K.E. = ½k(A² - x²), P.E. = ½kx²

At x = A/2, substitute

K.E.:P.E. = 3:1

Beat frequency = ν₁ - ν₂ = 4 Hz

Initial νA - 384 = 4, so νA = 388 Hz

After filing frequency reduces to 378 Hz

Water shows anomalous behavior between 0-4°C

Volume decreases from 0°C to 4°C

Volume increases from 4°C to 10°C

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

η = 1 - T₂/T₁

η = 1 - 300/500

η = 0.4 = 40%

Hydraulic stress = Force/Area

Strain = Change in volume/Original volume

Bulk modulus = -(p×V)/(ΔV)

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

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

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

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

Consider type of wave in air - compression and rarefaction

These are longitudinal waves

Due to reflection at closed ends, standing waves form

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

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

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

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

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

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

Write equation for closed pipe overtones

Write equation for open pipe harmonics

Equate and solve for ratio

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

Wavelength λ = 2(15) = 30 units

Distance between node and antinode = λ/4

Therefore distance = 30/4 = 7.5 units

Apply Newton's Law of Cooling

Calculate temperature differences ratio

Solve proportional equation

For contact, acceleration ≤ g at all points

Max acceleration = ω²A = (2πf)²A

Solve (2π×0.5)²A = g gives A = 1m

At constant P: V₁/T₁ = V₂/T₂

For V₂ = 2V₁, T₂ = 2T₁ = 600K

Q = nCp(T₂-T₁) = (7/2)R×300 = 1050R

Use Δλ/λ = v/c

Substitute values and solve for v

Convert units to km/s

Buoyancy force = ρgh where ρ is fluid density

Water density maximum at 4°C

Therefore buoyancy maximum at 4°C

For constant volume, ΔV/V = 0

ΔV/V = ΔL/L + 2(ΔR/R) = 0

Poisson's ratio = -(ΔR/R)/(ΔL/L) = 0.5

Consider surface area to volume ratio

Plate has highest ratio, sphere lowest

Rate of cooling ∝ surface area

For monoatomic gas, γ = 5/3

At constant P, work done = dW/dQ

dW/dQ = 1/(γ) = 2/5

Use PVᵞ = constant

During expansion, V increases

Therefore P must decrease more

Use ρᵢVᵢ = ρwVw

Vw/Vᵢ = ρᵢ/ρw

0.917/1 = 0.917

Consider definition: Y = stress/strain

For perfect rigid body, strain = 0

Y = F/A ÷ (Δl/l) = ∞ when Δl = 0

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

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

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.

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

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

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

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!

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!

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!

Check all equations for damped SHM

x=Ae⁻ᵇᵗ/²ᵐcos(ω't+φ) is correct form

Given option has wrong damping term

Apply heat equation mᵢLf+mᵢcΔT₁=mwcΔT₂

Substitute values and solve

Final temperature=10°C

Use Poisson's ratio to find longitudinal strain

Calculate stress using Young's modulus

Energy density=½×stress×strain

Apply Bernoulli's equation

Consider pressure difference and height

Flow rate=Area×velocity

Consider heating in solid phase

Add latent heat of fusion

Add heating in liquid phase and vaporization

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

Use continuity equation Q=Av

A=πr²=π(0.1)²=0.0314m²

v=Q/A=0.314/0.0314=10ms⁻¹

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

Consider both incident and reflected momentum

Total change in momentum is twice incident momentum

Force=2×momentum/time=2IA/c

Consider varying force along length

Integrate strain over length

Get extension=½ρgL²/Y

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