A solid metallic cube having total surface area 24 m 2 is uniformly heated. If its temperature is increased by 10° C, calculate the increase in volume of the cube (. Given: .α=5.0 × 10-4 ° C -1)

Question

A solid metallic cube having total surface area 24 m 2 is uniformly heated. If its temperature is increased by 10° C, calculate the increase in volume of the cube (. Given: .α=5.0 × 10-4 ° C -1) (A) 2.4 × 106 cm 3 (B) 1.2 × 105 cm 3 (C) 6.0 × 104 cm 3 (D) 4.8 × 105 cm 3

Tardigrade Admin · Accepted Answer

Increase in volume Δ V =&gamma; V 0 Δ T &gamma;=3 α So Δ V =(3 α) V 0 Δ T Total surface area =6 a 2, where a is side length 24=6 a 2 a =2 m Volume V 0=(2)3=8 m 3 Δ V =(3 × 5 × 10-4)(8) × 10 =1.2 × 105 cm 3

Q. A solid metallic cube having total surface area $24 m^{2}$ is uniformly heated. If its temperature is increased by $1 0^{\circ} C$ , calculate the increase in volume of the cube $($ Given : $α = 5.0 \times 1 0^{- 4}^{\circ} C^{- 1})$

$2.4 \times 1 0^{6} c m^{3}$

$1.2 \times 1 0^{5} c m^{3}$

$6.0 \times 1 0^{4} c m^{3}$

$4.8 \times 1 0^{5} c m^{3}$

Increase in volume $Δ V = γ V_{0} Δ T$
$γ = 3 α$
So $Δ V = (3 α) V_{0} Δ T$
Total surface area $= 6 a^{2}$ ,
where $a$ is side length $24 = 6 a^{2} a = 2 m$
Volume $V_{0} = (2)^{3} = 8 m^{3}$
$Δ V = (3 \times 5 \times 1 0^{- 4}) (8) \times 10$
$= 1.2 \times 1 0^{5} c m^{3}$

Q. A solid metallic cube having total surface area 24m2 is uniformly heated. If its temperature is increased by 10∘C, calculate the increase in volume of the cube ( Given : α=5.0×10−4∘C−1)

2.4×106cm3

1.2×105cm3

6.0×104cm3

4.8×105cm3