The coefficient of linear expansion of brass and steel are -1 and -2- If we take a brass rod of length l 1 and steel rod of length l 2 at 0- C- their difference in length l 2- l 1 will remain the same at a temperature ifA- -1 l2-2 l1B- -1 l1-2 l2C- -1 l22-2 l12D- -12 l1-22 l2

The correct option is B d) L_2=l_2(l+α_2θ) and L_1= l_1(l+α_1θ) ⇒ (L_2 L_1)=(l_2 l_1)+θ(l_2α_2 l_1α_1) Now (L_2 L1)=(l_2 l_1) so, l_2α_2 l_1α_1=0 ...

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Standard XII

Physics

Thermal Expansion

The coefficie...

Question

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 $l_{2} a t 0^{\circ} C,$ their difference in length $(l_{2} - l_{1})$ will remain the same at a temperature if

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Solution

The correct option is D
$d) L_{2} = l_{2} (l + α_{2} △ θ)$ and $L_{1} = l_{1} (l + α_{1} △ θ)$
$\Rightarrow (L_{2} - L_{1}) = (l_{2} - l_{1}) + △ θ (l_{2} α_{2} - l_{1} α_{1})$
Now $(L_{2} - L 1) = (l_{2} - l_{1})$ so, $l_{2} α_{2} - l_{1} α_{1} = 0$

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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

The correct option is D d) L2=l2(l+α2△θ) and L1=l1(l+α1△θ) ⇒ (L2−L1)=(l2−l1)+△θ(l2α2−l1α1) Now (L2−L1)=(l2−l1) so, l2α2−l1α1=0

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 $l_{2} a t 0^{\circ} C,$ their difference in length $(l_{2} - l_{1})$ will remain the same at a temperature if

The correct option is D
$d) L_{2} = l_{2} (l + α_{2} △ θ)$ and $L_{1} = l_{1} (l + α_{1} △ θ)$
$\Rightarrow (L_{2} - L_{1}) = (l_{2} - l_{1}) + △ θ (l_{2} α_{2} - l_{1} α_{1})$
Now $(L_{2} - L 1) = (l_{2} - l_{1})$ so, $l_{2} α_{2} - l_{1} α_{1} = 0$