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# The moment of inertia of the disc used in a torsional pendulum about the suspension wire is $0.2kg - {m^2}$ . It oscillates with a period of $2s$ . Another disc is placed over the first one and the time period of the system becomes $2.5s$ . Find the moment of inertia of the second disc about the wire. (A) 0.12(B) 0.13(C) 0.11(D) 0.07

Last updated date: 20th Jun 2024
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Hint: We know that the time period of the torsional pendulum is given by –
$T = 2\pi \sqrt {\dfrac{I}{K}}$ . Also, the second disc is placed over the first disc it means the moment of inertia about the axis passing through the torsional wire increases and this increases the time period of the torsional pendulum from $2s$ to $2.5s$ . We will find the value in the time period in both cases and divide both the equation to get the value of the moment of inertia of the second disc about the wire

It is given in the question that the moment of inertia of the disc used in the torsional pendulum about the suspension wire is $0.2kg - {m^2}$ , let’s say this as I. The time period of oscillation is $2s$ .
Another disc is placed over the first one and the time period of the system becomes $2.5s$ . Then, we have to find the moment of inertia of the second disc about the wire.

As the second disc is placed over the first disc it means the moment of inertia about the axis passing through the torsional wire increases and this increases the time period of the torsional pendulum form $2s$ to $2.5s$ .
Let us assume that the momentum of inertia of the second disc be ${I_1}$ and the torsional constant of the wire be K.
We know that the time period of the torsional pendulum is given by –
$T = 2\pi \sqrt {\dfrac{I}{K}}$ .
Here ‘I’ is the moment of inertia of the disc and K is the torsional constant of the wire.
As in the first case, the time period of the torsional pendulum is $2s$ and the moment of inertia of the disc used in torsional pendulum about the suspension wire is $0.2kg - {m^2}$ we get-
$2 = 2\pi \sqrt {\dfrac{{0.2}}{K}}$ … (1)
In case two we have the time period of the torsional pendulum is $2.5s$ and the moment of inertia of
the second disc be ${I_1}$ then it is expressed by-
$2.5 = 2\pi \sqrt {\dfrac{{0.2 + {I_1}}}{K}}$ … (2)
On dividing equation (2) with the equation (1), we get-
$\dfrac{{2.5}}{2} = \dfrac{{2\pi \sqrt {\dfrac{{0.2 + {I_1}}}{K}} }}{{2\pi \sqrt {\dfrac{{0.2}}{K}} }}$
Cancelling the similar terms both the sides we get-
$\dfrac{{2.5}}{2} = \dfrac{{\sqrt {0.2 + {I_1}} }}{{\sqrt {0.2} }}$
On squaring both the sides we get-
$\dfrac{{6.25}}{4} = \dfrac{{0.2 + {I_1}}}{{0.2}}$
On cross multiplying both the sides, we get-
$6.25 \times 0.2 = 4\left( {0.2 + {I_1}} \right)$
$1.25 = 0.8 + 4{I_1}$
Transposing 0.8 from R.H.S from L.H.S we get-
$1.25 - 0.8 = 4{I_1}$
${I_1} = \dfrac{{0.45}}{4}$
${I_1} = 0.1125$
So, the new moment of inertia of the second disc about the wire will be ${I_1} = 0.1125$ .

Therefore, option C is correct.

In general, the moment of inertia of a uniform disc about an axis through its centre and perpendicular to its plane is given by $I = 12M{R^2}$ . Here M is the mass of the disc and R is the distance of the centre of the disc from the axis.