Navbodh Physics Class 12

Explanation: A metal cube carries a positive charge Q. The question concerns the Electric Field and potential inside and on the surface of the cube.

In conductors, free charges move to the surface in electrostatic equilibrium. The Electric Field inside is zero, and the potential is uniform throughout the conductor. At the surface, the Electric Field is perpendicular to the surface because charges repel each other and distribute to maintain equilibrium.

Since the cube is a conductor, the charges are only on the outer faces. Inside, the Electric Field cancels out due to symmetry, keeping the potential constant. On the surface, the field is normal to each face, preventing any motion of charges inside.

Think of a balloon with static charges on its surface; the inside air feels no electric force, similar to potential remaining constant.

In summary, the interior of a conductor has zero Electric Field, uniform potential, and the surface field is perpendicular.

Explanation: Grounding a positively charged body allows electrons from the Earth to flow into it, affecting its electric potential.

The Earth is a vast reservoir of charges, so connecting an object provides a path for electrons to neutralize excess positive charge. Electric potential is the work required to bring a unit charge from infinity to a point. The flow continues until the potential of the object matches that of the Earth.

For a positively charged body, electrons move from the Earth, reducing its potential gradually. Once the potential equals that of the Earth, no further charge flows.

This is similar to a high-pressure water tank connected to a large reservoir; water flows until pressure equalizes.

Grounding stabilizes the potential of the body relative to the Earth.

Explanation: The question concerns the electric potential inside a hollow conducting sphere.

In conductors, charges reside on the outer surface. Inside a hollow conductor, the Electric Field is zero, so there is no change in potential throughout the interior. All points inside the sphere have the same potential as the surface because work done moving a unit charge inside the conductor requires no force.

The potential inside is uniform, and no point inside has higher or lower potential than another. This uniformity occurs regardless of the sphere’s size or total charge.

It’s like a completely sealed metal shell; the inside feels no electric influence from charges on the surface.

Hence, the potential at the center equals the surface potential.

Explanation: The potential gradient measures how electric potential changes over a distance.

It is defined as the rate of change of potential with respect to position, which involves direction as well as magnitude. Since it describes change in potential per unit length in a particular direction, it has both size and orientation, making it a Vector quantity.

Magnitude is calculated as ΔV/Δx, and the direction points in the direction of maximum decrease of potential. Scalars, by contrast, have only magnitude and no direction, so potential itself is a scalar, but its gradient is not.

For example, walking downhill, the steepness of the slope indicates the potential gradient; it points directly downhill.

Thus, the potential gradient is a Vector that indicates rate and direction of potential change.

Explanation: An equipotential surface is defined by constant electric potential across all points on it.

In Electrostatics, movement along such a surface requires no work since the electric potential difference between any two points is zero. The surface can take various shapes depending on the charge distribution. Electric Field lines always intersect equipotential surfaces at right angles.

For example, around a single point charge, spheres centered on the charge are equipotential surfaces. Moving a charge along one of these spheres requires no energy because potential is constant.

Hence, every point on an equipotential surface has the same electric potential.

Explanation: The amount of charge a conductor can hold at a given potential depends on its capacitance, which is proportional to its geometry.

Both spheres have the same radius, so their capacitances are identical. Capacitance determines the charge stored per unit potential, defined as Q = CV. The Solid or hollow nature does not affect the surface distribution of charge because only the outer surface participates in charge storage.

For example, a hollow chocolate sphere holds the same chocolate weight on the surface as a Solid one of the same radius, ignoring the inside.

Thus, charge storage depends on surface geometry, not Solid or hollow nature.

Explanation: Connecting two conductors allows charges to redistribute until both reach the same potential.

The potential on a sphere is V = Q/(4πε₀R). When a charge is placed on the larger sphere and connected to a smaller one, charge flows so that both spheres have identical potentials. Since the larger radius allows more charge at the same potential, the smaller sphere ends up with less charge, while the bigger retains more.

It’s similar to connecting two water tanks of different sizes at the bottom; water redistributes until the same water level is reached.

The final distribution depends on their radii to maintain equal potentials.

Explanation: The question refers to the work done in moving a test charge in an electric field.

