Q.A polythene piece rubbed with wool is found to have a negative charge of $3 \times 10^{-7}\,\text{C}$.

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.A potentiometer wire of length 10 m and resistance 9 $\Omega$ is connected to a cell of emf E and internal resistance 1 $\Omega$. If V is the maximum potential difference that can be measured using the potentiometer, then the condition for the potentiometer to work is (A) $E = V$ (B) $9E \geq 10V$ (C) $E < V$ (D) $9V = 10E$

   ›Reveal solutionSolution

   The key idea is that a potentiometer works only when the voltage across its entire wire (the maximum measurable voltage) is slightly greater than the voltage to be measured. Here, the maximum potential difference across the wire is $\frac{9}{10}E$, so the condition for it to measure a voltage $V$ is $9E \ge 10V$, which corresponds to option (B).

   ---

   Why this approach works

   A potentiometer is a device for measuring an unknown potential difference without drawing current from the source being measured. It works by balancing the unknown voltage against a known potential difference taken from a uniform wire. For the potentiometer to function, the unknown voltage $V$ must be less than or equal to the maximum potential difference that can be obtained across the entire length of the wire. If $V$ exceeds that maximum, the balance point cannot be reached — the slider would have to go beyond the wire.

   So the problem reduces to: What is the maximum voltage across the potentiometer wire? That depends on the driving cell's emf $E$, its internal resistance, and the wire's resistance.

   ---

   Step-by-step reasoning

   1. Find the current in the main circuit The potentiometer wire (resistance $9\,\Omega$) is connected in series with the driving cell (emf $E$, internal resistance $1\,\Omega$). The total resistance in the circuit is

   $R_{\text{total}} = 9 + 1 = 10\,\Omega.$

   Hence the current supplied by the cell is

   $I = \frac{E}{10}.$

   2. Find the potential difference across the whole wire The wire has resistance $9\,\Omega$, so the voltage drop from one end to the other is

   $V_{\text{wire}} = I \times 9 = \frac{E}{10} \times 9 = \frac{9E}{10}.$

   This is the maximum voltage that can be tapped from the wire (by using its full length).

   3. Condition for the potentiometer to work …

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2. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.A potentiometer wire of length 4 m and resistance $20~\Omega$ is connected in series to a resistor of resistance $28~\Omega$. If the emf of the cell is $5~\mathrm{V}$ and its internal resistance is $2~\Omega$, then the maximum potential difference that the potentiometer can measure is (A) $4~\mathrm{V}$ (B) $3~\mathrm{V}$ (C) $2~\mathrm{V}$ (D) $5~\mathrm{V}$

   ›Reveal solutionSolution

   The maximum potential difference a potentiometer can measure is the potential drop across its entire wire. By calculating the total current in the primary circuit and then the potential drop across the potentiometer wire, we find this maximum measurable potential difference to be $\boxed{2~\mathrm{V}}$.

   A potentiometer is a device used to measure an unknown electromotive force (EMF) or potential difference accurately. Its core principle relies on the concept of a uniform potential gradient along a wire.

   The primary circuit of a potentiometer consists of a cell (driver cell), a rheostat (or a fixed series resistor), and the potentiometer wire itself, all connected in series. A constant current flows through this primary circuit, establishing a potential drop across the potentiometer wire. This potential drop is distributed uniformly along the length of the wire, meaning there's a constant potential difference per unit length, known as the potential gradient.

   The maximum potential difference that a potentiometer can measure is limited by the total potential drop across its entire wire. If you try to measure a potential difference greater than this, you won't be able to find a null point on the wire. Therefore, to find the maximum measurable potential difference, we need to calculate the potential difference across the potentiometer wire when the primary circuit is operational.

   Here's how to calculate it:

   1. Calculate the total resistance of the primary circuit. The primary circuit consists of the potentiometer wire, the external series resistor, and the internal resistance of the driving cell, all connected in series. Let $R_p$ be the resistance of the potentiometer wire, $R_{ext}$ be the external series resistor, and $r$ be the internal resistance of the cell. The total resistance $R_{total}$ is:

   $R_{total} = R_p + R_{ext} + r$

   Given $R_p = 20~\Omega$, $R_{ext} = 28~\Omega$, and $r = 2~\Omega$.
   

