In a sonometer wire,the tension is maintained | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density. Since $l$ and $\mu$ are constant,$n \propto \sqrt{T}$.Initially,the tension $T_1 = M g = 50.7 	imes 10 = 507 \, N$.When the mass is submerged in water,the buoyant force $F_B = V ho_{water} g = 0.0075 	imes 1000 	imes 10 = 75 \, N$.The new tension $T_2 = M g - F_B = 507 - 75 = 432 \, N$.Using the ratio $\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}}$,we get $\frac{260}{n_2} = \sqrt{\frac{507}{432}}$.$\frac{260}{n_2} = \sqrt{1.1736} \approx 1.0833$.$n_2 = \frac{260}{1.0833} \approx 240 \, Hz$.

Similar Questions

$A$ segment of wire vibrates with a fundamental frequency of $450 \,Hz$ under a tension of $9 \,kg-wt$. The tension at which the fundamental frequency of the same wire becomes $900 \,Hz$ is:

$A$ string in a musical instrument is $50 \ cm$ long and its fundamental frequency is $800 \ Hz$. If a frequency of $1000 \ Hz$ is to be produced,then the required length of the string is ..... $cm$.

$A$ wire of length $0.4 \,m$ stretched at both ends vibrates $250$ times per second. If the length of the wire is increased by $0.1 \,m$ and the stretching force is reduced to $1/4$ of its original value, then the new frequency is (in $\,Hz$)

$A$ sonometer wire $49 \ cm$ long is in unison with a tuning fork of frequency '$n$'. If the length of the wire is decreased by $1 \ cm$ and it is vibrated with the same tuning fork,$6$ beats are heard per second. The value of '$n$' is (in $Hz$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module