The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half of its amplitude, the distance being measured from its equilibrium position will be
🥳 Wohoo! Correct answer
E = K.E. + P.E. = constant
K.E. = ½k(A² - x²), P.E. = ½kx²
At x = A/2, substitute
K.E.:P.E. = 3:1
Wrong energy equations
😢 Uh oh! Incorrect answer, Try again
Use energy conservation
E = K.E. + P.E. = constant
K.E. = ½k(A² - x²), P.E. = ½kx²
At x = A/2, substitute
K.E.:P.E. = 3:1
Wrong energy equations
When two tuning forks A and B are sounded together, 4 beats per second are heard. The frequency of fork B is 384 Hz. When one of the prongs of the fork A is filed and sounded with B, the beat frequency increases, then the frequency of the fork A is
🥳 Wohoo! Correct answer
νbeat = ν₁ - ν₂
Beat frequency = ν₁ - ν₂ = 4 Hz
Initial νA - 384 = 4, so νA = 388 Hz
After filing frequency reduces to 378 Hz
Not considering frequency decrease
😢 Uh oh! Incorrect answer, Try again
Beats occur due to frequency difference
νbeat = ν₁ - ν₂
Beat frequency = ν₁ - ν₂ = 4 Hz
Initial νA - 384 = 4, so νA = 388 Hz
After filing frequency reduces to 378 Hz
Not considering frequency decrease
Water is heated from 0°C to 10°C then its volume
🥳 Wohoo! Correct answer
Density max at 4°C
Water shows anomalous behavior between 0-4°C
Volume decreases from 0°C to 4°C
Volume increases from 4°C to 10°C
Assuming regular expansion
😢 Uh oh! Incorrect answer, Try again
Consider anomalous expansion
Density max at 4°C
Water shows anomalous behavior between 0-4°C
Volume decreases from 0°C to 4°C
Volume increases from 4°C to 10°C
Assuming regular expansion
1 gram of ice is mixed with 1 gram of steam. At thermal equilibrium, the temperature of the mixture is
🥳 Wohoo! Correct answer
Lf = 80 cal/g, Lv = 540 cal/g
Heat from steam = mLv + mL = 1×540 + 1×100 = 640 cal
Heat needed for ice = mLf + mL = 1×80 + 1×100 = 180 cal
Excess heat keeps temperature at 100°C
Not considering all phase changes
😢 Uh oh! Incorrect answer, Try again
Compare latent and specific heats
Lf = 80 cal/g, Lv = 540 cal/g
Heat from steam = mLv + mL = 1×540 + 1×100 = 640 cal
Heat needed for ice = mLf + mL = 1×80 + 1×100 = 180 cal
Excess heat keeps temperature at 100°C
Not considering all phase changes
The efficiency of a Carnot engine which operates between the two temperatures T₁ = 500 K and T₂ = 300 K is
🥳 Wohoo! Correct answer
η = 1 - T₂/T₁
η = 1 - T₂/T₁
η = 1 - 300/500
η = 0.4 = 40%
Using Celsius instead of Kelvin
😢 Uh oh! Incorrect answer, Try again
Use Kelvin temperatures
η = 1 - T₂/T₁
η = 1 - T₂/T₁
η = 1 - 300/500
η = 0.4 = 40%
Using Celsius instead of Kelvin
The ratio of hydraulic stress to the corresponding strain is known as
🥳 Wohoo! Correct answer
K = -p/(ΔV/V)
Hydraulic stress = Force/Area
Strain = Change in volume/Original volume
Bulk modulus = -(p×V)/(ΔV)
Confusing different moduli
😢 Uh oh! Incorrect answer, Try again
Identify elastic constants
K = -p/(ΔV/V)
Hydraulic stress = Force/Area
Strain = Change in volume/Original volume
Bulk modulus = -(p×V)/(ΔV)
Confusing different moduli
Four rods with different radii r and length l are used to connect to heat reservoirs at different temperatures. Which one will conduct most heat?
🥳 Wohoo! Correct answer
Q = KA(T₁-T₂)/l
Apply heat conduction equation: Q = [KAΔt]/l where A = πr²
Rate of heat flow dθ/dt ∝ [AΔT/l]
Compare ratios: max heat flow needs largest A and smallest l, so r=2cm, l=1/2m gives maximum
Students might overlook area dependency
😢 Uh oh! Incorrect answer, Try again
Look for largest radius and shortest length
Q = KA(T₁-T₂)/l
Apply heat conduction equation: Q = [KAΔt]/l where A = πr²
Rate of heat flow dθ/dt ∝ [AΔT/l]
Compare ratios: max heat flow needs largest A and smallest l, so r=2cm, l=1/2m gives maximum
Students might overlook area dependency
A Carnot engine working between 300 K and 400 K has 800 J of useful work. The amount of heat energy supplied to the engine from the source is
🥳 Wohoo! Correct answer
η = 1-(TL/TH) = W/QH
Apply Carnot efficiency formula: η = 1 - (TL/TH) = W/QH
Substitute values: (1-300/400) = 800/QH
Solve for QH: QH = 800/(0.25) = 3200 J
Students may confuse QH and QL
😢 Uh oh! Incorrect answer, Try again
Efficiency relates work and heat input
η = 1-(TL/TH) = W/QH
Apply Carnot efficiency formula: η = 1 - (TL/TH) = W/QH
Substitute values: (1-300/400) = 800/QH
Solve for QH: QH = 800/(0.25) = 3200 J
Students may confuse QH and QL
A pan filled with hot food from 94°C to 86°C in 2 minutes. When the room temperature is 20°C. How long will it cool from 74°C to 66°C?
