Question 272: Properties of Matter

Question

The length of a wire becomes $$l_{1}$$ and $$l_{2}$$ when $$100 \mathrm{~N}$$ and $$120 \mathrm{~N}$$ tensions are applied respectively. If $$10 ~l_{2}=11~ l_{1}$$, the natural length of wire will be $$\frac{1}{x} ~l_{1}$$. Here the value of $$x$$ is _____________.

Accepted Answer

Given:
When tension $$T_1 = 100 \mathrm{~N}$$, extension $$= l_1 - l_0$$.
When tension $$T_2 = 120 \mathrm{~N}$$, extension $$= l_2 - l_0$$.
Now, let's write the equations using Hooke's Law:
$$100 = k(l_1 - l_0)$$
$$120 = k(l_2 - l_0)$$
Divide the first equation by the second equation:
$$\frac{5}{6} = \frac{l_1 - l_0}{l_2 - l_0}$$
Given the relationship between $$l_1$$ and $$l_2$$:
$$10l_2 = 11l_1$$
Now, let's solve for $$l_0$$:
$$5l_2 - 5l_0 = 6l_1 - 6l_0$$
$$l_0 = 6l_1 - 5l_2$$
Substitute the relationship between $$l_1$$ and $$l_2$$:
$$l_0 = 6l_1 - 5\left(\frac{11l_1}{10}\right)$$
$$l_0 = 6l_1 - \frac{11l_1}{2}$$
$$l_0 = \frac{l_1}{2}$$
Therefore, the natural length of the wire is $$\frac{1}{x}l_1 = \frac{2}{1}l_1 = 2l_1$$.
The value of $$x$$ is $$2$$.

Properties of Matter

The length of a wire becomes $l_{1} and l_{2} when 100 \mathrm{~N} and 120 \mathrm{~N} tensions are applied respectively. If 10 ~l_{2}=11~ l_{1}, the natural length of wire will be \frac{1}{x} ~l_{1}. Here the value of x$ is _____________.

Integer / Numeric Answer Type