A sonometer wire supports a $4Kg$ load and vibrates in fundamental mode with a tuning fork of frequency $416Hz$ . The length of the wire between the bridges is now doubled. In order to maintain fundamental mode, the load should be changes to
A) $1Kg$
B) $2Kg$
C) $8Kg$
D) $16Kg$
Answer
299.7k+ views
Hint: Here in this question, we have to find the change of the load in fundamental mode for which the length of the wire is doubled between the bridges. This requires the usage of the fundamental frequency since it is at this frequency that the fundamental mode's load varies as the length is increased.
Formula Used:
Fundamental Frequency, ${f_ \circ } = \dfrac{V}{{4l}}$
Since, ${f_ \circ } = \dfrac{1}{{4l}}\sqrt {\dfrac{T}{\mu }} $
Formula of Sonometer, $T = mg$
Complete answer:
We know that, Fundamental Frequency,
${f_ \circ } = \dfrac{1}{{4l}}\sqrt {\dfrac{T}{\mu }} $
As according to the question, after increasing the length,
$l' = 2l$
As putting this value in Fundamental frequency equation, we get,
$f = \dfrac{1}{{4l'}}\sqrt {\dfrac{{T'}}{\mu }} $
After putting the value of $l' = 2l$ in the above equation, we get,
$f = \dfrac{1}{{4 \times 2l}}\sqrt {\dfrac{{T'}}{\mu }} $
In accordance with the query, it is implied that,
$f = {f_ \circ }$
By putting the value of both we get the following equation,
$\dfrac{1}{{4 \times 2l}}\sqrt {\dfrac{{T'}}{\mu }} = \dfrac{1}{{4l}}\sqrt {\dfrac{T}{\mu }} $
By further solution and using $T = mg$ in the above equation we get,
$\dfrac{1}{2}\sqrt {\dfrac{{m'g}}{\mu }} = \sqrt {\dfrac{{mg}}{\mu }} $
By doing solution by making them simple we get,
$\dfrac{1}{2}\sqrt {m'} = \sqrt m $
By taking square both side, we get,
$m' = 4 \times m$
By putting the value of m from the question we get,
$m' = 4 \times 4$
By completing the solution we get,
$m' = 16Kg$
Therefore, the correct answer for change of load in fundamental mode is $16Kg$ .
Hence, the correct option is (D).
Note: We all should know that, A stretched string's length, linear mass density, frequency, and tension are all essentially studied with a sonometer. Devices based on the resonance theory include sonometers. It is utilized to establish the tuning fork's frequency as well as the rules of vibration of stretched strings.
Formula Used:
Fundamental Frequency, ${f_ \circ } = \dfrac{V}{{4l}}$
Since, ${f_ \circ } = \dfrac{1}{{4l}}\sqrt {\dfrac{T}{\mu }} $
Formula of Sonometer, $T = mg$
Complete answer:
We know that, Fundamental Frequency,
${f_ \circ } = \dfrac{1}{{4l}}\sqrt {\dfrac{T}{\mu }} $
As according to the question, after increasing the length,
$l' = 2l$
As putting this value in Fundamental frequency equation, we get,
$f = \dfrac{1}{{4l'}}\sqrt {\dfrac{{T'}}{\mu }} $
After putting the value of $l' = 2l$ in the above equation, we get,
$f = \dfrac{1}{{4 \times 2l}}\sqrt {\dfrac{{T'}}{\mu }} $
In accordance with the query, it is implied that,
$f = {f_ \circ }$
By putting the value of both we get the following equation,
$\dfrac{1}{{4 \times 2l}}\sqrt {\dfrac{{T'}}{\mu }} = \dfrac{1}{{4l}}\sqrt {\dfrac{T}{\mu }} $
By further solution and using $T = mg$ in the above equation we get,
$\dfrac{1}{2}\sqrt {\dfrac{{m'g}}{\mu }} = \sqrt {\dfrac{{mg}}{\mu }} $
By doing solution by making them simple we get,
$\dfrac{1}{2}\sqrt {m'} = \sqrt m $
By taking square both side, we get,
$m' = 4 \times m$
By putting the value of m from the question we get,
$m' = 4 \times 4$
By completing the solution we get,
$m' = 16Kg$
Therefore, the correct answer for change of load in fundamental mode is $16Kg$ .
Hence, the correct option is (D).
Note: We all should know that, A stretched string's length, linear mass density, frequency, and tension are all essentially studied with a sonometer. Devices based on the resonance theory include sonometers. It is utilized to establish the tuning fork's frequency as well as the rules of vibration of stretched strings.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