🥳 Wohoo! Correct answer
dT/dt = -k(T-Ts)
Use Newton's law of cooling with approximation: dT/dt = -k/ms(T-Ts). For case 1: dt=2min, T₁=90°C avg, Ts=20°C
Calculate cooling constant: -k/ms = (8/2)×(1/[90-20]) = 0.05714
For case 2: T₂=70°C avg, solve (8/dt) = 0.05714×(70-20), dt = 2.8 min
Students often assume equal time intervals
😢 Uh oh! Incorrect answer, Try again
Rate of cooling depends on temperature difference
dT/dt = -k(T-Ts)
Use Newton's law of cooling with approximation: dT/dt = -k/ms(T-Ts). For case 1: dt=2min, T₁=90°C avg, Ts=20°C
Calculate cooling constant: -k/ms = (8/2)×(1/[90-20]) = 0.05714
For case 2: T₂=70°C avg, solve (8/dt) = 0.05714×(70-20), dt = 2.8 min
Students often assume equal time intervals
A spring is stretched by applying a load to its free end. The strain produced in the spring is
🥳 Wohoo! Correct answer
ε = ΔL/L, γ = tanθ
Analyze deformation: Spring extension causes change in length (longitudinal strain)
Spring coils undergo twisting motion during extension (shear strain)
Both types of strain occur simultaneously in a spring
Students often consider only longitudinal strain
😢 Uh oh! Incorrect answer, Try again
Consider both linear extension and angular deformation
ε = ΔL/L, γ = tanθ
Analyze deformation: Spring extension causes change in length (longitudinal strain)
Spring coils undergo twisting motion during extension (shear strain)
Both types of strain occur simultaneously in a spring
Students often consider only longitudinal strain
The waves set up in a closed pipe are
🥳 Wohoo! Correct answer
λ = 4L/(2n-1) for closed pipes
Consider type of wave in air - compression and rarefaction
These are longitudinal waves
Due to reflection at closed ends, standing waves form
Confusing with open pipe conditions
😢 Uh oh! Incorrect answer, Try again
Closed pipes have specific resonant frequencies
λ = 4L/(2n-1) for closed pipes
Consider type of wave in air - compression and rarefaction
These are longitudinal waves
Due to reflection at closed ends, standing waves form
Confusing with open pipe conditions
Two simple pendulums A and B are made to oscillate simultaneously and it is found that A completes 10 oscillations in 20 sec and B completes 8 oscillations in 10 sec. The ratio of the lengths of A and B is
🥳 Wohoo! Correct answer
T = 2π√(l/g)
Calculate time periods: TA = 20/10 = 2s, TB = 10/8 = 1.25s
Use formula T ∝ √l and square the ratio: (TA/TB)² = lA/lB
(2/1.25)² = 64/25
Not squaring the time period ratio
😢 Uh oh! Incorrect answer, Try again
Time period ratio squared gives length ratio
T = 2π√(l/g)
Calculate time periods: TA = 20/10 = 2s, TB = 10/8 = 1.25s
Use formula T ∝ √l and square the ratio: (TA/TB)² = lA/lB
(2/1.25)² = 64/25
Not squaring the time period ratio
The mean energy of a molecule of an ideal gas is
🥳 Wohoo! Correct answer
E = (3/2)kT
For ideal gas, energy is purely kinetic
Mean kinetic energy per molecule = (3/2)kT where k is Boltzmann constant
This comes from equipartition of energy principle
Forgetting factor of 3/2
😢 Uh oh! Incorrect answer, Try again
Remember factor 3/2 comes from three degrees of freedom
E = (3/2)kT
For ideal gas, energy is purely kinetic
Mean kinetic energy per molecule = (3/2)kT where k is Boltzmann constant
This comes from equipartition of energy principle
Forgetting factor of 3/2
For which combination of working temperatures, the efficiency of 'Carnot's engine' is the least?
🥳 Wohoo! Correct answer
η = 1 - T₁/T₂
Use Carnot efficiency formula: η = 1 - Tc/Th
Calculate efficiency for each case: (A) 1-40/60=0.33, (B) 1-20/40=0.5, (C) 1-60/80=0.25, (D) 1-80/100=0.2
Compare values: 0.33, 0.5, 0.25, 0.2; lowest is 0.2
Not converting temperatures to Kelvin
😢 Uh oh! Incorrect answer, Try again
Efficiency depends on ratio of temperatures, not difference
η = 1 - T₁/T₂
Use Carnot efficiency formula: η = 1 - Tc/Th
Calculate efficiency for each case: (A) 1-40/60=0.33, (B) 1-20/40=0.5, (C) 1-60/80=0.25, (D) 1-80/100=0.2
Compare values: 0.33, 0.5, 0.25, 0.2; lowest is 0.2
Not converting temperatures to Kelvin
The S.I. unit of specific heat capacity is
🥳 Wohoo! Correct answer
c = Q/(m × ΔT)
Analyze dimensions in specific heat capacity formula
Specific heat = heat energy/mass × temperature change
Units combine as Joules/(kg × Kelvin) = J kg⁻¹ K⁻¹
Confusing with molar specific heat units
😢 Uh oh! Incorrect answer, Try again
Remember specific heat is energy per unit mass per unit temperature change
c = Q/(m × ΔT)
Analyze dimensions in specific heat capacity formula
Specific heat = heat energy/mass × temperature change
Units combine as Joules/(kg × Kelvin) = J kg⁻¹ K⁻¹
Confusing with molar specific heat units
Young's modulus is defined as the ratio of
🥳 Wohoo! Correct answer
E = (F/A)/(ΔL/L)
Young's modulus relates stress and strain in same direction
Only tensile stress and longitudinal strain give Young's modulus
Other ratios define different elastic moduli
Confusing different elastic moduli
😢 Uh oh! Incorrect answer, Try again
Remember stress and strain must be along same direction
E = (F/A)/(ΔL/L)
Young's modulus relates stress and strain in same direction
Only tensile stress and longitudinal strain give Young's modulus
Other ratios define different elastic moduli
Confusing different elastic moduli
Hydraulic lift' works on the basis of
🥳 Wohoo! Correct answer
P₁ = P₂ in hydraulic systems
Hydraulic lift uses pressure transmission in enclosed fluid
Pascal's law states pressure is transmitted undiminished throughout fluid
This allows force multiplication based on area ratio
Confusing different fluid laws
😢 Uh oh! Incorrect answer, Try again
Pressure remains same but forces differ due to area difference
P₁ = P₂ in hydraulic systems
Hydraulic lift uses pressure transmission in enclosed fluid
Pascal's law states pressure is transmitted undiminished throughout fluid
This allows force multiplication based on area ratio
Confusing different fluid laws
Two wires A and B are stretched by the same load. If the area of cross-section of wire 'A' is double that of 'B', then the stress on 'B' is
🥳 Wohoo! Correct answer
σ = F/A
Write stress formula: σ = F/A where F is force and A is area
Compare stresses: σ_B/σ_A = (F/A_B)/(F/A_A) = A_A/A_B
Since A_A = 2A_B, σ_B = 2σ_A
Confusing stress and strain
😢 Uh oh! Incorrect answer, Try again
Stress is inversely proportional to area for same force
σ = F/A
Write stress formula: σ = F/A where F is force and A is area
Compare stresses: σ_B/σ_A = (F/A_B)/(F/A_A) = A_A/A_B
Since A_A = 2A_B, σ_B = 2σ_A
Confusing stress and strain
Pressure of an ideal gas is increased by keeping the temperature constant. The kinetic energy of molecules
🥳 Wohoo! Correct answer
E = (3/2)kT
Apply ideal gas equation PV = nRT where T is constant
Note that kinetic energy of molecules = (3/2)kT where k is Boltzmann constant
Since T is constant, kinetic energy must remain constant regardless of pressure change
Confusing pressure with kinetic energy
😢 Uh oh! Incorrect answer, Try again
Temperature is direct measure of average kinetic energy
E = (3/2)kT
Apply ideal gas equation PV = nRT where T is constant
Note that kinetic energy of molecules = (3/2)kT where k is Boltzmann constant
Since T is constant, kinetic energy must remain constant regardless of pressure change
Confusing pressure with kinetic energy
First overtone frequency of a closed pipe of length 'l₁' is equal to the 2nd harmonic frequency of an open pipe of length 'l₂'. The ratio l₁/l₂ =
🥳 Wohoo! Correct answer
f₁ = 3v/4L₁, f₂ = 2v/2L₂
Write equation for closed pipe overtones
Write equation for open pipe harmonics
Equate and solve for ratio
Confusing overtones and harmonics
😢 Uh oh! Incorrect answer, Try again
Remember fundamental frequencies differ for open and closed pipes
f₁ = 3v/4L₁, f₂ = 2v/2L₂
Write equation for closed pipe overtones
Write equation for open pipe harmonics
Equate and solve for ratio
Confusing overtones and harmonics
A Carnot engine takes 300 calories of heat from a source at 500 K and rejects 150 calories of heat to the sink. The temperature of the sink is
🥳 Wohoo! Correct answer
Q₁/Q₂ = T₁/T₂
Start with the equation of Carnot efficiency: η = 1 - T₂/T₁. Given T₁ = 500K, Q₁ = 300 cal, Q₂ = 150 cal
Calculate efficiency using heat: η = (Q₁-Q₂)/Q₁ = (300-150)/300 = 0.5
Use both equations: 0.5 = 1 - T₂/500, solve for T₂ to get 250K
Confusing heat and temperature ratios
😢 Uh oh! Incorrect answer, Try again
For Carnot engine, ratio of temperatures equals ratio of heats
Q₁/Q₂ = T₁/T₂
Start with the equation of Carnot efficiency: η = 1 - T₂/T₁. Given T₁ = 500K, Q₁ = 300 cal, Q₂ = 150 cal
Calculate efficiency using heat: η = (Q₁-Q₂)/Q₁ = (300-150)/300 = 0.5
Use both equations: 0.5 = 1 - T₂/500, solve for T₂ to get 250K
Confusing heat and temperature ratios
The equation of a stationary wave is y = 2sin(πx/15)cos(48πt). The distance between a node and its next antinode is
🥳 Wohoo! Correct answer
λ = 2L where L = 15
Wavelength λ = 2(15) = 30 units
Distance between node and antinode = λ/4
Therefore distance = 30/4 = 7.5 units
Not identifying wavelength from equation
😢 Uh oh! Incorrect answer, Try again
Consider wave equation form
λ = 2L where L = 15
Wavelength λ = 2(15) = 30 units
Distance between node and antinode = λ/4
Therefore distance = 30/4 = 7.5 units
Not identifying wavelength from equation
A cup of tea cools from 65.5°C to 62.5°C in 1 minute in a room at 22.5°C. How long will it take to cool from 46.5°C to 40.5°C in the same room?
