A glass flask of volume 400 cm3 is just filled with mercury at 0oC. How much mercury overflows when the temperature of the system is raised to 80oC? The coefficient of cubical expansivity of glass is 1.2 x 10-5 oC-1 and that of mercury is 1.8 x 10-5 oC-1.
Solution
For glass flask,
Volume (V1) = 400 cm3
Initial temperature [latex](\theta_1)[/latex] = 0oC
Final temperature [latex](\theta_2)[/latex] = 80oC
Change in temperature [latex](\Delta \theta) = \theta_2 – \theta_1[/latex] = 80 – 0 = 80oC
Coefficient of cubical expansivity of glass [latex](\gamma_g)[/latex] = 1.2 x 10-5 oC-1
Volume of glass flask at 80oC (V2) = [latex]V_1(1+\gamma_g \Delta \theta)[/latex]
= 400(1+ 1.2 x 10-5 x 80)
= 400.38 cm3
Now, for mercury,
Volume (V1) = 400 cm3 [∵ glass flask is just filled with mercury]
Coefficient of cubical expansivity of mercury [latex](\gamma_{Hg})[/latex] = 1.8 x 10-5 oC-1
Volume of mercury at 80oC (V2‘) = [latex]V_1(1+\gamma_{Hg}\Delta \theta)[/latex]
= 400(1+ 1.8 x 10-5 x 80)
= 400.58 cm3
So, Volume of mercury overflows = V2‘ – V2
= 400.58 – 400.38
= 0.2 cm3
= 2 x 10-7 m3
A glass flask with volume [latex]200\ cm^3[/latex] is filled to the brim with mercury at 20°C. How much mercury overflows when the temperature of the system is raised to 100°C? (Linear expansivity of glass = [latex]0.4\times10^{-5}\ K^{-1}[/latex], cubical expansivity of mercury = [latex]18\times10^{-5}\ K^{-1}[/latex]).
Solution
For glass flask,
Volume [latex](V_1) = 200\ cm^3[/latex]
Initial temperature [latex](\theta_1) = 20^oC = 20+273 = 293\ K[/latex]
Final temperature [latex](\theta_2) = 100^oC = 100+273 = 373\ K[/latex]
Change in temperature [latex](\Delta \theta) = \theta_2-\theta_1 = 373-293 = 80K[/latex]
Coefficient of linear expansivity of glass [latex](\alpha) = 0.4\times10^{-5}\ K^{-1}[/latex]
Volume of glass flask at 100°C [latex](V_2) = V_1(1+\gamma_g\Delta\theta)[/latex]
[latex]= V_1(1+3\alpha\Delta\theta)[/latex]
[latex]= 200(1+3\times0.4\times10^{-5}\times80)[/latex]
[latex]= 200.192\ cm^3[/latex]
Now, for mercury,
Volume [latex](V_1) = 200\ cm^3[/latex] [∵ glass flask is just filled with mercury]
Coefficient of cubical expansivity of mercury [latex](\gamma_m) = 18\times10^{-5}\ ^oK^{-1}[/latex]
Volume of mercury at 80°C [latex](V_4) = V_1(1+\gamma_{Hg}\Delta\theta)[/latex]
[latex]= 200(1+18\times10^{-5}\times80)[/latex]
[latex]= 202.88\ cm^3[/latex]
So, Volume of mercury overflows = [latex]V_4-V_2[/latex]
[latex]= 202.88-200.192[/latex]
[latex]= 2.688\ cm^3[/latex]
A glass vessel of volume 50 [latex]cm^3[/latex] is filled with mercury and is heated from 20°C to 60°C. What volume of mercury will overflow? Cubical expansivity of glass = [latex]1.8\times10^{-6}\ K^{-1}[/latex] and cubical expansivity of mercury = [latex]1.8\times10^{-4}\ K^{-1}[/latex].
