When a thin rod of length l is heated from t-1C to t-2 C length increases by 1

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Poisson's Effect

When a thin r...

Question

When a thin rod of length l is heated from $t_{1}^{\circ} C$ to $t_{2}^{\circ} C$ length increases by 1%. If plate of length 2 l and breadth l made of same material is heated form $t_{1}^{\circ} C$ to $t_{2}^{\circ} C$ , percentage increase in area is

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Solution

The correct option is D 2
We know, $l_{2} = l_{1} (1 + α d t)$ where dt = $t_{2} - t_{1}$ degree centigrade
Or, $(l_{2} - l_{1}) / l_{1} = α d t$
Percentage change in length = $[(l_{2} - l_{1}) / l_{1}]$ $\times 100 =$ $α$ $d t 100 = 1$
Or, $α$ $= 1 / (100 d t)$
Or $β$ $= 2 / (100 d t)$
Percentage change of area will be,
$[(A_{2} - A_{1}) / A_{1}] \times 100 = β$ $d t 100$
$= 2$ (putting value of $β$ )

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When a thin rod of length l is heated from t∘1C to t∘2C length increases by 1%. If plate of length 2 l and breadth l made of same material is heated form t∘1C to t∘2C , percentage increase in area is

The correct option is D 2We know, l2=l1(1+αdt) where dt = t2−t1 degree centigradeOr, (l2−l1)/l1=αdtPercentage change in length = [(l2−l1)/l1] ×100= α dt100=1Or, α =1/(100dt)Or β =2/(100dt)Percentage change of area will be,[(A2−A1)/A1]×100=βdt100 =2 (putting value of β)

When a thin rod of length l is heated from $t_{1}^{\circ} C$ to $t_{2}^{\circ} C$ length increases by 1%. If plate of length 2 l and breadth l made of same material is heated form $t_{1}^{\circ} C$ to $t_{2}^{\circ} C$ , percentage increase in area is