🥳 Wohoo! Correct answer
dT/dt = -k(T-T₀)
Apply Newton's Law of Cooling
Calculate temperature differences ratio
Solve proportional equation
Not considering proportionality constant
😢 Uh oh! Incorrect answer, Try again
Rate of cooling proportional to temperature difference with surroundings
dT/dt = -k(T-T₀)
Apply Newton's Law of Cooling
Calculate temperature differences ratio
Solve proportional equation
Not considering proportionality constant
A piston is performing S.H.M. in the vertical direction with a frequency of 0.5Hz. A block of 10kg is placed on the piston. The maximum amplitude of the system such that the block remains in contact with the piston is
🥳 Wohoo! Correct answer
a = ω²A ≤ g
For contact, acceleration ≤ g at all points
Max acceleration = ω²A = (2πf)²A
Solve (2π×0.5)²A = g gives A = 1m
Not considering acceleration limit
😢 Uh oh! Incorrect answer, Try again
Consider acceleration condition
a = ω²A ≤ g
For contact, acceleration ≤ g at all points
Max acceleration = ω²A = (2πf)²A
Solve (2π×0.5)²A = g gives A = 1m
Not considering acceleration limit
One mole of O₂ gas is heated at constant pressure starting at 27°C. How much energy must be added to the gas as heat to double its volume?
🥳 Wohoo! Correct answer
ΔQ = nCpΔT
At constant P: V₁/T₁ = V₂/T₂
For V₂ = 2V₁, T₂ = 2T₁ = 600K
Q = nCp(T₂-T₁) = (7/2)R×300 = 1050R
Not using correct heat capacity
😢 Uh oh! Incorrect answer, Try again
Use gas laws and heat capacity
ΔQ = nCpΔT
At constant P: V₁/T₁ = V₂/T₂
For V₂ = 2V₁, T₂ = 2T₁ = 600K
Q = nCp(T₂-T₁) = (7/2)R×300 = 1050R
Not using correct heat capacity
Due to Doppler's effect the shift in wavelength observed is 0.1Å for a star producing wavelength 6000Å. Velocity of recession of the star will be
🥳 Wohoo! Correct answer
v = c(Δλ/λ)
Use Δλ/λ = v/c
Substitute values and solve for v
Convert units to km/s
Unit conversion errors
😢 Uh oh! Incorrect answer, Try again
Use Doppler shift equation
v = c(Δλ/λ)
Use Δλ/λ = v/c
Substitute values and solve for v
Convert units to km/s
Unit conversion errors
An aluminium sphere is dipped into water. Which of the following is true?
🥳 Wohoo! Correct answer
FB = ρgh
Buoyancy force = ρgh where ρ is fluid density
Water density maximum at 4°C
Therefore buoyancy maximum at 4°C
Forgetting water density variation
😢 Uh oh! Incorrect answer, Try again
Consider water density anomaly
FB = ρgh
Buoyancy force = ρgh where ρ is fluid density
Water density maximum at 4°C
Therefore buoyancy maximum at 4°C
Forgetting water density variation
A wire is stretched such that its volume remains constant. The poission's ratio of the material of the wire is
🥳 Wohoo! Correct answer
μ = -(lateral strain)/(longitudinal strain)
For constant volume, ΔV/V = 0
ΔV/V = ΔL/L + 2(ΔR/R) = 0
Poisson's ratio = -(ΔR/R)/(ΔL/L) = 0.5
Forgetting volume conservation condition
😢 Uh oh! Incorrect answer, Try again
Use volume conservation equation for cylinder
μ = -(lateral strain)/(longitudinal strain)
For constant volume, ΔV/V = 0
ΔV/V = ΔL/L + 2(ΔR/R) = 0
Poisson's ratio = -(ΔR/R)/(ΔL/L) = 0.5
Forgetting volume conservation condition
A sphere, cube and thin circular plate of same material and mass initially at same temperature cool down. Which statement is true?
🥳 Wohoo! Correct answer
E ∝ A
Consider surface area to volume ratio
Plate has highest ratio, sphere lowest
Rate of cooling ∝ surface area
Ignoring surface area ratio
😢 Uh oh! Incorrect answer, Try again
Higher surface area means faster cooling
E ∝ A
Consider surface area to volume ratio
Plate has highest ratio, sphere lowest
Rate of cooling ∝ surface area
Ignoring surface area ratio
Heat energy supplied to monoatomic ideal gas expanding at constant pressure. Fraction converted to work?
