## Expansion Of Solid And Liquids – Practical Applications Of Thermal Expansion Of Solids

Solids expand on heating and contract on cooling. This may be advantageous in some cases, and in some other cases this is disadvantageous. A few examples are cited below.

**Practical Applications Of Thermal Expansion Of Solids Advantages**:

1. The iron rim of a bullock cart wheel has a diameter slightly smaller than that of the wheel. Hence, the rim cannot be fitted to the wheel at ordinary temperature. The rim, when heated, expands and thus can be fitted easily around the wheel. When cooled to normal temperature, the iron rim contracts and fits tightly around the wheel.

2. Some holes are drilled through two metal plates by placing them one above the other. A rivet is heated and inserted through each hole. Head of the rivet is then flattened by hammering.

This keeps the two plates tightly fitted. On cooling, the rivet contracts lengthwise, fixing the two plates even tighter. Iron bridges are made by riveting large plates in this way. The flattened heads of the rivets can be easily seen on this type of bridges.

**Read and Learn More: Class 11 Physics Notes**

3.** Thermostat:** Thermostats are devices used for automatic temperature control. These make use of the property of a bimetallic strip which bends when heated. Thermostats are used in electrical oven, refrigerators, thermostat condition at normal temperature electric iron, incubators, electric heater, etc. Thermostat is an automatic switch that turns on at a definite temperature and off at some other fixed temperature.

- Fig shows the working of a thermostat. The two metals in the bimetallic strip are brass and invar. The contact point A of the thermostat is kept adjacent to the brass part of the bimetallic strip. At ordinary temperature the metal strip remains straight.
- This keeps the thermostat switched on and current flows through the heater. Heat from the heater warms up the air. The strip gets heated up and bends.
- Coefficient of linear expansion of brass is more than that of invar and so the bimetallic strip bends away from the contact point A and disconnects the circuit, stopping current flow through the heater.
- Hence, temperature of the heater falls. This also cools the bimetallic strip. The strip straightens up again restoring the electrical connection, and the heater is on again.

Therefore, the heater cannot remain ‘ori above a fixed temperature as the thermostat controls the current flow. Thus, the temperature is also controlled.

4.** Fire alarm:** In a fire alarm, an electric bell is con¬nected through a thermostat to the power supply. Thermostat is essentially a bimetallic strip of brass and invar, invar piece having the contact point with the supply.

At ordinary temperatures, the thermostat remains disconnected from the power supply. When the strip gets heated up due to fire, it bends towards the contact point. This establishes current through the electric bell and the bell rings.

5. **Temperature measurement:** The curvature of a bime¬tallic strip increases, and so its radius of curvature decreases with the increase in temperature. From the straight upright stage, it bends as temperature increases. This moves the centre of curvature closer to the strip. If at temperature t_{0} the strip is upright, and bends to have a radius of curvature r at temperature t, then \(r \propto \frac{1}{t-t_0}, \text { i.e., } r\left(t-t_0\right)=\text { constant. }\)

Value of the constant can be found out by measuring r at a known temperature t. Any unknown temperature can be determined by the measurement of the corresponding value of r at that temperature. However, bimetallic strip is never used as a thermometer due to inconsistencies in its behaviour.

**Practical Applications Of Thermal Expansion Of Solids Disadvantages:**

1. Railway tracks are made by connecting pieces of rails using fishplates. A small gap is maintained between two consecutive rail pieces. Iron nuts and bolts connect and hold the fishplate with the rail.

- The bolt holes are slotted to allow free movement at the rail joints. Due to factors like sunlight or friction, the rails get heated up and expand.
- The gap between the two rails and the longitudinal slots for the bolts, allow the rail to expand lengthwise. Otherwise, the rails would have bent due to high thermal stress.
- But in case of tram lines no gap is maintained. The line is embedded on the earth surface by concrete and gran¬ite stones. This type of fixing can easily withstand the thermal stress developed, and the rails do not bend.

2. Iron or steel girders are used in the construction of a bridge. One end of the girder is rigidly fixed with bricks and concrete. The other end is not fixed.

Instead, the end is set on a roller over the support, as shown in Fig. When there is rise or fall in temperature due to seasonal changes, girders may expand or contract, without developing any thermal stress.

3. A thick-walled glass pot often cracks when hot water is poured into it. The inner surface of the thick glass, in contact with the hot water, warms up and expands. Glass is a bad conductor of heat and so the outer sur face remains colder, and hence expansion is less.

