Two rods P and Q of different metals- having the same area A and the same length L- are placed between two rigid walls as shown in figure- The coefficients of linear expansion of P and Q are -1 and -2 respectively- and their Young-s moduli are Y1 and Y2 - The temperature of both rods is now raised by T degrees- The new length of the rod P is-

The correct option is C L_1=L [ 1+α _1T F/AY_1 ] L_P=L_1=L+Δ L =L+LT/y_1+y_2[y_1α _1 y_2α _2] =L+LT/y_1+y_2[y_1α _1 α _2y_2] =L+LT[α ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Thermal Expansion

Two rods P an...

Question

Two rods P and Q of different metals, having the same area A and the same length L, are placed between two rigid walls as shown in figure. The coefficients of linear expansion of P and Q are $α_{1}$ and $α_{2}$ respectively, and their Young's moduli are $Y_{1}$ and $Y_{2}$ . The temperature of both rods is now raised by T degrees. The new length of the rod P is:

L_{1} = L [1 + α_{1} T + \frac{F}{A Y_{1}}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

L_{1} = L [1 - α_{1} T + \frac{F}{A Y_{1}}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

L_{1} = L [1 + α_{1} T - \frac{F}{A Y_{1}}]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

L_{1} = L [1 - α_{1} T - \frac{F}{A Y_{1}}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $L_{1} = L [1 + α_{1} T - \frac{F}{A Y_{1}}]$
$L_{P} = L_{1} = L + Δ L$
$= L + \frac{L T}{y_{1} + y_{2}} [y_{1} α_{1} - y_{2} α_{2}]$
$= L + \frac{L T}{y_{1} + y_{2}} [y_{1} α_{1} - α_{2} y_{2}]$
$= L + L T [\frac{α_{1} y_{1} + α_{1} y_{2} - y_{2} α_{1} - α_{2} y_{2}}{y_{1} + y_{2}}]$
$= L + L α_{1} T - \frac{L y_{2} T}{(y_{1} + y_{2})} [α_{1} + α_{2}]$
$= L + L α_{1} T - \frac{L}{A y_{1}} \frac{y_{1} y_{2}}{(y_{1} + y_{2})} A T [α_{1} + α_{2}]$
$= L [1 + α_{1} T - \frac{F}{A y_{1}}]$

Suggest Corrections

Similar questions

Q. Two rods P and Q of different metals having the same area A and the same length L are placed between two rigid walls as shown in figure. The coefficients of linear expansion of P and Q are

α_{1}

and

α_{2}

respectively and their Youngs modulii are

Y_{1}

and

Y_{2}

. The temperature of both rods is now raised by T degrees. The new length of rod Q is

Q. Two uniform metal rods of length

l_{1}

and

l_{2}

and the linear coefficients of expansion

α_{1}

and

α_{2}

respectively are connected to form a single rod of length

l_{1} + l_{2}

. When the temperature of the combined rod is raised by

t^{o} C

, the length of each rod increases by the same amount. Then

\frac{α_{2}}{α_{1} + α_{2}}

is :

Two rods of different materials, having coefficients of linear expansion $α_{1}$ and $α_{2}$ and Young's moduli $Y_{1}$ and $Y_{2}$ respectively, are fixed between two rigid walls. The rods are heated to the same temperature. There is no bending of the rods. If $α_{1} : α_{2} = 2 : 3$ , the thermal stress developed in the two rods will be equal provided $Y_{1} : Y_{2}$ is equal to

Q. Two rods of lengths

l_{1}

and

l_{2}

are made of materials whose coefficients of linear expansion are

α_{1}

and

α_{2}

. If the difference between the two lengths is independent of temperature, then

Join BYJU'S Learning Program

The correct option is C L1=L[1+α1T−FAY1]LP=L1=L+ΔL=L+LTy1+y2[y1α1−y2α2]=L+LTy1+y2[y1α1−α2y2]=L+LT[α1y1+α1y2−y2α1−α2y2y1+y2]=L+Lα1T−Ly2T(y1+y2)[α1+α2]=L+Lα1T−LAy1y1y2(y1+y2)AT[α1+α2]=L[1+α1T−FAy1]