Question

A metal cube of length of 10mm
at o'cis heated to 200°C. Given
a = 2x10ʻK. The % change in it's
volume is:
0.1
0.2.
0.4
1.2.​

Answer 1

Answer:

Here percentage change in volume is given as, ΔV=γ×t×100

where, γ=3α

So, ΔV=3α×(T1 −T2)×100

ΔV=3×2×10-⁵ ×(473−273)×100=1.2

Answer 2

Answer:

Please mark it as Brainliest.

=> 1.2

Solution

Here percentage change in volume is given as,

=> ΔV= γ × t × 100

where,

=> γ=3α

So,

ΔV = 3α × (T1 −T2) × 100

ΔV = 3 × 2 × 10^−5 × (473−273) × 100

∆V = 1.2

This solution is given in NCERT I think