- This unequal expansion sets up a thermal stress and the glass cracks. In case of a thin waliec glass pot, heat transfer to the outer surface is easier and there is almost eqUal expansion of belli the surfaces.
- The chance of cracking is reduced. Pyrex gloss has a low coefficient of expansion. Hence, beakers, test tube, etc. for laboratory use are usually made of pyrex glass.

4. A metal scale can measure true reading only at the temperature in which it has been graduated. At any other temperature, the interval between two consecutive graduations, increases or decreases, making the measurement inaccurate. Hence, the measured length has to be corrected using the value of a for the metal of the scale in use.

## Expansion Of Solid And Liquids – Practical Applications Of Thermal Expansion Of Solids Numerical Examples

**Example 1. The difference in length of two metal rods A and B is 25 cm at all temperatures. The coefficients of linear expansion of the materials of A and B are 1.28 x 10 ^{-5} °C^{-1} and 1.92 x 10^{-5 °}C^{-1} respectively. Find the length of each rod at 0°C.**

**Solution:**

Let l_{A} and l_{B} be the lengths of the rods A and B respectively at 0°C.

From the given condition, increase in length of rod A due to t°C rise in temperature = increase in length of rod B due to the same rise in temperature

∴ \(l_A \times 1.28 \times 10^{-5} \times t=l_B \times 1.92 \times 10^{-5} \times t\)

∴ \(2 l_A=3 l_B \quad \text { or, } l_A=\frac{3}{2} l_B\)

∴ \(l_A>l_B,\)

∴ \(l_A-l_B=25 \quad \text { or, } \frac{3}{2} l_B-l_B=25\)

or, \(l_B=50 \mathrm{~cm}\)

∴ \(l_A=\frac{3}{2} l_B=\frac{3}{2} \times 50=75 \mathrm{~cm}\)

∴ Length of rod A at 0°C is 75 cm, and that of rod B at 0°C is

**Example 2. The difference in length of an iron rod and a copper rod at 50°C is 2 cm. This difference remains the same at 450°C. What are the lengths of the rods at 0°C? Given, a for iron =12x 10 ^{-6} °C^{-1} and α for copper = 17 x 10^{-6} °C^{-1}.**

**Solution:**

Let x and y be the lengths of the iron and the copper rods at 50 °C respectively.

Since the difference in lengths of the two rods remains the same for any rise in temperature, both the rods will have the same expansion.

Increase in length of the iron rod = x x 12 x 10^{-6}(450 – 50) = x x 12 x 10^{-6} x 400

Increase in length of the copper rod = y x 17 x 10^{-6}(450 – 50) = y x 17 x 10^{-6} x 400

According to the question, x x 12 x 10^{-6} x 400 = y x 17 x 10^{-6} x 400

12x = 17y or, x = 17/12 y

∴ x > y

∴ x-y = 2 or, x = 2 +y

or, 2 + y = 17/12 y or, 5y = 24 or, y = 4.8 cm

∴ x = 2 + 4.8 = 6.8 cm

Now, suppose the lengths of the iron and the copper rods are x_{0} and y_{0} respectively at 0°C.

6.8 = x_{0}{1+ 12 x 10-6 x 50} or, x_{0} = 6.796 cm

4.8 = y_{0}{1 + 17 x 10-6 x 50} or, y_{0} = 4.796 cm

Hence, lengths of the iron and the copper rods at 0°C are 6.796 cm and 4.796 cm respectively.

**Example 3. A metal scale measures correct reading at 25°C. A rod is measured to be 80 cm by the scale at 15°C. What is the actual length of the rod? (α = 15 x 10 ^{-6 }C^{-1})**

**Solution:**

The actual length l of measured 1 cm, using the metal scale at 15°C, should be l =1{1-0.000015x(25-15)}

= 1 – 0.00015 = 0.99985 cm

∴ The correct length corresponding to the measured 80 cm = 0.99985 x 80 = 79.988 cm

**Example 4. A steel scale is errorless at 50°F. Using the scale, the length of a brass rod is found to be 1.5 m at 50°C. What is the true length of the rod at 100°C? (Coefficients of linear expansion of steel and brass are 11.2 x 10 ^{-6}°C^{-1} and 18x 10^{-6}°C^{-1} respectively.)**

**Solution:**

Let 50°F be equal to l°C, so that \(\frac{t}{5}=\frac{50-32}{9}\) i.e., t = 10. Hence, the steel scale is errorless at 10°C.

Therefore the true length of 1 m, measured at 50 °C, should be l =1(1 + 11.2 x 10-6 x (50 – 10)}

= 1 + 11.2 x 10-6 x 40 = 1.000448 m

∴ The true length of measured 1.5 m of the brass rod at 50°C = 1.5x 1.000448 = 1.5007 m.

∴ The true length of the brass rod at 100°C

= 1.5007(1 + 0.000018(100-50)}

= 1.5007(1 +0.000018×50} = 1.502 m.

**Example 5. The brass scale of a barometer has no error at 0°C. α of brass = 0.00002°C ^{-1}. The reading of the barometer at 27°C is 75 cm. What is the actual reading of the barometer?**

**Solution:**

As the scale is error-less at 0°C, its length wil increase at 27°C.

Actual length corresponding to the measured value of 75 cm at 27 °C in this scale

= 75(1 + 0.00002 x 27} = 75 x 1.00054 = 75.0405 cm

∴Actual reading of the barometer = 75.0405 cm.

**Example 6. A steel ring is heated up to 95°C to fit exactly on the outer surface of an iron cylinder of diameter 10 cm at 20°C. After fitting, the ring is cooled so that the system attains a temperature 20°C. What is the thermal stress of the ring? [Young’s modulus of steel =21 x 10 ^{5} kg · cm^{-2} and α for steel = 12 x 10^{-6 °}C^{-1}]**

**Solution:**

Thermal stress = \(Y \alpha t=21 \times 10^5 \times 12 \times 10^{-6} \times 75\)

= \(1890 \mathrm{~kg} \cdot \mathrm{cm}^{-2}\)

= \(1890 \times 1000 \times 980 \mathrm{dyn} \cdot \mathrm{cm}^{-2}\)

= \(1.8522 \times 10^9 \mathrm{dyn} \cdot \mathrm{cm}^{-2}\)

**Example 7. A metre scale made ofsteelis to be so graduated that at any temperature a reading of 1 item should be correct up to 0.0005 mm. What can be the maximum allowed change in temperature while marking the millimetre gaps? Coefficient of linear expansion of steel = 13.22 x 10 ^{-6 °}C^{-1}.**

**Solution:**

Suppose while marking the millimetre gaps, the maximum change in temperature allowed is 81.

If the temperature increases, then at the maximum temperature, the value of 1 mm will be 1 + 0.0005 = 1.0005 mm .

∴ \(1.0005=1+\alpha \cdot \delta t=1+13.22 \times 10^{-6} \times \delta t\)

or, \(\delta t=\frac{0.0005}{13.22 \times 10^{-6}}=37.82^{\circ} \mathrm{C}\).

**Example 8. At 0°C, three rods of equal length form an equilateral triangle. Among the three rods, one is made of invar (with negligible expansion) and the other two rods are made of some other metal. When the triangle is heated up to 100°C, the angle between the two rods of the same metal changes to \(\left(\frac{\pi}{3}-\theta\right)\) Show that the coefficient of linear expansion of the metal is \(\frac{\sqrt{3} \theta}{200}{ }^{\circ} \mathbf{C}^{-1}\)**

**Solution:**

Suppose at 0°C the lengths of the rods are Z and the coefficient of linear expansion of the metal of AD and BD is α. The inner rod AB has no expansion.

If l_{1} is the length of each of the metal rods AD and BD at 100 °C, l_{1} = l(1 + 100α).

A perpendicular DO is drawn from the vertex D on AB

From the triangle ADO, we get,

⇒ \(\frac{\frac{l}{2}}{\sin \left(\frac{\pi}{6}-\frac{\theta}{2}\right)}=\frac{l_1}{\sin 90^{\circ}}\)

or, \(\frac{l}{2 \sin \left(\frac{\pi}{6}-\frac{\theta}{2}\right)}=\frac{l(1+100 \alpha)}{1}\)

or, \(\sin \left(\frac{\pi}{6}-\frac{\theta}{2}\right)=\frac{1}{2(1+100 \alpha)}\)

or, \(\sin \frac{\pi}{6} \cos \frac{\theta}{2}-\cos \frac{\pi}{6} \sin \frac{\theta}{2}=\frac{1}{2(1+100 \alpha)}\)

or, \(\frac{1}{2}-\frac{\sqrt{3}}{2} \cdot \frac{\theta}{2}=\frac{1}{2(1+100 \alpha)}\)

