The length of a metallic rod is 5 m at 0° C and becomes 5.01 m, on heating upto 100° C. The linear expansion of the metal will be

Question

The length of a metallic rod is 5 m at 0° C and becomes 5.01 m, on heating upto 100° C. The linear expansion of the metal will be (A) 2.33 × 10-5 / ° C (B) 6.0 × 10-5 / ° C (C) 4.0 × 10-5 / ° C (D) 2.0 × 10-5 / ° C

Tardigrade Admin · Accepted Answer

ℓ=5 m t 1=0° C ℓ2=5.01 m t 2=100° C α=(ℓ2-ℓ1/ℓ1( t 2- t 1))=(5.01-5/5 × 100) =2 × 10-5 / ° C

Q. The length of a metallic rod is $5 m$ at $0^{\circ} C$ and becomes $5.01 m$ , on heating upto $10 0^{\circ} C$ . The linear expansion of the metal will be

$2.33 \times 1 0^{- 5} /^{\circ} C$

$6.0 \times 1 0^{- 5} /^{\circ} C$

$4.0 \times 1 0^{- 5} /^{\circ} C$

$2.0 \times 1 0^{- 5} /^{\circ} C$

$ℓ = 5 m$
$t_{1} = 0^{\circ} C$
$ℓ_{2} = 5.01 m$
$t_{2} = 10 0^{\circ} C$
$α = \frac{ℓ _{2} - ℓ _{1}}{ℓ _{1} ( t _{2} - t _{1} )} = \frac{5.01 - 5}{5 \times 100}$
$= 2 \times 1 0^{- 5} /^{\circ} C$

Q. The length of a metallic rod is 5m at 0∘C and becomes 5.01m, on heating upto 100∘C. The linear expansion of the metal will be

2.33×10−5/∘C

6.0×10−5/∘C

4.0×10−5/∘C

2.0×10−5/∘C