NEET 2022 Physics Simple Pendulum – Time Period & Phase MCQ Question

Chapter: Oscillations

Class: 11 Physics Topic: Simple Pendulum – Time Period & Phase Difficulty: 🟡 Medium

✅ Ans: C — 11 vibrations

For a simple pendulum:

$$T=2\pi\sqrt{\frac{L}{g}}$$

So,

$$T\propto\sqrt L$$

Given:

$$L_1=121\,cm,\qquad L_2=100\,cm$$

Therefore,

$$\frac{T_1}{T_2} =\sqrt{\frac{121}{100}} =\frac{11}{10}$$

So the time periods are in the ratio:

$$T_{\text{long}}:T_{\text{short}}=11:10$$

Step 2: Find when they are again in the same phase

In the same amount of time:

$$\text{Number of vibrations}\propto\frac{1}{T}$$

Therefore,

$$N_{\text{long}}:N_{\text{short}} =\frac{1}{11}:\frac{1}{10} =10:11$$

Thus, when the longer pendulum completes 10 vibrations, the shorter pendulum completes 11 vibrations.

$$\boxed{N_{\text{short}}=11}$$

❌ Why other options are wrong?

• 10 ❌ → the shorter pendulum completes 10 vibrations, but the longer completes $10\times10/11$, not an integer.
• 8 ❌ → they do not return to the same phase.
• 9 ❌ → they do not return to the same phase.
• 11 ✅ → longer completes 10 and shorter completes 11; both are again at the mean position in the same phase.

📌 NCERT Concept

$$\boxed{T=2\pi\sqrt{\frac{L}{g}}}$$

Hence, time period increases with the square root of length.

🧠 NEET Trick

For lengths 121 cm and 100 cm:

$$\sqrt{121}:\sqrt{100}=11:10$$

Time-period ratio = 11:10

Vibration ratio = 10:11

Therefore:

$$\boxed{\text{Shorter pendulum = 11 vibrations}}$$

Final Answer: $\boxed{C-11}$.