[Since θ is very small, \(\sin \frac{\theta}{2} \rightarrow \frac{\theta}{2} and \cos \frac{\theta}{2} \rightarrow 1] or, \quad \frac{\sqrt{3}}{2} \cdot \frac{\theta}{2}=\frac{1}{2}-\frac{1}{2(1+100 \alpha)}=\frac{100 \alpha}{2(1+100 \alpha)}\)]

or, \(\frac{\sqrt{3}}{2} \cdot \frac{\theta}{2}=\frac{1}{2}-\frac{1}{2(1+100 \alpha)}=\frac{100 \alpha}{2(1+100 \alpha)}\)

or, \(\frac{\sqrt{3} \theta}{2}=\frac{100 \alpha}{1+100 \alpha} \approx 100 \alpha\)

[As 100α is very small compared to 1, it is negligible]

or, \(\alpha=\frac{\sqrt{3} \theta^{\circ}}{200}{ }^{\circ} \mathrm{C}^{-1}\).

**Example 9. On each surface of a solid cube, a uniform pressure p is applied. Calculate the temperature rise of the solid cube so that its volume remains unaltered. Given, the coefficient of volume expansion of the material of the cube = γ and its bulk modulus of elasticity = B.**

**Solution:**

Let the initial volume of the cube be V and the change in volume due to the applied pressure be ΔV.

Hence, bulk modulus of elasticity,

B = \(\frac{p}{\frac{\Delta V}{V}} \quad \text { or, } \Delta V=\frac{p V}{B} \text {. }\)

∴ Due to the pressure applied the present volume of the cube = (V- ΔV).

Suppose with the rise in temperature by t, the reduced volume increases by ΔV so that the cube regains its initial volume.

∴ \(\Delta V=(V-\Delta V) \gamma t=\left(V-\frac{p V}{B}\right) \gamma t=\left(\frac{B-p}{B}\right) V \gamma t\)

or, \(\frac{p V}{B}=\left(\frac{B-p}{B}\right) \gamma V t \quad \text { or, } t=\frac{p}{(B-p) \gamma}\)

**Example 10. One end of a 100 cm long rod is fixed. At its free end a screw is attached and the pitch of the screw is 0.5 mm. The rod can move along its length on turning the screw. The screw has a circular scale with 100 divisions. It moves by one small-scale division of 0.5 mm per turn. At 20°C, the pitch scale reads a little over zero and the circular scale reads 92. When the temperature is increased to 100°C, the pitch scale reading changes to a little above 4 divisions and the circular scale reading is 72. Find the coefficient of linear expansion of the material of the rod.**

**Solution:**

Screw pitch = 0.5 mm and total number of circular scale divisions =100

∴ Least count of the screw = \(\frac{\text { screw pitch }}{\text { total no. of circular scale divisions }}\)=\(\frac{0.5}{100} \mathrm{~mm}\)

∴ The reading of the screw scale at 20° C = 0 x 0.5 + 92 x 0.005 = 0.46 mm ;

and the reading on that scale at 100°C = 4 x 0.5 + 72 x 0.005 = 2.36 mm

∴ The linear expansion of the rod = 2.36-0.46 = 1.9 mm = 0.19 cm

∴ Coefficient of linear expansion of the material of the rod = \(\frac{0.19}{100 \times(100-20)}\)=\(2.375 \times 10^{-5 \circ} \mathrm{C}^{-1}\)

**Example 11. Two rods of the same cross-sectional area are attached end to end forming a total length of 1 m, at 25°C. One rod in the combination is a 30 cm-long copper rod. The composite rod, Increases by 1.91 mm at 125°C. If the composite rod is rigidly fixed between two walls so that no change In length may occur even with the rise in temperature, find Young’s modulus (Y) and the coefficient of linear expansion (a) for the second rod. a for copper =1.7x 10 ^{-5} °C^{-1}, Y for copper = 1.3 x 10^{-11} N · m^{-2}.**

**Solution:**

Length of the 2nd rod at 25°C = 100 cm – length of the copper rod at 25 °C = 100-30 = 70 cm.

