Q.The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:

[!FORMULA] $R = R_0\left[1 + \alpha (T - T_0)\right]$

The resistance is $101.6\ \Omega$ at the triple-point of water $273.16\ \text{K}$, and $165.5\ \Omega$ at the normal melting point of lead ($600.5\ \text{K}$). What is the temperature when the resistance is $123.4\ \Omega$?

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If the errors in the measurements of diameter, length and electrical resistance of a wire are 1%, 0.5% and 2% respectively, then percentage error in the determination of the resistivity of material of the wire is (A) 3.5 (B) 4 (C) 4.5 (D) 2.5

   ›Reveal solutionSolution

   The percentage error in resistivity is the sum of the percentage errors in resistance, length, and twice the diameter error, giving 2% + 0.5% + 2×1% = 4.5%. The correct option is (C).

   The key idea is that resistivity $\rho$ is not measured directly but calculated from other measured quantities: resistance $R$, length $L$, and diameter $d$ (which gives cross-sectional area). When each measurement has an uncertainty, those uncertainties propagate into the final result. For multiplication and division, percentage errors simply add.

   Why this works:

   If a quantity $Q = \frac{A \cdot B}{C}$, then the relative error in $Q$ is the sum of the relative errors in $A$ and $B$ minus the relative error in $C$ (but since errors are usually given as positive magnitudes, we add absolute values). For powers, the error is multiplied by the exponent. Here, area depends on $d^2$, so the diameter error gets doubled.

   ---

   1. Write the formula for resistivity Resistivity $\rho$ is given by

   $\rho = \frac{R \cdot A}{L}$

   where $A$ is the cross-sectional area. For a wire of diameter $d$,

   $A = \frac{\pi d^2}{4}$

   So

   $\rho = \frac{R \cdot \pi d^2}{4L}$

   The constant $\pi/4$ has no error, so it doesn’t affect the percentage error.

   2. Apply the rule for error propagation For a product/quotient, the percentage error in $\rho$ is the sum of the percentage errors in each factor, with each error multiplied by the power of that factor:

   $\frac{\Delta \rho}{\rho} \times 100\% = \frac{\Delta R}{R} \times 100\% + 2 \cdot \frac{\Delta d}{d} \times 100\% + \frac{\Delta L}{L} \times 100\%$

   (Note: $L$ is in the denominator, but we still add the magnitude of its error.)

   3. Insert the given percentage errors

      • Error in resistance: $\frac{\Delta R}{R} \times 100\% = 2\%$
      • Error in diameter: $\frac{\Delta d}{d} \times 100\% = 1\%$ → multiplied by 2 gives $2\%$
      • Error in length: $\frac{\Delta L}{L} \times 100\% = 0.5\%$

      So total percentage error in $\rho$ is:

   $2\% + 2\% + 0.5\% = 4.5\%$

   Tip

   A common mistake is to forget that the diameter appears squared. If you only add 1% for diameter instead of 2%, you get 3.5% — which is option (A), a tempting distractor.

   Watch out

   Do not subtract the length error because it’s in the denominator. In error propagation, we always add absolute relative errors for multiplication and division — the sign of the operation doesn’t matter for the magnitude of uncertainty.

   4. Match with the options The result 4.5% corresponds to option (C).

   ✓Final answer

   The correct option is (C).

   ANSWER: C

2. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.If the pressure of the gas in a constant volume gas thermometer at ice point is 91 kPa, then the pressure of the gas in the thermometer at steam point is nearly (A) 144.3 kPa (B) 138.6 kPa (C) 172.6 kPa (D) 124.3 kPa

   ›Reveal solutionSolution

   Using the constant-volume gas law $P\propto T$ between the ice point (273 K) and steam point (373 K) gives a pressure of about 124.3 kPa.

   Concept and Intuition

   A constant-volume gas thermometer relies on the ideal-gas relation $PV=nRT$: with volume and moles fixed, pressure is directly proportional to absolute temperature. Measuring pressure at two known reference temperatures (ice point and steam point) lets us relate them directly through this proportionality.

   Step-by-Step Solution

   1. Ice point temperature: $T_1 = 0^\circ\text{C} = 273\ \text{K}$, with $P_1 = 91\ \text{kPa}$.
   2. Steam point temperature: $T_2 = 100^\circ\text{C} = 373\ \text{K}$.
   3. At constant volume, $\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2} \implies P_2 = P_1\times\dfrac{T_2}{T_1}$.
   4. Substitute: $P_2 = 91\times\dfrac{373}{273} = 91\times1.3663\approx124.3\ \text{kPa}$.

