Skip to content
Login Register

Practice question

MCQ

Question

A steel cube of side 20cm at 25∘C is heated to 125∘C. What is the increase in volume? (αl\=1.2×10−5K−1)

Options

Choose one · Correct answer highlighted

Explanation

Given: L0 = 20cm, V0 = 203 = 8000cm3, ΔT = 125−25 = 100∘C, αl = 1.2×10−5K−1. αv = 3αl = 3×1.2×10−5 = 3.6×10−5K−1. ΔV = V0αvΔT = 8000×3.6×10−5×100 = 28.8cm3.

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.

Related blogs

Guides and insights connected to this topic