Navbodh Physics Class 11

Explanation: A SET of dry cells with 1.5 V emf and internal resistance 0.5 Ω each is connected in series to a 20 Ω resistor to produce a specific current. In series circuits, total emf is the sum of individual emfs and total resistance is the sum of internal resistances plus external resistance. Using Ohm’s law, I = total emf / total resistance, one can relate the number of cells to the required current. The calculation involves balancing the combined voltage against the total series resistance, which includes both the internal and external contributions. Think of it like connecting small pumps in series to push water through a pipe: the number of pumps needed depends on the pipe’s resistance and the desired flow. By combining these relationships, the series configuration needed to achieve the SET current can be estimated.

Explanation: Maximum power transfer occurs when the external resistor matches the internal resistance of the source. The power delivered to the resistor is P = I²R, where I = V / (R + r) and r is the internal resistance. By expressing power as P = V²R / (R + r)², the condition for maximum power is obtained mathematically by differentiating P with respect to R and setting it to zero. This ensures the external load receives the highest possible energy output from the battery. A simple analogy is adjusting a garden hose nozzle to maximize water flow: the nozzle’s resistance must balance the hose’s internal resistance for peak flow. This principle applies universally to resistive electrical loads.

Explanation: The highest current from a battery occurs when the external resistance is negligible compared to the internal resistance. Ohm’s law (I = V / R_total) applies, with total resistance being the sum of internal and external resistances. As external resistance decreases, the total resistance approaches the internal resistance, resulting in maximum current. This scenario is like opening a wide water tap: the current is limited only by the internal friction of the battery, analogous to the pipe’s internal resistance. Careful consideration of internal resistance determines the limit of current that the source can provide without exceeding its capacity.

Explanation: When a battery supplies current to a resistor, the terminal voltage drops due to internal resistance. The voltage drop across the internal resistance equals the product of current and internal resistance. Using Ohm’s law and the change in terminal voltage, the internal resistance can be deduced. Conceptually, the internal resistance acts like friction inside a pipe: some of the battery’s potential is “lost” internally before reaching the load. By comparing unloaded and loaded voltages, the internal resistance’s effect is quantified.

Explanation: power dissipation in a resistor depends on P = V² / R. For identical resistors, total power changes with the configuration. In series, voltage divides among resistors; in parallel, each resistor receives the full supply voltage. The total dissipation is the sum of individual powers. Think of water flowing through multiple parallel pipes versus a single series path: each pipe in parallel carries the full pressure, increasing the total flow, just as parallel resistors increase total power dissipation. Understanding the relationship between voltage, resistance, and power is essential here.

Explanation: Resistance depends on length and cross-sectional area, R = ρL / A. Halving the wire length and doubling the cross-section by twisting ends together changes the resistance. The series and parallel combination concepts apply, as two halves effectively form a parallel path. Visualizing it as two short wires carrying current simultaneously helps understand why resistance decreases. This illustrates the interplay between geometric changes and electrical properties in conductors.

Explanation: Stretching a wire changes its length and cross-sectional area. Resistance is given by R = ρL / A. When the wire is stretched, its length doubles while the volume remains constant, so the cross-sectional area halves. This geometric change increases resistance, as a thinner, longer conductor offers more opposition to current. The effect is analogous to squeezing a hose: water faces more resistance in a longer, narrower pipe. Understanding how dimensional changes affect resistance is key.

Explanation: Carnot engine efficiency depends on source and sink temperatures: η = 1 − T_c / T_h. The work done per cycle is the difference between Heat absorbed from the source and Heat expelled to the sink. By expressing work as W = Q_h − Q_c and relating efficiency to temperatures, the Heat absorbed can be calculated conceptually. Imagine a perfect steam engine where the fraction of energy converted to work is limited by the temperature difference; the rest is expelled. Temperature ratio determines the portion of Heat converted to useful work.

Explanation: Efficiency of a Carnot engine is η = 1 − T_c / T_h, where T_c and T_h are absolute temperatures. Knowing the sink temperature and efficiency allows calculation of the source temperature. Conceptually, the engine can’t exceed the Carnot limit; the temperature difference dictates energy conversion. Think of it like a hill: the height difference determines how much water can flow downhill to generate energy. Using the efficiency ratio, the source temperature corresponding to a given efficiency is determined.