   Common Mistakes

   • Using Celsius temperatures directly in the ratio instead of converting to Kelvin (absolute temperature) first — this proportionality only holds for absolute temperature.
   • Rounding errors in the $373/273$ ratio leading to a value that doesn't match any option closely; keeping enough decimal places is needed to land near 124.3 kPa.

   ✓Final answer

   The correct option is (D) — 124.3 kPa.

   ANSWER: D

3. KCET 2025Set D-41 markMCQ
   Q.In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He unplugged a resistance of $5200\Omega$ in $R$. When $K_1$ is closed and $K_2$ is open, the deflection observed in the galvanometer is $26$ div. When $K_2$ is also closed and a resistance of $90\Omega$ is removed in $S$, the deflection between $13$ div. The resistance of galvanometer is nearly (A) $45.0\ \Omega$ (B) $103.0\ \Omega$ (C) $91.6\ \Omega$ (D) $116.0\ \Omega$

   ›Reveal solutionSolution

   Apply the standard half-deflection result $G = \dfrac{RS}{R - S}$, with $R = 5200\ \Omega$ (the series resistance) and $S = 90\ \Omega$ (the shunt that halves the deflection).

   Step 1 — What happens with $K_2$ open.

   Only the series resistance $R$ and the galvanometer $G$ carry the current. If the cell emf is $E$ (internal resistance negligible), the current through the galvanometer is

   $I_g = \frac{E}{R + G}$

   This produces the full deflection, $\theta = 26$ divisions. Since deflection $\propto$ current through the coil,

   $26 \propto \frac{E}{R+G} \qquad (1)$

   Step 2 — What happens when $K_2$ is closed.

   The shunt $S = 90\ \Omega$ is now in parallel with the galvanometer. Because $R \gg$ the parallel combination ($5200\ \Omega$ versus at most $90\ \Omega$), the total current drawn from the cell is essentially unchanged:

   $I \approx \frac{E}{R}$

   That current now splits between $G$ and $S$. By the current-divider rule the fraction through the galvanometer is

   $I_g' = I \cdot \frac{S}{G + S}$

   The observed deflection is halved, to $13$ divisions.

   Step 3 — Impose the half-deflection condition.

   Deflection halves $\Rightarrow$ the coil current halves:

   $\frac{I_g'}{I_g} = \frac{13}{26} = \frac{1}{2}$

   $\frac{S}{G+S} = \frac{1}{2}\ \text{(of the near-unchanged total)} \quad\Longrightarrow\quad \frac{E}{R}\cdot\frac{S}{G+S} = \frac{1}{2}\cdot\frac{E}{R+G}$

   Working this through (the standard derivation) yields the well-known result

   $\boxed{G = \frac{RS}{R - S}}$

   Why this form? In the ideal limit the shunt must carry exactly as much current as the coil for the coil current to halve — which requires $S = G$ if the total were fixed. The finite $R$ correction produces the $R/(R-S)$ factor, and because $R \gg S$ that factor is only slightly above 1, so $G$ comes out slightly larger than $S$. We should therefore expect an answer a little above $90\ \Omega$ — a useful check before computing.

   Step 4 — Substitute the numbers.

   $G = \frac{5200 \times 90}{5200 - 90} = \frac{468000}{5110}$

   $G = 91.585\ldots \approx 91.6\ \Omega$

   This sits just above $S = 90\ \Omega$, exactly as anticipated.

   Rejecting the distractors: (A) $45.0\ \Omega$ is $S/2$ (halving the wrong quantity); (B) $103\ \Omega$ and (D) $116\ \Omega$ do not follow from $RS/(R-S)$ with these values.

   ✓Final answer

   The correct option is (C) — $91.6\ \Omega$.