   $R_{total} = 20~\Omega + 28~\Omega + 2~\Omega = 50~\Omega$

   2. Calculate the current flowing through the primary circuit. The current $I$ in the primary circuit is determined by the EMF of the driving cell ($\mathcal{E}$) and the total resistance of the circuit. Using Ohm's Law:

   $I = \frac{\mathcal{E}}{R_{total}}$

   Given $\mathcal{E} = 5~\mathrm{V}$.
   

   $I = \frac{5~\mathrm{V}}{50~\Omega} = 0.1~\mathrm{A}$

   3. Calculate the potential difference across the potentiometer wire. …

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3. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.When monochromatic photons incident on a photosensitive material of cut-off wavelength 496 nm, photoelectrons are emitted with a maximum velocity of $8 \times 10^5$ ms$^{-1}$. Then the energy of the incident photons is nearly (Mass of the electron $= 9 \times 10^{-31}$ kg and charge of the electron $= 1.6 \times 10^{-19}$ C) (A) 6.1 eV (B) 3.7 eV (C) 4.3 eV (D) 5.2 eV

   ›Reveal solutionSolution

   Use the photoelectric equation: the incident photon energy equals the work function plus the maximum kinetic energy of the emitted electron. The work function is found from the cut-off wavelength. The result is about 4.3 eV, option (C).

   The photoelectric effect is a clean example of energy conservation at the quantum level. A single photon gives all its energy to one electron. Part of that energy is used to overcome the binding force that holds the electron in the metal — that's the work function. The rest appears as the electron's kinetic energy. The cut-off wavelength tells you the minimum photon energy needed to just barely eject an electron (with zero kinetic energy), so it directly gives the work function.

   Here, you're given the cut-off wavelength (496 nm) and the maximum speed of the emitted electrons ($8 \times 10^5$ m/s). The mass of the electron is also provided. The plan is straightforward: find the work function from the cut-off wavelength, find the maximum kinetic energy from the speed, and add them to get the incident photon energy. Then convert everything to electronvolts (eV) to match the options.

   1. Find the work function from the cut-off wavelength. The cut-off wavelength $\lambda_0 = 496$ nm is the longest wavelength that can still cause emission. At this wavelength, the photon energy exactly equals the work function $\phi$ (since the electron gets zero kinetic energy).

   $\phi = \frac{hc}{\lambda_0}$

   Use $hc = 1240$ eV·nm (a very handy constant for such problems).

   $\phi = \frac{1240 \text{ eV·nm}}{496 \text{ nm}} = 2.5 \text{ eV}$

   2. Find the maximum kinetic energy of the photoelectrons. The maximum speed $v = 8 \times 10^5$ m/s. The kinetic energy is

   $K_{\text{max}} = \frac{1}{2} m v^2$

   with $m = 9 \times 10^{-31}$ kg.

   $K_{\text{max}} = \frac{1}{2} \times (9 \times 10^{-31}) \times (8 \times 10^5)^2$

   First compute $v^2 = 64 \times 10^{10} = 6.4 \times 10^{11}$ m²/s².

   Then …

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.As shown in the figure, a dielectric of constant $K$ is placed between the plates of a parallel plate capacitor and is charged to a potential $V$ using a battery. If the dielectric is pulled out after disconnecting the battery from the capacitor, the final potential difference across the plates of the capacitor is (A) $\left(1 + \frac{1}{K}\right) 2V$ (B) $2KV$ (C) $\frac{2V}{\left(1 + \frac{1}{K}\right)}$ (D) $\frac{V}{2} \left(1 + \frac{1}{K}\right)$

   ›Reveal solutionSolution

   Charge is conserved once the battery is disconnected. With the dielectric slab shown occupying half the plate separation, the initial capacitance is a series combination of a dielectric-filled half-gap and a vacuum half-gap; once the dielectric is pulled out and the full gap is vacuum, working through $Q = C_iV = C_fV_f$ gives $V_f = \dfrac{2V}{1+\frac{1}{K}}$, option (C).

   Concept and intuition

   The key idea is charge conservation after the battery is disconnected: the charge $Q$ on the plates stays fixed no matter what happens to the dielectric afterward. Because the figure shows the dielectric slab filling only half the plate separation (not the whole gap), the capacitor must be modeled as two capacitors in series while it is being charged — one half-gap with the dielectric, one half-gap with vacuum. Once the dielectric is removed, the whole gap becomes vacuum and the capacitance drops to that of a plain parallel-plate capacitor of separation $d$; since $Q$ is unchanged, $V$ must rise to compensate for the smaller $C$.

