Q.Two moving coil meters, $M_1$ and $M_2$ have the following particulars:
$R_1 = 10\ \Omega$, $N_1 = 30$, $A_1 = 3.6 \times 10^{-3}\ \text{m}^2$, $B_1 = 0.25\ \text{T}$
$R_2 = 14\ \Omega$, $N_2 = 42$, $A_2 = 1.8 \times 10^{-3}\ \text{m}^2$, $B_2 = 0.50\ \text{T}$
(The spring constants are identical for the two meters). Determine the ratio of

Showing the 12 most recent of 53 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Two spherical black bodies A and B of equal radii are at absolute temperatures $2T$ and $3T$ respectively. If the absolute temperature of the surroundings is $T$, then the ratio of the radiant powers emitted by the bodies A and B is (A) $3 : 16$ (B) $3 : 8$ (C) $1 : 2$ (D) $2 : 3$

   ›Reveal solutionSolution

   The net radiant power of a body is the difference between its emitted power and the absorbed power from the surroundings. Using the Stefan-Boltzmann law, the ratio for A and B is $3:16$, so option (A) is correct.

   The key here is that "radiant power" in the presence of surroundings means net power — what the body actually radiates away after accounting for the radiation it absorbs from the environment. A black body both emits and absorbs perfectly, so its net power is $\sigma A (T^4 - T_0^4)$, where $T_0$ is the surrounding temperature. Many students mistakenly use just $T^4$, forgetting the absorption term, and get a wrong ratio.

   Let’s work through it step by step.

   1. Write the formula for net radiant power. For a black body of surface area $A$ at absolute temperature $T$, surrounded by a medium at temperature $T_0$, the net power radiated is

   $P = \sigma A (T^4 - T_0^4)$

   where $\sigma$ is the Stefan-Boltzmann constant. This is because the body emits $\sigma A T^4$ and absorbs $\sigma A T_0^4$ from the surroundings.

   2. Apply to body A. Body A has temperature $T_A = 2T$ and surroundings at $T_0 = T$. Its net power is

   $P_A = \sigma A \left[(2T)^4 - T^4\right] = \sigma A (16T^4 - T^4) = \sigma A (15T^4).$

   3. Apply to body B. Body B has temperature $T_B = 3T$ and the same surroundings $T_0 = T$. Its net power is

   $P_B = \sigma A \left[(3T)^4 - T^4\right] = \sigma A (81T^4 - T^4) = \sigma A (80T^4).$

   4. Find the ratio. The ratio $P_A : P_B$ is $\frac{P_A}{P_B} = \frac{15T^4}{80T^4} = \frac{15}{80} = \frac{3}{16}.$ …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.The magnetic field at a point A on the axis of a short bar magnet is 1500% more than the magnetic field at a point B on the normal bisector of the magnet. If the distance of point A from the centre of the magnet is 18 cm, then the distance of point B from the centre of the magnet is (A) 72 cm (B) 36 cm (C) 48 cm (D) 54 cm

   ›Reveal solutionSolution

   The axial field of a short bar magnet is twice the equatorial field at the same distance. The problem states the axial field is 1500% more (i.e., 16 times) the equatorial field, so the distances must adjust by the cube root of 16. The answer is 36 cm.

   The key concept here is the variation of the magnetic field of a short bar magnet along its axis and on its equatorial line. For a short magnet (length negligible compared to distance), the field at a point on the axis at distance $r$ is $B_{\text{axis}} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}$, and on the equatorial line at the same distance $r$ it is $B_{\text{eq}} = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3}$. Notice the axial field is exactly twice the equatorial field at the same distance.

   But here the distances are different. The problem says the axial field at point A is 1500% more than the equatorial field at point B. "1500% more" means the axial field is the equatorial field plus 1500% of it — that is, $B_A = B_B + 15 B_B = 16 B_B$. So $B_A = 16 B_B$.

   Now we write the expressions and solve for the unknown distance.