   ANSWER: C

4. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.A metal metre scale that is accurate up to 0.5 mm is made at a temperature of 25°C. The range of temperatures within which it can be used is (Coefficient of linear expansion of the metal $= 10^{-5}/^\circ \mathrm{C}$) (A) $+25^\circ \mathrm{C}$ to $+75^\circ \mathrm{C}$ (B) $+25^\circ \mathrm{C}$ to $+50^\circ \mathrm{C}$ (C) $-25^\circ \mathrm{C}$ to $+75^\circ \mathrm{C}$ (D) $0^\circ \mathrm{C}$ to $+50^\circ \mathrm{C}$

   ›Reveal solutionSolution

   The scale is accurate to 0.5 mm over 1 m, so the allowed expansion or contraction is ±0.5 mm. Using $\Delta L = L_0 \alpha \Delta T$, the allowable temperature change is ±50°C, giving a range from –25°C to +75°C. The correct option is (C).

   The key idea is that the scale is “accurate up to 0.5 mm” — meaning that over its full length of 1 metre, the error due to thermal expansion or contraction must not exceed 0.5 mm in either direction. Since the scale was made at 25°C, it reads correctly only at that temperature. At any other temperature, the metal expands or contracts, so the markings shift. We need the range of temperatures for which the shift is at most 0.5 mm.

   The coefficient of linear expansion $\alpha = 10^{-5}/^\circ\mathrm{C}$ tells us how much each metre of the metal changes per degree Celsius. The formula is:

   $\Delta L = L_0 \, \alpha \, \Delta T$

   where $L_0 = 1\ \mathrm{m} = 1000\ \mathrm{mm}$.

   1. Set the maximum allowed change in length.

      The scale is accurate to 0.5 mm, so $|\Delta L| \le 0.5\ \mathrm{mm}$.

   2. Solve for the maximum temperature change.

   $0.5 = 1000 \times 10^{-5} \times |\Delta T|$

   $0.5 = 0.01 \times |\Delta T|$

   $|\Delta T| = \frac{0.5}{0.01} = 50^\circ\mathrm{C}$

   3. Interpret the sign.

      The scale can be used when the temperature is up to 50°C above or below the calibration temperature of 25°C. So:

      • Lower bound: $25 - 50 = -25^\circ\mathrm{C}$
      • Upper bound: $25 + 50 = 75^\circ\mathrm{C}$

   4. Check the options.

      The range $-25^\circ\mathrm{C}$ to $+75^\circ\mathrm{C}$ matches option (C).

   Watch out

   A common mistake is to forget that the scale can contract as well as expand, so the range must extend below 25°C, not just above it. Options (A) and (B) only allow heating, which is incomplete.

   Tip

   Notice that the 0.5 mm accuracy is tiny compared to 1 m — it’s 0.05% of the length. The coefficient is also small, so the allowable temperature swing ends up being a generous 50°C in each direction.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

5. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.A metal metre scale that is accurate up to 0.5 mm is made at a temperature of $25^\circ \mathrm{C}$. The range of temperatures within which it can be used is (Coefficient of linear expansion of the metal $= 10^{-5}/^\circ \mathrm{C}$) (A) $0^\circ \mathrm{C}$ to $+50^\circ \mathrm{C}$ (B) $+25^\circ \mathrm{C}$ to $+75^\circ \mathrm{C}$ (C) $-25^\circ \mathrm{C}$ to $+75^\circ \mathrm{C}$ (D) $+25^\circ \mathrm{C}$ to $+50^\circ \mathrm{C}$

   ›Reveal solutionSolution

   The scale is accurate to 0.5 mm over 1 m, so the allowed expansion or contraction is ±0.5 mm. Using $\Delta L = L_0 \alpha \Delta T$, the allowable temperature change is ±50°C from the calibration temperature of 25°C, giving a range of –25°C to +75°C. The correct option is (C).

   The key idea here is that a "metre scale" is supposed to read exactly 1 m between its end marks. But if the temperature changes, the metal expands or contracts, so the physical distance between the marks changes. The scale is only accurate if that change stays within the stated tolerance of 0.5 mm.

   The coefficient of linear expansion tells you how much length changes per degree per unit length. For a 1 m scale, a temperature change $\Delta T$ produces a change $\Delta L = L_0 \alpha \Delta T$. We want $|\Delta L| \le 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}$.

   Let’s work it through.

   1. Set up the condition. The scale is calibrated at $25^\circ \mathrm{C}$. Let the temperature at which it is used be $T$. Then $\Delta T = T - 25^\circ \mathrm{C}$. The change in length is

   $\Delta L = (1 \text{ m}) \times (10^{-5}/^\circ \mathrm{C}) \times (T - 25^\circ \mathrm{C}).$

   2. Apply the accuracy limit. We need $|\Delta L| \le 0.5 \times 10^{-3} \text{ m}$. So

   $|10^{-5} (T - 25)| \le 0.5 \times 10^{-3}.$

   3. Solve for the temperature range. Divide both sides by $10^{-5}$:

   $|T - 25| \le \frac{0.5 \times 10^{-3}}{10^{-5}} = 50.$

   This means

   $-50 \le T - 25 \le 50,$

   so

   $-25^\circ \mathrm{C} \le T \le 75^\circ \mathrm{C}.$

   Watch out

   A common mistake is to forget that the scale can contract as well as expand. The problem says "accurate up to 0.5 mm", meaning the error in either direction (too long or too short) must not exceed that. So you must consider both positive and negative $\Delta T$, giving a symmetric range around 25°C.

   Tip

   Notice that the numbers work out neatly: $\alpha = 10^{-5}$ and the tolerance is 0.5 mm = $5 \times 10^{-4}$ m. Dividing gives $|\Delta T| \le 50^\circ \mathrm{C}$ directly. This is a quick mental check.

   ✓Final answer

   The correct option is (C), $-25^\circ \mathrm{C}$ to $+75^\circ \mathrm{C}$.

6. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.The change in moment of inertia of a solid sphere of mass ‘M’, radius ‘R’, for a small change in temperature ‘Δt’ is ($\alpha$ is coefficient of linear expansion) (A) $\dfrac{2}{5}MR^2\alpha\Delta t$ (B) $\dfrac{4}{5}MR^2\alpha\Delta t$ (C) $\dfrac{7}{5}MR^2\alpha\Delta t$ (D) $\dfrac{3}{5}MR^2\alpha\Delta t$

   ›Reveal solutionSolution

   When a solid sphere undergoes a small temperature change, its radius expands due to thermal expansion. This change in radius directly affects its moment of inertia. The change in moment of inertia is $\boxed{\dfrac{4}{5}MR^2\alpha\Delta t}$.

   The problem asks us to find the change in the moment of inertia of a solid sphere when its temperature changes by a small amount $\Delta t$. This involves two key physical concepts: thermal expansion and moment of inertia.

   Concept and Intuition

   1. Thermal Expansion: When the temperature of an object changes, its dimensions (length, area, volume) also change. For a solid sphere, an increase in temperature causes its radius to increase. This is known as linear thermal expansion. The change in radius is directly proportional to the original radius, the change in temperature, and the material's coefficient of linear expansion ($\alpha$).
   2. Moment of Inertia: The moment of inertia ($I$) of an object is a measure of its resistance to angular acceleration. For a solid sphere rotating about an axis passing through its center, the moment of inertia depends on its mass ($M$) and the square of its radius ($R$). Specifically, $I = \frac{2}{5}MR^2$.
   3. Connecting the Concepts: As the temperature changes, the radius $R$ of the sphere changes. Since the moment of inertia depends on $R^2$, a change in $R$ will lead to a change in $I$. The mass $M$ of the sphere, however, remains constant with temperature change. We need to calculate this change in $I$ for a small $\Delta t$.

   Let's work through the steps to find the change in moment of inertia.

   1. Initial Moment of Inertia

      The moment of inertia of a solid sphere of mass $M$ and radius $R$ about an axis passing through its center is given by:

      $I = \frac{2}{5}MR^2$

   2. Change in Radius due to Temperature

      When the temperature changes by $\Delta t$, the radius $R$ of the sphere changes. The change in radius, $\Delta R$, is given by the formula for linear thermal expansion:

   $\Delta R = R\alpha\Delta t$

   The new radius, $R'$, will be the original radius plus the change in radius:
   

   $R' = R + \Delta R = R + R\alpha\Delta t = R(1 + \alpha\Delta t)$

   3. New Moment of Inertia Now, we can find the new moment of inertia, $I'$, by substituting the new radius $R'$ into the formula for the moment of inertia:

   $I' = \frac{2}{5}M(R')^2$

   Substitute $R' = R(1 + \alpha\Delta t)$:
   

   $I' = \frac{2}{5}M[R(1 + \alpha\Delta t)]^2$

   $I' = \frac{2}{5}MR^2(1 + \alpha\Delta t)^2$

   4. Calculate Change in Moment of Inertia The change in moment of inertia, $\Delta I$, is the difference between the new moment of inertia and the initial moment of inertia:

   $\Delta I = I' - I$

   $\Delta I = \frac{2}{5}MR^2(1 + \alpha\Delta t)^2 - \frac{2}{5}MR^2$

   Factor out $\frac{2}{5}MR^2$:
   

   $\Delta I = \frac{2}{5}MR^2[(1 + \alpha\Delta t)^2 - 1]$

   5. Apply Small Change Approximation

      Since $\Delta t$ is a small change in temperature, $\alpha\Delta t$ will be a very small quantity. We can use the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$. In our case, $x = \alpha\Delta t$ and $n=2$.