Explanation: Carnot efficiency depends on η = 1 − T_c / T_h. Maximum efficiency occurs when the ratio T_c / T_h is smallest, i.e., a large difference between source and sink temperatures. Conceptually, a greater temperature gradient allows more energy to be converted into work. An analogy is using a steeper slope in a hydroelectric dam: more potential energy converts into Electricity with a higher drop. Comparing ranges, the one with the largest temperature difference yields the highest efficiency.

Explanation: The efficiency of a Carnot engine is determined by η = 1 − T_c / T_h. If the engine has the same efficiency in two different temperature ranges, the ratio of sink to source temperature must be equal in both cases. This allows setting up an equation relating T_c and T_h of the two scenarios. Conceptually, a Carnot engine’s performance is fully determined by the relative difference between source and sink temperatures, not their absolute values. By equating the temperature ratios, the unknown temperature can be calculated. This is similar to comparing slopes of two ramps: equal slopes produce the same energy conversion efficiency.

Explanation: Using η = 1 − T_c / T_h for Carnot engines, the equality of efficiencies in two temperature ranges implies the sink-to-source ratios are the same. Solving the resulting equation allows determination of the unknown temperature. This highlights that a Carnot engine’s efficiency depends solely on relative temperatures. Imagine water flowing down two channels: if the slopes are the same, the fraction of potential energy converted is identical. The temperature ratio dictates the fraction of Heat transformed into work.

Explanation: For adiabatic expansion, temperature and volume are related via T V^γ−1 = constant, where γ is the Heat capacity ratio (for diatomic gas, γ ≈ 1.4). Carnot efficiency depends on the ratio of sink to source temperature. Using the volume change, the temperatures at the start and end of expansion can be estimated, then applied to efficiency formula η = 1 − T_c / T_h. The process demonstrates how expansion affects thermal energy conversion. An analogy is a piston in a cylinder: as the gas expands, the temperature changes, affecting the work extractable per cycle.

Explanation: Efficiency η = 1 − T_c / T_h can be expressed in terms of Heat absorbed and rejected: η = W / Q_h = 1 − Q_c / Q_h. Given the absorbed and rejected Heat, the efficiency ratio can be calculated, which then allows finding the sink temperature. Conceptually, the engine converts part of the absorbed Heat into work, and the remainder is expelled to the sink. Think of it as a two-tank system: the energy not converted into work flows into the lower-temperature tank. Temperature ratio determines how much heat must be discarded.

Explanation: A refrigerator requires work input to transfer heat from a colder region to a warmer one. The minimum work is linked to the coefficient of performance (COP) of a reversible refrigerator: COP = Q_c / W. The latent heat of fusion for water gives Q_c. By combining these relationships, the minimum work to freeze the water can be estimated. Conceptually, the refrigerator moves energy against the natural flow, requiring external work. Analogous to lifting a weight uphill: energy must be provided to overcome natural gradients.

Explanation: The coefficient of performance for a Carnot refrigerator is COP = T_c / (T_h − T_c), with temperatures in Kelvin. Given COP and the cold reservoir temperature, the temperature of the surroundings can be found by rearranging the formula. Conceptually, COP represents how efficiently heat is moved from cold to hot regions. It’s like pumping water uphill: the efficiency depends on the height difference and the work input. Using the relationship, the hot reservoir temperature is determined.

Explanation: Work input is related to the heat removed and COP: W = Q_c / COP. Given the heat extracted and COP, the minimum energy required can be calculated. Conceptually, a refrigerator uses work to shift heat from the cold region to the surroundings. Analogous to lifting water: the energy needed depends on how much water is moved and the efficiency of the pump. By applying the formula, the required work is determined.

Explanation: For an ideal (Carnot) refrigerator, COP = T_c / (T_h − T_c), with temperatures in Kelvin. Converting the freezer temperature to Kelvin, the surrounding temperature can be calculated by rearranging the COP formula. Conceptually, the refrigerator moves heat against the natural gradient: the colder it is inside, the more work required to expel energy to the warmer surroundings. Think of it as lifting water from a deep well to a tank at a higher level. The relationship ensures energy conservation and maximum efficiency.

