Answer 1

Given:
Initial pressure on the steel ball = 1.0 × 10⁵ Pa 
The ball is heated from 20 °C to 120 °C.
So, change in temperature, Δθ = 100°C.

Coefficient of linear expansion of steel, α = 12 ×10^-6°C^-1

Bulk modulus of steel ,B = 1.6 × 10¹¹ Nm^–2

Pressure is given as,

⇒ P = B × γ Δθ

⇒ P =B × 3 αΔθ  (∵ γ = 3 α )

⇒ P = 1.6 × 10¹¹ ×3 × 12 × 10^-6 ×(120-20)

=1.6 ×  3 × 12 ×10¹¹ ×10^-6 ×10²

= 57.6 × 10⁷ 

⇒ P =5.8 × 10⁸ Pa

Therefore, the pressure inside the ball is 5.8 × 10⁸ Pa.