      Tip

      For a small value $x$ (i.e., $|x| \ll 1$), the binomial approximation is $(1+x)^n \approx 1+nx$.

      Here, $\alpha\Delta t$ is very small, so $(1 + \alpha\Delta t)^2 \approx 1 + 2(\alpha\Delta t)$.

      Substitute this approximation into the expression for $\Delta I$:

   $\Delta I \approx \frac{2}{5}MR^2[(1 + 2\alpha\Delta t) - 1]$

   $\Delta I \approx \frac{2}{5}MR^2[2\alpha\Delta t]$

   $\Delta I \approx \frac{4}{5}MR^2\alpha\Delta t$

   This result represents the change in the moment of inertia of the solid sphere due to the small change in temperature $\Delta t$.

   The correct option is (B).

   ✓Final answer

   The change in moment of inertia of the solid sphere is $\boxed{\dfrac{4}{5}MR^2\alpha\Delta t}$.

7. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.A pendulum clock loses 10.8 seconds a day when the temperature is 38°C and gains 10.8 seconds a day when the temperature is 18°C. The coefficient of linear expansion of the metal of the pendulum clock is (A) $7 \times 10^{-5}~^\circ \mathrm{C}^{-1}$ (B) $1.25 \times 10^{-5}~^\circ \mathrm{C}^{-1}$ (C) $5 \times 10^{-5}~^\circ \mathrm{C}^{-1}$ (D) $2.5 \times 10^{-5}~^\circ \mathrm{C}^{-1}$

   ›Reveal solutionSolution

   Equal loss and gain fix the correct-time temperature at $28^\circ$C, and the daily error over $\Delta\theta = 10^\circ$C gives $\alpha = 2.5\times10^{-5}~^\circ\text{C}^{-1}$.

   The fractional change in the time period of a pendulum with temperature is

   $\frac{\Delta T}{T} = \frac{1}{2}\,\alpha\,\Delta\theta,$

   so the time lost or gained per day is

   $\Delta t = \frac{1}{2}\,\alpha\,\Delta\theta \times 86400~\text{s}.$

   The clock loses time at $38^\circ$C and gains time at $18^\circ$C by the same amount ($10.8$ s), so the temperature $\theta_0$ at which it keeps correct time is symmetric between them:

   $38 - \theta_0 = \theta_0 - 18 \implies \theta_0 = 28^\circ\text{C}.$

   Thus the temperature swing from correct time is $\Delta\theta = 38 - 28 = 10^\circ$C. Substituting:

   $10.8 = \frac{1}{2}\,\alpha\,(10)\,(86400)$

   $10.8 = 432000\,\alpha \implies \alpha = \frac{10.8}{432000} = 2.5\times10^{-5}~^\circ\text{C}^{-1}.$

   ✓Final answer

   The coefficient of linear expansion is $2.5\times10^{-5}~^\circ\text{C}^{-1}$ — option (D).

8. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
   Q.'n' conducting wires of same dimensions but having resistivities 1, 2, 3...n respectively are connected in series. The equivalent resistivity of the combination is (A) $\dfrac{n(n+1)}{2}$ (B) $\dfrac{n+1}{2}$ (C) $\dfrac{n+1}{2n}$ (D) $\dfrac{2n}{n+1}$

   ›Reveal solutionSolution

   When wires of equal length and cross-section are connected in series, the total resistance is the sum of individual resistances. Since each wire’s resistance is proportional to its resistivity, the equivalent resistivity becomes the average of the given resistivities — which for $1, 2, 3, \dots, n$ is $\frac{n+1}{2}$.

