Question

Two metre scales, one of steel and the other of aluminium, agree at 20°C. Calculate the ratio aluminium-centimetre/steel-centimetre at (a) 0°C, (b) 40°C and (c) 100°C. α for steel = 1.1 × 10–5 °C–1 and for aluminium = 2.3 × 10–5°C–1.

Answer 1

The ratios aluminium-centimetre/steel-centimetre are given below.

(a) Lo st Lo al =  0.999759

(b) L 40 A x l/L 40 x st  =  1.0002496

(c) L-100 Al/L-100 st =  1.00096

Explanation:

Lst = La lat 20°C  

αal = 2.3×10−5^ 0°C

αst = 1.1×10−5^ 0°C

so Lost (1−αst×20)

=Lo al (1−αal×20)

(a) Lo st Lo al = 1−αal×20 / 1−αst×20

= 1−2.3×10−5 × 20 / 1−1.1×10−5×20

= 0.99954 / 0.99978

= 0.999759

(b) L40 A x l/L 40 x st

= Lo Al 1 + αal × 40/  Lo st(1+αst×40)

L_40 Al/L_40 st=L_0 Al/Lost xx (1+2.3 x 10^-5 xx 40)/(1+1.1 x 10^5 xx 40)

=0.99977 x 1.00092/1.00044

= 1.0002496

(c) L- 100 Al/L- 100 st

=Lo Al(1+alpha_Al xx 100)/Lo st(1+alpha_st xx 100)

=0.99977 x 1.0023/1.0023

= 1.00096

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Answer 2

The ratio aluminium-centimetre/steel-centimetre

a) at 0°C is  0.999759

b) at 40°C is 0.99951

c)at 100°C is 1.00096

Explanation:

Given:

$L_{st} = L_{al}$  at 20°C  

$\alpha{al} = 2.3 \times 10^{-5} /^{\circ} \mathrm{C}$

$\alpha{st} = 1.1\times 10^{−5} /^{\circ}C$

Solution:

We know that,

$Lo_{st} (1-\alpha{st}\times 20) = Lo_{al} (1- \alpha{al}\times 20)$

(a) The ratio aluminium-centimetre/steel-centimetre at 0°C,

$\frac{\mathrm{Lo}_{\mathrm{St}}}{\mathrm{Lo}_{\mathrm{al}}}=\frac{\left(1-\alpha_{\mathrm{a} 1} \times 20\right)}{\left(1-\alpha_{\mathrm{St}} \times 20\right)}$

$\frac{\mathrm{Lo}_{\mathrm{St}}}{\mathrm{Lo}_{\mathrm{al}}}=\frac{\left(1-2.3 \times 10^{-5} \times 20\right)}{\left(1-1.1 \times 10^{-5} \times 20\right)}$

$\frac{\mathrm{Lo}_{\mathrm{St}}}{\mathrm{Lo}_{\mathrm{al}}}=\frac{\left(1-4.6 \times 10^{-4}\right)}{\left(1-2.2 \times 10^{-4}\right)}$

$\frac{\mathrm{Lo}_{\mathrm{St}}}{\mathrm{Lo}_{\mathrm{al}}}=\frac{0.99954}{0.99978}$

$\frac{\mathrm{Lo}_{\mathrm{St}}}{\mathrm{Lo}_{\mathrm{al}}}=0.999759$

b) The ratio aluminium-centimetre/steel-centimetre at 40°C,

$\frac{\mathrm{Lo}_{40 \mathrm{St}}}{\mathrm{Lo}_{40 \mathrm{al}}}=\frac{\left(1-\alpha_{\mathrm{al}} \times 40\right)}{\left(1-\alpha_{\mathrm{st}} \times 40\right)}$

$\frac{\mathrm{Lo}_{40 \mathrm{St}}}{\mathrm{Lo}_{40 \mathrm{al}}}=\frac{\left(1-2.3 \times 10^{-5} \times 40\right)}{\left(1-1.1 \times 10^{-5} \times 40\right)}$

$\frac{\mathrm{Lo}_{40 \mathrm{st}}}{\mathrm{Lo}_{40 \mathrm{al}}}=\frac{\left(1-9.2 \times 10^{-4}\right)}{\left(1-4.4 \times 10^{-4}\right)}$

$\frac{\mathrm{Lo}_{40 \mathrm{St}}}{\mathrm{Lo}_{40 \mathrm{al}}}=\frac{0.99908}{0.99956}$

$\frac{\mathrm{Lo}_{40 \mathrm{St}}}{\mathrm{Lo}_{40 \mathrm{al}}}=0.99951$

C) The ratio aluminium-centimetre/steel-centimetre at 100°C

$\frac{\mathrm{Lo}_{100 \mathrm{St}}}{\mathrm{Lo}_{100 \mathrm{al}}}=\frac{\left(1+\alpha_{\mathrm{al}} \times 100\right)}{\left(1+\alpha_{\mathrm{st}} \times 100\right)}$

$\frac{\mathrm{Lo}_{100 \mathrm{st}}}{\mathrm{Lo}_{100 \mathrm{al}}}=\frac{\left(1+2.3 \times 10^{-5} \times 100\right)}{\left(1+1.1 \times 10^{-5} \times 100\right)}$

$\frac{\mathrm{Lo}_{100 \mathrm{st}}}{\mathrm{Lo}_{100 \mathrm{al}}}=\frac{\left(1+2.3 \times 10^{-3}\right)}{\left(1+1.1 \times 10^{-3}\right)}$

$\frac{\mathrm{Lo}_{100 \mathrm{st}}}{\mathrm{Lo}_{100 \mathrm{al}}}=\frac{0.99977 \times 1.00092}{1.00011}$

$\frac{\mathrm{Lo}_{100 \mathrm{st}}}{\mathrm{Lo}_{100 \mathrm{al}}}=\frac{(1.0023)}{1.0011}$

$\frac{\mathrm{Lo}_{100 \mathrm{st}}}{\mathrm{Lo}_{100 \mathrm{al}}}=1.00096$