🥳 Wohoo! Correct answer
γ = Cp/Cv = 5/3
For monoatomic gas, γ = 5/3
At constant P, work done = dW/dQ
dW/dQ = 1/(γ) = 2/5
Not using correct γ value
😢 Uh oh! Incorrect answer, Try again
Remember specific heat ratios
γ = Cp/Cv = 5/3
For monoatomic gas, γ = 5/3
At constant P, work done = dW/dQ
dW/dQ = 1/(γ) = 2/5
Not using correct γ value
In adiabatic expansion of ideal gas, product of pressure and volume
🥳 Wohoo! Correct answer
PVᵞ = constant
Use PVᵞ = constant
During expansion, V increases
Therefore P must decrease more
Confusing with isothermal
😢 Uh oh! Incorrect answer, Try again
Remember adiabatic process equations
PVᵞ = constant
Use PVᵞ = constant
During expansion, V increases
Therefore P must decrease more
Confusing with isothermal
Iceberg floats with part submerged. Fraction of volume submerged if ice density = 0.917 g/cm³?
🥳 Wohoo! Correct answer
ρᵢVᵢ = ρwVw
Use ρᵢVᵢ = ρwVw
Vw/Vᵢ = ρᵢ/ρw
0.917/1 = 0.917
Not using density ratio
😢 Uh oh! Incorrect answer, Try again
Archimedes principle application
ρᵢVᵢ = ρwVw
Use ρᵢVᵢ = ρwVw
Vw/Vᵢ = ρᵢ/ρw
0.917/1 = 0.917
Not using density ratio
Young's modulus of a perfect rigid body is
🥳 Wohoo! Correct answer
Y = stress/strain = F/A ÷ Δl/l
Consider definition: Y = stress/strain
For perfect rigid body, strain = 0
Y = F/A ÷ (Δl/l) = ∞ when Δl = 0
Confusing zero and infinity
😢 Uh oh! Incorrect answer, Try again
Perfect rigid body has no deformation
Y = stress/strain = F/A ÷ Δl/l
Consider definition: Y = stress/strain
For perfect rigid body, strain = 0
Y = F/A ÷ (Δl/l) = ∞ when Δl = 0
Confusing zero and infinity
To propagate both longitudinal and transverse waves, a material must have
🥳 Wohoo! Correct answer
v_l = √(B/ρ), v_t = √(G/ρ)
Let's understand what each wave needs: Longitudinal waves involve compression and expansion, requiring bulk modulus (B)
Transverse waves involve shearing motion, requiring shear modulus (G)
Therefore, material needs both moduli to support both types of waves. Option A is correct
Confusing moduli requirements
😢 Uh oh! Incorrect answer, Try again
Wave type depends on elastic properties
v_l = √(B/ρ), v_t = √(G/ρ)
Let's understand what each wave needs: Longitudinal waves involve compression and expansion, requiring bulk modulus (B)
Transverse waves involve shearing motion, requiring shear modulus (G)
Therefore, material needs both moduli to support both types of waves. Option A is correct
Confusing moduli requirements
A pendulum oscillates simple harmonically and only if [statements]
🥳 Wohoo! Correct answer
T = 2π√(L/g)
Statement i: Bob size must be small compared to length for point mass approximation
Statement ii: Small angle approximation (θ < 10°) needed for SHM as sinθ ≈ θ
Both conditions necessary for pendulum to exhibit SHM - making option A correct
Overlooking assumptions
😢 Uh oh! Incorrect answer, Try again
Small angle and point mass needed
T = 2π√(L/g)
Statement i: Bob size must be small compared to length for point mass approximation
Statement ii: Small angle approximation (θ < 10°) needed for SHM as sinθ ≈ θ
Both conditions necessary for pendulum to exhibit SHM - making option A correct
Overlooking assumptions
LC oscillations are similar and analogous to the mechanical oscillations of a block attached to a spring. The electrical equivalent of the force constant of the spring is
🥳 Wohoo! Correct answer
ω = √(k/m) = √(1/LC) ∴ k ∝ 1/C
Let's approach this by comparing the differential equations of both systems. For a spring-mass system, we have d²x/dt² + (k/m)x = 0, where k is the spring constant.
In an LC circuit, we have d²q/dt² + (1/LC)q = 0, where q is charge. Looking at these equations, we can see they have the same form. The term k/m in mechanical oscillation corresponds to 1/LC in electrical oscillation.
Therefore, k corresponds to 1/C (not C or XC). This makes sense because a stiffer spring (larger k) is like a smaller capacitance (harder to store charge). The frequency equations ω = √(k/m) and ω = √(1/LC) also confirm this.
Confusing capacitive reactance with capacitance
😢 Uh oh! Incorrect answer, Try again
Match the coefficients of similar terms in both differential equations
ω = √(k/m) = √(1/LC) ∴ k ∝ 1/C
Let's approach this by comparing the differential equations of both systems. For a spring-mass system, we have d²x/dt² + (k/m)x = 0, where k is the spring constant.
In an LC circuit, we have d²q/dt² + (1/LC)q = 0, where q is charge. Looking at these equations, we can see they have the same form. The term k/m in mechanical oscillation corresponds to 1/LC in electrical oscillation.
Therefore, k corresponds to 1/C (not C or XC). This makes sense because a stiffer spring (larger k) is like a smaller capacitance (harder to store charge). The frequency equations ω = √(k/m) and ω = √(1/LC) also confirm this.
