent resistance. Therefore, taking the arithmetic mean of the endpoint currents would give an incorrect result. \(\textbf{Final answer:}\) \[ \boxed{\alpha_{20}=5.0\times10^{-3}\,\text{K}^{-1},\quad I_{70}=2.4\,\text{A}} \]

Explanation: \(\textbf{Resistance of equal wire sections at the reference temperature:}\) The complete wire has resistance \[ R_0=20\,\Omega \] Its two sections have equal lengths, equal cross-sectional areas and the same material. At the common reference temperature of \(0^\circ\text{C}\), they therefore have equal resistance. \[ R_{P,0}=R_{Q,0}=\frac{20}{2} \] \[ R_{P,0}=R_{Q,0}=10\,\Omega \] \(\textbf{Cold section P:}\) Section P remains at \(0^\circ\text{C}\). Therefore, \[ R_P=10\,\Omega \] \(\textbf{Hot section Q:}\) Section Q is maintained at \(100^\circ\text{C}\). Using \[ R_T=R_0[1+\alpha_0(T-T_0)], \] we obtain \[ R_Q=10[1+(4.0\times10^{-3})(100)] \] \[ R_Q=10(1.40) \] \[ R_Q=14\,\Omega \] \(\textbf{Total resistance and current:}\) The sections are connected in series, so their resistances add. \[ R_{\text{total}}=R_P+R_Q \] \[ R_{\text{total}}=10+14 \] \[ R_{\text{total}}=24\,\Omega \] The source voltage is \(12\,\text{V}\). Therefore, \[ I=\frac{12}{24} \] \[ I=0.50\,\text{A} \] \(\textbf{Power dissipated in the hot section:}\) The same steady current flows through both sections. Hence, \[ P_Q=I^2R_Q \] Substituting, \[ P_Q=(0.50)^2(14) \] \[ P_Q=(0.25)(14) \] \[ P_Q=3.5\,\text{W} \] The hotter section dissipates more electrical power because it has the greater resistance while carrying the same current. Its voltage drop is also greater than that of the colder section. The temperature coefficient must be applied to Q's own reference resistance of \(10\,\Omega\), not to the resistance of the entire wire. Treating the complete wire as though it were uniformly at \(100^\circ\text{C}\) would overestimate its total resistance. \(\textbf{Final answer:}\) \[ \boxed{I=0.50\,\text{A},\quad P_Q=3.5\,\text{W}} \]

Explanation: \(\textbf{Temperature dependence of the two resistors:}\) For the stated linear interval, \[ R_T=R_0(1+\alpha_0\Delta T) \] The temperature change is \[ \Delta T=100\,\text{K} \] The coefficients have opposite signs, so the two resistances change in opposite directions. \(\textbf{Resistance of P at }100^\circ\textbf{C:}\) \[ R_P=12[1+(0.004)(100)] \] \[ R_P=12(1.40) \] \[ R_P=16.8\,\Omega \] Thus, P's resistance increases by \(4.8\,\Omega\). \(\textbf{Resistance of Q at }100^\circ\textbf{C:}\) \[ R_Q=8.0[1-(0.006)(100)] \] \[ R_Q=8.0(0.40) \] \[ R_Q=3.2\,\Omega \] Thus, Q's resistance decreases by \(4.8\,\Omega\). \(\textbf{Total series resistance:}\) \[ R_{\text{total}}=R_P+R_Q \] \[ R_{\text{total}}=16.8+3.2 \] \[ R_{\text{total}}=20.0\,\Omega \] The increase in P's resistance exactly compensates for the decrease in Q's resistance. Consequently, their total series resistance remains equal to its value at \(0^\circ\text{C}\). \(\textbf{Determine the circuit current:}\) \[ I=\frac{V}{R_{\text{total}}} \] \[ I=\frac{10}{20.0} \] \[ I=0.50\,\text{A} \] \(\textbf{Potential difference across P:}\) \[ V_P=IR_P \] \[ V_P=(0.50)(16.8) \] \[ V_P=8.4\,\text{V} \] The remaining potential difference across Q is \[ V_Q=(0.50)(3.2)=1.6\,\text{V} \] The drops add to \(10.0\,\text{V}\), consistent with the source voltage. Although the total resistance and current are unchanged, the individual voltage drops change substantially. This demonstrates that unchanged equivalent resistance does not require the resistances of the individual components to remain unchanged. \(\textbf{Final answer:}\) \[ \boxed{R_{\text{total}}=20.0\,\Omega,\quad V_P=8.4\,\text{V}} \]

