A 0.1 ,kg mass is suspended from a wire of ne | vedclass

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

Question

(C) The angular frequency $\omega$ of a mass $m$ suspended from a wire of length $L$, area $A$, and Young's modulus $Y$ is given by $\omega = \sqrt{\f

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The angular frequency $\omega$ of a mass $m$ suspended from a wire of length $L$, area $A$, and Young's modulus $Y$ is given by $\omega = \sqrt{\frac{YA}{mL}}$.Given: $m = 0.1 \,kg$, $L = 1 \,m$, $A = 4.9 	imes 10^{-7} \,m^2$, $\omega = 140 \,rad \,s^{-1}$.Squaring both sides: $\omega^2 = \frac{YA}{mL}$.Rearranging for $Y$: $Y = \frac{\omega^2 mL}{A}$.Substituting the values: $Y = \frac{(140)^2 	imes 0.1 	imes 1}{4.9 	imes 10^{-7}} = \frac{19600 	imes 0.1}{4.9 	imes 10^{-7}} = \frac{1960}{4.9 	imes 10^{-7}} = 400 	imes 10^7 = 4 	imes 10^9 \,N \,m^{-2}$.Comparing this with $n 	imes 10^9 \,N \,m^{-2}$, we get $n = 4$.

Similar Questions

In the given figure,if the dimensions of the two wires are same but materials are different,then Young's modulus is ........

Young's modulus of rubber is $10^4 \ N/m^2$ and the area of cross-section is $2 \ cm^2$. If a force of $2 \times 10^5 \ dynes$ is applied along its length,then its initial length $L$ becomes:

$A$ load $W$ produces an extension of $1 \ mm$ in a thread of radius $r$. If the load is increased to $4W$ and the radius is increased to $2r$,while all other factors remain the same,what will be the new extension in $mm$?

$A$ load of $2 \,kg$ produces an extension of $1 \,mm$ in a wire of $3 \,m$ in length and $1 \,mm$ in diameter. The Young's modulus of the wire will be .......... $Nm^{-2}$.

Write the equation of bending in a rod. Write the unit of bending and its dimensional formula.

Vedclass Test Series

Exam Paper Generator

Online Exam Module