Electric fields are conservative; the work done around a closed path in a static electric field is zero. For a stationary point charge +q, the field is radial, and moving a unit charge along a circular path around it involves no change in potential energy. work depends on potential difference, and in a closed loop, this difference is zero.

For example, walking around a hill at constant height requires no work to maintain height, similar to moving along an equipotential.

Thus, the work done around a closed loop in an electrostatic field is zero.

Explanation: Black body radiation depends on temperature according to Stefan–Boltzmann law: E = σT⁴, where T is absolute temperature.

Halving the temperature reduces the emitted power by a factor of (1/2)⁴ = 1/16. Radiation is very sensitive to temperature changes due to the fourth power dependence. Absolute temperature must be used in Kelvin to apply the law correctly.

It’s like a small fire glowing less brightly than a hotter one; reducing temperature drastically lowers energy emission.

Therefore, when the temperature is halved, the emitted radiation drops to a sixteenth of the original.

Explanation: Stefan–Boltzmann law applies here: E ∝ T⁴, with T in Kelvin.

Convert temperatures to Kelvin: 227 °C = 500 K, 727 °C = 1000 K. Doubling absolute temperature increases radiation by a factor of (1000/500)⁴ = 16. The original rate is 5 cal/s cm², so multiplying by 16 gives the new rate.

This demonstrates the extreme sensitivity of radiated power to temperature. For example, doubling a stove’s temperature drastically increases the Heat output.

Thus, higher absolute temperature produces significantly more radiated energy.

Explanation: Stefan–Boltzmann law states that the radiated energy E ∝ T⁴ with absolute temperature T in Kelvin.

Convert temperatures to Kelvin: 227 °C = 500 K, 27 °C = 300 K. The ratio of radiated power is (500/300)⁴. This shows that small changes in absolute temperature can significantly affect the rate of radiation.

It is similar to comparing two lamps, one much hotter than the other; the hotter one radiates energy far more intensely.

Hence, the ratio depends strongly on the fourth power of absolute temperatures.

Explanation: Using Stefan–Boltzmann law, E ∝ T⁴, we calculate radiated energy ratio using absolute temperatures.

Convert to Kelvin: 427 °C = 700 K, 127 °C = 400 K. The ratio of radiation is (700/400)⁴. This demonstrates the strong sensitivity of emission rate to temperature differences.

It’s like comparing a hot oven to a warm stove; the hotter oven radiates vastly more energy due to the fourth power dependence.

Therefore, the ratio is determined by the fourth power of the ratio of absolute temperatures.

Explanation: Radiated power depends on surface area, P ∝ A = 4πR², for spheres.

Given the radii ratio 2:1, the areas are proportional to R², giving a ratio of (2²)/(1²) = 4:1. Temperature and material being equal ensures that emissivity and T⁴ factors are identical for both.

It’s like comparing two spherical lamps of different sizes; the larger one radiates more due to greater surface area.

Thus, the rate of radiation is proportional to the square of the radius ratio.

Explanation: Stefan–Boltzmann law: E ∝ T⁴ with T in Kelvin.

0 °C = 273 K, 273 °C = 546 K. Doubling absolute temperature increases the radiated energy by (546/273)⁴ = 16 times the original E. This shows how radiation increases dramatically with temperature.

Like heating a metal plate; doubling its temperature increases energy emission exponentially.

Hence, radiated power depends on the fourth power of absolute temperature.

Explanation: Convert temperatures to Kelvin: 727 °C = 1000 K, 1727 °C = 2000 K.

Using Stefan–Boltzmann law, radiated energy ∝ T⁴. Doubling temperature doubles radiated energy by a factor of (2000/1000)⁴ = 16 times. Radiated energy increases sharply with temperature.

It’s similar to comparing two ovens, one at double the absolute temperature of the other; it emits far more Heat.

Thus, higher absolute temperature leads to substantially increased energy radiation.

Explanation: Stefan–Boltzmann law: E ∝ T⁴.

Reducing temperature to T/2 lowers radiated energy by (1/2)⁴ = 1/16 of the original E. Energy emission depends extremely sensitively on absolute temperature, so halving T drastically reduces output.

