Question

A number of holes are drilled along a diameter of a disc of radius $R$ . To get minimum time period of oscillations the disc should be suspended from a horizontal axis passing through a hole whose distance from the centre should be

• A. $\frac { R } { 2 }$
• B. $\frac { R } { \sqrt { 2 } }$
• C. $\frac { R } { 2 \sqrt { 2 } }$
• D. Zero

Answer 1

Answer: (B)

$\frac { R } { \sqrt { 2 } }$

Answer 2

$T = 2 \pi \sqrt { \frac { L } { g } }$ where $L = \frac { l ^ { 2 } + k ^ { 2 } } { l }$

Here $k ^ { 2 } = \frac { R ^ { 2 } } { 2 }$ $\therefore$ $L = \frac { l ^ { 2 } + \frac { R ^ { 2 } } { 2 } } { l }$ $= l + \frac { R ^ { 2 } } { 2 l }$

For minimum time period $L$ should be minimum $\frac { d L } { d l } = 0$

⇒ $\frac { d } { d l } \left( l + \frac { R ^ { 2 } } { 2 l } \right) = 0$

⇒ $1 + \frac { R ^ { 2 } } { 2 } \left( \frac { - 1 } { l ^ { 2 } } \right) =$ $1 - \frac { R ^ { 2 } } { 2 l ^ { 2 } } = 0$ ⇒ $l = \frac { R } { \sqrt { 2 } }$.