Contents
Some of the most important Physics Topics include energy, motion, and force.
What are the Advantages and Disadvantages of Solid Expansion?
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.
Advantages:
i) 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 tighdy around the wheel.
ii) Some holes are drilled through two metal plates by placing them one above the other. A rivet is heated and inserted through each hole [Fig.], 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.
iii) Thermostat: Thermostats are devices used for auto-matic temperature control. These make use of the prop-erty of a bimetallic strip which bends when heated. Thermostats are used in electrical oven, refrigerator,
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 ‘on’ above a fixed temperature as the thermostat controls the current flow. Thus, the temperature is also controlled.
iv) Fire alarm: In a fire alarm, an electric bell is connected 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 temperature, 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.
v) Temperature measurement: The curvature of a bimetallic 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 t0 the strip is upright, and bends to have a radius of curvature r at temperature t, then r ∝ \(\frac{1}{t-t_0}\), i.e., r(t – t0) = 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.
Disadvantages:
i) Railway tracks are made by connecting pieces of rails using fishplates [Fig.], 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 granite stones. This type of fixing can easily withstand the thermal stress developed, and the rails do not bend.
ii) 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.
iii) 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 surface remains colder, and hence expansion is less. This unequal expansion sets up a thermal stress and the glass cracks. In case of a thin wallec glass pot, heat transfer to the outer surface is easier and there is almost equal expansion of both the surfaces. The chance of cracking is reduced. Pyrex glass has a low coefficient of expansion. Hence, beaker, test tube, etc. for laboratory use are usually made of pyrex glass.
iv) A metal scale can measure true reading only at the tem-perature in which it has been graduated. At any other temperature, the interval between two consecutive graduations, increases or decreases, making the mea-surement inaccurate. Hence, the measured length has to be corrected using the value of α for the metal of the scale in use.
Numerical Examples
Example 1.
The difference in length of two metal rods A and B is 25 cm at all temperatures. Coefficients of linear expansion of the materials of A and B are 1.28 × 10-5°C and 1.92 × 10-5°C-1 respectively. Find the length of each rod at 0°C. [HS 01]
Solution:
Let lA and lB 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
∴ Length of rod A at 0°C is 75 cm, and that of rod B at 0°C is 50 cm.
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 also at 450°C. What are the lengths of the rods at 0°C? Given, α for iron = 12 × 10-6°C-1 and a for copper = 17 × 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 × 12 × 10-6(450 – 50) = x × 12 × 10-6 × 400
Increase in length of the copper rod
= y × 17 × 10-6(450 – 50) = y × 17 × 10-6 × 400
According to the question,
x × 12 × 10-6 × 400 = y × 17 × 10-6 × 400
or, 12x = 17y or, x = \(\frac{17}{12}\)y
∴ x – y = 2 or, x = 2 + y
or, 2 + y = \(\frac{17}{12}\)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 x0 and y0 respectively at 0°C.
∴ 6.8 = x0{1 + 12 × 10-6 × 50} or, x0 = 6.796 cm
∴ 4.8 = y0{ 1 + 17 × 10-6 × 50} or, y0 = 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 × 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 – 0.000015 × (25 – 15))
= 1 – 0.00015 = 0.99985 cm
∴ The correct length corresponding to the measured 80 cm
= 0.99985 × 80 =79.988 cm.
Example 4.
A steel scale is error-less 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 × 10-6°C-1 and 18 × 10-6°C-1 respectively.) [HS 2000]
Solution:
Let 50°F be equal to t°C, so that \(\frac{t}{5}\) = \(\frac{50-32}{9}\) i.e., t = 10. Hence, the steel scale is error-less at 10°C.
Therefore the true length of 1 m, measured at 50 °C, should be l = 1(1 + 11.2 × 10-6 × (50 – 10)}
= 1 + 11.2 × 10-6 × 40 = 1.000448 m
∴ The true length of measured 1.5 m of the brass rod at 50°C = 1.5 × 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 will increase at 27°C. Actual length corresponding to the measured value of 75 cm at 27°C in this scale
= 75{1 + 0.00002 × 27} = 75 × 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 × 105 kg ᐧ cm-2 and a for steel = 12 × 10-6°C-1.]
Solution:
Thermal stress = Yαt = 21 × 105 × 12 × 10-6 × 75
= 1890 kg ᐧ cm-2 = 1890 × 1000 × 980 dyn ᐧ cm-2
= 1.8522 × 109 dyn ᐧ cm-2
Example 7.
A metre scale made of steel is to be so graduated that at any temperature a reading of 1 mm 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 × 10-6°C-1.
Solution:
Suppose while marking the millimetre gaps, the maximum change in temperature allowed is δt. 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 + α ᐧ δt = 1 + 13.22 × 10-6 × δt
or, δt = \(\frac{0.0005}{13.22 \times 10^{-6}}\) = 37.82°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 (\(\frac{\pi}{3}\) – θ). Show that the coefficient of linear expansion of the metal is \(\frac{\sqrt{3} \theta}{200}\)°C-1.
Solution:
Suppose at 0°C the lengths of the rods are l and the coefficient of linear expansion of the metal of AD and BD is α. The inner rod AB has no expansion.
If l1 is the length of each of the metal rods AD and BD at 100 °C, l1 = l(1 + 100α).
A perpendicular DO is drawn from the vertex D on AB [Fig.],
From the triangle ADO, we get,
or, \(\frac{\sqrt{3} \theta}{2}\) = \(\frac{100 \alpha}{1+100 \alpha}\) \(\simeq\) 100α
[As 100α is very small compared to 1, it is negligible]
or, α = \(\frac{\sqrt{3} \theta}{200}\)°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 Δ V.
Hence, bulk modulus of elasticity,
B = \(\frac{\frac{p}{\Delta V}}{V}\) or, ΔV = \(\frac{p V}{B}\).
∴ 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 Δ V so that the cube regains its initial volume.
∴ ΔV = (V – ΔV)γt = (V – \(\frac{p V}{B}\))γt = \(\left(\frac{B-p}{B}\right)\)Vγt
or, \(\frac{p V}{B}\) = \(\left(\frac{B-p}{B}\right)\)γVt 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
\(=\frac{\text { screw pitch }}{\text { total no. of circular scale divisions }}\) = \(\frac{0.5}{100}\)mm
∴ The reading of the screw scale at 20° C = 0 × 0.5 + 92 × 0.005 = 0.46 mm;
and the reading on that scale at 100°C
= 4 × 0.5 + 72 × 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 × 10-5°C-1