The coefficient of linear expansion for a certain metal varies with temperature as α( T ). If L 0 is the initial length of the metal and the temperature of metal is changed from T 0 to T ( T 0 T ), then

Question

The coefficient of linear expansion for a certain metal varies with temperature as α( T ). If L 0 is the initial length of the metal and the temperature of metal is changed from T 0 to T ( T 0> T ), then (A) L=L0 ∫ limitsT0T α(T) d T (B) L=L0[1+∫ limitsT0T α(T) d T] (C) L=L0[1-∫ limitsT0T α(T) d T] (D) L >L0

Tardigrade Admin · Accepted Answer

As , ( dL / L 0)=α( T ) dT ∫ limits L 0 L dL = L 0 ∫ limits T 0 T α( T ) dT L - L 0= L 0 ∫ limits T 0 T α( T ) dT L = L 0[1+∫ limits T 0 T α( T ) dT ]

Q. The coefficient of linear expansion for a certain metal varies with temperature as $α (T)$ . If $L_{0}$ is the initial length of the metal and the temperature of metal is changed from $T_{0}$ to $T$ $(T_{0} > T)$ , then

$L = L_{0} T_{0} \int T α (T) d T$

$L = L_{0} [1 + T_{0} \int T α (T) d T]$

$L = L_{0} [1 - T_{0} \int T α (T) d T]$

$L > L_{0}$

$A s, \frac{d L}{L _{0}} = α (T) d T L_{0} \int L d L = L_{0} T_{0} \int T α (T) d T$
$L - L_{0} = L_{0} T_{0} \int T α (T) d T L = L_{0} [1 + T_{0} \int T α (T) d T]$

Q. The coefficient of linear expansion for a certain metal varies with temperature as α(T). If L0​ is the initial length of the metal and the temperature of metal is changed from T0​ to T (T0​>T), then

L=L0​T0​∫T​α(T)dT

L=L0​[1+T0​∫T​α(T)dT]

L=L0​[1−T0​∫T​α(T)dT]

L>L0​