   1. Let $d_A = 18\ \text{cm}$ be the distance of point A from the centre along the axis. Let $d_B$ be the distance of point B from the centre on the normal bisector (equatorial line). For a short bar magnet:

   $B_A = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d_A^3}, \quad B_B = \frac{\mu_0}{4\pi} \cdot \frac{M}{d_B^3}.$

   2. The given condition is $B_A = 16 B_B$. Substitute:

   $\frac{\mu_0}{4\pi} \cdot \frac{2M}{d_A^3} = 16 \cdot \frac{\mu_0}{4\pi} \cdot \frac{M}{d_B^3}.$

   3. Cancel the common factor $\frac{\mu_0}{4\pi} M$ (non-zero):

   $\frac{2}{d_A^3} = \frac{16}{d_B^3}.$

   4. Rearrange: $\frac{d_B^3}{d_A^3} = \frac{16}{2} = 8.$ …

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3. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.When two resistances P and Q are connected in the left and right gaps of a meter bridge respectively, then the balancing point is obtained at a distance greater than 25 cm. If the resistance P is increased by $5~\Omega$, the balancing point shifts by 10 cm. After that if the resistance Q is halved, the balancing point further shifts by 15 cm, then the initial value of P is (A) $5~\Omega$ (B) $15~\Omega$ (C) $10~\Omega$ (D) $20~\Omega$

   ›Reveal solutionSolution

   The key idea is to use the meter bridge balance condition $P/Q = l/(100-l)$ and track how the balancing length changes with each modification. Solving the system of equations yields the initial value $P = 10\,\Omega$.

   The meter bridge works on the principle of a Wheatstone bridge. When the bridge is balanced, the ratio of the two resistances in the left and right gaps equals the ratio of the lengths of the wire on either side of the balance point. If the balance point is at $l$ cm from the left end, then:

   $\frac{P}{Q} = \frac{l}{100 - l}$

   We are told the initial balance point is greater than 25 cm. Then we apply two successive changes, each shifting the balance point by a known amount. The trick is to carefully track the direction of each shift — whether the balance point moves left or right — because that tells us the sign of the change in length.

   ---

   1. Set up the initial condition. Let the initial balance point be at $l$ cm (with $l > 25$). Then:

   $\frac{P}{Q} = \frac{l}{100 - l} \quad \text{(1)}$

   2. First change: increase $P$ by $5\,\Omega$. New left gap resistance = $P + 5$. The balance point shifts by 10 cm. Since increasing $P$ makes the left side stronger, the balance point must move to the right (longer left arm to compensate). So the new balance length is $l + 10$.

   $\frac{P + 5}{Q} = \frac{l + 10}{90 - l} \quad \text{(2)}$

   3. Second change: then halve $Q$. Now left gap = $P + 5$, right gap = $Q/2$. The balance point shifts further by 15 cm. But careful: halving $Q$ makes the right side weaker, so the balance point again moves to the right. So from the previous position, it moves another 15 cm to the right. The new balance length = $(l + 10) + 15 = l + 25$.

   $\frac{P + 5}{Q/2} = \frac{l + 25}{75 - l} \quad \text{(3)}$

   4. Simplify equation (3). Multiply numerator and denominator:

   $\frac{2(P + 5)}{Q} = \frac{l + 25}{75 - l}$

   So:

   $\frac{P + 5}{Q} = \frac{l + 25}{2(75 - l)} \quad \text{(3')}$

   5. Equate expressions for $(P+5)/Q$ from (2) and (3'). From (2): $\frac{P+5}{Q} = \frac{l+10}{90-l}$ From (3'): $\frac{P+5}{Q} = \frac{l+25}{2(75-l)}$ Set them equal:

   $\frac{l+10}{90-l} = \frac{l+25}{2(75-l)}$

   Cross-multiply:

   $2(l+10)(75-l) = (l+25)(90-l)$

   6. Solve for $l$. Expand left: $2[(l)(75-l) + 10(75-l)] = 2[75l - l^2 + 750 - 10l] = 2[-l^2 + 65l + 750] = -2l^2 + 130l + 1500$ Expand right: $(l)(90-l) + 25(90-l) = 90l - l^2 + 2250 - 25l = -l^2 + 65l + 2250$ So equation: $-2l^2 + 130l + 1500 = -l^2 + 65l + 2250$ …

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.When a load is attached at the midpoint of a steel bar A of length 80 cm, breadth 2.5 cm and thickness 2 mm, it sags by 1.2 mm. If the same load is attached at the midpoint of another steel bar B of length 120 cm, breadth 3 cm and thickness 3 mm, the bar B sags by (In both the cases, the bars are supported at the two ends) (A) 0.18 cm (B) 0.10 cm (C) 0.12 cm (D) 0.15 cm

   ›Reveal solutionSolution

   For a centrally loaded beam supported at both ends, $\delta \propto \dfrac{L^3}{b\,d^3}$ for the same load and material; this gives $\delta_B = 1.0\text{ mm} = 0.10$ cm — option (B).