   Step-by-step solution

   1. Initial capacitance (dielectric fills half the gap, thickness $d/2$) Two capacitors in series, each with the full plate area $A$ but half the separation:

   $C_1 = \frac{K \varepsilon_0 A}{d/2} = \frac{2K \varepsilon_0 A}{d}, \qquad C_2 = \frac{\varepsilon_0 A}{d/2} = \frac{2 \varepsilon_0 A}{d}$

   $\frac{1}{C_i} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{2\varepsilon_0 A}\left( \frac{1}{K} + 1 \right) \quad\Rightarrow\quad C_i = \frac{2\varepsilon_0 A}{d} \cdot \frac{K}{K+1}$

   2. Charge stored while connected to the battery

   $Q = C_i V = \frac{2\varepsilon_0 A}{d} \cdot \frac{K}{K+1} \cdot V$

   3. After disconnecting the battery and removing the dielectric The gap is now entirely vacuum, so

   $C_f = \frac{\varepsilon_0 A}{d}$

   Charge is unchanged (no battery connected), so $Q = C_f V_f$:

   $$$$

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.Photons of energy 4.5 eV are incident on a photosensitive material of work function 3 eV. The de Broglie wavelength associated with the photoelectrons emitted with maximum kinetic energy is nearly (A) 10 $\text{\AA}$ (B) 5 $\text{\AA}$ (C) 20 $\text{\AA}$ (D) 15 $\text{\AA}$

   ›Reveal solutionSolution

   Use the photoelectric equation to find the maximum kinetic energy of the emitted electrons, then convert that energy to momentum and apply de Broglie’s relation to get the wavelength. The result is about 8.7 Å, so the closest option is 10 Å — option (A).

   ---

   The problem combines two classic ideas: the photoelectric effect and the de Broglie wavelength of a particle.

   The key insight: the maximum kinetic energy of the photoelectrons comes from the photon energy minus the work function. Once we know that kinetic energy, we can find the electron’s momentum, and then the de Broglie wavelength is simply $\lambda = h / p$.

   ---

   1. Find the maximum kinetic energy of the photoelectrons The photoelectric equation:

   $K_{\text{max}} = h\nu - \phi$

   Here $h\nu = 4.5\ \text{eV}$ and $\phi = 3\ \text{eV}$.

   So

   $K_{\text{max}} = 4.5 - 3 = 1.5\ \text{eV}.$

   2. Convert kinetic energy to joules Since $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$,

   $K_{\text{max}} = 1.5 \times 1.602 \times 10^{-19} = 2.403 \times 10^{-19}\ \text{J}.$

   3. Find the momentum of the electron For a non‑relativistic electron (1.5 eV is tiny compared to its rest energy 511 keV),

   $p = \sqrt{2 m_e K_{\text{max}}}.$

   Electron mass $m_e = 9.109 \times 10^{-31}\ \text{kg}$.

   $p = \sqrt{2 \times 9.109 \times 10^{-31} \times 2.403 \times 10^{-19}}.$

   First compute the product inside:

   $2 \times 9.109 \times 10^{-31} \times 2.403 \times 10^{-19} = 4.378 \times 10^{-49}.$

   Then

   $p = \sqrt{4.378 \times 10^{-49}} = 6.617 \times 10^{-25}\ \text{kg·m/s}.$

   4. Apply de Broglie’s relation

   $\lambda = \frac{h}{p},$

   with Planck’s constant $h = 6.626 \times 10^{-34}\ \text{J·s}$.

   $$$$

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6. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.If the wavelength of the incident radiation on a photosensitive metal surface is decreased from $3100~\text{\AA}$ to $1550~\text{\AA}$, the maximum kinetic energy of the emitted photoelectrons is tripled. The work function of the metal surface is (A) $3~\text{eV}$ (B) $4~\text{eV}$ (C) $2~\text{eV}$ (D) $6~\text{eV}$

   ›Reveal solutionSolution

   When the wavelength of incident light decreases, the photon energy increases, leading to higher kinetic energy of emitted photoelectrons. By applying Einstein's photoelectric equation for two different wavelengths and using the given relationship between kinetic energies, we find the work function of the metal. The work function is $\boxed{2~\text{eV}}$.