Explanation: \(\textbf{Separate material and geometrical effects:}\) The resistance of a uniform wire is \[ R=\frac{\rho L}{A} \] The two specimens differ in temperature, length and cross-sectional area. Therefore, their measured resistance ratio cannot be identified directly with their resistivity ratio. \(\textbf{Geometrical comparison:}\) For X, \[ R_X=\frac{\rho_{20}L}{A} \] For Y at \(120^\circ\text{C}\), \[ R_{Y,120}=\frac{\rho_{120}(2L)}{A/2} \] Simplifying, \[ R_{Y,120}=\frac{4\rho_{120}L}{A} \] Therefore, \[ \frac{R_{Y,120}}{R_X} =4\frac{\rho_{120}}{\rho_{20}} \] \(\textbf{Infer the resistivity ratio:}\) The measured resistance ratio is \[ \frac{R_{Y,120}}{R_X}=\frac{30}{5.0} \] \[ \frac{R_{Y,120}}{R_X}=6 \] Hence, \[ 6=4\frac{\rho_{120}}{\rho_{20}} \] \[ \frac{\rho_{120}}{\rho_{20}}=\frac32 \] The resistivity at \(120^\circ\text{C}\) is \(1.5\) times its value at \(20^\circ\text{C}\). \(\textbf{Determine the temperature coefficient:}\) Using the linear relation, \[ \rho_{120}=\rho_{20}[1+\alpha_{20}(120-20)] \] Dividing by \(\rho_{20}\), \[ \frac32=1+100\alpha_{20} \] Therefore, \[ 100\alpha_{20}=0.50 \] \[ \alpha_{20}=5.0\times10^{-3}\,\text{K}^{-1} \] \(\textbf{Resistivity at }70^\circ\textbf{C:}\) The temperature rise above \(20^\circ\text{C}\) is \(50\,\text{K}\). \[ \rho_{70}=\rho_{20}[1+(0.005)(50)] \] \[ \rho_{70}=1.25\rho_{20} \] \(\textbf{Resistance of Y at }70^\circ\textbf{C:}\) Specimen Y retains its geometry, so its resistance is proportional to the material resistivity. \[ \frac{R_{Y,70}}{R_{Y,120}} =\frac{\rho_{70}}{\rho_{120}} \] Substituting, \[ \frac{R_{Y,70}}{30}=\frac{1.25}{1.50} \] Therefore, \[ R_{Y,70}=30\left(\frac56\right) \] \[ R_{Y,70}=25\,\Omega \] The essential step is removing the geometrical factor before inferring the material's temperature dependence. Comparing only the measured resistances would incorrectly attribute the entire sixfold difference to temperature. \(\textbf{Final answer:}\) \[ \boxed{\alpha_{20}=5.0\times10^{-3}\,\text{K}^{-1},\quad R_{Y,70}=25\,\Omega} \]

Explanation: \(\textbf{Distinguish wire and contact resistance:}\) The measured resistance is the sum of the wire resistance and the constant contact resistance. \[ R_m=R_{\text{wire}}+R_c \] Here, \[ R_c=1.0\,\Omega \] The temperature coefficient must be calculated using the wire's own resistance, not the combined measured resistance. \(\textbf{Wire resistance at }0^\circ\textbf{C:}\) \[ R_{\text{wire},0}=R_{m,0}-R_c \] Substituting, \[ R_{\text{wire},0}=9.0-1.0 \] \[ R_{\text{wire},0}=8.0\,\Omega \] \(\textbf{Wire resistance at }100^\circ\textbf{C:}\) \[ R_{\text{wire},100}=13.0-1.0 \] \[ R_{\text{wire},100}=12.0\,\Omega \] \(\textbf{Determine the temperature coefficient:}\) Using \[ R_T=R_0(1+\alpha_0\Delta T), \] we obtain \[ 12.0=8.0[1+\alpha_0(100)] \] Dividing by \(8.0\), \[ 1.5=1+100\alpha_0 \] Therefore, \[ 100\alpha_0=0.5 \] \[ \alpha_0=5.0\times10^{-3}\,\text{K}^{-1} \] \(\textbf{Wire resistance at }50^\circ\textbf{C:}\) \[ R_{\text{wire},50}=8.0[1+(0.005)(50)] \] \[ R_{\text{wire},50}=8.0(1.25) \] \[ R_{\text{wire},50}=10.0\,\Omega \] \(\textbf{Restore the contact contribution:}\) The voltmeter-based measurement includes the same contact resistance. Therefore, \[ R_{m,50}=R_{\text{wire},50}+R_c \] \[ R_{m,50}=10.0+1.0 \] \[ R_{m,50}=11.0\,\Omega \] The measured resistance rises linearly from \(9.0\,\Omega\) to \(13.0\,\Omega\), but its temperature coefficient differs from that of the wire alone. This happens because the fixed contact resistance contributes to the reference resistance without contributing to its temperature-dependent change. Subtracting the contact resistance before calculating the coefficient is therefore necessary. \(\textbf{Final answer:}\) \[ \boxed{\alpha_0=5.0\times10^{-3}\,\text{K}^{-1},\quad R_{m,50}=11.0\,\Omega} \]