For example, a cooled stove emits much less infrared energy than when it is hot.

Hence, energy radiation decreases to a sixteenth when absolute temperature is halved.

Explanation: Radiated energy of a sphere: E ∝ R²T⁴, with radius R and absolute temperature T in Kelvin.

Convert temperatures: 127 °C = 400 K, 527 °C = 800 K. Surface areas ∝ R², so ratio E_P/E_Q = (8² * 400⁴) / (2² * 800⁴). Calculating shows a very small fraction because Q is much hotter and smaller but T⁴ dominates.

This is like comparing a large warm ball to a small hot one; the smaller one can radiate more intensely due to higher temperature.

The rate depends on both size and the fourth power of absolute temperature.

Explanation: Stefan–Boltzmann law: E ∝ T⁴.

Let initial T₁ = 400 °C = 673 K. If E₂ = 2E₁, then T₂⁴ / T₁⁴ = 2 → T₂ = T₁ * 2^1/4. This gives new absolute temperature in Kelvin.

This illustrates that to double radiation, the temperature increase is less than double in Kelvin scale due to the fourth root relation.

Hence, radiated energy doubling requires T₂ slightly higher than T₁.

Explanation: Radiant energy ∝ T⁴.

Doubling the Sun’s absolute temperature increases energy by (2)⁴ = 16 times. Thus, incident radiation rises sharply. Energy received depends strongly on the fourth power of temperature.

It’s like comparing a normal stove to one four times hotter; the hotter one radiates vastly more energy per unit area.

Therefore, radiation increases drastically with doubled source temperature.

Explanation: Convert to Kelvin: 27 °C = 300 K, 51.2 °C = 324.2 K.

Using Stefan–Boltzmann law, the ratio of emitted energy is (324.2/300)⁴ ≈ 1.46–1.5 times. This shows even a small rise in absolute temperature significantly increases energy emission.

For example, a slightly hotter stove element radiates noticeably more infrared energy.

Hence, radiated power increases by roughly 1.46–1.5 times with this temperature rise.

Explanation: The energy radiated by the Sun follows Stefan–Boltzmann law: E ∝ T⁴.

Doubling the Sun’s absolute temperature increases emitted energy per unit area by (2)⁴ = 16 times. The energy reaching Earth depends on this emission and distance, so the factor of increase remains 16.

It’s like turning a Light bulb to four times its effective brightness; the emitted energy increases sharply.

Thus, doubling the source temperature greatly amplifies energy received on Earth.

Explanation: Stefan–Boltzmann law: E ∝ T⁴.

Initial temperature T₁ = 400 K, final T₂ = 1200 K. The ratio of radiated energy = (T₂/T₁)⁴ = (1200/400)⁴ = 81. Multiply by 5 W gives the new energy.

This shows that radiated energy increases very rapidly with absolute temperature.

For instance, heating a metal plate from 400 K to 1200 K results in an enormous increase in radiated power.

Explanation: Stefan–Boltzmann law states the total energy radiated per unit area ∝ T⁴, where T is the absolute temperature.

The fourth power dependence means even small increases in temperature lead to large increases in radiation. The material’s emissivity also influences actual emission, but the basic law relates energy to T⁴.

Like comparing a warm stove to a red-hot one; the hotter surface radiates exponentially more Heat.

Hence, radiation is proportional to the fourth power of absolute temperature.

Explanation: Using Stefan–Boltzmann law, E ∝ T⁴.

Doubling absolute temperature gives (2)⁴ = 16 times the radiated energy. The rate of radiation is extremely sensitive to temperature due to the fourth power relationship.

Like heating a metal plate from warm to very hot; the energy emitted increases dramatically.

Thus, doubling T increases emission sixteenfold.

Explanation: The Stefan–Boltzmann law states that the power emitted per unit area of a black body ∝ T⁴.

It is a fundamental law of thermal radiation and applies to ideal black bodies. This law relates absolute temperature to radiant energy output and does not depend on material for a perfect black body.

For example, hotter objects like the Sun radiate much more energy than cooler ones.

Hence, this principle describing radiated energy versus temperature is known as Stefan’s law.

