The diameter of a metal ring is D and the coefficient of linear expansion is α. If the temperature of the ring is increased by 10C, the circumference and the area of the ring will increase by

Question

The diameter of a metal ring is D and the coefficient of linear expansion is α. If the temperature of the ring is increased by 10C, the circumference and the area of the ring will increase by (A) π D α ,2π Dα (B) 2π D α ,2π D2α (C) π Dα,(π Dα/2) (D) π Dα,(π D2α/2)

Tardigrade Admin · Accepted Answer

Increase in Diameter, Δ D = D α So, increase in circumference =π Δ D =π D α Also, we note that coefficient of area expansion =2 α So change in area = initial area × 2 × α=(π D 2/4) × 2 × α=(π D 2 α/2)

Q. The diameter of a metal ring is $D$ and the coefficient of linear expansion is $α$ . If the temperature of the ring is increased by $1^{0} C$ , the circumference and the area of the ring will increase by

$π D α, 2 π D α$

$2 π D α, 2 π D^{2} α$

$π D α, \frac{π D α}{2}$

$π D α, \frac{π D ^{2} α}{2}$

Increase in Diameter, $Δ D = D α$
So, increase in circumference $= π Δ D = π D α$
Also, we note that coefficient of area expansion $= 2 α$
So change in area $=$ initial area $\times 2 \times α = \frac{π D ^{2}}{4} \times 2 \times α = \frac{π D ^{2} α}{2}$

Q. The diameter of a metal ring is D and the coefficient of linear expansion is α. If the temperature of the ring is increased by 10C, the circumference and the area of the ring will increase by

πDα,2πDα

2πDα,2πD2α

πDα,2πDα​

πDα,2πD2α​