MathJax Widget Testin
An Al-Fatah missile has an initial velocity ( vec{u} ). It is subjected to a constant force ( vec{F} ) for t seconds, causing a constant acceleration ( vec{a} ). The force is not in the same direction as the initial velocity. A vector diagram is drawn to find the final velocity ( vec{v} ).
\[ \frac{1}{2}mv^2 \]
\[ \frac{1}{2}mv^2 \]
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
\[ \frac{1}{2}mv^2 \]
Detailed Explanation for Question 1
Question:
A brass pendulum clock is calibrated at 20°C. If the temperature rises to 40°C, how much time will the clock lose per day? (Coefficient of linear expansion of brass = 19×10⁻⁶/°C)
Step-by-Step Solution:
1. Understand the relationship between temperature and pendulum period:
The period \( T \) of a simple pendulum is given by:
where \( L \) is the length of the pendulum.
- When temperature increases, the brass pendulum expands, increasing its length \( L \).
- Since \( T \propto \sqrt{L} \), the period \( T \) also increases.
- A longer period means the clock runs slower (loses time).
2. Calculate the change in length (\( \Delta L \)):
The change in length due to thermal expansion is:
where:
- \( L_0 \) = original length,
- \( \alpha = 19 \times 10^{-6} \, \text{°C}^{-1} \) (coefficient of linear expansion for brass),
- \( \Delta T = 40°C – 20°C = 20°C \).
3. Relate the change in length to the change in period (\( \Delta T \)):
Since \( T \propto \sqrt{L} \), the fractional change in period is:
This means the period increases by \( 1.9 \times 10^{-4} \) of its original value.
4. Calculate time lost per day:
- In one day, there are \( 86,400 \) seconds.
- If the clock runs slower, the time lost per day is:
\[ \text{Time lost} \approx 16.4 \, \text{seconds} \]
Final Answer: (b) 16.4 seconds
Key Concept:
- Thermal expansion increases the pendulum’s length, slowing its period.
- The fractional time lost is half the fractional length increase (because \( T \propto \sqrt{L} \)).
- This is a classic problem combining thermal physics and oscillations.
Note: For competitive exams, memorizing that a pendulum loses \( \approx 0.5 \alpha \Delta T \times \text{time interval} \) can save calculation time.