Confusing capacitive reactance with capacitance
A gas mixture contains monoatomic and diatomic molecules of 2 moles each. Total internal energy is
🥳 Wohoo! Correct answer
U = (f/2)nRT
For monoatomic gas, U = (3/2)RT per mole. For 2 moles, U₁ = 3RT
For diatomic gas, U = (5/2)RT per mole. For 2 moles, U₂ = 5RT
Total U = U₁ + U₂ = 3RT + 5RT = 8RT
Forgetting mole numbers
😢 Uh oh! Incorrect answer, Try again
Add energies for different molecules
U = (f/2)nRT
For monoatomic gas, U = (3/2)RT per mole. For 2 moles, U₁ = 3RT
For diatomic gas, U = (5/2)RT per mole. For 2 moles, U₂ = 5RT
Total U = U₁ + U₂ = 3RT + 5RT = 8RT
Forgetting mole numbers
Two capillary tubes P and Q are dipped vertically in water. The height of water level in capillary tube P is 2rd/3 of the height in capillary tube Q. The ratio of their diameter is
🥳 Wohoo! Correct answer
h = 2Tcosθ/ρgr
Use capillary rise formula: h = 2Tcosθ/ρgr, where r is radius.
For same liquid, T, θ, ρ, g are constant. So h ∝ 1/r
If hₚ/hq = 2/3, then rₚ/rq = 3/2 (inverse relation). Since diameter = 2r, ratio is 3:2
Inverse relationship confusion
😢 Uh oh! Incorrect answer, Try again
Height inversely proportional to radius
h = 2Tcosθ/ρgr
Use capillary rise formula: h = 2Tcosθ/ρgr, where r is radius.
For same liquid, T, θ, ρ, g are constant. So h ∝ 1/r
If hₚ/hq = 2/3, then rₚ/rq = 3/2 (inverse relation). Since diameter = 2r, ratio is 3:2
Inverse relationship confusion
In an oscillating LC circuit, L=3.00mH and C=2.70µF. At t=0 the charge on the capacitor is zero and the current is 2.00A. The maximum charge that will appear on the capacitor will be
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Energy equations: EL = ½LI², EC = ½Q²/C
First, let's understand what's happening in the circuit. At t=0, all energy is in the form of magnetic energy in the inductor (since charge is zero). We can find this energy using ½LI².
As the oscillation proceeds, this energy will completely transfer to electrical energy in the capacitor at some point. At this moment, current will be zero and charge will be maximum. Using energy conservation: ½LI² = ½Q²/C
Substituting our values: ½(3×10⁻³)(2)² = ½Q²/(2.70×10⁻⁶). Solving for Q: Q = 2√(LC) = 2√(3×10⁻³×2.70×10⁻⁶) = 18×10⁻⁵C = 18×10⁵µC
Not considering energy conservation
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Energy oscillates between L and C, with total energy remaining constant
Energy equations: EL = ½LI², EC = ½Q²/C
First, let's understand what's happening in the circuit. At t=0, all energy is in the form of magnetic energy in the inductor (since charge is zero). We can find this energy using ½LI².
As the oscillation proceeds, this energy will completely transfer to electrical energy in the capacitor at some point. At this moment, current will be zero and charge will be maximum. Using energy conservation: ½LI² = ½Q²/C
Substituting our values: ½(3×10⁻³)(2)² = ½Q²/(2.70×10⁻⁶). Solving for Q: Q = 2√(LC) = 2√(3×10⁻³×2.70×10⁻⁶) = 18×10⁻⁵C = 18×10⁵µC
Not considering energy conservation
The displacement x=3sin(2πt+π/4) where x in meters, t in seconds. Amplitude and maximum speed:
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v = Aω
Let's break down this SHM equation. The amplitude is the coefficient of sine, so A=3m. For velocity, we need to differentiate x with respect to t.
When we differentiate, we get v=-3(2π)cos(2πt+π/4). The maximum speed occurs when cosine = ±1.
So, maximum speed = 3×2π = 6π m/s. Notice how ω=2π rad/s appears in our calculation - this is the angular frequency!
Forgetting chain rule
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Remember: amplitude is the coefficient of sine, and velocity is dx/dt
v = Aω
Let's break down this SHM equation. The amplitude is the coefficient of sine, so A=3m. For velocity, we need to differentiate x with respect to t.
When we differentiate, we get v=-3(2π)cos(2πt+π/4). The maximum speed occurs when cosine = ±1.
So, maximum speed = 3×2π = 6π m/s. Notice how ω=2π rad/s appears in our calculation - this is the angular frequency!
Forgetting chain rule
A metallic rod breaks when strain produced is 0.2%. The Young's modulus is 7×10⁹ N/m². Area of cross-section to support 4×10⁴ N is:
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Y = F/(A×ε) where ε is strain
Let's break this down step by step. We know strain = 0.2% = 0.002 (always convert percentage to decimal!). Young's modulus (Y) relates stress and strain: Y = stress/strain. Since we know Y and strain, we can find stress.
Now, stress is force per unit area (F/A). We can rearrange our equation: Y = (F/A)/strain. Therefore, F/A = Y × strain. We know F = 4×10⁴ N, Y = 7×10⁹ N/m², strain = 0.002
Solving for A: A = F/(Y × strain) = (4×10⁴)/(7×10⁹ × 0.002) = 7.1×10⁻⁴ m². Always check your units match!