   Depression formula. With central load $W$, Young's modulus $Y$, length $L$, breadth $b$, thickness $d$:

   $\delta = \frac{W L^3}{4 Y b\, d^3}\ \Rightarrow\ \delta \propto \frac{L^3}{b\, d^3}\quad(\text{same } W,\ Y).$

   Ratio.

   $\frac{\delta_B}{\delta_A} = \left(\frac{L_B}{L_A}\right)^3 \cdot \frac{b_A}{b_B} \cdot \left(\frac{d_A}{d_B}\right)^3 .$

   Data: $L_A=80,\ L_B=120$ cm; $b_A=2.5,\ b_B=3$ cm; $d_A=2,\ d_B=3$ mm. …

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5. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.When a $15~\Omega$ resistor is connected in parallel to a galvanometer, its deflection becomes one-fourth. Then the resistance of the galvanometer is (A) $45~\Omega$ (B) $60~\Omega$ (C) $15~\Omega$ (D) $30~\Omega$

   ›Reveal solutionSolution

   A shunt resistor in parallel diverts most of the current away from the galvanometer. When the deflection drops to one-fourth, the current through the galvanometer is one-fourth of the total current, so the shunt carries three-fourths. Using the parallel current-divider rule gives the galvanometer resistance as $45~\Omega$.

   The key idea here is that a galvanometer’s deflection is proportional to the current passing through it. When you connect a resistor in parallel (called a shunt), some of the total current bypasses the galvanometer, reducing its deflection. The problem tells you the deflection becomes one-fourth of its original value — that means the current through the galvanometer is now one-fourth of the total current entering the parallel combination.

   Let’s work through it step by step.

   1. Understand what “deflection becomes one-fourth” means.

      The original deflection corresponds to the full current $I_g$ (when no shunt is connected). After adding the shunt, the new deflection is one-fourth, so the current through the galvanometer is now $I_g' = \frac{I}{4}$, where $I$ is the total current entering the parallel circuit. The remaining current $\frac{3I}{4}$ flows through the shunt resistor $R_s = 15~\Omega$.

   2. Apply the parallel current-divider rule.

      In a parallel circuit, the current divides inversely with resistance. For two resistors in parallel, the current through one branch is:

   $I_{\text{branch}} = I_{\text{total}} \times \frac{\text{other resistance}}{\text{sum of resistances}}$

   Here, the galvanometer (resistance $G$) and the shunt ($15~\Omega$) are in parallel. The current through the galvanometer is:

   $I_g' = I \times \frac{15}{G + 15}$

   But we already know $I_g' = \frac{I}{4}$. So:

   $\frac{I}{4} = I \times \frac{15}{G + 15}$

   3. Cancel $I$ (assuming $I \neq 0$) and solve for $G$.

   $\frac{1}{4} = \frac{15}{G + 15}$

   Cross-multiply:

   $G + 15 = 60$

   $G = 45~\Omega$ …

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6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Two identical circular coils each of radius $R$ and number of turns $N$ are arranged coaxially with a separation of $R$ between their centers. If the current through each coil is $I$, then the maximum resultant magnetic field induced at the midpoint of the line joining the centers of the two coils is (A) $\dfrac{2\mu_0 NI}{R}$ (B) $\dfrac{4\mu_0 NI}{\sqrt{5}R}$ (C) $\dfrac{8\mu_0 NI}{5\sqrt{5}R}$ (D) $\dfrac{\mu_0 NI}{\pi R}$

   ›Reveal solutionSolution

   The maximum resultant field at the midpoint occurs when the currents are in the same direction, and the field from each coil is found using the Biot–Savart law for a circular loop. Adding the two equal axial fields gives $\frac{8\mu_0 N I}{5\sqrt{5}\,R}$, which corresponds to option (C).

   ---

   The key idea is that two coaxial coils produce magnetic fields along their common axis. At the midpoint, the fields from each coil are parallel (if currents are in the same sense), so they simply add. The field on the axis of a single circular coil is a standard result, and we apply it at a distance $R/2$ from each coil’s center (since the separation is $R$ and the midpoint is halfway).

