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  • We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm.

We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length would change in such a way that difference between their lengths remain constant. If \alpha _{iron }= \frac{1.2 \times 10^{-5}}{K} and \alpha _{brass}= \frac{1.8 \times 10^{-5}}{K}what should we take as the length of each strip?

Answers (1)

We can use iron and brass to make the required scale. Here, one end will be connected with brass and the other end will be only of iron at a distance of 10 cm at any temperature. Let us consider the initial length of iron and brass to be-

\alpha _{iron }= \frac{1.2 \times 10^{-5}}{K} &

\alpha _{brass}= \frac{1.8 \times 10^{-5}}{K}

Now, L_{11 }- L_{1B }= 10cm      …… (i)

Thus, \alpha =\frac{ \Delta L}{L_{0}\Delta T }or\frac{ L_{2}-L_{1}}{L_{1}\Delta T}

Now, L_{2} = L_{1} + L_{1}\alpha \Delta T

                        = L_{1} (1+\alpha _{B}\Delta T)

If the rod is heated then, the length will become,

L_{21} and L_{2B}

Now, L_{21}- L_{2B}=10cm

L_{11}(1 + \alpha _{1}\Delta T) - L_{1B} (1+\alpha _{B}\Delta T) = 10

L_{11} + \alpha _{1}L_{11}\Delta T - L_{1B} - L_{1B} \alpha _{B}\Delta T = 10

Thus, from (i),

10 + (\alpha _{1}L_{11} -\alpha _{B}L_{1B}) \Delta T = 10 or \alpha _{1}L_{11} - \alpha _{B}L_{1B} = 0

\alpha _{1}L_{11} = \alpha _{B}L_{1B}

\frac{L_{11}}{L_{1B}}= \frac{\alpha _{B}}{\alpha _{1}}

          = \frac{1.8 \times 10^{-5}}{1.2 \times 10^{-5}}

            = \frac{3}{2}

Now, letL_{11} = 3x and L_{1B} = 2x

L_{11} - L_{1} = 10

3x – 2x = 10

Thus, x = 10.

Therefore, length of iron rod, (3x) = 30 & length of brass rod, (2x) = 20

Thus, the difference between the second ends will be 10 cm.

Posted by

infoexpert24

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