The rods of length L_1 and L_2 are made of ma | vedclass

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

Question

(B) Let the lengths of the rods at temperature $T$ be $L_1$ and $L_2$. At a temperature $T + \Delta T$,the new lengths are $L_1' = L_1(1 + \alpha_1 \D

Vedclass · Accepted Answer

$\frac{L_1}{L_2} = \frac{\alpha_2}{\alpha_1}$

Vedclass · Answer

(B) Let the lengths of the rods at temperature $T$ be $L_1$ and $L_2$. At a temperature $T + \Delta T$,the new lengths are $L_1' = L_1(1 + \alpha_1 \Delta T)$ and $L_2' = L_2(1 + \alpha_2 \Delta T)$. The difference between the lengths is given as independent of temperature,so $L_1' - L_2' = L_1 - L_2$. Substituting the expressions: $L_1 + L_1 \alpha_1 \Delta T - (L_2 + L_2 \alpha_2 \Delta T) = L_1 - L_2$. This simplifies to $L_1 \alpha_1 \Delta T = L_2 \alpha_2 \Delta T$. Since this holds for any $\Delta T$,we have $L_1 \alpha_1 = L_2 \alpha_2$. Therefore,$\frac{L_1}{L_2} = \frac{\alpha_2}{\alpha_1}$.

The rods of length $L_1$ and $L_2$ are made of materials whose coefficients of linear expansion are $\alpha_1$ and $\alpha_2$. If the difference between the two lengths is independent of temperature,then:

Similar Questions

Lactose is composed of .......

$A$ stationary charge produces . . . . . . .

Let $P$ be a plane passing through the points $(2,1,0)$,$(4,1,1)$ and $(5,0,1)$ and $R$ be the point $(2,1,6)$. Then the image of $R$ in the plane $P$ is

If $[2,6]$ is divided into four intervals of equal length,then the approximate value of $\int_2^6 \frac{1}{x^2-x} dx$ using Simpson's rule is

The centre of the circle $r^2-4r(\cos \theta+\sin \theta)-4=0$ in Cartesian coordinates is

Vedclass Test Series

Exam Paper Generator

Online Exam Module