Question 1 Report
(a) Define stable equilibrium as applied to a rigid body.
(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.
(c) At the beginning of a race, a tyre of volume \(8.0 \times 10^{-4}\) at 20°C-has a gas pressure of \(4.5 \times 10^5\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to \(4.6 \times 1 0^5\) Pa.
(d)(i)
Ice point 273k |
373k Steam point |
|
Resistance/\(\Omega\) |
5.67 | 7.75 |
Pressure/Pa |
\(7.13 \times 10^4\) | \(9.74 \times 10^4\) |
The table above shows readings of the resistance and pressure of a platinum resistance thermometer and a constant-volume gas thermometer respectively, when immersed in the same liquid bath. Use this data to determine the temperature of the bath on the: (\(\alpha\)) resistance thermometer; (\(\beta\)) gas thermometer
(ii) By what percentage is the temperature measured on the platinum resistance thermometer in error?
(a) Stable equilibrium
A rigid body is in stable equilibrium if, when it is slightly displaced and released, it tends to return to its original position. Its centre of gravity rises on displacement, so that its potential energy is a minimum at the equilibrium position.
(b) Block and tackle of velocity ratio 3
The rope is fixed to the movable block. It then passes over the left fixed pulley, round the movable pulley, and over the right fixed pulley. Thus, the movable block is supported by three tensioned sections of the rope, giving:
\[\text{V.R.}=3\]
(c) Temperature of the gas at the end of the race
Since the volume of the tyre is constant,
\[\frac{P_1}{T_1}=\frac{P_2}{T_2}\]
\[P_1=4.5\times10^5\ \text{Pa},\qquad P_2=4.6\times10^5\ \text{Pa}\]
\[T_1=20^\circ\text{C}+273=293\ \text{K}\]
\[T_2=\frac{P_2T_1}{P_1}=\frac{(4.6\times10^5)(293)}{4.5\times10^5}=299.5\ \text{K}\]
\[\boxed{T_2=299.5\ \text{K}\approx 26.5^\circ\text{C}}\]
(d)(i) Temperature scales
The calibration readings are:
| Fixed point | Temperature | Resistance | Gas pressure |
|---|---|---|---|
| Ice point | 273 K | 5.67 \(\Omega\) | \(7.13\times10^4\) Pa |
| Steam point | 373 K | 7.75 \(\Omega\) | \(9.74\times10^4\) Pa |
Let \(R_\theta\) be the resistance of the platinum thermometer in the liquid bath and \(P_\theta\) be the pressure of the gas thermometer in the same bath.
(α) Platinum resistance thermometer
\[\frac{R_\theta-5.67}{7.75-5.67}=\frac{T_R-273}{373-273}\]
\[\boxed{T_R=273+\frac{100(R_\theta-5.67)}{2.08}\ \text{K}}\]
(β) Constant-volume gas thermometer
\[\frac{P_\theta-7.13\times10^4}{(9.74-7.13)\times10^4}=\frac{T_G-273}{373-273}\]
\[\boxed{T_G=273+\frac{100(P_\theta-7.13\times10^4)}{2.61\times10^4}\ \text{K}}\]
(d)(ii) Percentage error of the platinum resistance thermometer
Taking the constant-volume gas thermometer temperature as the reference temperature,
\[\boxed{\%\,\text{error}=\frac{T_R-T_G}{T_G}\times100\%}\]
where \(T_R\) and \(T_G\) are obtained from the two boxed calibration equations above.
Everything you need to excel in your exams