A wooden wheel of radius R is made of two semicircular parts (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area S and length L. L is slightly less than 2R. To fit the ring on the wheel, it is heated so that its temperature rises by T and it just steps over the wheel. As it cools down to surroundifng temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is R, and its Young's modulus is Y, the force that one part of the wheel applies on the other part is :

Question

A wooden wheel of radius R is made of two semicircular parts (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area S and length L. L is slightly less than 2R. To fit the ring on the wheel, it is heated so that its temperature rises by T and it just steps over the wheel. As it cools down to surroundifng temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is R, and its Young's modulus is Y, the force that one part of the wheel applies on the other part is : (A) 2π SY αΔ T (B) SY αΔ T (C) π SY αΔ T (D) 2SY αΔ T

Tardigrade Admin · Accepted Answer

L ⇒ S Δ L=L propto Δ T (F/A)=(Δ L/L)=Y F=αΔ T YS So, T=2F T=2αΔ TYS

$2 π S Y α Δ T$

$S Y α Δ T$

$π S Y α Δ T$

$2 S Y α Δ T$

$L \Rightarrow S$
$Δ L = L \propto Δ T$
$\frac{F}{A} = \frac{Δ L}{L} = Y$
$F = α Δ T Y S < b r / > S o, T = 2 F$
$T = 2 α Δ T Y S$

2πSYαΔT

SYαΔT

πSYαΔT

2SYαΔT

L⇒S ΔL=L∝ΔT AF​=LΔL​=Y F=αΔTYS<br/>So,T=2F T=2αΔTYS

$2 π S Y α Δ T$

$S Y α Δ T$

$π S Y α Δ T$

$2 S Y α Δ T$

$L \Rightarrow S$
$Δ L = L \propto Δ T$
$\frac{F}{A} = \frac{Δ L}{L} = Y$
$F = α Δ T Y S < b r / > S o, T = 2 F$
$T = 2 α Δ T Y S$