Explanation: The statements concern properties of thermal radiation and Stefan–Boltzmann law.

Stefan’s constant is universal, thermal radiation travels at Light speed, and Diffraction occurs, but Stefan’s law applies specifically to black bodies, not all heated objects. Ordinary bodies may deviate due to emissivity < 1. It’s like assuming all lamps radiate the same energy per unit temperature, which is not true; only ideal black bodies strictly follow the law. Hence, identifying the incorrect statement requires knowing the limits of Stefan’s law. [/explain] Show Answer

A metal cube of each side 0.05 m long emits 0.6 kcal in 80 seconds. Then the emissive power of its surface in kcal/s m² 2 is :

(A) 0.5

(B) 0.05

(C) 5

(D) 5 x 10²

[explain]Explanation: Emissive power is the rate of energy radiated per unit area.

Total surface area of cube: A = 6 * (0.05)² m². Energy emitted per unit time = 0.6 kcal / 80 s. Divide by surface area to get emissive power in kcal/s m².

This shows how geometry and time influence surface energy emission.

It’s like calculating how much Heat per square meter a small heater panel emits.

Explanation: Total energy radiated = emissive power × surface area × time.

E = 0.5 kcal/s m² × 0.02 m² × 20 s = 0.2 kcal.

This shows the relationship between surface area, emissive power, and radiation over time.

Similar to determining total heat output of a lamp over a period given its surface emission rate.

Explanation: Emissive power is defined as the energy emitted per unit area per unit time at a given temperature.

It is a measure of the radiation intensity from a surface and depends on temperature and material properties. Total energy emitted over entire surface = emissive power × area × time.

For example, a stove plate’s heat output per square meter per second defines its emissive power.

Hence, it quantifies radiation per unit area and time at a specified temperature.

Explanation: Coefficient of emission (emissivity) = energy absorbed / energy incident.

Here, absorbed energy = 224 cal, incident = 512 cal → ε = 224/512. Emissivity indicates how efficiently a surface absorbs and emits radiation.

It’s like comparing a black cloth that absorbs most Light versus a shiny metal reflecting most Light.

Thus, emissivity quantifies absorption efficiency relative to perfect black body.

Explanation: A black body is an idealized object that absorbs all incident radiation and radiates energy across all wavelengths.

Unlike colored or reflective surfaces, a black body emits radiation continuously throughout the Spectrum. Its emission depends solely on temperature, following Planck’s law and Stefan–Boltzmann law.

For example, the Sun approximates a black body, radiating across visible, infrared, and ultraviolet ranges.

Thus, a black body emits radiation of all wavelengths, determined only by its temperature.

Explanation: An ideal absorber takes in 100% of incident radiation.

Such a body also radiates energy efficiently and is called a black body in Physics. Real objects may approximate this behavior to varying degrees depending on color and material.

For instance, a perfectly black-painted metal surface absorbs nearly all sunlight.

Hence, a body that absorbs all incident radiation is called a black body.

Explanation: Emissivity (coefficient of emission) measures how efficiently a surface radiates energy compared to a perfect black body.

For a perfectly black body, emissivity = 1, meaning it radiates the maximum energy possible at a given temperature. Real materials have emissivity < 1. Like a blackened oven plate radiating more heat than a shiny one at the same temperature. Thus, the coefficient of emission of a perfectly black body is unity. [/explain] Show Answer

SI unit of emissive power is :

(A) J/s

(B) J/m²

(C) J/s m²

(D) W/m

[explain]Explanation: Emissive power is energy radiated per unit area per unit time.

Energy is in Joules (J), time in seconds (s), and area in square meters (m²). Hence, the SI unit is J/(s·m²) or W/m².

For example, a heater emitting 100 W over 2 m² has an emissive power of 50 W/m².

Thus, the proper SI unit for emissive power is J/s·m².

Explanation: Materials that block the passage of radiation are called athermanous.

They do not transmit thermal radiation and often reflect or absorb it. Transparent materials like glass transmit radiation, while opaque Metals block it.

Like a metal sheet blocking infrared radiation from a heater.

Hence, substances that prevent radiation from passing through are called athermanous.

Explanation: Coefficient of transmission = fraction of incident radiation transmitted.