Explanation: \(\textbf{Relate resistance to resistivity:}\) For a uniform specimen, \[ R=\frac{\rho L}{A} \] The specimen dimensions remain unchanged. Therefore, \[ R\propto\rho \] The measured resistance ratio is \[ \frac{R_{120}}{R_{20}}=\frac{18}{12} \] \[ \frac{R_{120}}{R_{20}}=\frac32 \] Hence, \[ \frac{\rho_{120}}{\rho_{20}}=\frac32 \] \(\textbf{Relate resistivity to relaxation time:}\) In the elementary free-electron model, \[ \sigma=\frac{ne^2\tau}{m} \] Since \[ \rho=\frac1\sigma, \] we obtain \[ \rho=\frac{m}{ne^2\tau} \] The carrier number density remains unchanged, so \[ \rho\propto\frac1\tau \] Therefore, \[ \frac{\tau_{120}}{\tau_{20}} =\frac{\rho_{20}}{\rho_{120}} \] \[ \frac{\tau_{120}}{\tau_{20}}=\frac23 \] The shorter relaxation time at higher temperature is consistent with increased scattering in the metallic specimen. \(\textbf{Use the fixed-current condition:}\) The total current and cross-sectional area are unchanged. Therefore, the current density also remains unchanged: \[ j=\frac IA \] Microscopic Ohm's law gives \[ E=\rho j \] At fixed current density, \[ E\propto\rho \] Hence, \[ \frac{E_{120}}{E_{20}} =\frac{\rho_{120}}{\rho_{20}} \] \[ \frac{E_{120}}{E_{20}}=\frac32 \] The average electron drift speed remains unchanged because \(I=neAv_d\), with \(I,n,e,A\) all unchanged. Nevertheless, a greater electric field is required at the higher temperature to maintain that same drift speed. The increased field compensates for the reduced relaxation time. The result connects measured resistance, microscopic scattering and the field required under a fixed-current constraint. \(\textbf{Final answer:}\) \[ \boxed{\frac{\tau_{120}}{\tau_{20}}=\frac23,\quad\frac{E_{120}}{E_{20}}=\frac32} \]

Explanation: Statement I is correct because the resistance of a uniform conductor is related to resistivity through \(R=\rho L/A\). When length and cross-sectional area remain effectively unchanged, resistance changes in the same proportion as resistivity. For ordinary metallic conductors over suitable temperature ranges, increasing temperature generally increases electron scattering and therefore increases resistivity. Statement II is correct because the temperature dependence of semiconductors differs from that of ordinary metals. In many semiconductors, increasing temperature produces a substantial increase in the concentration of available charge carriers. This increase can outweigh the reduction in carrier mobility, causing conductivity to increase and resistance to decrease. Statement III is correct because the commonly used linear relation is an approximation valid over a suitable temperature range. It is not a universal description of material resistance at every possible temperature. A large extrapolation can give inaccurate or physically inappropriate predictions when the actual material dependence is nonlinear. Statement IV is incorrect because conductivity is the reciprocal of resistivity. For an isotropic material, \(\sigma=1/\rho\), so increasing resistivity necessarily decreases conductivity. This reciprocal relation does not depend on the specimen's dimensions. Dimensions determine the resistance or conductance of a particular specimen, while resistivity and conductivity are material properties. Thus, Statements I, II and III are correct, and Statement IV is incorrect.

Explanation: \(\textbf{Infer conductivity from the current-density measurement:}\) For an ohmic isotropic material, \[ j=\sigma E \] Therefore, \[ \sigma=\frac jE \] At \(120^\circ\text{C}\), \[ \sigma_{120}=\frac{8.0\times10^6}{0.25} \] \[ \sigma_{120}=3.2\times10^7\,\text{S m}^{-1} \] The measured conductivity is smaller than its value at \(20^\circ\text{C}\). \(\textbf{Determine the resistivity ratio:}\) Resistivity is the reciprocal of conductivity. \[ \rho=\frac1\sigma \] Therefore, \[ \frac{\rho_{120}}{\rho_{20}} =\frac{\sigma_{20}}{\sigma_{120}} \] Substituting, \[ \frac{\rho_{120}}{\rho_{20}} =\frac{4.0\times10^7}{3.2\times10^7} \] \[ \frac{\rho_{120}}{\rho_{20}}=\frac54 \] The resistivity increases by \(25\%\) over the \(100\,\text{K}\) temperature rise. \(\textbf{Determine the temperature coefficient:}\) Using the reference at \(20^\circ\text{C}\), \[ \rho_{120}=\rho_{20}[1+\alpha_{20}(120-20)] \] Dividing by \(\rho_{20}\), \[ \frac54=1+100\alpha_{20} \] Therefore, \[ 100\alpha_{20}=\frac14 \] \[ \alpha_{20}=2.5\times10^{-3}\,\text{K}^{-1} \] \(\textbf{Compare resistance under unchanged geometry:}\) Since \[ R=\frac{\rho L}{A} \] and \(L,A\) remain unchanged, \[ \frac{R_{120}}{R_{20}}=\frac54 \] \(\textbf{Power ratio under fixed voltage:}\) At constant applied voltage, \[ P=\frac{V^2}{R} \] Therefore, \[ \frac{P_{120}}{P_{20}} =\frac{R_{20}}{R_{120}} \] \[ \frac{P_{120}}{P_{20}}=\frac45 \] The hotter specimen dissipates less power at the same voltage because its resistance is greater. The current-density measurement reveals conductivity, which must first be converted into resistivity before applying the temperature-coefficient relation. The final power comparison additionally requires identifying that the voltage, rather than the current, is held fixed. \(\textbf{Final answer:}\) \[ \boxed{\alpha_{20}=2.5\times10^{-3}\,\text{K}^{-1},\quad\frac{P_{120}}{P_{20}}=\frac45} \]

Explanation: \(\textbf{Determine the resistivity change:}\) The resistivity varies with temperature according to \[ \rho_T=\rho_{20}[1+\alpha_{20}(T-20)] \] The temperature rise for experiment Q is \[ \Delta T=120-20=100\,\text{K} \] Therefore, \[ \rho_{120}=\rho_{20}[1+(4.0\times10^{-3})(100)] \] \[ \rho_{120}=1.40\rho_{20} \] \(\textbf{Experiment P — Current along }3a\textbf{:}\) For a uniform conductor, \[ R=\frac{\rho L}{A} \] The current path length is \[ L_P=3a \] The area normal to current is \[ A_P=2a^2 \] Thus, \[ R_P=\frac{\rho_{20}(3a)}{2a^2} \] \[ R_P=\frac{3\rho_{20}}{2a} \] \(\textbf{Experiment Q — Current along }a\textbf{:}\) The current path length is \[ L_Q=a \] The area normal to current is \[ A_Q=6a^2 \] At the higher temperature, \[ \rho_Q=1.40\rho_{20} \] Hence, \[ R_Q=\frac{(1.40\rho_{20})a}{6a^2} \] \[ R_Q=\frac{1.40\rho_{20}}{6a} \] Since \(1.40=7/5\), \[ R_Q=\frac{7\rho_{20}}{30a} \] \(\textbf{Compare the measured resistances:}\) \[ \frac{R_Q}{R_P} = \frac{7\rho_{20}/(30a)}{3\rho_{20}/(2a)} \] Cancelling the common factors, \[ \frac{R_Q}{R_P}=\frac7{30}\times\frac23 \] \[ \frac{R_Q}{R_P}=\frac7{45} \] The material resistivity is greater in experiment Q because the block is hotter. However, its measured resistance is much smaller because current travels along the shortest dimension through the largest cross-sectional area. A greater resistivity does not guarantee a greater measured resistance when the geometry of current flow also changes. The result requires combining a material-property change with the three-dimensional orientation of the conducting path. \(\textbf{Final answer:}\) \[ \boxed{\frac{R_Q}{R_P}=\frac7{45}} \]