The difference in length between an iron rod | vedclass

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

Question

(A) Given that the difference in length is constant at all temperatures,the change in length for both rods must be equal for any change in temperature

Vedclass · Accepted Answer

$28.3 \ cm, 18.3 \ cm$

Vedclass · Answer

(A) Given that the difference in length is constant at all temperatures,the change in length for both rods must be equal for any change in temperature $\Delta T$.Let $L_{Fe}$ and $L_{Cu}$ be the lengths of the iron and copper rods respectively.We are given $L_{Fe} - L_{Cu} = 10 \ cm$ ..... $(i)$Since the difference is constant,$\Delta L_{Fe} = \Delta L_{Cu}$.Using the formula for linear expansion $\Delta L = L \alpha \Delta T$,we get:$L_{Fe} \alpha_{Fe} \Delta T = L_{Cu} \alpha_{Cu} \Delta T$$L_{Fe} \alpha_{Fe} = L_{Cu} \alpha_{Cu}$$\frac{L_{Fe}}{L_{Cu}} = \frac{\alpha_{Cu}}{\alpha_{Fe}} = \frac{17 	imes 10^{-6}}{11 	imes 10^{-6}} = \frac{17}{11}$ ..... $(ii)$From $(i)$,$L_{Fe} = L_{Cu} + 10$.Substituting this into $(ii)$:$\frac{L_{Cu} + 10}{L_{Cu}} = \frac{17}{11}$$11(L_{Cu} + 10) = 17 L_{Cu}$$11 L_{Cu} + 110 = 17 L_{Cu}$$6 L_{Cu} = 110$$L_{Cu} = \frac{110}{6} \approx 18.33 \ cm$$L_{Fe} = 18.33 + 10 = 28.33 \ cm$.Thus,the lengths are approximately $28.3 \ cm$ and $18.3 \ cm$.

The difference in length between an iron rod and a copper rod is $10 \ cm$ at all temperatures. If ${\alpha _{Fe}} = 11 \times {10^{ - 6}} \, ^\circ C^{ - 1}$ and ${\alpha _{Cu}} = 17 \times {10^{ - 6}} \, ^\circ C^{ - 1}$,what are their lengths respectively?

Similar Questions

$A$ solid ball of metal has a concentric spherical cavity within it. If the ball is heated,the volume of the cavity will

$A$ cylindrical metal rod of length $L_0$ is shaped into a ring with a small gap as shown. On heating the system,

An iron sphere having diameter $D$ and mass $M$ is immersed in hot water so that the temperature of the sphere increases by $\delta T$. If $\alpha$ is the coefficient of linear expansion of the iron,then the change in the surface area of the sphere is:

When the temperature of a body increases from $T$ to $T+\Delta T$, its moment of inertia increases from $I$ to $I+\Delta I$. If $\alpha$ is the coefficient of linear expansion of the material of the body, then $\frac{\Delta I}{I}$ is (neglect higher orders of $\alpha$ ).

Vedclass Test Series

Exam Paper Generator

Online Exam Module