The coefficient of linear expansion of brass and steel are α1 and α2. If we take a brass rod of length l1 and steel rod of length l2 at 0° c , their difference in length (l2-l1) will remain the same at a temperature if.

Question

The coefficient of linear expansion of brass and steel are α1 and α2. If we take a brass rod of length l1 and steel rod of length l2 at 0° c , their difference in length (l2-l1) will remain the same at a temperature if. (A) α1 l1=α2 l2 (B) α1 l12=α2 l22 (C) α12 l1=α22 l2 (D) α1 l2=α2 l1

Tardigrade Admin · Accepted Answer

L2=ℓ2(1+α2 Δ θ) ldots(1) L1=ℓ1(1+α1 Δ θ) ldots (2) Subtract eq (2) by (1), we get (L2-L1)=(ℓ2-ℓ1)+Δ θ(ℓ2 α2-ℓ1 α1) Δ θ(ℓ2 α2-ℓ1 α1)=0 α2 ℓ2=α1 ℓ1

Q. The coefficient of linear expansion of brass and steel are $α_{1}$ and $α_{2}$ . If we take a brass rod of length $l_{1}$ and steel rod of length $l_{2}$ at $0^{\circ} c,$ their difference in length $(l_{2} - l_{1})$ will remain the same at a temperature if.

$α_{1} l_{1} = α_{2} l_{2}$

$α_{1} l_{1}^{2} = α_{2} l_{2}^{2}$

$α_{1}^{2} l_{1} = α_{2}^{2} l_{2}$

$α_{1} l_{2} = α_{2} l_{1}$

$L_{2} = ℓ_{2} (1 + α_{2} Δ θ) \dots$ (1)
$L_{1} = ℓ_{1} (1 + α_{1} Δ θ) \dots$ (2)
Subtract eq (2) by $(1),$ we get
$(L_{2} - L_{1}) = (ℓ_{2} - ℓ_{1}) + Δ θ (ℓ_{2} α_{2} - ℓ_{1} α_{1})$
$Δ θ (ℓ_{2} α_{2} - ℓ_{1} α_{1}) = 0$
$α_{2} ℓ_{2} = α_{1} ℓ_{1}$

Q. The coefficient of linear expansion of brass and steel are α1​ and α2​. If we take a brass rod of length l1​ and steel rod of length l2​ at 0∘c, their difference in length (l2​−l1​) will remain the same at a temperature if.

α1​l1​=α2​l2​

α1​l12​=α2​l22​

α12​l1​=α22​l2​

α1​l2​=α2​l1​