Forgetting to convert % to decimal
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First find stress from strain and Young's modulus, then use force to find area
Y = F/(A×ε) where ε is strain
Let's break this down step by step. We know strain = 0.2% = 0.002 (always convert percentage to decimal!). Young's modulus (Y) relates stress and strain: Y = stress/strain. Since we know Y and strain, we can find stress.
Now, stress is force per unit area (F/A). We can rearrange our equation: Y = (F/A)/strain. Therefore, F/A = Y × strain. We know F = 4×10⁴ N, Y = 7×10⁹ N/m², strain = 0.002
Solving for A: A = F/(Y × strain) = (4×10⁴)/(7×10⁹ × 0.002) = 7.1×10⁻⁴ m². Always check your units match!
Forgetting to convert % to decimal
"Heat cannot itself flow from lower to higher temperature" corresponds to:
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ΔS(universe) ≥ 0
Think about your everyday experience - have you ever seen a cold object spontaneously make a hot object hotter? Never! This is a fundamental observation about nature. Let's understand why this is related to the Second Law.
The First Law only talks about energy conservation, but doesn't specify direction. The Second Law introduces the concept of entropy and tells us about the natural direction of processes. It's like water flowing downhill - it happens spontaneously in one direction.
This statement is specifically the Clausius statement of the Second Law. It's equivalent to saying entropy always increases in natural processes. It's not about mass or momentum conservation - those are different laws entirely!
Confusing First and Second Laws
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Think about why your hot coffee cools down but a cold drink doesn't spontaneously get hot
ΔS(universe) ≥ 0
Think about your everyday experience - have you ever seen a cold object spontaneously make a hot object hotter? Never! This is a fundamental observation about nature. Let's understand why this is related to the Second Law.
The First Law only talks about energy conservation, but doesn't specify direction. The Second Law introduces the concept of entropy and tells us about the natural direction of processes. It's like water flowing downhill - it happens spontaneously in one direction.
This statement is specifically the Clausius statement of the Second Law. It's equivalent to saying entropy always increases in natural processes. It's not about mass or momentum conservation - those are different laws entirely!
Confusing First and Second Laws
The speed of sound in an ideal gas at a given temperature T is v. The rms speed of gas molecules at that temperature is vᵣₘₛ. The ratio of velocities v and vᵣₘₛ for helium and oxygen gases are X and X' respectively. Then X/X' is equal to
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v/vᵣₘₛ=√(γ/3)
Use v=√(γRT/M) and vᵣₘₛ=√(3RT/M)
Calculate ratio for each gas
Compare ratios
Gas properties
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Consider molecular mass
v/vᵣₘₛ=√(γ/3)
Use v=√(γRT/M) and vᵣₘₛ=√(3RT/M)
Calculate ratio for each gas
Compare ratios
Gas properties
Pressure of ideal gas at constant volume is proportional to
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P∝KE(avg)
Use ideal gas equation PV=nRT
P∝T at constant V
T∝KE(avg)
Energy relations
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Consider molecular motion
P∝KE(avg)
Use ideal gas equation PV=nRT
P∝T at constant V
T∝KE(avg)
Energy relations
A block of mass m is connected to a light spring of force constant k. The system is placed inside a damping medium of damping constant b. The instantaneous values of displacement, acceleration and energy of the block are x, a and E respectively. The initial amplitude of oscillation is A and ω is the angular frequency of oscillations. The incorrect expression related to the damped oscillations is
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ω'=√(k/m-b²/4m²)
Check all equations for damped SHM
x=Ae⁻ᵇᵗ/²ᵐcos(ω't+φ) is correct form
Given option has wrong damping term
Damping effect
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Consider damping factor
ω'=√(k/m-b²/4m²)
Check all equations for damped SHM
x=Ae⁻ᵇᵗ/²ᵐcos(ω't+φ) is correct form
Given option has wrong damping term
Damping effect
100 g of ice at 0°C is mixed with 100 g of water at 100°C. The final temperature of the [Take Lf=3.36×10⁵ J kg⁻¹ and Sw=4.2×10³ J kg⁻¹ k⁻¹]
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Q=mL+mcΔT
Apply heat equation mᵢLf+mᵢcΔT₁=mwcΔT₂
Substitute values and solve
Final temperature=10°C
Heat exchange
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Consider latent heat
Q=mL+mcΔT
Apply heat equation mᵢLf+mᵢcΔT₁=mwcΔT₂
Substitute values and solve
Final temperature=10°C
Heat exchange
A stretched wire of material whose Young's modulus Y=2×10¹¹Nm⁻² has Poisson's ratio 0.25. Its lateral strain εₗ=10⁻³. The elastic energy density of the wire is
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U=½Yε²
Use Poisson's ratio to find longitudinal strain
Calculate stress using Young's modulus
Energy density=½×stress×strain
Strain relations
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Consider strain energy
U=½Yε²
Use Poisson's ratio to find longitudinal strain
Calculate stress using Young's modulus
Energy density=½×stress×strain
Strain relations
A closed water tank has cross-sectional area A. It has a small hole at a depth of h from the free surface of water. The radius of the hole is r so that r≪√(A/π). If P₀ is the pressure inside the tank above water level, and Pₐ is the atmospheric pressure, the rate of flow of water coming out of the hole is [ρ is the density of water]
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Q=Av
Apply Bernoulli's equation
Consider pressure difference and height
Flow rate=Area×velocity
Pressure effects
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Consider continuity
Q=Av
Apply Bernoulli's equation
Consider pressure difference and height
Flow rate=Area×velocity
Pressure effects
A solid cube of mass m at temperature θ₀ heated at constant rate...
