Consider two uniform discs of the same thickness and different radii $R_{1} = R$ and $R_{2} = α R$ made of the same material. If the ratio of their moments of inertia $I_{1}$ and $I_{2}$ , respectively, about their axes is $I_{1} : I_{2} = 1 : 16$ then the value of $α$ is:

Option 1 - 4

Option 2 - <math> <mroot> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> </math>

Option 3 - 2

Option 4 - <math> <mn>2</mn> <mroot> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> </math>  (Rotational Mechanics)

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Alok Kumar Singh | Contributor-Level 10

8 months ago

Correct Option - 3

Detailed Solution:

$\begin{array}{r} \Rightarrow V^{2} = U^{2} + 2 g S \\ \Rightarrow V^{2} = 0 + 2 g (h - y) \\ \Rightarrow V^{2} = 2 g h - 2 g y \\ \Rightarrow V = \sqrt[]{2 g h - 2 g y} \end{array}$

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A $\sqrt[]{34} m$ long ladder weighing 10kg leans on a frictionless wall. Its feet rest on the floor 3m away from the wall as shown in the figure. If F_f and F_w are the reaction forces of the floor and the wall, then ratio of F_w / F_f will be:

(Use g = 10 m/s²)

Alok Kumar Singh

mg = 10 × 10

= 100 of

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By NLM 1

Translational fB = NA

NB = mg = 100N

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Alok Kumar Singh

a = 6t² – 2t, of t 20, w = 10 rad/sec

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Consider two uniform discs of the same thickness and different radii R 1 = R and R 2 = α R made of the same material. If the ratio of their moments of inertia I 1 and I 2 , respectively, about their axes is I 1 : I 2 = 1 : 16 then the value of α is: