Q.At room temperature ($27.0\ ^\circ\text{C}$) the resistance of a heating element is $100\ \Omega$. What is the temperature of the element if the resistance is found to be $117\ \Omega$, given that the temperature coefficient of the material of the resistor is $1.70 \times 10^{-4}\ ^\circ\text{C}^{-1}$.

A real electrical cell is not an ideal voltage source; it possesses an inherent internal resistance, denoted by $r$. When this cell is connected to an external resistor $R$, the total resistance in the circuit becomes $R+r$. This internal resistance causes a voltage drop within the cell itself, meaning the terminal voltage across the external resistor is less than the cell's electromotive force (EMF), $E$.

The current flowing through the circuit is determined by Ohm's Law applied to the entire circuit: $I = \frac{E}{R+r}$. The power consumed by the external resistor $R$ is given by $P = I^2 R$. The problem states that the power consumption is the same for two different external resistors. This allows us to set up an equation and solve for the unknown internal resistance $r$.

Here's how to solve the problem:

1. Define the circuit parameters and current: Let $E$ be the electromotive force (EMF) of the cell and $r$ be its internal resistance. When an external resistor $R$ is connected, the total resistance in the circuit is $R+r$. The current $I$ flowing through the circuit is given by Ohm's Law:

2. Express power consumed in the external resistor: The power $P$ consumed by the external resistor $R$ is given by the formula $P = I^2 R$. Substituting the expression for $I$:

3. Apply the condition of equal power consumption: We are given two external resistors, $R_1 = 2~\Omega$ and $R_2 = 4.5~\Omega$. The power consumption is the same in both cases. Let $P_1$ be the power for $R_1$ and $P_2$ for $R_2$.

Since $P_1 = P_2$:

4. Solve for the internal resistance $r$: We can cancel $E^2$ from both sides (assuming $E \neq 0$):

Rearrange the terms:

Take the square root of both sides. Since $R_1, R_2, r$ are all positive resistances, $R_1+r$ and $R_2+r$ must be positive, so we take the positive square root:

Cross-multiply:

Group terms involving $r$:

Factor out $r$ on the left side and $\sqrt{R_1 R_2}$ on the right side:

Since $R_1 \neq R_2$, we know that $\sqrt{R_1} \neq \sqrt{R_2}$, so $(\sqrt{R_2} - \sqrt{R_1}) \neq 0$. We can divide both sides by this term:

> [!IMPORTANT]
> This is a useful general result: if a cell delivers the same power to two different external resistors $R_1$ and $R_2$, its internal resistance $r$ is the geometric mean of $R_1$ and $R_2$.

5. Substitute the given values:

Given $R_1 = 2~\Omega$ and $R_2 = 4.5~\Omega$.

The internal resistance of the cell is $3~\Omega$.

The key idea is that in a semiconductor, both electrons and holes contribute to current flow. Conductivity isn't just about one carrier — it's the sum of each carrier's contribution, where each contribution is the product of charge, concentration, and mobility.

The formula is straightforward: $\sigma = e(n\mu_n + p\mu_p)$, where $e = 1.6 \times 10^{-19}\ \mathrm{C}$ is the elementary charge, $n$ and $p$ are electron and hole concentrations, and $\mu_n$, $\mu_p$ are their mobilities.

1. Identify the given values

   $n = 5 \times 10^{18}\ \mathrm{m}^{-3}$

   $p = 8 \times 10^{19}\ \mathrm{m}^{-3}$

   $\mu_n = 0.24\ \mathrm{m}^2\,\mathrm{V}^{-1}\,\mathrm{s}^{-1}$

   $\mu_p = 0.01\ \mathrm{m}^2\,\mathrm{V}^{-1}\,\mathrm{s}^{-1}$

   $e = 1.6 \times 10^{-19}\ \mathrm{C}$

2. Compute the electron contribution

   $e n \mu_n = (1.6 \times 10^{-19}) \times (5 \times 10^{18}) \times 0.24$

   First, $1.6 \times 10^{-19} \times 5 \times 10^{18} = 1.6 \times 5 \times 10^{-1} = 8 \times 10^{-1} = 0.8$

   Then $0.8 \times 0.24 = 0.192\ \mathrm{S\,m^{-1}}$

3. Compute the hole contribution

   $e p \mu_p = (1.6 \times 10^{-19}) \times (8 \times 10^{19}) \times 0.01$

   First, $1.6 \times 10^{-19} \times 8 \times 10^{19} = 1.6 \times 8 = 12.8$

   Then $12.8 \times 0.01 = 0.128\ \mathrm{S\,m^{-1}}$

4. Add the contributions

   $\sigma = 0.192 + 0.128 = 0.32\ \mathrm{S\,m^{-1}}$

The key idea is Curie’s law: for a paramagnetic material, the magnetic susceptibility $\chi$ is inversely proportional to the absolute temperature. This means as temperature increases, the random thermal motion of atomic magnetic moments increases, reducing their alignment with an external field — hence susceptibility drops. The law is simple: $\chi = C/T$, where $C$ is the Curie constant. So if you know $\chi$ at one temperature, you can find it at another by scaling with the ratio of absolute temperatures.

Why this works: The susceptibility measures how easily a material magnetizes. In paramagnets, thermal energy fights the aligning effect of the field. Higher temperature → more disorder → lower susceptibility. The inverse proportionality is a direct consequence of statistical mechanics for non-interacting magnetic moments.

Now, step by step:

1. Convert temperatures to Kelvin.

   Curie’s law uses absolute temperature.

   $T_1 = 23^\circ\mathrm{C} + 273 = 296\,\mathrm{K}$

   $T_2 = 467^\circ\mathrm{C} + 273 = 740\,\mathrm{K}$

2. Apply Curie’s law.

   Since $\chi \propto 1/T$, we have $\chi_1 T_1 = \chi_2 T_2$.

   Given $\chi_1 = 2.3 \times 10^{-5}$, solve for $\chi_2$:

3. Simplify the fraction.

   $\frac{296}{740} = \frac{296 \div 4}{740 \div 4} = \frac{74}{185}$

   Further simplify: divide numerator and denominator by 37 → $\frac{2}{5} = 0.4$

   (Check: $74 \div 37 = 2$, $185 \div 37 = 5$)

4. Compute the result.

   $\chi_2 = 2.3 \times 10^{-5} \times 0.4 = 0.92 \times 10^{-5} = 9.2 \times 10^{-6}$

The key concept here is the drift velocity formula for electrons in a conductor under an electric field. Drift speed $v_d$ is the average velocity electrons acquire due to the field, and it depends linearly on both the relaxation time $\tau$ (the average time between collisions) and the applied electric field $E$. The formula is:

where $e$ is the electron charge and $m$ is its mass. Since $e$ and $m$ are constants for a given material, any change in $v_d$ comes only from changes in $E$ and $\tau$.

Now, let’s work through the problem step by step.

1. Write the initial drift speed. Let the initial relaxation time be $\tau_0$ and the initial electric field be $E_0$. Then the initial drift speed is:

2. Apply the given changes.

   The relaxation time is doubled: $\tau = 2\tau_0$.

   The electric field is tripled: $E = 3E_0$.

3. Write the new drift speed.

   Substitute these into the formula:

4. Interpret the result. The new drift speed is exactly six times the initial value. This matches option (B).

The key idea is to distinguish between resistance (a property of a particular object) and resistivity (a property of the material). Resistance changes when you alter the wire’s dimensions; resistivity changes only if you change the material or its temperature. Here, the material (iron) and temperature are unchanged, so resistivity is fixed. We just need to see how the geometry affects $R$.

1. Recall the formula for resistance For a wire of uniform cross-section,

where $\rho$ is the resistivity (specific resistance), $L$ is the length, and $A$ is the cross-sectional area. For a cylindrical wire, $A = \pi r^2$.

2. Identify what changes

   The problem says: length and radius are both doubled.

   So new length $L' = 2L$ and new radius $r' = 2r$.

3. Compute the new cross-sectional area

The area becomes four times larger.

4. Find the new resistance Since resistivity $\rho$ is unchanged (same material, same temperature),

The resistance is halved.

5. What about specific resistance? Specific resistance is just another name for resistivity $\rho$. It depends only on the material and temperature — not on the wire’s shape or size. Since we are still using iron at the same temperature, $\rho$ remains unchanged.

Concept & Intuition

A galvanometer is a sensitive current-measuring device that can only handle a small fraction of the total current in a circuit. To measure larger currents, we connect a low-resistance shunt in parallel with the galvanometer. The shunt “steals” most of the current, leaving only a safe, small current through the galvanometer.

The key principle: In a parallel circuit, the voltage across each branch is the same. So the voltage across the galvanometer (due to its small current) equals the voltage across the shunt (due to the large current). This voltage equality gives us a direct relationship between the resistances and the currents.

Step-by-step solution

1. Interpret the given percentage “5% of the total current passes through the galvanometer” means:

where $I_g$ is the current through the galvanometer and $I_{\text{total}}$ is the total current entering the parallel combination.

2. Find the current through the shunt The shunt carries the rest of the current:

3. Apply the parallel voltage condition The voltage across the galvanometer is $V_g = I_g G$. The voltage across the shunt is $V_s = I_s S$. Since they are in parallel:

4. Substitute the currents

Cancel $I_{\text{total}}$ (it is non-zero):

5. Solve for $S$

So $S = \dfrac{G}{19}$.

The concept first

When a network is not a plain series–parallel ladder, don't force it. Do two things instead: (i) collapse any obvious parallel pairs, and (ii) write node equations (Kirchhoff's current law). Injecting $1$ A at $A$ and taking it out at $B$ makes $R_{AB}$ simply equal to $V_A-V_B$ — that is the definition of resistance and it never fails, even on a bridge.

Step-by-step

1. Label the nodes. $A$; the intermediate node $M$ where the two $10\,\Omega$ resistors meet; the right-hand node which is joined by plain wire to $B$ (so it is $B$); and the bottom rail $G$ that all the vertical resistors hang onto.

2. Collapse the obvious parallels.

• Two $10\,\Omega$ between $A$ and $M$: $\ 10\parallel10=5\,\Omega$.
• Two $6\,\Omega$ from $B$ (and from the node shorted to $B$) down to $G$: $\ 6\parallel6=3\,\Omega$.

3. The reduced network.

This is a bridge: $M$ and $G$ are the two "middle" nodes, linked by the $8\,\Omega$.

4. Inject $1$ A at $A$, extract at $B$, set $V_B=0$. Conductances: $A\!-\!M:0.2$, $A\!-\!G:0.2$, $M\!-\!B:\tfrac13$, $M\!-\!G:0.125$, $B\!-\!G:\tfrac13$.

KCL at $A$: $\;0.2(V_A-V_M)+0.2(V_A-V_G)=1$

KCL at $M$: $\;0.2(V_M-V_A)+\tfrac13 V_M+0.125(V_M-V_G)=0$

KCL at $G$: $\;0.2(V_G-V_A)+\tfrac13 V_G+0.125(V_G-V_M)=0$

5. Exploit the symmetry. The equations for $M$ and $G$ are identical in form: each has $5\,\Omega$ to $A$, $8\,\Omega$ to the other, and $3\,\Omega$ to $B$. Hence $V_M=V_G=v$ and the $8\,\Omega$ link carries no current (a balanced bridge!).

6. Solve. With $V_M=V_G=v$, the $M$-equation gives

and the $A$-equation gives

7. Read off the resistance. With $1$ A flowing and $V_A-V_B=4$ V,

Equivalently, once you see the bridge is balanced you can drop the $8\,\Omega$: then $A\!\to\! M\!\to\! B$ is $5+3=8\,\Omega$ and $A\!\to\! G\!\to\! B$ is $5+3=8\,\Omega$, and $8\parallel8=4\,\Omega$. Same answer, in one line.

ANSWER: B

Concept and Intuition

This problem tests our understanding of fundamental concepts in current electricity:

1. Temperature dependence of resistance in metals: How the electrical resistance of a material changes with temperature, characterized by the temperature coefficient of resistance.
2. Drift velocity of electrons: The average velocity attained by charge carriers in a material due to an electric field, which is directly related to the current flowing through the material.
3. Ohm's Law and its limitations: The conditions under which a material obeys Ohm's law, and which types of materials (like semiconductors) deviate from it.

We will evaluate each statement individually based on these principles.

Step-by-step Evaluation

1. Evaluating Statement (I)

Statement (I) says: "The temperature coefficient of resistance for most of metals in pure form is positive."

• Concept: In pure metals, electrical conduction occurs due to the movement of free electrons. As temperature increases, the thermal vibrations of the lattice atoms become more vigorous. These increased vibrations lead to more frequent collisions between the free electrons and the vibrating lattice atoms.
• Effect on resistance: These collisions impede the smooth flow of electrons, effectively increasing the resistivity of the metal. Since resistance $R$ is directly proportional to resistivity $\rho$ ($R = \rho L/A$), an increase in resistivity leads to an increase in resistance.
• Temperature coefficient: The resistance of a material at temperature $T$ is given by $R_T = R_0 (1 + \alpha (T - T_0))$, where $R_0$ is the resistance at reference temperature $T_0$, and $\alpha$ is the temperature coefficient of resistance. For metals, since $R_T > R_0$ when $T > T_0$, the value of $\alpha$ must be positive.

• Conclusion for Statement (I): This statement is True.

2. Evaluating Statement (II)

Statement (II) says: "A metal wire 2 mm in diameter carries a charge of $360\pi$ C in two hours. If the metal contains $5 \times 10^{22}$ free electrons / cm$^3$, then drift velocity of the electrons in the wire is $6.25 \times 10^{-6}$ m/s."

• Concept: The relationship between current ($I$), number density of charge carriers ($n$), cross-sectional area ($A$), drift velocity ($v_d$), and elementary charge ($e$) is given by the formula:

  $I = n A v_d e$

• Given values and unit conversions:

  • Diameter $d = 2$ mm $\implies$ Radius $r = 1$ mm $= 1 \times 10^{-3}$ m.
  • Charge $Q = 360\pi$ C.
  • Time $t = 2$ hours $= 2 \times 3600$ s $= 7200$ s.
  • Number density $n = 5 \times 10^{22}$ electrons/cm$^3$. We need to convert this to electrons/m$^3$: $n = 5 \times 10^{22} \text{ electrons/cm}^3 \times (100 \text{ cm/m})^3 = 5 \times 10^{22} \times 10^6 \text{ electrons/m}^3 = 5 \times 10^{28} \text{ electrons/m}^3$.
  • Elementary charge $e = 1.6 \times 10^{-19}$ C.

• Calculations:

  1. Calculate the current ($I$):

     $I = \frac{Q}{t} = \frac{360\pi \text{ C}}{7200 \text{ s}} = \frac{\pi}{20} \text{ A}$

  2. Calculate the cross-sectional area ($A$):

     $A = \pi r^2 = \pi (1 \times 10^{-3} \text{ m})^2 = \pi \times 10^{-6} \text{ m}^2$

  3. Calculate the drift velocity ($v_d$):

     Rearranging the formula $I = n A v_d e$, we get $v_d = \frac{I}{n A e}$.

     Substitute the calculated values:

     $v_d = \frac{\frac{\pi}{20} \text{ A}}{(5 \times 10^{28} \text{ m}^{-3}) \times (\pi \times 10^{-6} \text{ m}^2) \times (1.6 \times 10^{-19} \text{ C})}$

     $v_d = \frac{\pi/20}{\pi \times (5 \times 1.6) \times 10^{28 - 6 - 19}}$

     $v_d = \frac{1/20}{8 \times 10^3}$

     $v_d = \frac{1}{20 \times 8 \times 10^3}$

     $v_d = \frac{1}{160 \times 10^3}$

     $v_d = \frac{1}{1.6 \times 10^5}$

     $v_d = 0.625 \times 10^{-5} \text{ m/s}$

     $v_d = 6.25 \times 10^{-6} \text{ m/s}$

• Comparison: The calculated drift velocity ($6.25 \times 10^{-6}$ m/s) matches the value given in the statement.

• Conclusion for Statement (II): This statement is True.