   Why this approach works:

   The Biot–Savart law for a circular loop gives the axial field at a point a distance $x$ from the center. The formula is symmetric and depends only on $x$, $R$, and the current. Since the two coils are identical and symmetrically placed, the fields at the midpoint are equal in magnitude. The maximum resultant occurs when they reinforce each other — i.e., currents in the same direction.

   ---

   1. Recall the axial field of a single circular coil For a coil of radius $R$, $N$ turns, and current $I$, the magnetic field at a point on its axis at distance $x$ from the center is

   $B_{\text{single}} = \frac{\mu_0 N I R^2}{2\,(R^2 + x^2)^{3/2}}.$

   This comes from integrating the Biot–Savart law; the perpendicular components cancel, leaving only the axial component.

   2. Find the distance from each coil to the midpoint The coils are separated by $R$ (center-to-center). The midpoint is therefore at a distance

   $x = \frac{R}{2}$

   from each coil’s center.

   3. Compute the field from one coil at the midpoint Substitute $x = R/2$ into the formula:

   $B_1 = \frac{\mu_0 N I R^2}{2\left(R^2 + \left(\frac{R}{2}\right)^2\right)^{3/2}} = \frac{\mu_0 N I R^2}{2\left(R^2 + \frac{R^2}{4}\right)^{3/2}} = \frac{\mu_0 N I R^2}{2\left(\frac{5R^2}{4}\right)^{3/2}}.$

   Simplify the denominator:

   $\left(\frac{5R^2}{4}\right)^{3/2} = \left(\frac{5}{4}\right)^{3/2} R^3 = \frac{5^{3/2}}{4^{3/2}} R^3 = \frac{5\sqrt{5}}{8} R^3,$

   since $4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$.

   Thus

   $$$$

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7. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Two metal wires A and B of lengths 80 cm and 120 cm respectively having equal cross-sectional area are connected to form a long wire. When the combination is subjected to a tension, the elongations of the wires A and B are 0.8 mm and 1.5 mm respectively. If the Young’s modulus of the material of the wire A is $1.6 \times 10^{11}$ Nm$^{-2}$, then the Young’s modulus of the material of the wire B (in $10^{11}$ Nm$^{-2}$) is (A) 1.28 (B) 1.44 (C) 1.76 (D) 1.92

   ›Reveal solutionSolution

   Wires A and B are joined end-to-end, so they carry the same tension and (equal area) the same stress; $Y_B = Y_A\cdot\dfrac{\Delta L_A}{L_A}\cdot\dfrac{L_B}{\Delta L_B} = 1.28\times10^{11}$ N m$^{-2}$.

   Because A and B are connected in series and pulled by one tension $F$ with equal cross-sectional area $A$, the stress $\sigma = F/A$ is identical in both wires.

   $Y = \frac{\sigma}{\text{strain}} = \frac{\sigma\,L}{\Delta L}\quad\Rightarrow\quad \sigma = \frac{Y_A\,\Delta L_A}{L_A} = \frac{Y_B\,\Delta L_B}{L_B}$

   Equating the common stress and solving for $Y_B$:

   $Y_B = Y_A\cdot\frac{\Delta L_A}{L_A}\cdot\frac{L_B}{\Delta L_B}$ …

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8. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.If the magnification of a compound microscope when the final image forms at near point is 14% greater than its magnification when the final image forms at infinity, then the ratio of the focal length of the eyepiece and least distance of distinct vision is (A) $3 : 25$ (B) $3 : 50$ (C) $7 : 50$ (D) $7 : 25$

   ›Reveal solutionSolution

   The key idea is to use the standard magnification formulas for a compound microscope for the two cases (image at near point vs. at infinity) and set up the given percentage relation. The required ratio is $3 : 50$.

   Concept and Intuition

   In a compound microscope, the total magnification is the product of the magnification produced by the objective lens and the magnification produced by the eyepiece. The objective's magnification is the same in both cases because the object is placed such that its image is formed at the same position relative to the objective (just inside the focal point of the eyepiece). The difference arises only in how the eyepiece is used.

   When the final image is at infinity, the eyepiece is adjusted so that the intermediate image lies exactly at its focal point. The angular magnification of the eyepiece in this case is $D/f_e$, where $D$ is the least distance of distinct vision (usually 25 cm).

   When the final image is at the near point (the closest distance for clear vision, $D$), the eyepiece is adjusted so that the intermediate image lies just inside its focal point, producing a virtual image at $D$. The angular magnification of the eyepiece in this case is $1 + D/f_e$.