Athermanous substances block radiation entirely, so the coefficient of transmission = 0. Other bodies allow partial or full transmission depending on transparency.

For example, a metal plate does not let infrared heat pass through, giving zero transmission.

Thus, the transmission coefficient of athermanous substances is zero.

Explanation: Perfect black bodies absorb all incident radiation and do not reflect any.

Coefficient of reflection = fraction of incident radiation reflected. For a perfect black body, this value = 0. Real materials may reflect partially.

Like a matte black surface absorbing all Light without reflection.

Hence, the reflection coefficient of a perfect black body is zero.

Explanation: Energy conservation: incident = absorbed + reflected + transmitted.

Absorbed = 0.31 × 200 = 62 cal, reflected = 0.41 × 200 = 82 cal. Transmission = 200 – (62+82) = 56 cal.

It’s like splitting a Light beam into absorbed, reflected, and transmitted portions.

Hence, transmitted energy depends on subtracting absorbed and reflected energy from incident energy.

Explanation: Athermanous body: no transmission.

Reflection coefficient = 1 – absorption = 1 – 0.3 = 0.7. This shows that any incident energy not absorbed is reflected.

Example: A shiny metal plate reflects the rest of the radiation it does not absorb.

Thus, reflection coefficient = 0.7 for an athermanous body absorbing 30%.

Explanation: Transmission coefficient = fraction of incident energy passing through.

Total incident = 10 J, absorbed = 2 J, reflected = 6 J → transmitted = 10 – (2+6) = 2 J.

This demonstrates energy conservation: incident energy splits among absorption, reflection, and transmission.

Hence, the transmission coefficient = 2/10 = 0.2.

Explanation: Energy conservation: incident energy = absorbed + reflected + transmitted.

Absorbed energy = 0.68 × 100 = 68 cal, reflected = 27 cal. Transmission = 100 – (68 + 27) = 5 cal.

This shows that even if a surface absorbs a large fraction, some energy may still pass through depending on transparency.

Thus, the coefficient of transmission = 5/100 = 0.05.

Explanation: Athermanous body transmits no radiation.

Energy conservation: absorption + reflection + transmission = 1 → reflection = 1 – absorption – transmission = 1 – 0.65 – 0 = 0.35.

Example: A metal sheet absorbs some infrared and reflects the rest, letting none pass through.

Hence, the reflection coefficient = 0.35.

Explanation: Reflected energy = incident × (1 – absorption).

Emissivity = absorptivity = 0.8 → absorbed fraction = 0.8 → reflected fraction = 1 – 0.8 = 0.2.

Incident power = 10 W → reflected power = 10 × 0.2 = 2 W. Over 1 minute: energy = 2 W × 60 s = 120 J.

Like a reflective surface where only a small portion of heat is absorbed, the rest is reflected.

Hence, reflected energy in 1 minute = 120 J.

Explanation: Transparent to thermal radiation → diathermanous.

Such materials allow radiation to transmit through without significant absorption. Examples: glass for infrared or sunlight.

Like a greenhouse allowing sunlight to enter, diathermanous substances permit heat to pass.

Thus, materials that let radiation pass are called diathermanous.

Explanation: All Matter with temperature above absolute zero emits electromagnetic radiation.

It is not limited to hot bodies or specific antennas; emission depends on temperature and material properties. Even room-temperature objects emit infrared radiation.

For example, a warm cup radiates infrared waves, and the Sun emits visible Light, infrared, and ultraviolet.

Hence, all bodies emit electromagnetic radiation.

Explanation: This is conduction, where kinetic energy of molecules transfers to neighboring particles without bulk motion.

It requires a medium, typically Solid, for energy transfer via vibrating atoms or free electrons.

Like a metal spoon getting hot from its handle when placed in hot water.

Thus, heat transfer by Molecular collisions occurs via conduction.

Explanation: Conduction requires molecules to vibrate about fixed positions, transferring kinetic energy to neighbors.

Medium should not require free movement; Solids are ideal since atoms are closely packed and can vibrate.

For example, Metals conduct heat effectively due to closely spaced atoms and free electrons.

Hence, conduction occurs via Molecular vibrations in Solids.

