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ra1l [238]
3 years ago
10

A steel bar of 2 m length is fixed at both ends at 20°C. The coefficient of thermal expansion is 11 x 10-6/°C and the modulus of

the elasticity is 2 x 106 kg/cm2. If the temperature is changed to 18°C, then the bar will be experience a stress of
A. 22 kg/cm2 (tensile) B. 22 kg/cm2 (compressive) C. 44 kg/cm2 (compressive) D. 44 kg/cm2 (tensile)

Physics
2 answers:
Anna007 [38]3 years ago
8 0

Answer:

C. 44 kg/cm2 (compressive).

Explanation:

The main difference between tensile and compressive stress is that tensile stress results in elongation whereas compressive stress results in shortening.

Below is an attachment containing the solution.

Daniel [21]3 years ago
5 0

Answer:

A. 22 kg/cm2 (tensile)

Explanation:

If the temperature drops to 18C degrees (by 2 degrees), it would shrink by an amount of

\Delta L = 11\times10^{-6}*2 = 22\times10^{-6} m

The strain would be:

\epsilon = \Delta L / L = 22\times10{-6} / 2 = 11\times10^{-6}

We can calculate the stress from strain and modulus of the elasticity:

\sigma = E\epsilon = 2\times10^6 * 11\times10^{-6} = 22 kg/cm^2

This stress would be tensile because the bar shrinks

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One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00 ∘C by an ice–water mixture. Th
lozanna [386]

Answer:

k=105.0359\times 10^4\,W.m^{-1}.K^{-1}

Explanation:

Given:

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length of rod through which the heat travels, dx=0.7\,m

cross-sectional area of rod, A=1.1\times 10^{-4}\,cm^2

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time taken for the melting of ice, t=15\times60=900\,s

thermal conductivity k=?

By Fourier's Law of conduction we have:

\dot{Q}=k.A.\frac{dT}{dx}......................................(1)

where:

\dot{Q}=rate of heat transfer

dT= temperature difference across the length dx

Now, we need the total heat transfer according to the condition:

we know the latent heat of fusion of ice,  L = 334000\,J.kg^{-1}

Q=m.L

Q=8.7\times 10^{-3}\times 334000

Q=2905.8\,J

Now the heat rate:

\dot{Q}=\frac{Q}{t}

\dot{Q}=\frac{2905.8}{900}

\dot{Q}=3.2287\,W

Now using eq,(1)

3.2287=k\times 1.1\times 10^{-4} \times \frac{100}{0.7}

k=205.4606\,W.m^{-1}.K^{-1}

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