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BITSAT 2020 Paper

Section: Physics

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Question : 38 of 150

Marks: +1, -0

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

1: 1
2: 1
1: 2
4: 1

Solution:

∵ Both wires are same materials so both will have same Young's modulus, and let it be Y.

Y=\frac{\text{stress}}{\text{strain}}=\frac{F}{A.\left(\frac{\Delta L}{L}\right)},

F = applied force

A = area of cross-section of wire

Now,

Y_{1}=Y_{2} \Rightarrow \frac{F L}{(A_{1})(\Delta L_{1})}

= \frac{F L}{(A_{2})(\Delta L_{2})}

Since load and length are same for both

\Rightarrow r_1^2 \Delta L_1 = r_2^2 \Delta L_2

,

\left( \frac{\Delta L_1}{\Delta L_2} \right) = \left( \frac{r_2}{r_1} \right)^2 = 4

\Delta L_1 : \Delta L_2 = 4 : 1

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