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NEET PHYSICSWAVE OPTICSMedium

Question

Each of the two strings of lengths 51.6 cm51.6 \text{ cm} and 49.1 cm49.1 \text{ cm} is tensioned separately by 20 N20 \text{ N} of force. The mass per unit length of both strings is the same and equals 1 g/m1 \text{ g/m}. When both the strings vibrate simultaneously, the number of beats is:

A

5

B

7

C

8

D

3

Step-by-Step Solution

Given: Lengths of the strings, l1=51.6 cm=0.516 ml_1 = 51.6 \text{ cm} = 0.516 \text{ m} and l2=49.1 cm=0.491 ml_2 = 49.1 \text{ cm} = 0.491 \text{ m} Tension in the strings, T=20 NT = 20 \text{ N} Mass per unit length, μ=1 g/m=1×103 kg/m\mu = 1 \text{ g/m} = 1 \times 10^{-3} \text{ kg/m}

The velocity of a transverse wave in a stretched string is given by: v=Tμv = \sqrt{\frac{T}{\mu}} v=201×103=20000141.42 m/sv = \sqrt{\frac{20}{1 \times 10^{-3}}} = \sqrt{20000} \approx 141.42 \text{ m/s}

The fundamental frequency of a vibrating string of length ll is f=v2lf = \frac{v}{2l}. The frequencies of the two strings are: f1=v2l1=141.422×0.516=141.421.032137.03 Hzf_1 = \frac{v}{2l_1} = \frac{141.42}{2 \times 0.516} = \frac{141.42}{1.032} \approx 137.03 \text{ Hz} f2=v2l2=141.422×0.491=141.420.982144.01 Hzf_2 = \frac{v}{2l_2} = \frac{141.42}{2 \times 0.491} = \frac{141.42}{0.982} \approx 144.01 \text{ Hz}

The number of beats produced per second is the difference between the two frequencies: Number of beats=f2f1=144.01137.03=6.987\text{Number of beats} = f_2 - f_1 = 144.01 - 137.03 = 6.98 \approx 7

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE OPTICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVE OPTICSstringslengthstensionedseparatelylength

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