AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A 25 cm long string is fixed at both ends. It has a mass of 2.5 g and vibrates in its first overtone under tension. Also given is a closed pipe which is 30 cm long which vibrates in its fundamental frequency. When both of these are sounded simultaneously, 8 beats per second are heard. It is also observed that decreasing the tension in the string also decreases the beat frequency. What is the tension (in N) in the string? [Given, speed of sound in air is $360 {ms}^{- 1}$ ]

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

55.28 N

52.11 N

59.29 N

56.66 N

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$μ = \frac{2 .5}{25} = 0 .1 g / cm = 10^{- 2} kg / m$

$1^{st} overtone, λ_{s} = 25 cm = 0 .25 m \Rightarrow f_{s} = \frac{1}{λ_{s}} \sqrt{\frac{T}{μ}}$
Pipe in fundamental frequency,

$\frac{λ_{p}}{4} = 0 .3 \Rightarrow λ_{p} = 1 .2 m \Rightarrow f_{p} = \frac{v_{s}}{λ_{p}} = \frac{360}{1 .2} = 300 Hz$
$∵$ By decreasing the tension, beat frequency is decreased $∴ f_{s} > f_{p}$

$\Rightarrow f_{s} - f_{p} = 8 \Rightarrow \frac{1}{0 .25} \sqrt{\frac{T}{10^{- 2}}} - 300 = 8$

$\Rightarrow \frac{10 \times 100}{25} \sqrt{T} - 300 = 8 \Rightarrow \sqrt{T} = \frac{308 \times 25}{1000} \Rightarrow T = 59 .29 N$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!