InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
We would like to perpare a scale whose length does not change with temperature. It is proposed to prepare a unit scale f this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length (both components) would change in such a way that differnece between theri lenght rermain constant. If `alpha_(iron) = 1.2 xx 10^(-5)//K and alpha_(brass) = 1.8 xx 10^(-5)//K,` what should we take as lenght of each strip ? |
|
Answer» Let `l_i and l_b` be the length of iron scale and brass at temper4ature `Delta theta .^@C.` Let `l_(io)` and `l_(bo)` be the legnth of iron scale and brass scale at temperature `0^@C.` Then as per question, `l_(omega) - l_(bo) = 10 cm =l_i - l_b` ...... (i) `l_i = l_(io)(1 + alpha_i Delta theta)` `l_b = l_(bo) (1 +alpha_b Delta theta)` `:. l_i - l_b = (l_(io) - l_(bo) + Delta theta (I_(io) alpha_i - I_(bo) alpha b)` Since `I_i - I_b = I_(io) - I_(bo) = `a constant as per question hence `I_(io) alpha_i - I_(bo) alpha_b = 0` or `I_(io) alpha_i = I_(bo) alpha_b` or `(I_(io))/(I_(bo)) = (alpha_b)/(alph_i) = (1.8 xx 10^(-5))/(1.2 xx 10^(-5)) = (3)/(2)` ..... (ii) As per question `(I_(io))/(I_(bo)) = (3)/(2)` or `I_(io) = (3)/(2) I_(bo)` Putting this value in (i), we have `(3)/(2)I_(bo) - I_(bo) = 10` or `I_(bo) = 20cm` and `I_(io) = (3)/(2) xx 20 = 30 cm` |
|