Explanation

For a string wave: v = √(T/μ) and from wave equation Y = A sin(kx + ωt), v = ω/k

1. Find wave speed from equation  

   Y = 0.021 sin(x + 30t) → compare with sin(kx + ωt)  

   So k = 1 rad/m, ω = 30 rad/s  

   v = ω/k = 30/1 = 30 m/s

2. Use v = √(T/μ) → T = μv²  

   μ = 1.3 × 10⁻³ kg/m  

   T = 1.3 × 10⁻³ × (30)² = 1.3 × 10⁻³ × 900  

   T = 1.17 N ≈ 1.20 N

Last verified on 05-06-2026

Explanation

For an ideal gas, average translational KE ∝ T. It doesn’t depend on the gas type, only on temperature.

KE_avg = (3/2)kT, so KE₂/KE₁ = T₂/T₁

Given:  

KE₁ = 6 × 10⁻²⁰ J at T₁ = 300 K  

T₂ = 750 K  

KE₂ = KE₁ × (T₂/T₁) = 6 × 10⁻²⁰ × (750/300)  

= 6 × 10⁻²⁰ × 2.5 = 1.5 × 10⁻¹⁹ J

Last verified on 05-06-2026

Explanation

Collision frequency = average speed / mean free path

ν = v / λ

Given:  

v = 600 m/s  

λ = 3 × 10⁻⁷ m  

ν = 600 / (3 × 10⁻⁷) = 200 × 10⁷ = 2 × 10⁹ s⁻¹

So each N₂ molecule bumps into another about 2 billion times every second at 27°C.

Last verified on 05-06-2026

Explanation

Work done = Change in rotational kinetic energy = ½Iω₂² - ½Iω₁²

1. Moment of inertia of disc I = ½mr²  

   m = 2 kg, diameter = 2m → radius r = 1m  

   I = ½ × 2 × 1² = 1 kg·m²

2. Convert rpm to rad/s  

   ω₁ = 300 rpm = 300 × 2π/60 = 10π rad/s  

   ω₂ = 600 rpm = 600 × 2π/60 = 20π rad/s  

3. ΔKE = ½(1)[(20π)² - (10π)²]  

   = 0.5 [400π² - 100π²]  

   = 0.5 × 300π² = 150π²  

   π² ≈ 9.87 → 150 × 9.87 ≈ 

1479 J

So you need ∼1479 J of work to double the rpm from 300 to 600.

Last verified on 05-06-2026

Explanation

• The unforced rising of hot air in the atmosphere is due to buoyancy.
• Hot air becomes less dense than surrounding cold air.
• The surrounding denser air pushes it upward.
• This upward force is called buoyant force.

Last verified on 05-06-2026