Solution
For glass flask,
Volume [latex](V_1) = 50\ cm^3[/latex]
Initial temperature [latex](\theta_1) = 20^oC = 20+273 = 293\ K[/latex]
Final temperature [latex](\theta_2) = 60^oC = 60+273 = 333\ K[/latex]
Change in temperature [latex](\Delta \theta) = \theta_2-\theta_1 = 333-293 = 40K[/latex]
Coefficient of cubical expansivity of glass [latex](\gamma_g) = 1.8\times10^{-6}\ K^{-1}[/latex]
Volume of glass flask at 100°C [latex](V_2) = V_1(1+\gamma_g\Delta\theta)[/latex]
[latex]= 50(1+1.8\times10^{-6}\times40)[/latex]
[latex]= 50.0036\ cm^3[/latex]
Now, for mercury,
Volume [latex](V_1) = 50\ cm^3[/latex] [∵ glass flask is just filled with mercury]
Coefficient of cubical expansivity of mercury [latex](\gamma_m) = 1.8\times10^{-4}\ ^oK^{-1}[/latex]
Volume of mercury at 80°C [latex](V_4) = V_1(1+\gamma_{Hg}\Delta\theta)[/latex]
[latex]= 50(1+1.8\times10^{-4}\times40)[/latex]
[latex]= 50.36\ cm^3[/latex]
So, Volume of mercury overflows = [latex]V_4-V_2[/latex]
[latex]= 50.36-50.0036[/latex]
[latex]= 0.3564\ cm^3[/latex]
A glass flask of volume [latex]500\ cm^3[/latex] is just filled with mercury at 0°C. how much mercury overflows when the temperature of the system is raised to 80°C?
Solution
For glass flask,
Volume latex = 500\ cm^3[/latex]
Initial temperature [latex](\theta_1) = 0^oC = 0+273 = 273\ K[/latex]
Final temperature [latex](\theta_2) = 80^oC = 80+273 = 353\ K[/latex]
Change in temperature [latex](\Delta \theta) = \theta_2-\theta_1 = 353-273 = 80K[/latex]
Coefficient of cubical expansivity of glass [latex](\gamma_g) = 1.2\times10^{-5}\ K^{-1}[/latex]
Volume of glass flask at 80°C [latex](V_2) = V_1(1+\gamma_g\Delta\theta)[/latex]
[latex]= 500(1+1.2\times10^{-5}\times80)[/latex]
[latex]= 500.48\ cm^3[/latex]
Now, for mercury,
Volume [latex](V_1) = 500\ cm^3[/latex] [∵ glass flask is just filled with mercury]
Coefficient of cubical expansivity of mercury [latex](\gamma_m) = 1.8\times10^{-4}\ ^oK^{-1}[/latex]
Volume of mercury at 80°C [latex](V_4) = V_1(1+\gamma_{Hg}\Delta\theta)[/latex]
[latex]= 500(1+1.8\times10^{-4}\times80)[/latex]
[latex]= 507.2\ cm^3[/latex]
So, Volume of mercury overflows = [latex]V_4-V_2[/latex]
[latex]= 507.2-500.48[/latex]
[latex]= 6.72\ cm^3[/latex]
A copper vessel with a volume of exactly [latex]100\ m^3[/latex] at a temperature of 15°C is filled with glycerine. If the temperature rises to 25°C, how much glycerine will spill out?
