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PART - 2 CH - 8 Mechanical Properties of Solids — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 8 Mechanical Properties of Solids5 Marks
Question
A $5 ~ m$ long steel wire hanging from the ceiling of a room has a mass of 25 kg at a radius of 10 cm . The height of the ceiling is 5.21 m . When the body is made to oscillate, it passes through touching the floor. If $Y$ is $2 \times 10^{11} N / m ^2$ and the radius of the wire is 0.05 cm . If so, what will be the velocity of the body at its lowest point?
✓

Answer

Depth of the centre of mass of the moving body (which is a sphere of radius 10 cm and 4 kg)
from the ceiling $=$ Length of the wire + Radius of the object
$\begin{array}{l}=5 m+10 cm \\=5+\frac{10}{100}=5.1 m\end{array}$
Depth of the lowest point of the object from the ceiling $=$ Length of the wire + Diameter of the object
$\begin{array}{l}=5 m+2 \times 10 cm \\=5.2 m\end{array}$
Height of the lowest point of the object from the floor $=$ Height of the ceiling from the floor - (Length of wire + Diameter of the object)
$=5.21-5.20=0.01 m$
$\because$ While oscillating, the body moves touching the floor at the lowest point.
$\therefore$ Increase in length of wire at lowest point $=$ 0.01 m and depth from the ceiling of the centre of mass of the body $=5.1+0.01=5.11 m$
Force applied on the body due to increase in length of wire by 0.01 m
$F=\frac{YAl}{L}$
Here, $\Delta L =$ Increase in length, $A =$ Cross-sectional area, $Y=$ Young's Modulus of elasticity of material of a wire, $L =$ Original length.
$\begin{aligned}
\therefore \quad F & =\frac{Y \times \pi r^2 l}{L} \\& =\frac{2 \times 10^{11} \times 3.14 \times\left(5 \times 10^{-4}\right)^2 \times(0.01)}{5} \\F & =\frac{2 \times 10^{11} \times 3.14 \times 25 \times 10^{-8} \times 10^{-2}}{5} \\
F & =\frac{2 \times 3.14 \times 25 \times 10^1}{5}=314 N\end{aligned}$
The second force on the object is its weight
$W=mg=25 \times 9.8$
$=245 N$, this force is applied downwards
Resultant force on the object
$F=314-245=69 N \text { upwards. }$
$\therefore$ The motion of the body is circular motion and provides resultant centripetal force on the body, if $v$ is the velocity of the body at lowest point then
$\begin{aligned}\frac{m v^2}{r} & =F^{\prime} \\\frac{25 \times v^2}{5.11} & =F^{\prime}=69 \\v^2 & =\frac{69 \times 5.11}{25} \\v & =\sqrt{\frac{69 \times 5.11}{25}} \\& =3.76 m / s\end{aligned}$

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