Explanation: For a reversible refrigerator, COP = Q_c / W = T_c / (T_h − T_c). Given W = 1 J, the heat expelled to the hot reservoir is Q_h = W + Q_c. Using the temperature ratio, Q_c can be determined first, then Q_h calculated. Conceptually, the refrigerator moves heat from cold to hot using work, with the expelled heat being the sum of input work and extracted energy. This is similar to raising water to a higher tank: the total energy delivered equals the lifted portion plus input energy.

Explanation: Cooling rate depends on surface area to volume ratio and material properties. Objects with larger surface area relative to volume lose heat faster. Among a sphere, cube, and thin plate of same Mass and material, the thin plate has the highest surface area per unit volume, so it cools fastest. Conceptually, more exposed area allows more energy to radiate or convect. Think of spreading butter thin: it cools faster than a Solid lump due to greater exposure. This principle applies universally to objects losing heat to the Environment.

Explanation: Heat loss by a body to its surroundings is governed by Newton’s law of cooling: rate of heat loss ∝ temperature difference between object and surroundings. At 80 °C, ΔT = 80 − 20 = 60 °C, giving 60 cal/s. At 40 °C, ΔT = 40 − 20 = 20 °C. By proportionality, the rate of heat loss decreases with smaller temperature difference. Conceptually, the driving force for heat transfer is the difference between the body and ambient temperatures. Think of it as water flowing from a hotter tank to a cooler one: flow slows as temperatures equalize.

Explanation: RMS speed of gas molecules is given by v_rms = √(3RT / M), where R is gas constant, T is temperature, and M is molar Mass. Air density allows calculation of molar Mass or confirmation of known composition. The root-mean-square speed indicates the average Molecular motion contributing to kinetic energy. Conceptually, molecules move faster at higher temperatures, and heavier molecules move slower. Think of it like a crowd of people: lighter individuals move faster on average than heavier ones at the same energy.

Explanation: RMS speed v_rms = √(3kT / m) for a gas Molecule, where k is Boltzmann constant and m is Molecular Mass. To achieve v_rms equal to Earth’s escape velocity, the temperature T can be found by rearranging the formula. Conceptually, extremely high temperatures are required for Molecular speeds to match planetary escape velocities. Analogous to giving a ball enough kinetic energy to escape Earth’s gravity.

Explanation: RMS speed is the square root of the average of the squares of individual speeds: v_rms = √((v₁² + v₂² + … + vₙ²)/n). Each Molecule’s speed is squared, summed, divided by the total number, and square-rooted. Conceptually, RMS gives an effective speed that relates to kinetic energy, not just an arithmetic mean. Think of it as calculating an average “energy speed” rather than simple velocity average.

Explanation: Boyle’s law states P₁V₁ = P₂V₂ for constant temperature. If volume decreases by 10%, V₂ = 0.9V₁. Solving for P₂ gives the required pressure. Conceptually, reducing volume at constant temperature increases Molecular collisions, thereby increasing pressure. Analogously, compressing a balloon makes it harder to push inward, reflecting the increased pressure.

Explanation: At constant pressure, Charles’s law applies: V₁/T₁ = V₂/T₂, with temperatures in Kelvin. The new volume is proportional to the ratio of absolute temperatures. Conceptually, gas expands when heated at constant pressure as molecules move faster and occupy more space. Like inflating a balloon in a warm room: the volume increases with temperature.

Explanation: The pressure of a moving gas combines contributions from random Molecular motion (RMS speed) and bulk flow. Pressure due to Molecular collisions is proportional to ρC² / 3. Bulk velocity adds directional kinetic energy but doesn’t directly contribute to isotropic pressure. Conceptually, pressure is created by random impacts of molecules on container walls. Like balls bouncing in a box: random motion increases the overall force on the walls.

Explanation: Ideal gas law PV = nRT gives n = PV / RT. For both containers, calculate n using respective P, V, T values. The ratio n_A / n_B gives the required Molecule count ratio. Conceptually, doubling pressure or halving volume alters the number of particles needed to maintain the ideal gas relationship. Analogous to filling two balloons of different sizes and pressures: particle counts adjust accordingly.

Explanation: The speed of sound c in a gas is related to RMS Molecular speed by c = √(γ / 3) v_rms. Given γ and c, v_rms can be calculated. Conceptually, sound speed depends on how fast molecules transmit momentum; RMS speed represents the average Molecular motion contributing to kinetic energy. Like a crowd: how fast vibrations pass through depends on individuals’ typical speeds.

