A sphere of mass $2 \mathrm{~kg}$ and diameter $4.5 \mathrm{~cm}$ is attached to the lower end of a steel wire of $2 \mathrm{~m}$ length and area of cross-section $0.24 imes 10^{-6} \mathrm{~m}^2$. The wire is suspended from $205 \mathrm{~cm}$ high ceiling of a room. When the system is made to…

Question

A sphere of mass $2 \mathrm{~kg}$ and diameter $4.5 \mathrm{~cm}$ is attached to the lower end of a steel wire of $2 \mathrm{~m}$ length and area of cross-section $0.24 	imes 10^{-6} \mathrm{~m}^2$. The wire is suspended from $205 \mathrm{~cm}$ high ceiling of a room. When the system is made to oscillate as a simple pendulum, the sphere just grazes the floor at its lowest position. The velocity of the sphere at the lowest position is (Young's modulus of steel $=2 	imes 10^{11} \mathrm{Nm}^{-2}$ and acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Aicharya · Accepted Answer

Extension in wire due to load and centrifugal force $ =\Delta l=(205-204.5) \mathrm{cm}=0.5 \mathrm{~cm} $ If velocity of sphere at lowest point is $v$, then $ \begin{gathered} Y=\frac{\left(M g+\frac{M v^2}{R}\right) L}{A \Delta l} \\ \Rightarrow \quad M g+\frac{M v^2}{R}=\frac{Y A \Delta l}{L} \\ \text { where, } \quad R=202.25 \mathrm{~cm} \\ \Rightarrow \quad 2 \times 10+\frac{2 \times v^2}{202.25 \times 10^{-2}} \\ =\frac{2 \times 10^{11} \times 0.24 \times 10^{-6} \times 0.5 \times 10^{-2}}{2} \\ =0.12 \times 10^3 \quad 2 \times v^2 \\ \Rightarrow \quad \frac{202.25 \times 10^{-2}}{202.00} \Rightarrow v^2=101.125 \\ \Rightarrow \quad v=10.05 \mathrm{~ms}^{-1} \text { or } v=10 \mathrm{~ms}^{-1} \end{gathered} $

A sphere of mass $2 \mathrm{~kg}$ and diameter $4.5 \mathrm{~cm}$ is attached to the lower end of a steel…

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