Increase in length of the copper rod for the rise in temperature

= 30 x 1.7 x 10^{-5} x (125-25) cm = 0.051 cm =0.51 mm

Increase in length of the 2nd rod for the same rise in temperature = 1.91-0.51 = 1.4 mm =0.14 cm

If the coefficient of linear expansion of the material of the second rod is α,

0.14 = \(70 \times \alpha \times(125-25)\)

∴ \(\alpha=\frac{0.14}{70 \times(125-25)}=2.0 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}p\)

Again, thermal stress = Yα(t_{2} – t_{1}). For copper rod, thermal stress =1.3 x 10^{11} x 1.7 x 10^{-5} x (t_{2}– t_{1}) and for the 2nd rod, the thermal stress = Y x 2 x 10^{-5 }x (t_{2}-t_{1}).

[Y = Young’s modulus for the second rod]

Since the two rods have the same area of cross-section, the thermal stresses should be equal.

∴ \(1.3 \times 10^{11} \times 1.7 \times 10^{-5} \times\left(t_2-t_1\right)\)

= \(Y \times 2 \times 10^{-5} \times\left(t_2-t_1\right)\)

or, \(Y=\frac{1.3 \times 10^{11} \times 1.7 \times 10^{-5}}{2 \times 10^{-5}}=1.105 \times 10^{11} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

**Example 12. An aluminium sphere at 20 °C Is kept at 1 standard atmosphere pressure, In a pressure chamber covered with oil. What should be the rise In pressure on the sphere to keep Its volume unchanged even at 35°C? Coefficient of linear expansion of aluminium, α = 23 x 10 ^{-6 }°C^{-1} and Its bulk modulus of elasticity, k = 7.7 x 10^{10} N · m^{-2}.**

**Solution:**

Let the volume of the sphere be V_{0} at 20°C and V at 35°C.

∴ V = \(V_0\{1+3 \alpha(35-20)\}=V_0\left\{1+3 \times 23 \times 10^{-6} \times 15\right\}\)

or, \(\frac{V}{V_0}=1+1035 \times 10^{-6} \quad \text { or, } \frac{V}{V_0}-1=1035 \times 10^{-6}\)

∴ \(\frac{V-V_0}{V_0}=1035 \times 10^{-6}\)

Let the required increase in pressure be p.

∴ Bulk modulus of elasticity, K = \(\frac{p}{\frac{V-V_0}{V_0}}\)

= \(\frac{7.7 \times 10^4 \times 1035}{101300}\) standard atmosphere

= 786.7 standard atmosphere

**Example 13. Two rods, each of coefficient of linear expansion α _{2} and length l_{2}, form the two sides of an isosceles triangle and the base is formed by another rod of length l_{1} and coefficient of linear expansion α_{1}. The base is fixed horizontally at its mid-point What should be the relationship between l_{1} and l_{2} so that the distance of the vertex from the mid-point of the base, does not change for any increase in temperature?**

**Solution:**

ABC is the isosceles triangle and D is the fixed point on the base.

From the figure,

AD = \(\sqrt{l_2^2-\left(\frac{l_1}{2}\right)^2}=x\) (say)

∴ \(x^2=l_2^2-\left(\frac{l_1}{2}\right)^2\)

Suppose with the rise in temperature by t, the length of AB and AC change to l’_{2}, BC to l’_{1} and AD to y.

∴ \(l_2^{\prime}=l_2\left(1+\alpha_2 t\right), l_1^{\prime}=l_1\left(1+\alpha_1 t\right)\)

∴ \(y^2 =l_2^{\prime 2}-\left(\frac{l_1^{\prime}}{2}\right)^2=\left\{l_2\left(1+\alpha_2 t\right)\right\}^2-\frac{l_1^2\left(1+\alpha_1 t\right)^2}{4}\)\(\approx l_2^2\left(1+2 \alpha_2 t\right)-l_1^2\left(\frac{1+2 \alpha_1 t}{4}\right)\)

[neglecting higher powers of α_{1} and α_{2}]

By condition, length AD should remain the same,

i.e. x² = y²

or, \(l_2^2-\frac{l_1^2}{4}=l_2^2\left(1+2 \alpha_2 t\right)-\frac{1}{4} l_1^2\left(1+2 \alpha_1 t\right)\)

or, \(2 l_2^2 \alpha_2 t=\frac{1}{2} l_1^2 \alpha_1 t\)

or, \(\frac{l_2^2}{l_1^2}=\frac{1}{4} \cdot \frac{\alpha_1}{\alpha_2} \quad or, \frac{l_2}{l_1}=\frac{1}{2} \sqrt{\frac{\alpha_1}{\alpha_2}}\).