Solution
For copper vessel,
Volume [latex](V_1) = 100\ cm^3[/latex]
Initial temperature [latex](\theta_1) = 15^oC = 15+273 = 288\ K[/latex]
Final temperature [latex](\theta_2) = 25^oC = 25+273 = 298\ K[/latex]
Change in temperature [latex](\Delta \theta) = \theta_2-\theta_1 = 298-288 = 10K[/latex]
Coefficient of cubical expansivity of copper [latex](\gamma_c) = 5.1\times10^{-5}\ K^{-1}[/latex]
Volume of copper vessel at 25°C [latex](V_2) = V_1(1+\gamma_c\Delta\theta)[/latex]
[latex]= 100(1+5.1\times10^{-5}\times10)[/latex]
[latex]= 100.051\ m^3[/latex]
Now, for glycerine,
Volume [latex](V_1) = 100\ cm^3[/latex] [∵ copper vessel is just filled with glycerine]
Coefficient of cubical expansivity of glycerine [latex](\gamma_{Hg}) = 5\times10^{-4}\ ^oK^{-1}[/latex]
Volume of mercury at 25°C [latex](V_4) = V_1(1+\gamma_{Hg}\Delta\theta)[/latex]
[latex]= 100(1+5\times10^{-4}\times10)[/latex]
[latex]= 100.5\ m^3[/latex]
So, Volume of mercury overflows = [latex]V_4-V_2[/latex]
[latex]= 100.5-100.051[/latex]
[latex]= 0.449\ m^3[/latex]
An iron rod of length 100 m at 10°C is used to measure a distance of 2 km on a day when the temperature is 40°C. Calculate the error in measuring the distance. ([latex]\alpha[/latex] for iron = [latex]12\times10^{-6}\ ^oC^{-1}[/latex]).
Solution
Length of iron rod [latex](l_1) = 100\ m[/latex]
Initial temperature [latex](\theta_1) = 10^oC[/latex]
Final temperature [latex](\theta_2) = 40^oC[/latex]
Change in temperature [latex](\Delta \theta) = \theta_2-\theta_1 = 40-10 = 30^oC[/latex]
Total distance to be measured = 2 km = 2000 m
Error in measuring the distance = ?
Length of the rod at 40°C [latex](l_2) = l_1(1+\alpha\Delta\theta)[/latex]
[latex]= 100(1+12\times10^{-6}\times30)[/latex]
[latex]= 100.036\ m[/latex]
So, error in measuring 100m distance = [latex]l_2-l_1[/latex]
[latex]= 100.036-100[/latex]
[latex]= 0.036\ m[/latex]
So, error in measuring 2 km (= 2000m) distance [latex]= \dfrac{0.036}{100}\times2000[/latex]
[latex]= 0.72\ m = 72\ cm[/latex]
The marking on an aluminium ruler and a brass ruler are perfectly aligned at 0°C. How far apart will the 20.0 cm marks be on the two rulers at 100°C, if precise alignment of the left-hand ends of the rulers is maintained? Coefficient of linear expansion of aluminium and brass are [latex]2.4\times10^{-5}\ K^{-1}[/latex] and [latex]2.0\times10^{-5}\ K^{-1}[/latex] respectively.
Solution
Aluminium and brass scales are perfectly aligned at 0°C.
i.e., [latex]\theta_1 = 0^oC = 273\ K[/latex]
Also, length of both scale [latex](l_1) = 20\ cm[/latex]
Final temperature [latex](\theta_2) = 100^oC = 100+273 = 373\ K[/latex]
Change in temperature [latex](\Delta \theta) = \theta_2-\theta_1 = 373-273 = 100K[/latex]
Linear expansion of aluminium [latex](\alpha_a) = 2.4\times10^{-5}\ K^{-1}[/latex]
Linear expansion of brass [latex](\alpha_b) = 2.0\times10^{-5}\ K^{-1}[/latex]
Now, for aluminium scale,
Length at 100°C [latex](l_2) = l_1(1+\alpha_a\Delta\theta)[/latex]
[latex]= 20(1+2.4\times10^{-5}\times100)[/latex]
[latex]= 20.048\ cm[/latex]
Again, for brass scale,
Length at 100°C [latex](l_2) = l_1(1+\alpha_b\Delta\theta)[/latex]
[latex]= 20(1+2.0\times10^{-5}\times100)[/latex]
[latex]= 20.040\ cm[/latex]
So, difference in their scales = [latex]l_2-l_2′[/latex]
[latex]= 20.048-20.040[/latex]
[latex]= 0.008\ cm[/latex]
The length of an iron rod is measured by a brass scale. When both of them are at 10°C, the measured length is 50 cm. What is the length of the rod of 40°C when measured by the brass scale at 40°C? ([latex]\alpha[/latex] for brass = [latex]24\times10^{-6}\ ^oC^{-1}[/latex], [latex]\alpha[/latex] for iron = [latex]16\times10^{-6}\ ^oC^{-1}[/latex]).