   The key here is to separate two ideas that students often mix up: resistivity (a material property) and resistance (a property of the whole wire). The wires are identical in length and cross-section — only their resistivities differ. When you connect them in series, you’re adding resistances, not resistivities directly. But because the dimensions are the same, the total resistance can be re-expressed as an equivalent resistivity for the combined wire.

   Let’s walk through it.

   1. Resistance of a single wire For a wire of length $L$, cross-sectional area $A$, and resistivity $\rho$, the resistance is

   $R = \rho \frac{L}{A}.$

   Since all $n$ wires have the same $L$ and $A$, the resistance of the $i$-th wire (with resistivity $i$) is

   $R_i = i \cdot \frac{L}{A}.$

   2. Total resistance in series In series, resistances add directly:

   $R_{\text{total}} = R_1 + R_2 + \cdots + R_n = \frac{L}{A} (1 + 2 + 3 + \cdots + n).$

   The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$, so

   $R_{\text{total}} = \frac{L}{A} \cdot \frac{n(n+1)}{2}.$

   3. What does “equivalent resistivity” mean? The combination is a single conductor of length $nL$ (since series connection adds lengths) and the same cross-section $A$. If this combined wire had an equivalent resistivity $\rho_{\text{eq}}$, its resistance would be

   $R_{\text{total}} = \rho_{\text{eq}} \frac{nL}{A}.$

   4. Equate and solve Set the two expressions for $R_{\text{total}}$ equal:

   $\rho_{\text{eq}} \frac{nL}{A} = \frac{L}{A} \cdot \frac{n(n+1)}{2}.$

   Cancel $\frac{L}{A}$ (non-zero) and divide both sides by $n$:

   $\rho_{\text{eq}} = \frac{n+1}{2}.$

   Watch out

   A common mistake is to average the resistivities as $\frac{1+2+\cdots+n}{n} = \frac{n+1}{2}$ and stop there — which gives the same number, but the reasoning is wrong if you don’t account for the change in total length. Here it works because the wires have identical dimensions, so the factor of $n$ in length cancels neatly. In general, always go through resistance first.

   Tip

   Notice that the equivalent resistivity is simply the arithmetic mean of the individual resistivities. This happens because each wire contributes equally to the total length and has the same cross-section. If the dimensions differed, the mean would be weighted.

   ✓Final answer

   The equivalent resistivity is $\boxed{\frac{n+1}{2}}$, which corresponds to option (B).

9. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.A blacksmith fixes circular iron frame on the wooden wheel of a bullock cart. The diameter of wooden wheel and circular iron frame are $5.012\,\text{m}$ and $5\,\text{m}$ respectively at $27\,^\circ\text{C}$. The temperature (in $^\circ\text{C}$) to which the iron ring must be heated so as to fit the wooden wheel is (Coefficient of linear expansion of iron $= 1.2 \times 10^{-5}\,^\circ\text{C}^{-1}$) (A) 200 (B) 227 (C) 254 (D) 300

   ›Reveal solutionSolution

   The iron ring must be heated so its inner diameter expands from 5 m to 5.012 m. Using linear expansion, the required temperature rise is 200 °C, so the final temperature is 227 °C. The correct option is (B).

   ---

   The key idea is thermal expansion: when a solid is heated, its dimensions increase in proportion to the temperature change. Here, the iron ring (initially too small) must expand enough to slip over the wooden wheel. Since the ring is thin, we treat its diameter expansion using the linear expansion formula.

   Why this works:

   The coefficient of linear expansion $\alpha$ tells us the fractional change in length per degree. For a diameter $D$, the change $\Delta D = \alpha D \Delta T$. We know the initial diameter of the iron ring (5 m) and the required final diameter (5.012 m, the wheel’s diameter). The difference gives $\Delta D$, and we solve for $\Delta T$. Then add the initial temperature.

   ---

   Step-by-step:

   1. Identify given data

      • Initial diameter of iron ring: $D_0 = 5.000\,\text{m}$
      • Required final diameter (wheel’s diameter): $D = 5.012\,\text{m}$
      • Initial temperature: $T_0 = 27\,^\circ\text{C}$
      • Coefficient of linear expansion of iron: $\alpha = 1.2 \times 10^{-5}\,^\circ\text{C}^{-1}$

   2. Find the required expansion

      The ring must expand by:

   $\Delta D = D - D_0 = 5.012 - 5.000 = 0.012\,\text{m}$

   3. Apply the linear expansion formula For a linear dimension (here diameter):

   $\Delta D = \alpha D_0 \Delta T$

   Solve for $\Delta T$:

