A rod BC of negligible mass is fixed at end B | vedclass

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

Question

(A) The thermal contraction of the rod is $\Delta L_{thermal} = L \alpha \Delta 	heta = 4 	imes (2.2 	imes 10^{-4}) 	imes 100 = 0.088 \ m = 8.8 \

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The thermal contraction of the rod is $\Delta L_{thermal} = L \alpha \Delta 	heta = 4 	imes (2.2 	imes 10^{-4}) 	imes 100 = 0.088 \ m = 8.8 \ cm$.Let $x$ be the actual decrease in length of the rod. The spring is stretched by $x$ due to the contraction of the rod.The force exerted by the spring is $F = Kx$.The stress in the rod is $\sigma = \frac{F}{A} = \frac{Kx}{A}$.The strain in the rod is $\epsilon = \frac{\Delta L_{thermal} - x}{L}$.Using Hooke's Law,$Y = \frac{\sigma}{\epsilon} = \frac{Kx/A}{(\Delta L_{thermal} - x)/L} = \frac{KxL}{A(\Delta L_{thermal} - x)}$.Rearranging for $x$: $Y A (\Delta L_{thermal} - x) = KxL \implies Y A \Delta L_{thermal} = x(YA + KL)$.$x = \frac{Y A \Delta L_{thermal}}{YA + KL} = \frac{(10^{11}) 	imes (4 	imes 10^{-4}) 	imes 0.088}{(10^{11} 	imes 4 	imes 10^{-4}) + (10^4 	imes 4)} = \frac{3520}{40000 + 40000} = \frac{3520}{44000} = 0.08 \ m = 8 \ cm$.Wait,re-evaluating the calculation: $YA = 4 	imes 10^7$,$KL = 4 	imes 10^4$. $x = \frac{4 	imes 10^7 	imes 0.088}{4 	imes 10^7 + 4 	imes 10^4} = \frac{3520000}{40040000} \approx 0.0879 \ m = 8.79 \ cm$.The closest integer is $9 \ cm$.

Similar Questions

When a weight of $10 \,kg$ is suspended from a copper wire of length $3 \,m$ and diameter $0.4 \,mm$, its length increases by $2.4 \,cm$. If the diameter of the wire is doubled, then the extension in its length will be (in $\,cm$)

Two copper wires $A$ and $B$ of lengths in the ratio $1: 2$ and diameters in the ratio $3: 2$ are stretched by forces in the ratio $3: 1$. The ratio of the elastic potential energies stored per unit volume in the wires $A$ and $B$ is

The unit of Young's modulus is

There are two wires of the same material and same length. If the diameter of the second wire is two times the diameter of the first wire,then the ratio of the extension produced in the wires by applying the same load will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module