(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 x 10\(^{-4}\) at 20°C-has a gas pressure of 4.5 x 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 x 1 0\(^5\) Pa.
(d)(i)
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?
(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 x 10\(^{-4}\) at 20°C-has a gas pressure of 4.5 x 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 x 1 0\(^5\) Pa.
(d)(i)
| Ice point273k | 373kSteam point | |
| Resistance/\(\Omega\) | 5.67 | 7.75 |
| Pressure/Pa | 7.13 x 10\(^4\) | 9.74 x 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?
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
(a) A body is said to be in stable equillibrium if it tends to return to its original position after being slightly displaced.
(b)
i.e. V.R = Number of pulleys in the system
V.R = 3
(c) \(P_1 = 4.5 \times 10^5\)Pa; \(_2 = 4.6 \times 10^5\)Pa
T\(_1\) = 20\(^{o}\) + 273 = 293k
T\(_2\) = ?
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}; \frac{4.5 \times 10^5}{293} = \frac{4.6 \times 10^5}{T_2}\)
\(T_2 = \frac{4.6 \times 10^5 \times 293}{4.5 \times 10^5}\)
\(T_2\) = 299.5k or 26.5\(^o\)C
(d) \(\alpha\) Resistance thermometer
\(\frac{a}{b} = \frac{c}{d}; \frac{R \theta - 5.67}{7.75 - 5.67} = \frac{R - 273}{373 - 273}\)
\(\frac{R \theta - 5.67}{1,88} = \frac{R - 273}{100}\)
\(\beta\) Gas thermometer
\(\frac{a}{b} = \frac{c}{d}; \frac{G\theta - (7.13) x 10^4}{(9.74 - 7.13) \times 10^4} = \frac{G - 273}{373 - 273}\)
\(\frac{G\theta - 7.13}{2.16} = \frac{G - 273}{100}\)
(a) A body is said to be in stable equillibrium if it tends to return to its original position after being slightly displaced.
(b)
i.e. V.R = Number of pulleys in the system
V.R = 3
(c) \(P_1 = 4.5 \times 10^5\)Pa; \(_2 = 4.6 \times 10^5\)Pa
T\(_1\) = 20\(^{o}\) + 273 = 293k
T\(_2\) = ?
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}; \frac{4.5 \times 10^5}{293} = \frac{4.6 \times 10^5}{T_2}\)
\(T_2 = \frac{4.6 \times 10^5 \times 293}{4.5 \times 10^5}\)
\(T_2\) = 299.5k or 26.5\(^o\)C
(d) \(\alpha\) Resistance thermometer
\(\frac{a}{b} = \frac{c}{d}; \frac{R \theta - 5.67}{7.75 - 5.67} = \frac{R - 273}{373 - 273}\)
\(\frac{R \theta - 5.67}{1,88} = \frac{R - 273}{100}\)
\(\beta\) Gas thermometer
\(\frac{a}{b} = \frac{c}{d}; \frac{G\theta - (7.13) x 10^4}{(9.74 - 7.13) \times 10^4} = \frac{G - 273}{373 - 273}\)
\(\frac{G\theta - 7.13}{2.16} = \frac{G - 273}{100}\)