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Q.

Circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4 R is made from an iron plate of thickness. $\frac{t}{4}$ .Then, the relation between the moment of inertia I_X, and Iy is

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a

$I_{Y} = 32 I_{X}$

b

$I_{Y} = 16 I_{X}$

c

$I_{Y} = 64 I_{X}$

d

$I = I_{X}$

answer is A.

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Detailed Solution

The mass of the disc = Volume x density
$= {πR}^{2} t \times ρ$
For disc $X, I_{X} = \frac{1}{2} M_{1} R_{1}^{2}$
$= \frac{1}{2} {πR}^{4} tρ$
For disc Y,I_Y $= \frac{1}{2} π (4 R)^{2} (\frac{t}{4}) ρ (4 R)^{2}$
$\begin{aligned} = 32 {πR}^{4} tρ \\ ∴ I_{Y} = 64 I_{X} \end{aligned}$

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Circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4 R is made from an iron plate of thickness.t4.Then, the relation between the moment of inertia IX, and Iy is