Concepts Of Physics MCQ Edition [Volume 2]
- Exam
- Concepts Of Physics MCQ Edition [Volume 2]
- Subject
- Physics
- Chapter
- Heat and Temperature
Concepts Of Physics MCQ Edition [Volume 2] · Physics · Heat and Temperature
A horizontal aluminium plate has a hole with a diameter of 2.000 cm...
A horizontal aluminium plate has a hole with a diameter of . A steel sphere having a diameter of rests on this hole. These lengths are measured at an initial temperature of 10^{}^{{}^{\circ}} \text{C}. The temperature of the entire system is then gradually increased. Determine the temperature at which the ball will fall through the hole. The coefficient of linear expansion for aluminium is 23 \times 10^{-6} ^{}^{{}^{\circ}}\textC^{-1} and for steel is 11 \times 10^{-6} ^{}^{{}^{\circ}}\textC^{-1}.
Options
- A219^{}^{{}^{\circ}} \text{C}
- B157^{}^{{}^{\circ}} \text{C}
- C209^{}^{{}^{\circ}} \text{C}
- D199^{}^{{}^{\circ}} \text{C}
Correct answer
A. 219^{}^{{}^{\circ}} \text{C}
Step-by-step solution
Let the initial temperature be T_1 = 10^{}^{{}^{\circ}} \text{C} and the final temperature be .
The ball will fall through the hole when the diameter of the hole becomes equal to the diameter of the sphere.
\Delta T = \dfrac{0.005}{23.945 \times 10^{-6}} \approx 208.8^{}^{{}^{\circ}} \text{C}
T_2 = T_1 + \Delta T = 10^{}^{{}^{\circ}} \text{C} + 208.8^{}^{{}^{\circ}} \text{C} = 218.8^{}^{{}^{\circ}} \text{C} \approx 219^{}^{{}^{\circ}} \text{C}
The ball will fall through the hole when the diameter of the hole becomes equal to the diameter of the sphere.
\Delta T = \dfrac{0.005}{23.945 \times 10^{-6}} \approx 208.8^{}^{{}^{\circ}} \text{C}
T_2 = T_1 + \Delta T = 10^{}^{{}^{\circ}} \text{C} + 208.8^{}^{{}^{\circ}} \text{C} = 218.8^{}^{{}^{\circ}} \text{C} \approx 219^{}^{{}^{\circ}} \text{C}