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Q=mcΔθ+mL
Consider heating in solid phase
Add latent heat of fusion
Add heating in liquid phase and vaporization
Phase transition
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Account all phase changes
Q=mcΔθ+mL
Consider heating in solid phase
Add latent heat of fusion
Add heating in liquid phase and vaporization
Phase transition
The ratio of molar specific heats of oxygen is
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γ = Cp/Cv = 1 + 2/f
For diatomic oxygen (O₂): First identify degrees of freedom: 3 translational (movement in x,y,z directions) + 2 rotational (rotation around two perpendicular axes) = 5 total degrees
Calculate γ using the formula γ = 1 + 2/f, where f=5. This comes from equipartition of energy theorem where each degree contributes ½kT to internal energy
Substitute f=5 into equation: γ = 1 + 2/5 = 1.4. This ratio represents Cp/Cv
Confusing with monatomic gases (f=3)
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Check molecular structure for degrees of freedom
γ = Cp/Cv = 1 + 2/f
For diatomic oxygen (O₂): First identify degrees of freedom: 3 translational (movement in x,y,z directions) + 2 rotational (rotation around two perpendicular axes) = 5 total degrees
Calculate γ using the formula γ = 1 + 2/f, where f=5. This comes from equipartition of energy theorem where each degree contributes ½kT to internal energy
Substitute f=5 into equation: γ = 1 + 2/5 = 1.4. This ratio represents Cp/Cv
Confusing with monatomic gases (f=3)
Water flows through horizontal pipe... rate 0.314m³s⁻¹... radius 10cm...
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Q=Av
Use continuity equation Q=Av
A=πr²=π(0.1)²=0.0314m²
v=Q/A=0.314/0.0314=10ms⁻¹
Area calculation
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Consider flow conservation
Q=Av
Use continuity equation Q=Av
A=πr²=π(0.1)²=0.0314m²
v=Q/A=0.314/0.0314=10ms⁻¹
Area calculation
For a particle executing simple harmonic motion (SHM), at its mean position
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v²=ω²(A²-y²), a=-ω²y
At mean position y=0: First understand that mean position is where displacement is zero. The oscillating particle passes through this point with maximum speed
Apply velocity equation v²=ω²(A²-y²). At mean position y=0, so v²=ω²A², giving maximum velocity
Apply acceleration equation a=-ω²y. At y=0, acceleration=0. This shows velocity is maximum while acceleration is zero
Confusing mean position with extreme positions
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Remember phase relationships in SHM
v²=ω²(A²-y²), a=-ω²y
At mean position y=0: First understand that mean position is where displacement is zero. The oscillating particle passes through this point with maximum speed
Apply velocity equation v²=ω²(A²-y²). At mean position y=0, so v²=ω²A², giving maximum velocity
Apply acceleration equation a=-ω²y. At y=0, acceleration=0. This shows velocity is maximum while acceleration is zero
Confusing mean position with extreme positions
Electromagnetic waves are incident normally on a perfectly reflecting surface...
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F=2IA/c
Consider both incident and reflected momentum
Total change in momentum is twice incident momentum
Force=2×momentum/time=2IA/c
Momentum transfer
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Consider momentum change
F=2IA/c
Consider both incident and reflected momentum
Total change in momentum is twice incident momentum
Force=2×momentum/time=2IA/c
Momentum transfer
A thick metal wire of density ρ and length 'L' hung from rigid support...
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ΔL=∫(F/AY)dl
Consider varying force along length
Integrate strain over length
Get extension=½ρgL²/Y
Force variation
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Consider variable force
ΔL=∫(F/AY)dl
Consider varying force along length
Integrate strain over length
Get extension=½ρgL²/Y
Force variation
A motor-cyclist moving towards a huge cliff with a speed of 18 kmh⁻¹, blows a horn of source frequency 325 Hz. If the speed of the sound in air is 330 ms⁻¹, the number of beats heard by him is
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Doppler effect and beat frequency formulas
Convert speed: 18 km/h = 5 m/s. Consider both approaching wave and reflected wave from cliff. Source is moving towards cliff, causing apparent frequency increase
Use Doppler effect formula for approaching source: f'=f[(v+vs)/(v-vs)]. Original wave frequency increases due to approach, reflected wave frequency further increases due to reflection
Calculate beat frequency by finding difference between incident and reflected frequencies: Δf=325[(5)/(330-5)]=5 Hz
Forgetting reflected wave contribution
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Consider both incident and reflected waves
Doppler effect and beat frequency formulas
Convert speed: 18 km/h = 5 m/s. Consider both approaching wave and reflected wave from cliff. Source is moving towards cliff, causing apparent frequency increase
Use Doppler effect formula for approaching source: f'=f[(v+vs)/(v-vs)]. Original wave frequency increases due to approach, reflected wave frequency further increases due to reflection
Calculate beat frequency by finding difference between incident and reflected frequencies: Δf=325[(5)/(330-5)]=5 Hz
Forgetting reflected wave contribution