We are asked to find the extension of a vertically hanging uniform rod caused solely by its own weight. The data given are:
Weight of the rod $$W = 10\;{\rm kg\,m\,s^{-2}} = 10\;{\rm N}$$
Length $$L = 20\;{\rm cm} = 0.20\;{\rm m}$$
Cross-sectional area $$A = 100\;{\rm cm^2}$$
Because $$1\;{\rm cm} = 10^{-2}\;{\rm m},$$ we have
$$A = 100\;(10^{-2}\,{\rm m})^{2} = 100 \times 10^{-4}\;{\rm m^2} = 10^{-2}\;{\rm m^2} = 0.01\;{\rm m^2}.$$
Young’s modulus of the material is given as
$$Y = 2 \times 10^{11}\;{\rm N\,m^{-2}}.$$
Let us denote by $$x$$ the distance from the fixed support down the rod.
At a section situated at this distance, the portion of the rod lying below that section has a length $$(L - x)$$ and therefore weighs
$$P(x) = \frac{W}{L}\;(L - x).$$
The factor $$\dfrac{W}{L}$$ is simply the weight per unit length, so we first evaluate it:
$$\frac{W}{L} = \frac{10\;{\rm N}}{0.20\;{\rm m}} = 50\;{\rm N\,m^{-1}}.$$
Hence the tensile force acting at a distance $$x$$ from the top is
$$P(x) = 50\,(L - x)\;{\rm N}.$$
For a small element of thickness $$dx$$ situated at that point, the corresponding elongation $$d\ell$$ is obtained from Hooke’s (Young’s) law in the form
$$d\ell = \frac{P(x)}{A\,Y}\;dx.$$
Substituting $$P(x)$$, we get
$$d\ell = \frac{50\,(L - x)}{A\,Y}\;dx.$$
To find the total extension of the rod, we integrate $$d\ell$$ from the top end $$x = 0$$ to the bottom end $$x = L$$:
$$\Delta L = \int_{0}^{L} d\ell
= \int_{0}^{L} \frac{50\,(L - x)}{A\,Y}\;dx.$$
The constants $$\dfrac{50}{A\,Y}$$ can be taken outside the integral:
$$\Delta L = \frac{50}{A\,Y}\int_{0}^{L}(L - x)\;dx.$$
We now evaluate the integral:
$$\int_{0}^{L}(L - x)\;dx
= \bigg[Lx - \frac{x^{2}}{2}\bigg]_{0}^{L}
= \left(L\cdot L - \frac{L^{2}}{2}\right)
= \frac{L^{2}}{2}.$$
Therefore
$$\Delta L = \frac{50}{A\,Y}\,\frac{L^{2}}{2}
= \frac{50\,L^{2}}{2\,A\,Y}.$$
Substituting the numerical values now:
$$\Delta L = \frac{50\,(0.20\,{\rm m})^{2}}{2 \times 0.01\,{\rm m^{2}} \times 2 \times 10^{11}\,{\rm N\,m^{-2}}}.$$
We first compute $$L^{2}$$:
$$L^{2} = (0.20)^{2} = 0.04\;{\rm m^{2}}.$$
Multiplying by 50:
$$50 \times 0.04 = 2.0.$$
Now the denominator:
$$2 \times 0.01 \times 2 \times 10^{11} = 0.02 \times 2 \times 10^{11} = 0.04 \times 10^{11} = 4 \times 10^{9}.$$
Finally,
$$\Delta L = \frac{2.0}{4 \times 10^{9}}
= 0.5 \times 10^{-9}\;{\rm m}
= 5 \times 10^{-10}\;{\rm m}.$$
Hence, the correct answer is Option A.
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