Explanation: Using Charles’s law at constant pressure, V₁ / T₁ = V₂ / T₂ (absolute temperatures). Tripling volume implies T₂ = 3 × T₁. Conceptually, heating a gas at constant pressure allows molecules to spread apart, increasing both volume and temperature proportionally. Analogous to stretching a balloon slowly while heating: the final volume and temperature scale together.

Explanation: RMS speed depends on Molecular Mass and temperature: v_rms = √(3kT / m). At the same temperature, lighter molecules move faster. Since hydrogen has 1/16th the Mass of oxygen, its RMS speed is higher by a factor of √16 = 4. Conceptually, lighter particles move more quickly to maintain the same kinetic energy. This is like comparing a ping-pong ball and a bowling ball: under equal energy, the lighter ball moves much faster.

Explanation: At constant volume, pressure of an ideal gas is proportional to temperature in Kelvin (P ∝ T). Converting the temperature rise to Kelvin, ΔT = 1°C = 1 K. The percentage rise in pressure = (ΔP / P) × 100% = (ΔT / T) × 100%. Conceptually, increasing temperature raises Molecular motion, causing more frequent collisions with the container walls. Think of it as stirring a confined gas: molecules collide faster as energy increases.

Explanation: Boyle’s law applies at constant temperature: P₁V₁ = P₂V₂. Solving for V₂ gives V₂ = P₁V₁ / P₂. Conceptually, increasing pressure compresses the gas, reducing volume. Like squeezing a balloon slowly: the air inside occupies less space as external pressure increases, maintaining constant temperature.

Explanation: For an ideal gas, average kinetic energy per Molecule depends only on temperature: KE_avg = (3/2)kT. Since the gases are at the same temperature, their average kinetic energies are equal regardless of molecular Mass or proportion. Conceptually, the energy per particle is determined by thermal motion, not composition. Analogous to people running at the same speed: energy per individual is the same despite their number in the group.

Explanation: Using the ideal gas law, PV = nRT, and knowing masses, the number of moles differs due to molar Mass. Absolute temperature T ∝ PV / nR. With equal P and V, temperature ratio is inversely proportional to number of moles. Conceptually, heavier molecules require fewer moles for the same mass, so temperature adjusts to maintain pressure. This is like balancing two containers with different numbers of balls to maintain equal pressure inside.

Explanation: Charles’s law (V ∝ T at constant pressure) gives T₂ / T₁ = V₂ / V₁. Doubling volume implies doubling absolute temperature in Kelvin. Conceptually, heating a gas at constant pressure allows molecules to move faster, pushing against the walls and increasing volume proportionally. Think of a piston: when the gas is heated, it expands to maintain pressure.

Explanation: Using the ideal gas law PV = nRT, T ∝ PV. If P and V both double, T₂ = 2 × 2 × T₁ = 4T₁. Conceptually, the temperature scales with the product of pressure and volume. Think of it as inflating a balloon while increasing external pressure: the gas temperature must rise to maintain the relation between PV and nRT.

Explanation: Density ρ = mass/volume = (n × m) / V. Using n = PV / kT, density ρ = mP / kT. Conceptually, density depends on how much mass is contained in a volume at a given temperature and pressure. Higher pressure compresses molecules, increasing density; higher temperature spreads them out, reducing density. Like packing a suitcase: more items increase “density,” while expanding space reduces it.

Explanation: Use ideal gas law: PV = nRT. First, calculate moles n = mass / molar mass. Then substitute P, n, R, and T (in Kelvin) to find volume V. Conceptually, gas occupies volume proportional to the number of moles at given pressure and temperature. Like filling balloons: the number of molecules and their temperature determine how much space they take.

Explanation: At constant volume, P ∝ T. Percentage increase in pressure = ΔP / P = ΔT / T. Convert 1°C to Kelvin, SET up proportion ΔP/P = ΔT/T, solve for T. Conceptually, small temperature changes produce proportional pressure changes. Like heating a sealed can: the hotter it gets, the higher the internal pressure rises.

Explanation: Using the ideal gas law PV = nRT and knowing equal masses, the number of moles changes with molar mass. Pressure is proportional to the number of moles at constant volume and temperature. Helium has a lower molar mass than hydrogen, so there are more moles for the same mass, resulting in higher pressure. Conceptually, lighter molecules contribute more to collisions per unit volume at constant mass. Like comparing two boxes of balls: smaller balls allow more balls inside, creating more impacts per second.