Solution
Length of brass at 10°C [latex](l_{10}) = 50\ cm[/latex]
Length of brass at 40°C [latex](l_{40})[/latex] =?
Initial temperature [latex](\theta_1) = 10^oC[/latex]
Final temperature [latex](\theta_2) = 40^oC[/latex]
Linear expansivity of brass [latex](\alpha_b) = 24\times10^{-6}\ ^oC^{-1}[/latex]
Linear expansivity of iron [latex](\alpha_i) = 16\times10^{-6}\ ^oC^{-1}[/latex]
Now, Length of iron rod at 40°C = [latex]50[1+16\times10^{-6}(40-10)] = 50.024\ cm[/latex]
Length of one cm division of the brass scale at 40°C
[latex]= 1[1+24\times10^{-6}(40-10)] = 1.00072\ cm[/latex]
Length of the rod as measured by the brass scale at 40°C = [latex]\dfrac{50.024}{1.0072}[/latex]
[latex]= 49.988\ cm.[/latex]
A second pendulum made of brass keeps correct time at 10°C. How many seconds it will lose or gain per day when the temperature of its surrounding rises to 35°C? ([latex]\alpha[/latex] for brass [latex]= 2\times10^{-5}\ ^oC^{-1}[/latex]).
Solution
Initial temperature [latex](\theta_1) = 10^oC[/latex]
Final temperature [latex](\theta_2) = 35^oC[/latex]
Time period of second pendulum (T) = 2 s
Time (t) = 1 day = 86400 s
Time loss or gain per day =?
Let, [latex]T_1[/latex] and [latex]T_2[/latex] be the time period of the pendulum at 10°C and 35°C respectively.
At 10°C,
[latex]T_1 = 2\pi\sqrt{\dfrac{l_1}{g}}[/latex]
Also, at 35°C,
[latex]T_2 = 2\pi\sqrt{\dfrac{l_2}{g}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \dfrac{2\pi\sqrt{\frac{l_2}{g}}}{2\pi\sqrt{\frac{l_1}{g}}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{g}\times\dfrac{g}{l_1}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{l_1}}[/latex]
Or, [latex]T_2 = \sqrt{\dfrac{l_1(1+\alpha\Delta\theta)}{l_1}}\times T_1[/latex]
Or, [latex]T_2 = \sqrt{1+2\times10^{-5}\times25}\times2[/latex]
[latex]= 2.00049\ s[/latex]
Thus, time lost in 2 s = [latex]2.00049-2 = 0.00049\ s[/latex]
So, time lost in 1 day [latex]= \dfrac{0.00049}{2}\times86400[/latex]
[latex]= 21.168\ s[/latex]
A brass pendulum clock keeps correct time at 15°C. How many seconds per day it will lose or gain at 0°C? ([latex]\alpha[/latex] for brass [latex]= 2\times10^{-5}K^{-1}[/latex]).
Solution
Initial temperature [latex](\theta_1) = 0^oC[/latex]
Final temperature [latex](\theta_2) = 15^oC[/latex]
Time period of second pendulum (T) = 2 s
Time (t) = 1 day = 86400 s
Time lose or gain per day = ?
Let, [latex]T_1[/latex] and [latex]T_2[/latex] be the time period of the pendulum at 0°C and 15°C respectively.