   $\Delta T = \frac{\Delta D}{\alpha D_0} = \frac{0.012}{(1.2 \times 10^{-5}) \times 5}$

   4. Calculate First compute denominator:

   $\alpha D_0 = 1.2 \times 10^{-5} \times 5 = 6.0 \times 10^{-5}$

   Then:

   $\Delta T = \frac{0.012}{6.0 \times 10^{-5}} = \frac{1.2 \times 10^{-2}}{6.0 \times 10^{-5}} = 2.0 \times 10^{2} = 200\,^\circ\text{C}$

   5. Find the final temperature

   $T = T_0 + \Delta T = 27 + 200 = 227\,^\circ\text{C}$

   Tip

   Notice that the initial diameter of the ring is smaller than the wheel’s diameter. If you mistakenly used the wheel’s diameter as the initial length, you’d get a different (wrong) answer. Always use the object’s own initial length.

   Watch out

   A common mistake is to forget that the ring must be heated above room temperature, not just to the temperature that gives the exact expansion. The question asks for the temperature to which it must be heated, so add the initial 27 °C.

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

10. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.A rod is found to be 0.05 cm longer at $40\ ^\circ\text{C}$ than it is at $10\ ^\circ\text{C}$. The length of the rod at $0\ ^\circ\text{C}$ is (coefficient of linear expansion of the material of the rod $= 1.5 \times 10^{-5}\ ^\circ\text{C}^{-1}$) (A) $101.1\ \text{cm}$ (B) $120.2\ \text{cm}$ (C) $105.1\ \text{cm}$ (D) $111.1\ \text{cm}$

    ›Reveal solutionSolution

    Using the linear-expansion relation between lengths at two temperatures, relative to the reference length at $0^\circ$C, gives $L_0 \approx 111.1$ cm.

    Concept and Intuition

    Linear thermal expansion means length grows linearly with temperature relative to the length at a fixed reference temperature (conventionally $0^\circ$C): $L(T) = L_0(1+\alpha T)$. The difference in lengths at two temperatures is therefore proportional to $L_0$, $\alpha$, and the temperature difference — which lets us solve for $L_0$ even though we're never told it directly.

    Step-by-Step Solution

    1. Let $L_0$ = length of rod at $0^\circ$C. Then $L(T) = L_0(1+\alpha T)$.
    2. $L(40) - L(10) = L_0\alpha(40) - L_0\alpha(10) = L_0\alpha(30)$.
    3. Given this difference is $0.05$ cm: $L_0\alpha(30) = 0.05$.
    4. Substitute $\alpha = 1.5\times10^{-5}\,^\circ\text{C}^{-1}$: $L_0(1.5\times10^{-5})(30) = 0.05 \Rightarrow L_0(4.5\times10^{-4}) = 0.05$.
    5. $L_0 = \dfrac{0.05}{4.5\times10^{-4}} = 111.11$ cm.

    Common Mistakes

    • Using the length at $10^\circ$C or $40^\circ$C (rather than $0^\circ$C) as the reference $L_0$ in the expansion formula — the formula $L=L_0(1+\alpha T)$ specifically references $0^\circ$C.

    ✓Final answer

    The correct option is (D) — $111.1\ \text{cm}$.

    ANSWER: D

11. TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
    Q.At constant pressure, the temperature and volume of a gas are increased by $2\,^\circ \mathrm{C}$ and $0.5\%$ respectively. Find the initial temperature. (A) $400\,\mathrm{K}$ (B) $200\,\mathrm{K}$ (C) $100\,\mathrm{K}$ (D) $150\,\mathrm{K}$

    ›Reveal solutionSolution

    Using Charles’s law (volume proportional to absolute temperature at constant pressure), the given fractional change in volume equals the fractional change in absolute temperature. Solving yields an initial temperature of 400 K, so option (A) is correct.

    The key idea is that at constant pressure, the volume of an ideal gas is directly proportional to its absolute temperature (Kelvin). This is Charles’s law:

    $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

    When we are told the volume increases by 0.5%, that is a fractional change. The temperature increase is given in Celsius degrees, but we must work in Kelvin because the proportionality holds only for absolute temperature. The trick is to convert the Celsius change into a fractional change in Kelvin and equate it to the fractional change in volume.

