Two wires W1 and W2 have the same radius r and respective densities ρ1 and ρ2 such that ρ2=4ρ1.  They are joined together at the point O, as shown in the figure.  The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges.  When a stationary wave is set up in the composite wire, the joint is found to be a node.  The ratio of the number of antinodes formed in W1 to W2 is :  
Option: 1  1 : 1
Option: 2  1 : 2
Option: 3  1 : 3
Option: 4 4 : 1
 

Answers (1)

When the joint is found to be a node.

then the frequency is given as 

n=\frac{p}{2 l} \sqrt{\frac{T}{\pi r^{2} d}}

Where p is also equal to the number of antinodes formed  in the wire

As

\begin{aligned} &n_{1}=n_{2}\\ &T \rightarrow \text { same }\\ &r \rightarrow \text { same }\\ &l \rightarrow \text { same } \end{aligned}

So

\begin{aligned} &\frac{p_{1}}{\sqrt{d_{1}}}=\frac{p_{2}}{\sqrt{d_{2}}}\\ &\frac{p_{1}}{p_{2}}=\frac{1}{2} \end{aligned}

 

 

Most Viewed Questions

Preparation Products

Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Buy Now
Test Series JEE Main 2024

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 7999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2023

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, Faculty Support.

₹ 9999/- ₹ 6999/-
Buy Now
Knockout JEE Main (One Month Subscription)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, Faculty Support.

₹ 7999/- ₹ 4999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions