Question

A steel rod and brass rod when measured at 0०c are 50.1cm and 50.0cm respectively. At what temperature will their lengths be equal?(alpha of brass= 18*10^-6 and of steel= 12*10^-6)


plz help me only if you know it
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Answer 1

Given :

• A steel rod and brass rod when measured at 0 degree

• The length of steel rod =50.1cm

• The length of brass rod =50.0cm

$\bullet \rm \alpha(brass) = 18 \times {10}^{ - 6 } \: \: and \: \: \alpha(steel) = 12 \times {10}^{ - 6}$

To find :

• At what temperature will their length be equal =?

Formula :

$\red\bigstar\boxed{\sf{\delta L=L_{1}(1+\alpha \:\delta t)}}$

Solution :

• As we know that

$\rm{L_{1}-L_{2}= \alpha \: L_{1} \: t}$

$\rm{\bullet \: L_{1}= initial \: length}$

$\rm{\bullet \: L_{2} =Final \: length}$

$\rm{\bullet \: t= temperature \: difference }$

• As initial temperature is 0

Now final length for steel and brass rod will be same.

Generally written as,

$\\ \implies\sf{ L_{2}=L_{1} (1+\alpha \: \delta t) }$

$\\ \implies\sf{L_{1} \: for \: steel=L_{2} \: for \: brass}$

$\\ \implies \sf \: 50.1(1 + 1.8 \times {10}^{ - 6} t) = 50.0(1 + 1.2 \times {10}^{ - 6}t )$

$\\ \implies \sf \: 50.1 + 50.1 \times 1.8 \times {10}^{ - 6} t = 50.0 + 50.0 \times 1.2 \times {10}^{ - 6} t$

$\\ \\ \implies \sf \: 50.1 + 90.18 \times {10}^{ - 6} t= 50.0 + 60 \times {10}^{ - 6} t \\ \\ \implies \sf \: 50.1 - 50.0 + 90.18 \times {10}^{ - 6} t - 60.0 \times {10}^{ - 6} t \\ \\ \implies \sf \: 0.1 + (90.18 \times {10}^{ - 6} - 60.00 \times {10}^{ - 6} )t = 0 \\ \\ \implies \sf \: 0.1 + 30.18 \times {10}^{ - 6} t = 0 \\ \\ \implies \sf \: 30.18 \times {10}^{ - 6} t = - 0.1 \\ \\ \implies \sf \: t = - \frac{0.1 \times {10}^{ - 6} }{30.18} \\ \\ \implies \sf \: t = \frac{ - 1 \times {10}^{ - 6} }{301.8} \\ \\ \implies \sf \: t = - 0.00 33 \times {10}^{ - 6} \\ \\ \implies \sf \: \: t = 33 \times {10}^{ - 10}$

Answer will be approximately -33 Celsius