A thin steel ring of inner radius r and cross-sectional area A is fitted on to a wooden disc of radius R(R r). If Young's modulus be Y, then the tension in the steel ring is A Y((R-r/n r)). Find n.

Question

A thin steel ring of inner radius r and cross-sectional area A is fitted on to a wooden disc of radius R(R > r). If Young's modulus be Y, then the tension in the steel ring is A Y((R-r/n r)). Find n. (A)  (B)  (C)  (D)

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/90b9b65e1b67474726584190800fc020-.png" /> Strain =(Δ l/l)=((2 π R-2 π r/2 π r)) ⇒ (Δ l/l)=((R-r/r)) Y=(F / A/(Δ l)l) ⇒ Y (Δ l/l) A=F ⇒ F=A Y((R-r/r))

Q. A thin steel ring of inner radius $r$ and cross-sectional area $A$ is fitted on to a wooden disc of radius $R (R > r)$ . If Young's modulus be $Y$ , then the tension in the steel ring is $A Y (\frac{R - r}{n r})$ . Find $n$ .

Strain $= \frac{Δ l}{l} = (\frac{2 π R - 2 π r}{2 π r})$
$\Rightarrow \frac{Δ l}{l} = (\frac{R - r}{r})$
$Y = \frac{F / A}{\frac{Δ l}{l}}$
$\Rightarrow Y \frac{Δ l}{l} A = F$
$\Rightarrow F = A Y (\frac{R - r}{r})$

Q. A thin steel ring of inner radius r and cross-sectional area A is fitted on to a wooden disc of radius R(R>r). If Young's modulus be Y, then the tension in the steel ring is AY(nrR−r​). Find n.

Strain =lΔl​=(2πr2πR−2πr​) ⇒lΔl​=(rR−r​) Y=lΔl​F/A​ ⇒YlΔl​A=F ⇒F=AY(rR−r​)

Q. A thin steel ring of inner radius $r$ and cross-sectional area $A$ is fitted on to a wooden disc of radius $R (R > r)$ . If Young's modulus be $Y$ , then the tension in the steel ring is $A Y (\frac{R - r}{n r})$ . Find $n$ .

Strain $= \frac{Δ l}{l} = (\frac{2 π R - 2 π r}{2 π r})$
$\Rightarrow \frac{Δ l}{l} = (\frac{R - r}{r})$
$Y = \frac{F / A}{\frac{Δ l}{l}}$
$\Rightarrow Y \frac{Δ l}{l} A = F$
$\Rightarrow F = A Y (\frac{R - r}{r})$