At 0°C,
[latex]T_1 = 2\pi\sqrt{\dfrac{l_1}{g}}[/latex]
Also, at 15°C,
[latex]T_2 = 2\pi\sqrt{\dfrac{l_2}{g}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \dfrac{2\pi\sqrt{\frac{l_2}{g}}}{2\pi\sqrt{\frac{l_1}{g}}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{g}\times\dfrac{g}{l_1}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{l_1}}[/latex]
Or, [latex]T_2 = \sqrt{\dfrac{l_1(1+\alpha\Delta\theta)}{l_1}}\times T_1[/latex]
Or, [latex]T_2 = \sqrt{1+2\times10^{-5}\times15}\times2[/latex]
[latex]= 2.00030\ s[/latex]
Thus, time gained in 2 s = [latex]2.00030-2 = 0.00030\ s[/latex]
So, time gained in 1 day [latex]= \dfrac{0.00030}{2}\times86400[/latex]
[latex]= 12.96\ s[/latex]
An iron pendulum clock keeps correct time at 20°C. How many seconds will it gain or lose per day when temperature rises to 30°C?
Solution
Initial temperature [latex](\theta_1) = 20^oC[/latex]
Final temperature [latex](\theta_2) = 30^oC[/latex]
Time period of pendulum (T) = 2 s (Let)
Time (t) = 1 day = 86400 s
Time loss or gain per day =?
Let, [latex]T_1[/latex] and [latex]T_2[/latex] be the time period of the pendulum at 20°C and 30°C respectively.
At 20°C,
[latex]T_1 = 2\pi\sqrt{\dfrac{l_1}{g}}[/latex]
Also, at 30°C,
[latex]T_2 = 2\pi\sqrt{\dfrac{l_2}{g}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \dfrac{2\pi\sqrt{\frac{l_2}{g}}}{2\pi\sqrt{\frac{l_1}{g}}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{g}\times\dfrac{g}{l_1}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{l_1}}[/latex]
Or, [latex]T_2 = \sqrt{\dfrac{l_1(1+\alpha\Delta\theta)}{l_1}}\times T_1[/latex]
Or, [latex]T_2 = \sqrt{1+12\times10^{-6}\times10}\times2[/latex]
[latex]= 2.00012\ s[/latex]
Thus, time lost in 2 s = [latex]2.00012-2 = 0.00012\ s[/latex]
So, time lost in 1 day [latex]= \dfrac{0.00012}{2}\times86400[/latex]
[latex]= 5.18\ s[/latex]
The pendulum of a clock is made of brass. If the clock keeps correct time at 15°C, how many seconds per day will it lose at 20°C?
Solution
Initial temperature [latex](\theta_1) = 15^oC[/latex]
Final temperature [latex](\theta_2) = 20^oC[/latex]
Time period of second pendulum (T) = 2 s
Time (t) = 1 day = 86400 s
Time loss or gain per day =?
Let, [latex]T_1[/latex] and [latex]T_2[/latex] be the time period of the pendulum at 15°C and 20°C respectively.
At 15°C,
[latex]T_1 = 2\pi\sqrt{\dfrac{l_1}{g}}[/latex]
Also, at 20°C,
[latex]T_2 = 2\pi\sqrt{\dfrac{l_2}{g}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \dfrac{2\pi\sqrt{\frac{l_2}{g}}}{2\pi\sqrt{\frac{l_1}{g}}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{g}\times\dfrac{g}{l_1}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{l_1}}[/latex]
Or, [latex]T_2 = \sqrt{\dfrac{l_1(1+\alpha\Delta\theta)}{l_1}}\times T_1[/latex]
Or, [latex]T_2 = \sqrt{1+2\times10^{-5}\times5}\times2[/latex]
[latex]= 2.000099\ s[/latex]
Thus, time lost in 2 s = [latex]2.000099-2 = 0.000099\ s[/latex]
So, time lost in 1 day [latex]= \dfrac{0.000099}{2}\times86400[/latex]
[latex]= 4.32\ s[/latex]
A pendulum clock made of iron keeps correct time at 15°C. How many seconds will it lose or gain per day when the temperature rises to 35°C?
Solution
Initial temperature [latex](\theta_1) = 15^oC[/latex]
Final temperature [latex](\theta_2) = 35^oC[/latex]
Time period of second pendulum (T) = 2 s
Time (t) = 1 day = 86400 s
Time loss or gain per day =?
Let, [latex]T_1[/latex] and [latex]T_2[/latex] be the time period of the pendulum at 15°C and 35°C respectively.
At 15°C,
[latex]T_1 = 2\pi\sqrt{\dfrac{l_1}{g}}[/latex]
Also, at 35°C,
[latex]T_2 = 2\pi\sqrt{\dfrac{l_2}{g}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \dfrac{2\pi\sqrt{\frac{l_2}{g}}}{2\pi\sqrt{\frac{l_1}{g}}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{g}\times\dfrac{g}{l_1}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{l_1}}[/latex]
Or, [latex]T_2 = \sqrt{\dfrac{l_1(1+\alpha\Delta\theta)}{l_1}}\times T_1[/latex]
Or, [latex]T_2 = \sqrt{1+12\times10^{-6}\times20}\times2[/latex]
[latex]= 2.00024\ s[/latex]
Thus, time lost in 2 s = [latex]2.00024-2 = 0.00024\ s[/latex]
So, time lost in 1 day [latex]= \dfrac{0.00024}{2}\times86400[/latex]
[latex]= 10.37\ s[/latex]
A clock which has a brass pendulum beats seconds correctly when the temperature of the room is 30°C. How many seconds will it gain or lose per day when the temperature of the room falls to 10°C? ([latex]\alpha[/latex] for brass = 0.000018 [latex]^oC^{-1}[/latex]).
Solution
Initial temperature [latex](\theta_1) = 30^oC[/latex]
Final temperature [latex](\theta_2) = 10^oC[/latex]
Time period of pendulum (T) = 2 s (Let)
Time (t) = 1 day = 86400 s
Time lose or gain per day =?
Let, [latex]T_1[/latex] and [latex]T_2[/latex] be the time period of the pendulum at 30°C and 10°C respectively.
At 30°C,
[latex]T_1 = 2\pi\sqrt{\dfrac{l_1}{g}}[/latex]
Also, at 10°C,
[latex]T_2 = 2\pi\sqrt{\dfrac{l_2}{g}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \dfrac{2\pi\sqrt{\frac{l_2}{g}}}{2\pi\sqrt{\frac{l_1}{g}}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{g}\times\dfrac{g}{l_1}}[/latex]
Or, [latex]\dfrac{T_2}{T_1} = \sqrt{\dfrac{l_2}{l_1}}[/latex]
Or, [latex]T_2 = \sqrt{\dfrac{l_1(1+\alpha\Delta\theta)}{l_1}}\times T_1[/latex]
Or, [latex]T_2 = \sqrt{1+0.000018\times20}\times2[/latex]
[latex]= 2.000359\ s[/latex]
Thus, time gained in 2 s = [latex]2.000359-2 = 0.000359\ s[/latex]
So, time gained in 1 day [latex]= \dfrac{0.000359}{2}\times86400[/latex]
[latex]= 15.55\ s[/latex]
A steel wire 8m long and 4 mm in diameter is fixed to two rigid supports. Calculate the increase in tension when the temperature falls by 10°C. (Young’s modulus of elasticity of steel = [latex]2\times10^{11}\ N/m^2[/latex] and linear expansivity of steel = [latex]1.2\times10^{-5}\ K^{-1}[/latex]).
Solution
Length of steel wire (l) = 8m
Diameter (d) = 4 mm = [latex]4\times10^{-3}\ m[/latex]
Young’s modulus of elasticity of steel [latex](Y) = 2\times10^{11}\ N/m^2[/latex]
Linear expansivity of steel [latex](\alpha) = 1.2\times10^{-5}\ K^{-1}[/latex]
Increase in tension (T) = ?
Area of cross section [latex](A) = \dfrac{\pi d^2}{4}[/latex]
[latex]= \dfrac{22}{7}\times\dfrac{(4\times10^{-3})^2}{4}[/latex]
[latex]= 1.26\times10^{-5}\ m^2[/latex]
Fall in temperature [latex](\Delta \theta) = 10^oC[/latex]
We know that,
Young’s modulus [latex](Y) = \dfrac{Stress}{Strain}[/latex]
Or, [latex]Y = \dfrac{F/A}{e/l}[/latex]
Or, [latex]Y = \dfrac{F}{A}\times\dfrac{l}{e}[/latex]
Or, [latex]F = \dfrac{Y\times A\times e}{l}[/latex]
Or, [latex]F = \dfrac{Y\times A\times l_1\alpha\Delta\theta}{l_1}[/latex] [∵ l = [latex]l_1[/latex] and, e = [latex]l_1\alpha\Delta\theta[/latex]]
[latex]= YA \alpha\Delta\theta[/latex]
[latex]= (2\times10^{11})\times(1.263\times10^{-5})\times(1.2\times10^{-5})\times10[/latex]
[latex]= 302.4\ N[/latex]
So, tension in the wire (T) = F
[latex]= 302.4\ N[/latex]
A copper wire of diameter 0.5 mm is stretched between two points at 25°C. Calculate the increase in tension in the wire if the temperature falls to 0°C. (Young’s modulus for copper = [latex]1.2\times10^{11}\ Nm^{-2}[/latex], linear expansivity for copper = [latex]18\times10^{-6}\ K^{-1}[/latex]).
Solution
Diameter (d) = 0.5 mm = [latex]0.5\times10^{-3}\ m[/latex]
Young’s modulus of elasticity of copper [latex](Y) = 1.2\times10^{11}\ N/m^2[/latex]
Linear expansivity of copper [latex](\alpha) = 18\times10^{-6}\ K^{-1}[/latex]
Increase in tension (T) =?
Area of cross section [latex](A) = \dfrac{\pi d^2}{4}[/latex]
[latex]= \dfrac{22}{7}\times\dfrac{(0.5\times10^{-3})^2}{4}[/latex]
[latex]= 1.963\times10^{-7}\ m^2[/latex]
Fall in temperature [latex](\Delta \theta) = 25^oC[/latex]
We know that,
Young’s modulus [latex](Y) = \dfrac{Stress}{Strain}[/latex]
Or, [latex]Y = \dfrac{F/A}{e/l}[/latex]
Or, [latex]Y = \dfrac{F}{A}\times\dfrac{l}{e}[/latex]
Or, [latex]F = \dfrac{Y\times A\times e}{l}[/latex]
Or, [latex]F = \dfrac{Y\times A\times l_1\alpha\Delta\theta}{l_1}[/latex] [∵ l = [latex]l_1[/latex] and, e = [latex]l_1\alpha\Delta\theta[/latex]]
[latex]= Y\times A\times\alpha\Delta\theta[/latex]
[latex]= (1.2\times10^{11})\times(1.963\times10^{-7})\times(18\times10^{-6})\times25[/latex]
[latex]= 10.6\ N[/latex]
So, tension in the wire (T) = F
[latex]= 10.6\ N[/latex]
The density of silver at 0°C is [latex]10310\ kgm^3[/latex] and the coefficient of linear expansion is 0.000019/°C. Calculate its density at 100°C.
Solution
Given,
Density of silver [latex](\rho_1) = 10310\ kgm^{-3}[/latex]
Temperature [latex](\theta_1) = 0^oC[/latex]
Coefficient of linear expansivity [latex](\alpha) = 0.000019/^oC[/latex]
Density of silver [latex](\rho_2)[/latex] =?