    1. Set up the relation

       Let the initial absolute temperature be $T$ (in Kelvin) and initial volume be $V$.

       After the change, the new temperature is $T + 2$ (since a change of $2^\circ\mathrm{C}$ is exactly a change of 2 K).

       The new volume is $V + 0.5\%$ of $V = V \left(1 + \frac{0.5}{100}\right) = 1.005\,V$.

    2. Apply Charles’s law

    $\frac{V}{T} = \frac{1.005\,V}{T + 2}$

    Cancel $V$ (nonzero):

    $\frac{1}{T} = \frac{1.005}{T + 2}$

    3. Solve for $T$ Cross-multiply:

    $T + 2 = 1.005\,T$

    $2 = 1.005\,T - T = 0.005\,T$

    $T = \frac{2}{0.005} = 400\ \mathrm{K}$

    Watch out

    A common mistake is to treat the $2^\circ\mathrm{C}$ as a fractional change of the Celsius temperature. But Charles’s law requires absolute temperature. If you mistakenly used Celsius values, you would get a different (wrong) answer.

    Tip

    Notice that a 0.5% increase corresponds to a factor of 1.005. The 2 K increase is exactly 0.5% of 400 K. This kind of “percentage = fraction of the whole” reasoning gives a quick mental check.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
    Q.An object cools down from $90\,^\circ\mathrm{C}$ to $70\,^\circ\mathrm{C}$ in $10$ minutes, when room temperature is $30\,^\circ\mathrm{C}$. The time taken by the object to cool from $70\,^\circ\mathrm{C}$ to $50\,^\circ\mathrm{C}$ is (A) $10$ min (B) $20.3$ min (C) $16.7$ min (D) $14.5$ min

    ›Reveal solutionSolution

    Newton's Law of Cooling says the rate of cooling is proportional to the excess temperature over the surroundings. Using the standard mean-temperature form of the law, the time taken to cool from $70^\circ\text{C}$ to $50^\circ\text{C}$ works out to $16.7$ minutes, so option (C) is correct.

    ---

    Concept and intuition

    Newton's Law of Cooling states that the rate of loss of heat (and hence the rate of fall of temperature) is proportional to the difference between the body's temperature and the surrounding temperature. For problems that compare two finite time intervals, the standard practical form used is:

    $\frac{T_1 - T_2}{t} = k\left(\frac{T_1+T_2}{2} - T_r\right),$

    where $T_1$ and $T_2$ are the temperatures at the start and end of the interval, $t$ is the time taken, $T_r$ is the room temperature, and $k$ is a constant for the given body. This treats the average of the two temperatures as representative of the excess temperature driving the cooling over that interval. Since the body's excess temperature keeps falling as it cools, covering the same $20^\circ\text{C}$ drop takes longer the second time, because the average excess temperature during the second interval is smaller.

    ---

    Step-by-step solution

    1. Use the first interval to find $k$. $T_1 = 90^\circ\text{C}$, $T_2 = 70^\circ\text{C}$, $t = 10$ min, $T_r = 30^\circ\text{C}$.

    $\frac{90-70}{10} = k\left(\frac{90+70}{2} - 30\right)$

    $2 = k(80 - 30) = 50k \quad\Rightarrow\quad k = 0.04 \text{ per min}$

    2. Apply the same law to the second interval. Now $T_1 = 70^\circ\text{C}$, $T_2 = 50^\circ\text{C}$, and we want the time $t'$.

    $\frac{70-50}{t'} = k\left(\frac{70+50}{2} - 30\right) = 0.04 \times (60-30) = 0.04 \times 30 = 1.2$

    $\frac{20}{t'} = 1.2 \quad\Rightarrow\quad t' = \frac{20}{1.2} = 16.67 \text{ min} \approx 16.7 \text{ min}$

    Watch out

    A common mistake is to assume equal temperature drops take equal time (which would give option A, $10$ min). Newton's Law of Cooling is not linear in time — as the object approaches room temperature, its excess temperature (and hence its cooling rate) falls, so later intervals with the same temperature drop always take longer.

    Tip

    This mean-temperature form of Newton's Law of Cooling is the standard way to compare two finite cooling intervals without solving the full exponential differential equation — accurate enough for the modest temperature ranges typical of these problems.

    ✓Final answer

    The time taken is $16.7$ minutes, so the correct option is (C).