Explanation
Normally, we use Searle's method to measure Young's modulus of some material. Young's modulus is independent of the shape of the material, so we can utilize any shape for its calculation. Searle’s apparatus is used for the measurement of Young’s modulus experimentally. This apparatus consists of two equal length wires that are attached to a rigid support.
Consider the expression for Young’s modulus.
$$ Y=\dfrac{FL}{A\Delta L} $$
$$ \Delta L=\dfrac{FL}{AY} $$
Where A is the area, F is the applied force, $$\Delta L$$ is the increase in length.
So,
$$ \Delta L=\dfrac{FL}{\pi {{r}^{2}}Y} $$
$$ \Delta L\propto \dfrac{1}{{{r}^{2}}} $$
$$ \dfrac{{{l}_{2}}}{{{l}_{1}}}={{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}} $$
$$ ={{\left( \dfrac{{{d}_{1}}/2}{{{d}_{2}}/2} \right)}^{2}} $$
$$ ={{(n)}^{2}} $$
$$ {{l}_{2}}=n^2\,{{l}_{1}} $$
Hence, on applying the same load, the length is increased by a factor of $${{n}^{2}}$$.
A wire of length L can support a load W. If the wire is broken in to two equal parts , then how much load can be suspended by one of those cut wires?
Breaking force area of cross-section of wire i.e. load hold by the wire does not depend upon the length of the wire.
The correct option is (b)
A rubber cord 10 m long is suspended vertically. How much does it stretch under its own weight. ( [Density of rubber is $$1500 (kg / m^3), Y = 5 \times 10^8 N/m^2)$$
We know that,
When there is no change in liquid level in vessel then,
$$\gamma _{real}^{'}=\gamma _{vessel}^{'}$$
Change in volume in liquid relative to vessel
$$ \Delta {{V}_{app}}=V\gamma _{app}^{'}\Delta \theta $$
$$ \Delta {{V}_{app}}=V\left( \gamma _{real}^{'}-\gamma _{vessel}^{'} \right) $$
Hence, the fraction is $$1:4$$
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