**Example 14. Two metal plates of length l _{0} and width x are joined**

**together at temperature t by riveting them in such a way that the edges of the plates coincide. The coefficients of linear expansion of the materials of the plates are α**

_{1}and α_{2 }(α_{1}> α_{2}). When the bimetallic strip is heated to (t+ δ1) it bends and forms an arc of a circle. Find the radius of curvature of the strip.**Solution:**

A bimetallic strip bends on heating due to the dif-ference in values of a for the two metal plates forming the strip.

With the increase in temperature, suppose the lengths of the metal plates AB and CD change to l_{1} and l_{2}, and the radii of curvature are r_{1 }and r_{2} respectively. Let the angle subtended at the centre by the arc be ø.

∴ \(l_1=l_0\left(1+\alpha_1 \delta t\right)=r_1 \phi\) ….(1)

and \(l_2=l_0\left(1+\alpha_2 \delta t\right)=r_2 \phi\)…(2)

∴ \(\phi\left(r_1-r_2\right)=l_0\left(\alpha_1-\alpha_2\right) \delta t\)

∴ \(\phi=\frac{l_0\left(\alpha_1-\alpha_2\right) \delta t}{r_1-r_2 0}=\frac{l_0\left(\alpha_1-\alpha_2\right) \delta t}{2 x}\)….(3)

[because \(r_1-r_2\) =2x]

Adding (1) anil (2), \(\left(r_1+r_2\right) \phi=2 l_0+l_0\left(\alpha_1+\alpha_2\right) \delta \hat{t}\)

Let the average radius of curvature be r.

r = \(\frac{r_1+r_2}{2}=\frac{2 l_0+l_0\left(\alpha_1+\alpha_2\right) \delta t}{2 \phi}\)

= \(\frac{2 l_0+l_0\left(\alpha_1+\alpha_2\right) \delta t}{2 l_0\left(\alpha_1-\alpha_2\right) \delta t} \times 2 x\)=\(\frac{\left\{2+\left(\alpha_1+\alpha_2\right) \delta t\right\} x}{\left(\alpha_1-\alpha_2\right) \delta t}\)

**Example 15. One end of a 600 cm long steel bar Is fixed rigidly and the other endrests on a lever, 10 cm away from Its fulcrum. The lever turns through 2° when the bar is heated up to 50°C. Find the increase In length of the bar and Its coefficient of linear expansion.**

**Solution:**

The increase in length of the bar,

l = \(r \theta=10 \times \frac{2 \times \pi}{180}=\frac{\pi}{9}=0.349 \mathrm{~cm}\)

∴ Coefficient of linear expansion of the material of the bar,

α = \(\frac{\text { increase in length }}{\text { initial length } \times \text { rise in temperature }}\)

= \(\frac{0.349}{600 \times 50}=11.6 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1} .\)

**Example 16. Three rods of equal length at 0°C, are connected with one another to form an equilateral triangle ABC. The coefficient of linear expansion for the rod AB is a and that for the other two rods is β. What will be the Increment in the measure of the angle at C if the triangle is heated to t°C?**

**Solution:**

At 0°C, \(\angle C=60^{\circ}=\frac{\pi}{3} .\)

Suppose due to rise in temperature to t°C, ∠C = 2Φ where \(\phi=\frac{\pi}{6}+\theta\) (assuming that ∠C has increased by 2θ; θ is very small)

If l is the length of each rod at 0°C, then at t°C the lengths of AC and BC become l(1 +βt) and the length of AD becomes 1/2(1 + αt).

So for ΔACD in Fig.

⇒ \(\frac{l / 2(1+\alpha t)}{\sin \phi}=\frac{l(1+\beta t)}{\sin 90^{\circ}}\)

or, \(\sin \phi=\frac{l / 2(1+\alpha t)}{l(1+\beta t)}\)

or, \(\sin \left(\frac{\pi}{6}+\theta\right)=\frac{1+\alpha t}{2(1+\beta t)}\)

or, \(\sin \frac{\pi}{6} \cos \theta+\cos \frac{\pi}{6} \sin \theta\)

= \(\frac{1+\alpha t}{2(1+\beta t)}\)

or, \(\frac{1}{2} \cos \theta+\frac{\sqrt{3}}{2} \sin \theta=\frac{1}{2}\left(\frac{1+\alpha t}{1+\beta t}\right) \)

or, \(\frac{1}{2}+\frac{\sqrt{3}}{2} \cdot \theta=\frac{1}{2}\left(\frac{1+\alpha t}{1+\beta t}\right) \)

