Question-74384

Question Number 74384 by jagannath19 last updated on 23/Nov/19

Commented by jagannath19 last updated on 23/Nov/19

p l e a s e e x p l a i n

Answered by Tanmay chaudhury last updated on 23/Nov/19

m o m e n t o f i n e r t i a o f r o d = \frac{{m L}^{2}}{12}

L = l e n g t h o f r o d

l = m o m e n t o f i n e r t i a

T = t e m p a r a t u r e

l = \frac{{m L}^{2}}{12} ⇛ s o \frac{d l}{d T} = \frac{1}{12} \times m \times 2 L \times \frac{d L}{d T}

\frac{\frac{d l}{d T}}{l} = \frac{\frac{1}{12} \times m \times 2 L \times \frac{d L}{d T}}{\frac{{m L}^{2}}{12}} = \frac{2 \frac{d L}{d T}}{L}

\frac{d l}{l} = \frac{2 d L}{L} [n o w α = \frac{d L}{L d T}]

\frac{d l}{l} = 2 α d T