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Grade Upto college level Engineering Entrance Exams

The radius of gyration of a disc of mass 100 gm and radius 5 cm about an axis passing through its centre of gravity and perpendicular to the plane is (a) 0.5 cm (b) 2.5 cm (c) 3.54 cm (d) 6.54 cm

Profile image of sumit kumar
12 Years agoGrade Upto college level
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2 Answers

Profile image of Saurabh Koranglekar
6 Years ago

To solve this problem, we use the formula for the radius of gyration (k), which is related to the moment of inertia (I) and the mass (M) of the object:

Formula:
I = M * k²
=> k = sqrt(I / M)

Step 1: Moment of Inertia of a Disc
The moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is:

I = (1/2) * M * R²

where,
M = 100 gm = 0.1 kg (converted to kg),
R = 5 cm = 0.05 m (converted to meters).

Substituting the values:

I = (1/2) * (0.1) * (0.05)²
= (0.05) * (0.0025)
= 0.000125 kg·m²

Step 2: Calculate Radius of Gyration (k)
Using the formula:

k = sqrt(I / M)

Substituting values:

k = sqrt(0.000125 / 0.1)
= sqrt(0.00125)
= 0.0354 m
= 3.54 cm

Answer:
The correct option is (c) 3.54 cm.

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Profile image of Vikas TU
6 Years ago
the moment of inertia of a disc passing thrugh the centre of the axis is given by
i=mr^2/2
I=100*25/2
i=50*25
i=1250
radius of gyration is given by k=under root i/m
therefore the radius of a gyration is
k=under root 1250/100
k=under root 12.5
 

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