A uniform heavy rod of weight 10 kg m s$$^{-2}$$, cross-sectional area 100 cm$$^2$$ and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is $$2 \times 10^{11}$$ N m$$^{-2}$$. Neglecting the lateral contraction, find the elongation of rod due to its own weight:

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.