   The problem tells us that the near-point magnification is 14% greater than the infinity magnification. This gives a direct equation relating $f_e$ and $D$.

   Step-by-Step Solution

   1. Define the variables.

      Let $f_e$ be the focal length of the eyepiece and $D$ be the least distance of distinct vision. The objective magnification $m_o$ is the same in both cases, so we only need to compare the eyepiece contributions.

   2. Write the eyepiece magnification for the two cases.

      • When the final image is at infinity: $M_{e,\infty} = \frac{D}{f_e}$
      • When the final image is at the near point: $M_{e,np} = 1 + \frac{D}{f_e}$

   3. Express the total magnifications.

      Let $m_o$ be the objective magnification. Then:

      • Total magnification at infinity: $M_\infty = m_o \cdot \frac{D}{f_e}$
      • Total magnification at near point: $M_{np} = m_o \cdot \left(1 + \frac{D}{f_e}\right)$

   4. Use the given percentage relation.

      The problem states that $M_{np}$ is 14% greater than $M_\infty$. That means:

      $M_{np} = M_\infty + 0.14\, M_\infty = 1.14\, M_\infty$ …

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9. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.A coil of area $3 \times 10^{-4} \ \mathrm{m}^2$ and 80 turns of a moving coil galvanometer is suspended in a uniform radial magnetic field of 20 mT. If the resistance of the galvanometer is 50 $\Omega$ and its voltage sensitivity is 200 $\mathrm{rad \ V^{-1}}$, then the torsional constant of the spring of the galvanometer (in $10^{-8} \ \mathrm{Nm \ rad^{-1}}$) is (A) 3.6 (B) 4.8 (C) 2.4 (D) 1.2

   ›Reveal solutionSolution

   Voltage sensitivity $=\dfrac{NBA}{kR}$, so $k=\dfrac{NBA}{V_s\,R}=4.8\times10^{-8}\ \mathrm{N\,m\,rad^{-1}}$.

   For a moving-coil galvanometer the current sensitivity is $\dfrac{\theta}{I}=\dfrac{NBA}{k}$, and the voltage sensitivity is

   $V_s=\frac{\theta}{V}=\frac{\theta}{IR}=\frac{NBA}{kR}.$

   Rearranging for the torsional constant:

   $k=\frac{NBA}{V_s\,R}.$

   Substituting $N=80$, $B=20\ \mathrm{mT}=0.02\ \mathrm{T}$, $A=3\times10^{-4}\ \mathrm{m^2}$, $V_s=200\ \mathrm{rad\,V^{-1}}$, $R=50\ \Omega$:

   $NBA = 80\times0.02\times3\times10^{-4}=4.8\times10^{-4},$ …

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10. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
    Q.The drift speed of electrons in a material is found to be $0.3\, \mathrm{ms}^{-1}$ when an electric field of $2\, \mathrm{Vm}^{-1}$ is applied across it. The electron mobility (in $\mathrm{m}^2\, \mathrm{V}^{-1}\, \mathrm{s}^{-1}$) in the material is (A) $60 \times 10^{-2}$ (B) $15 \times 10^{-2}$ (C) $1350 \times 10^{6}$ (D) $5400 \times 10^{6}$

    ›Reveal solutionSolution

    Electron mobility is the ratio of drift speed to electric field: $\mu = v_d / E$.

    With $v_d = 0.3\ \mathrm{m/s}$ and $E = 2\ \mathrm{V/m}$, we get $\mu = 0.15\ \mathrm{m^2 V^{-1} s^{-1}} = 15 \times 10^{-2}\ \mathrm{m^2 V^{-1} s^{-1}}$, so the correct option is (B).

    Concept & Intuition

    Mobility measures how quickly charge carriers (here electrons) can move through a material in response to an applied electric field. It’s defined as the drift speed per unit electric field:

    $\mu = \frac{v_d}{E}$

    The units naturally come out as $\mathrm{(m/s) / (V/m)} = \mathrm{m^2 V^{-1} s^{-1}}$. The problem gives both $v_d$ and $E$ directly, so it’s a straightforward division — no hidden physics, just careful handling of the numbers and units.

    Step-by-step solution

    1. Write down the definition Electron mobility $\mu$ is given by

    $\mu = \frac{v_d}{E}$

    where $v_d$ is the drift speed and $E$ is the electric field strength.

