$A$ cuboid $ABCDEFGH$ is anisotropic with $\alpha_x = 1 \times 10^{-5} /^{\circ}C$,$\alpha_y = 2 \times 10^{-5} /^{\circ}C$,$\alpha_z = 3 \times 10^{-5} /^{\circ}C$. The coefficient of superficial expansion of the faces can be:

  • A
    $\beta_{ABCD} = 5 \times 10^{-5} /^{\circ}C$
  • B
    $\beta_{BCGH} = 4 \times 10^{-5} /^{\circ}C$
  • C
    $\beta_{CDEH} = 3 \times 10^{-5} /^{\circ}C$
  • D
    $\beta_{EFGH} = 2 \times 10^{-5} /^{\circ}C$

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Similar Questions

$A$ metal tape is calibrated at $25^{\circ} C$. On a cold day when the temperature is $-15^{\circ} C$,the percentage error in the measurement of length is (Coefficient of linear expansion of metal $= 1 \times 10^{-5} {}^{\circ} C^{-1}$) (in $\%$)

The density of a substance at $0 \, ^\circ C$ is $10 \, g/cm^3$ and at $100 \, ^\circ C$ is $9.7 \, g/cm^3$. The coefficient of linear expansion of the substance is ..... $^\circ C^{-1}$.

$A$ unit scale is to be prepared whose length does not change with temperature and remains $20\,cm$,using a bimetallic strip made of brass and iron each of different length. The length of both components would change in such a way that the difference between their lengths remains constant. If the length of brass is $40\,cm$,what is the length of iron in $cm$?
($\alpha_{\text{iron}} = 1.2 \times 10^{-5} K^{-1}$ and $\alpha_{\text{brass}} = 1.8 \times 10^{-5} K^{-1}$)

When a bimetallic strip is heated, it . . . . . . .

$A$ hole of diameter $5 \ cm$ is drilled in a metal sheet at $30^{\circ} C$. The coefficient of linear expansion of the metal is $2 \times 10^{-5} K^{-1}$. The diameter of the hole when the temperature is raised to $230^{\circ} C$ is equal to: (in $cm$)

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