Explanation: At constant volume, pressure varies directly with temperature in Kelvin (P ∝ T). Convert both temperatures to Kelvin and calculate P₂ = P₁ × (T₂ / T₁). Conceptually, heating gas increases molecular motion, producing more frequent and forceful collisions with the container walls. Like inflating a balloon inside a rigid box: higher temperature pushes molecules harder against walls, raising pressure proportionally.

Explanation: From PV = nRT, number of moles n = mass / molar mass. With equal masses and pressure, T ∝ n. Therefore, T_N₂ / T_O₂ = n_N₂ / n_O₂ = M_O₂ / M_N₂. Conceptually, fewer moles of heavier gas require a proportionally lower temperature to maintain the same pressure. Analogy: a smaller number of larger balls needs less energy to exert equal force on a surface.

Explanation: Reynolds number Re = (ρvD)/η. Solving for critical velocity v = Re × η / (ρD). Substituting values gives the threshold for laminar to turbulent flow. Conceptually, higher viscosity or smaller diameter lowers critical velocity. Analogy: stirring thick syrup slowly avoids turbulence, but water flows easily before turbulence sets in.

Explanation: Terminal velocity v_t = √((2r²g(ρ_sphere − ρ_fluid)) / (9η)) for a sphere in viscous Fluid. For same radius and Fluid, v_t ∝ √(ρ_sphere − ρ_fluid). Lighter silver sphere experiences smaller NET force, reducing terminal speed. Conceptually, terminal speed balances gravity and viscous drag. Analogy: heavier objects fall faster in syrup, lighter ones slower, until drag equals weight.

Explanation: Critical velocity for laminar to turbulent flow: v_c ∝ η / (ρD) × Re. With equal diameter, ratio v₁ / v₂ = (η₁ / η₂) × (ρ₂ / ρ₁). Conceptually, denser fluids slow turbulence onset, while more viscous fluids resist motion, increasing critical speed. Analogy: stirring thick honey vs. water in identical tubes; honey can maintain laminar flow longer.

Explanation: Pressure at the bottom = P_water + P_oil = ρ_water g h_water + ρ_oil g h_oil. Convert densities and heights to SI units. Conceptually, pressure increases with the weight of the column above a point. Analogy: stacking two liquids in a cylinder increases the pressure at the bottom due to combined weights.

Explanation: Viscous force F = η A v / d → d = η A v / F. Area A = (0.1)². Convert poise to SI units. Conceptually, thinner gaps increase resistance, so less separation allows same viscous force at given velocity. Analogy: dragging a card over syrup: closer plates make it harder to slide.

Explanation: Torricelli’s law: flow rate ∝ √h. For half height, rate reduces, so time increases inversely with √h. If full vessel drains in 10 min, half-filled drains in 10 × √2 ≈ 14.14 min. Conceptually, flow slows as water column height decreases, because gravitational potential energy driving the flow reduces. Analogy: a shorter hose ejects water more slowly.

Explanation: Boyle’s law: P₁V₁ = P₂V₂. Pressure at depth = atmospheric + water column. Doubling radius → volume increases by factor 8, so initial pressure = 8 × atmospheric. Solve for depth: P_total = P_atm + ρ g h. Conceptually, as bubble rises, external pressure decreases, allowing expansion. Analogy: squeezing a balloon underwater and releasing it: it grows as pressure lessens.

Explanation: Using Boyle’s law, P₁V₁ = P₂V₂, the volume triples → V₂ = 3V₁. Pressure at depth = P_atm + ρ g h. Solve for h. Conceptually, as a bubble rises, decreasing pressure allows it to expand. Analogous to a balloon released from water: the lower surrounding pressure allows it to inflate.

Explanation: Use Torricelli’s theorem: v = √(2 g h). Height h = 20 m. Conceptually, water flows out under gravity, converting potential energy into kinetic energy. Analogy: water rushing from a tank mimics free-fall motion, gaining speed proportional to height of water above the hole.

Explanation: Viscosity η = F d / (A v), where F is force, d is separation, A is contact area, v is velocity. Convert units consistently. Conceptually, viscosity quantifies internal resistance of Fluid to shear. Analogy: sliding a plate over honey requires more force than over water due to higher viscosity.