Explanation: Heat radiation is electromagnetic radiation, primarily in the infrared region of the Spectrum.

Wavelengths longer than visible Light are emitted by objects at typical temperatures. Hotter bodies emit shorter wavelengths as well.

For instance, a heater emits mostly infrared waves, which we feel as warmth.

Thus, heat radiation corresponds to the infrared Spectrum.

Explanation: Absorptive power = energy absorbed / energy incident.

It quantifies the ability of a body to absorb radiation; ranges from 0 (no absorption) to 1 (perfect absorption).

Example: Black surfaces have high absorptive power, while shiny Metals have low.

Hence, the ratio of absorbed to incident energy is the absorptive power.

Explanation: In vacuum, no conduction or convection is possible; only radiation can transfer heat.

Hot water will lose energy via thermal radiation to surroundings. The vacuum prevents air molecules from transferring heat.

Like a thermos bottle reducing heat loss primarily via radiation.

Thus, the calorimeter will cool by radiation in vacuum.

Explanation: In vacuum, heat transfer can only occur via radiation.

The hot body emits thermal radiation, losing energy, while the cold body absorbs some of this radiation. Over time, the temperature of the hot body decreases until thermal equilibrium is approached.

This is similar to feeling the warmth of a hot object in space where no air exists—cooling happens only via emitted radiation.

Hence, the hot body’s temperature will decrease due to radiation.

Explanation: Hot air rises due to being less dense, creating convection currents.

Ventilators at the top allow warm, stale air to escape, maintaining air circulation and improving ventilation. They also help regulate temperature and remove carbon dioxide.

Like an exhaust fan removing warm air in a kitchen.

Thus, top ventilators maintain convection currents to keep the air fresh.

Explanation: Sensation of temperature depends on heat transfer rate, not actual temperature.

Objects of different thermal conductivity feel differently: Metals conduct heat faster than wood. A special temperature exists (close to body temperature) where both conduct heat at similar rates, making them feel the same.

For example, touching metal and wood at room temperature: metal feels colder due to rapid heat conduction.

Hence, there is a temperature at which both feel equally hot or cold.

Explanation: Convection requires gravity to allow Fluid movement, which is absent in space.

Thus, heat transfer occurs only via conduction through Molecular vibrations and radiation from the liquid surface.

For example, heating water in zero gravity: the liquid does not circulate, so convection is eliminated.

Hence, heat transfer occurs by conduction and radiation under zero gravity.

Explanation: Heat radiation is electromagnetic radiation, which travels at the speed of light in vacuum.

It does not require a medium and moves at 3 × 10⁸ m/s. This is consistent for all electromagnetic waves.

Example: Infrared radiation from the Sun reaches Earth at light speed.

Hence, the velocity of heat radiation in vacuum equals the speed of light.

Explanation: Thermal conduction is transfer of heat via Molecular collisions, most effective in Solids with closely packed atoms.

Convection and radiation involve bulk movement or electromagnetic waves, less efficient in Solids. Metals conduct heat best due to free electrons and lattice vibrations.

Like a metal rod heating faster than water under the same conditions.

Hence, thermal conduction is maximum in conduction process.

Explanation: Water is a Fluid, and heating it causes density differences, generating convection currents.

Hot water rises and cold water sinks, transferring heat throughout the liquid. Conduction is slower, and radiation is minor.

Example: Boiling water in a pan creates circulating currents distributing heat evenly.

Thus, water is primarily heated by convection.

Explanation: Metals have free electrons that can move easily, transferring kinetic energy quickly between atoms.

Atoms themselves vibrate, but electrons dominate heat transfer efficiency. Non-Metals lack free electrons and conduct poorly.

Like copper pan handles heating rapidly due to free electrons carrying energy.

Hence, Metals are excellent heat conductors due to mobile electrons.

Explanation: Heating reduces the density of liquid locally, making it buoyant relative to surrounding cooler liquid.

This density difference creates upward motion, forming convection currents. No external force is required; the Fluid moves naturally.

Example: Boiling water forms rising bubbles due to less dense hot liquid.

Thus, heated liquid rises because its density is lower than the surrounding Fluid.