   The problem describes a scenario involving the photoelectric effect, where light incident on a metal surface causes electrons to be ejected. The core concept here is Einstein's photoelectric equation, which relates the energy of the incident photon, the work function of the metal, and the maximum kinetic energy of the emitted photoelectrons.

   Concept and Intuition

   When light shines on a metal surface, if the photons have sufficient energy, they can eject electrons from the metal. This phenomenon is called the photoelectric effect.

   • Photon Energy ($E$): Light consists of discrete packets of energy called photons. The energy of a single photon is directly proportional to its frequency ($\nu$) and inversely proportional to its wavelength ($\lambda$).

     $E = h\nu = \frac{hc}{\lambda}$

     where $h$ is Planck's constant and $c$ is the speed of light.

   • Work Function ($\phi$): This is the minimum energy required to remove an electron from the surface of a particular metal. It's a characteristic property of the metal. Electrons deeper inside the metal require more energy to escape, but the work function refers to the minimum energy for the most loosely bound electrons at the surface.
   • Maximum Kinetic Energy ($K_{max}$): When a photon strikes an electron, it transfers its energy. If this energy is greater than the work function, the electron is ejected. Any excess energy beyond the work function is converted into the kinetic energy of the emitted electron. Since the work function is the minimum energy to escape, the maximum kinetic energy is achieved by electrons that require only this minimum energy to escape.

   Einstein's photoelectric equation elegantly combines these ideas:

   $K_{max} = E - \phi = \frac{hc}{\lambda} - \phi$

   This equation tells us that the maximum kinetic energy of the emitted photoelectron is the energy of the incident photon minus the work function of the metal. If the photon energy is less than the work function, no photoemission occurs, regardless of the intensity of the light.

   In this problem, we are given two different wavelengths of incident radiation and the corresponding relationship between the maximum kinetic energies. We can set up two equations based on Einstein's photoelectric equation and then solve them simultaneously to find the unknown work function.

   Step-by-step Solution

   1. Identify the given information and constants:

      • Initial wavelength, $\lambda_1 = 3100~\text{\AA}$
      • Final wavelength, $\lambda_2 = 1550~\text{\AA}$
      • Relationship between maximum kinetic energies: $K_{max,2} = 3 K_{max,1}$
      • We need to find the work function, $\phi$.
      • For calculations involving photon energy, it's convenient to use the product $hc$. Since the wavelengths are in Angstroms and the options for work function are in electron volts (eV), we use the value $hc \approx 12400~\text{eV}\cdot\text{\AA}$.

   2. Apply Einstein's photoelectric equation for the first case:

      When the wavelength is $\lambda_1 = 3100~\text{\AA}$, let the maximum kinetic energy be $K_{max,1}$.

      The photon energy is $E_1 = \frac{hc}{\lambda_1}$.

      Using Einstein's equation: …

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7. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.Two photons of energies $2.5\, \text{eV}$ and $5.5\, \text{eV}$ incident on a metal surface of work function $1.5\, \text{eV}$. The ratio of the maximum speeds of the photoelectrons emitted from the metal surface is (A) $1:4$ (B) $1:2$ (C) $1:1$ (D) $5:11$

   ›Reveal solutionSolution

   The maximum kinetic energy of a photoelectron is the photon energy minus the work function; since kinetic energy scales with speed squared, the ratio of maximum speeds is the square root of the ratio of the excess energies, giving $1:2$.

   The key concept here is the photoelectric effect: a photon’s energy is used first to overcome the work function (the minimum energy needed to free an electron), and any leftover energy becomes the electron’s kinetic energy. The maximum speed corresponds to the case where no energy is lost to internal collisions, so all excess energy goes into kinetic energy: $K_{\text{max}} = hf - \phi$. Since kinetic energy is $\frac12 m v^2$, the speed is proportional to the square root of the excess energy.

   Let’s work through it step by step.

   1. Write the photoelectric equation For a photon of energy $E$, the maximum kinetic energy of an emitted electron is

   $K_{\text{max}} = E - \phi,$

   where $\phi = 1.5\,\text{eV}$ is the work function.