Temperature [latex](\theta_2) = 100^oC[/latex]
We know that,
[latex]\rho_2 = \rho_1(1-\gamma\Delta\theta)[/latex]
[latex]= \rho_1(1-3\alpha\Delta\theta)[/latex]
[latex]= 10310(1-3\times0.000019\times100)[/latex]
[latex]= 10310\times0.9943[/latex]
[latex]= 10251.233\ Kg/m^3.[/latex]
Using the following data, determine the temperature at which wood will just sink in benzene. Density of benzene at 0°C = [latex]9.0\times10^2\ Kgm^{-3}[/latex], Density of wood at 0°C = [latex]8.8\times10^2\ Kgm^{-3}[/latex]. (Cubical expansivity of benzene = [latex]1.2\times10^{-3}\ K^{-1}[/latex], Cubical expansivity of wood = [latex]1.5\times10^{-4}\ K^{-1}[/latex]).
Solution
Density of benzene at 0°C [latex](\rho_1) = 9.0\times10^2\ Kgm^{-3}[/latex]
Density of wood at 0°C [latex](\rho_2) = 8.8\times10^2\ Kgm^{-3}[/latex]
Initial temperature [latex](\theta_1) = 0^oC[/latex]
Final temperature [latex](\theta_2) = t^oC[/latex] (Let)
Cubical expansivity of benzene [latex](\gamma_b) = 1.2\times10^{-3}\ K^{-1}[/latex]
Cubical expansivity of wood [latex](\gamma_w) = 1.5\times10^{-4}\ K^{-1}[/latex]
For benzene,
Density at [latex]\theta_2\ (\rho_3) = \rho_1(1-\gamma_b\Delta\theta)[/latex]
[latex]= 9.0\times10^2(1-1.2\times10^{-3}\times t)[/latex]
Now, for wood,
Density at [latex]\theta_2(\rho_4) = \rho_2(1-\gamma_w\Delta\theta)[/latex]
[latex]= 8.8\times10^2(1-1.5\times10^{-4}\times t)[/latex]
For the wood just sink in benzene, [latex]\rho_3 = \rho_4[/latex]
Or, [latex]9.0\times10^2(1-1.2\times10^{-3}\times t) = 8.8\times10^2(1-1.5\times10^{-4}\times t)[/latex]
Or, [latex]\dfrac{9.0\times10^2}{8.8\times10^2} = \dfrac{1-1.5\times10^{-4}t}{1-1.2\times10^{-3}t}[/latex]
Or, [latex]1.0227 = \dfrac{1-1.5\times10^{-4}t}{1-1.2\times10^{-3}t}[/latex]
Or, [latex]1.0227 – 1.227\times10^{-3}t = 1 – 1.5\times10^{-4}t[/latex]
Or, [latex]1.0227-1 = -1.5\times10^{-4}t + 1.227\times10^{-3}t[/latex]
Or, [latex]0.0227 = 1.077\times10^{-3}t[/latex]
Or, [latex]t = \dfrac{0.0227}{1.077\times10^{-3}}[/latex]
[latex]= 21.07^oC[/latex]
A brass rod of length 0.40 m and steel rod of length 0.60 m, both are initially at 0°C are heated to 75°C. If the increase in length is the same for both the rods, calculate the linear expansivity of brass. The linear expansivity of steel is [latex]12\times10^{-6}\ ^oC^{-1}[/latex].
Solution
Length of brass rod [latex](l_1) = 0.4\ m[/latex]
Length of steel rod [latex](l_2) = 0.6\ m[/latex]
Change in temperature [latex](\Delta \theta) = 75^oC[/latex]
Linear expansivity of brass [latex](\alpha_1)[/latex] =?
Linear expansivity of steel [latex](\alpha_2) = 12\times10^{-6}\ ^oC^{-1}[/latex]
We have,
From differential expansion,
[latex]\dfrac{\alpha_1}{\alpha_2} = \dfrac{l_2}{l_1}[/latex]
Or, [latex]\alpha_1 = \dfrac{0.6}{0.4}\times12\times10^{-6} = 1.8\times10^{-5}\ ^oC^{-1}[/latex]