[because θ is small, sinθ=θ, cosθ=1]

or, \(\frac{\sqrt{3}}{2} \theta=\frac{1}{2}\left(\frac{1+\alpha t}{1+\beta t}-1\right) \)

or, \( \theta=\frac{1}{\sqrt{3}}\left(\frac{1+\alpha t-1-\beta t}{1+\beta t}\right)=\frac{(\alpha-\beta) t}{\sqrt{3}(1+\beta t)} \)

Increase in \(\angle C is 2 \theta=\frac{2(\alpha-\beta) t}{\sqrt{3}(1+\beta t)} \)

**Example 17. A steel wire with a cross-sectional area 0.4 mm² is fixed tightly at 20°C between two rigid supports. If the temperature falls to 0°C, what will be the tension on the string? Given, for steel, Y = 2 x 10 ^{12} dyn · cm^{-2} and α = 12 x 10^{-6} °C^{-1}.**

**Solution:**

The tension developed =YAαt

A = area of cross-section = 0.4 mm² = 0.004 cm²

Y = Young’s modulus = 2 x 10^{12} dyn • cm^{-2}

α = coefficient of linear expansion = 12 x 10^{-6} °C^{-1}

t = fall in temperature = 20-0 = 20°C

∴ Tension developed = 2 x 10^{12} x 0.004 x 12 x 10^{-6} x 20 = 19.2 x 10^{5} dyn.

**Example 16. Structure of an equilateral triangle ABC is made using three thin rods. Another rod AD connects the vertex A with D, the mid-point of BC. Coefficient of linear expansion for AB and AC is a and that for the base BC is β. Show that, if the coefficient of lin-ear expansion for the rod AD is 1/3 (4α – β), the arms of the system will not show any tendency to bend, for a small rise in temperature.**

**Solution:**

Let the initial length of each side of the triangle ABC be 2l

Hence, BD = CD = l

∴ AD = \(\sqrt{A B^2-B D^2}\)

= \(\sqrt{4 l^2-l^2}=\sqrt{3l}\)

Let γ be the coefficient of linear expansion for rod AD.

For a rise in temperature t,

changed length of AB = 2l(1 + αt),

changed length of BD = l(1 + βt) and

changed length of AD = √3(1 +γt)

If the sides do not bend, the triangle ABD continues to be a right-angled triangle,

i.e., \({[2 l(1+\alpha t)]^2=[l(1+\beta t)]^2+[\sqrt{3} l(1+\gamma t)]^2}\)

or, \(4 R^2\left(1+2 \alpha t+\alpha^2 t^2\right)=\mathbb{R}^2\left(1+2 \beta t+\beta^2 t^2\right)\)

+ \(3 l^2\left(1+2 \gamma t+\gamma^2 t^2\right)\)

or, \(4(1+2 \alpha t)=1+2 \beta t+3(1+2 \gamma t)\)

(Neglecting the terms containing α², β² and γ² for their very small values)

or, \(4+8 \alpha t=1+2 \beta t+3+6 \gamma t\) or, \(8 \alpha t=2 \beta t+6 \gamma t or, 4 \alpha=\beta+3 \gamma\) or, \(\gamma=\frac{1}{3}(4 \alpha-\beta).\)

**Example 19. A thin square metal plate has side of length i0. When the temperature of the plate is raised from 0°C to 100°C, its length increases by 1%. What is the percentage increase in area of the plate in this case?**

**Solution:**

At 100°C, the length of a side of the square metal plate, \(l=l_0 \times \frac{101}{100}\)

∴ \(l^2=l_0^2\left(\frac{101}{100}\right)^2\)

∴ The area of the metal plate at 100°C,

A = \(A_0\left(\frac{101}{100}\right)^2\left[A_0=\text { area of the plate at } 0^{\circ} \mathrm{C}\right]\)

∴ Increase in area, \(\Delta A=A-A_0=A_0\left(\frac{101}{100}\right)^2-A_0\)

= \(\frac{A_0}{(100)^2}\left[(101)^2-(100)^2\right]=\frac{201}{(100)^2} \cdot A_0\)

∴ Percentage increase, \(\frac{\Delta A}{A_0} \times 100=\frac{201}{(100)^2} \times 100=2.01\)

∴ Increase in area of the plate = 2.01%