   2. Compute the excess energies
      • For the $2.5\,\text{eV}$ photon:

   $K_1 = 2.5 - 1.5 = 1.0\,\text{eV}.$

   • For the $5.5\,\text{eV}$ photon:

   $K_2 = 5.5 - 1.5 = 4.0\,\text{eV}.$

   3. Relate kinetic energy to speed Since $K = \frac12 m v^2$, we have $v = \sqrt{\frac{2K}{m}}$. The mass $m$ of the electron is the same for both, so the ratio of maximum speeds is $\frac{v_1}{v_2} = \sqrt{\frac{K_1}{K_2}}.$ …

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8. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.The work function of a photosensitive metal surface is $1.1~\text{eV}$. Two light beams of energies $1.5~\text{eV}$ and $2~\text{eV}$ incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is (A) $3:4$ (B) $1:1$ (C) $2:3$ (D) $4:9$

   ›Reveal solutionSolution

   The maximum kinetic energy of a photoelectron is the photon energy minus the work function, and since kinetic energy scales with velocity squared, the ratio of velocities is the square root of the ratio of the excess energies. The correct ratio is $2:3$, so option (C).

   The key concept here is the photoelectric effect: a photon gives all its energy to an electron, but the electron must first overcome the binding energy (work function $\phi$) to escape. The leftover energy becomes kinetic energy. Since kinetic energy $K = \frac12 m v^2$, the velocity $v$ is proportional to $\sqrt{K}$. So the ratio of velocities is the square root of the ratio of the kinetic energies, not the photon energies themselves.

   Let’s work it through:

   1. Find the kinetic energy for each photon. For the first beam: photon energy $E_1 = 1.5~\text{eV}$, work function $\phi = 1.1~\text{eV}$. Maximum kinetic energy:

   $K_1 = E_1 - \phi = 1.5 - 1.1 = 0.4~\text{eV}.$

   For the second beam: $E_2 = 2.0~\text{eV}$, so

   $K_2 = 2.0 - 1.1 = 0.9~\text{eV}.$

   2. Relate kinetic energy to velocity. Since $K = \frac12 m v^2$, we have $v \propto \sqrt{K}$. Therefore,

   $\frac{v_1}{v_2} = \sqrt{\frac{K_1}{K_2}}.$

   3. Plug in the numbers. …

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9. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
   Q.An electric charge (q) moves along a metal tube with a variable cross-section as shown in the figure. It's velocity at the wider position A is $V_o$. Then the velocity of the charge when it approaches the neck of tube (position B) is (A) greater than $V_o$ (B) equal to $V_o$ (C) less than $V_o$ (D) less than or equal to $V_o$

   ›Reveal solutionSolution

   Inside a metallic tube the charge feels no force (a conductor screens the field), so it moves at constant speed — the velocity at the neck B equals $V_o$.

   Concept: The tempting (wrong) analogy is fluid flow, where the equation of continuity ($A_1v_1 = A_2v_2$) forces the speed to increase where the cross-section narrows. But a single charge is not an incompressible fluid; its motion is governed by Newton's laws, i.e. by the forces on it.

   Reasoning:

   • The tube is metal (a conductor). The electrostatic field inside the body of a conductor is zero, and a neutral conducting enclosure screens the interior. …

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10. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
    Q.Metal detector works on the principle of (A) Ohm’s law (B) Coulomb’s law (C) Electromagnetic induction (D) Stefan’s law of radiation

    ›Reveal solutionSolution

    A metal detector works by generating a changing magnetic field that induces currents in metal objects; this is the principle of electromagnetic induction, making option (C) correct.

    The key concept here is electromagnetic induction — the production of an electric current in a conductor when it is exposed to a changing magnetic field. A metal detector doesn’t simply “sense” metal; it actively creates a magnetic field that changes over time. When this field encounters a metal object, it induces tiny electric currents (called eddy currents) in the metal. Those currents then create their own magnetic field, which the detector picks up. The other options are unrelated: Ohm’s law deals with voltage, current, and resistance in a circuit; Coulomb’s law describes electrostatic forces between charges; Stefan’s law is about thermal radiation.

    Let’s break down why each option is or isn’t correct:

    1. Ohm’s law (A) — This law states $V = IR$, relating voltage, current, and resistance. While a metal detector contains circuits that obey Ohm’s law, the operating principle of detection is not Ohm’s law. It doesn’t explain how the detector senses metal.