    2. Insert the given values $v_d = 0.3\ \mathrm{m/s}$, $E = 2\ \mathrm{V/m}$.

    $\mu = \frac{0.3}{2} = 0.15\ \mathrm{m^2 V^{-1} s^{-1}}$

    3. Express in the form used by the options The options are given in terms of $10^{-2}$ or $10^{6}$. $0.15 = 15 \times 10^{-2}$, so

    $\mu = 15 \times 10^{-2}\ \mathrm{m^2 V^{-1} s^{-1}}$

    4. Match with the choices …

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11. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.Two wires A and B made of same material and areas of cross-section in the ratio 1:2 are stretched by same force. If the masses of the wires A and B are in the ratio 2:3, then the ratio of the elongations of the wires A and B is (A) $1:2$ (B) $8:3$ (C) $1:3$ (D) $4:3$

    ›Reveal solutionSolution

    The key is to relate elongation to length, area, and force via Young’s modulus, then use the given mass and area ratios to find the length ratio, and finally compute the elongation ratio. The result is 8:3.

    Concept & Intuition

    When the same force stretches two wires of the same material, the elongation depends on the wire’s length and cross-sectional area. But we aren’t given lengths directly — we’re given masses and areas. Since mass = density × volume = density × (area × length), and density is the same for both wires, the mass ratio tells us the (area × length) ratio. From that, we can extract the length ratio, and then use the elongation formula.

    Step-by-step solution

    1. Write the elongation formula For a wire of length $L$, area $A$, Young’s modulus $Y$, under force $F$:

    $\Delta L = \frac{F L}{A Y}$

    Since $F$ and $Y$ are the same for both wires,

    $\frac{\Delta L_A}{\Delta L_B} = \frac{L_A / A_A}{L_B / A_B} = \frac{L_A}{L_B} \cdot \frac{A_B}{A_A}$

    2. Use the given area ratio $A_A : A_B = 1 : 2$ means $\frac{A_B}{A_A} = 2$. So

    $\frac{\Delta L_A}{\Delta L_B} = \frac{L_A}{L_B} \times 2$

    3. Find the length ratio from the mass ratio Mass $m = \text{density} \times \text{volume} = \rho \times (A L)$. Given $m_A : m_B = 2 : 3$ and $\rho$ same for both: $$$$

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12. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.An alpha particle and a proton are accelerated from rest in a uniform electric field. The ratio of the times taken by proton and alpha particle to attain equal displacements is (A) $\sqrt{2}:1$ (B) $1:2$ (C) $1:\sqrt{2}$ (D) $2:1$

    ›Reveal solutionSolution

    For particles starting from rest in a uniform electric field, the time to cover a given displacement scales as the square root of mass divided by charge. The proton-to-alpha time ratio is $1:\sqrt{2}$, so the correct option is (C).

    ---

    Concept and Intuition

    When a charged particle is placed in a uniform electric field, it experiences a constant force $F = qE$. Since the field is uniform, the force is constant, so the acceleration is constant. That means the motion is uniformly accelerated from rest. For such motion, the displacement after time $t$ is $s = \frac{1}{2} a t^2$.

    The key insight: the acceleration depends on both the charge $q$ (which determines the force) and the mass $m$ (which resists acceleration). So different particles will have different accelerations, and thus different times to cover the same distance. We need to compare the proton and the alpha particle.

    Tip

    An alpha particle is a helium nucleus: it has charge $+2e$ and mass $4m_p$ (where $m_p$ is the proton mass). The proton has charge $+e$ and mass $m_p$.

    ---

    Step-by-step reasoning

    1. Write the equation of motion For a particle of mass $m$ and charge $q$ in a uniform electric field $E$, the force is $F = qE$. By Newton’s second law, acceleration is:

    $a = \frac{F}{m} = \frac{qE}{m}.$

    Since the particle starts from rest, the displacement after time $t$ is:

    $s = \frac{1}{2} a t^2 = \frac{1}{2} \left( \frac{qE}{m} \right) t^2.$

    2. Solve for time in terms of displacement Rearranging for $t$:

    $t = \sqrt{ \frac{2s m}{qE} }.$

    For a fixed displacement $s$ and fixed field $E$, the time depends on the ratio $\sqrt{m/q}$:

    $t \propto \sqrt{\frac{m}{q}}.$

    3. Apply to the proton For the proton: $m_p$, $q_p = e$. So: …

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