    2. Coulomb’s law (B) — This law describes the force between stationary electric charges: $F = k \frac{q_1 q_2}{r^2}$. Metal detectors do not rely on static electric charges; they rely on changing magnetic fields, so this is incorrect. …

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11. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
    Q.At very high frequencies, the current $(i)$ in the given circuit is [FIGURE] (A) $4~\mathrm{A}$ (B) $0.4~\mathrm{A}$ (C) $44~\mathrm{A}$ (D) $4.4~\mathrm{A}$

    ›Reveal solutionSolution

    At $\omega\to\infty$ capacitors short and inductors open. The ladder collapses to $1\,\Omega + (4\parallel4) + 2\,\Omega = 5\,\Omega$, giving $i=220/5=\mathbf{44\ A}$ — option (C).

    The concept first

    A frightening-looking AC network becomes trivial once you use the two limiting behaviours of reactive elements:

    $X_C=\frac{1}{\omega C}\ \xrightarrow[\ \omega\to\infty\ ]{}\ 0 \qquad\text{(capacitor} \to \textbf{short circuit)}$

    $X_L=\omega L\ \xrightarrow[\ \omega\to\infty\ ]{}\ \infty \qquad\text{(inductor} \to \textbf{open circuit)}$

    Why? A capacitor opposes a change of voltage; at very high frequency the voltage barely has time to build up across it, so it offers almost no opposition — a wire. An inductor opposes a change of current; at very high frequency the back-emf is enormous, so essentially no current gets through — a break.

    Once every capacitor is a wire and every inductor is a gap, all the reactances vanish, the impedance is purely resistive, and Ohm's law $i=V/R$ finishes the job. (The values $6\,\mu\text{F}$, $0.8\,\mu\text{F}$, $0.5\,\mu\text{F}$, $20$ mH, $50$ mH are deliberate red herrings — none of them matters in the limit.)

    Step-by-step

    1. Kill the inductors (open circuits).

    • The upper row begins $8\,\Omega \to 20\,\text{mH} \to \dots$. With the $20\,\text{mH}$ open, the $8\,\Omega$ leads nowhere: it is a dead branch carrying zero current. Delete it.
    • Similarly, near the right the $50\,\text{mH}$ opens, orphaning the last upper $8\,\Omega$ (which only connects to the right terminal through that broken inductor). Delete it too.

    2. Short every capacitor.

    • The three $0.8\,\mu\text{F}$ capacitors in the lower row become wires, so the lower row reduces to just its resistors: $1\,\Omega$, then $4\,\Omega$, then $2\,\Omega$ in series from the left terminal to the right terminal.
    • The $6\,\mu\text{F}$ capacitor in the upper row becomes a wire, so the upper $4\,\Omega$ is now directly connected between the two cross-link nodes.
    • The two $0.5\,\mu\text{F}$ cross-links become wires, welding each upper node to the lower node beneath it. …

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12. TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
    Q.When the brakes are applied to two cars moving with initial velocities V and 2V and the cars stop at distances $d_1$ and $d_2$ respectively. Assuming the work done on the cars is the same, then the value of ratio $\dfrac{d_2}{d_1}$ is (A) 1 (B) 2 (C) 0.5 (D) 4

    ›Reveal solutionSolution

    The key idea is that the work done by the brakes equals the change in kinetic energy, and since the work is the same for both cars, the stopping distance is proportional to the square of the initial speed. The ratio $\frac{d_2}{d_1}$ is 4, so the correct option is (D).

    Concept and Intuition

    When brakes are applied, the force of friction (assumed constant) does work to bring the car to rest. The work-energy theorem tells us that the net work done on an object equals its change in kinetic energy. Here, the work done by the brakes is the same for both cars, so the initial kinetic energy of each car is entirely dissipated by the same braking force over different distances. Since kinetic energy depends on the square of velocity, a car starting at twice the speed has four times the kinetic energy, and thus needs four times the stopping distance if the braking force is constant.

    Step-by-step solution

    1. Apply the work-energy theorem

       For each car, the work done by the brakes (which is the net work, since other forces are balanced) equals the change in kinetic energy. The car starts with kinetic energy $\frac{1}{2} m v^2$ and ends at rest, so the work done is $W = \frac{1}{2} m v^2$.

    2. Express work in terms of force and distance

       If the braking force $F$ is constant (same for both cars, as the problem implies identical braking conditions), then the work done is also $W = F \cdot d$, where $d$ is the stopping distance. Thus:

    $F d = \frac{1}{2} m v^2$

    3. Write equations for both cars For the first car (initial speed $V$, stopping distance $d_1$):

    $F d_1 = \frac{1}{2} m V^2$

    For the second car (initial speed $2V$, stopping distance $d_2$):

    $$$$

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