two circular loops of radius r and nr are made from the same wire. the moment of inertia about the axis passing through the centre and perpendicular to the plane of the loop of the larger loop is 8 times that of the smaller loop what is the value of n.
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Answer : n = 2.
Circular loops (rings) of the wire:
Radii: r₁ = r , r₂ = n * r
Mass of the wire per unit length = d
Mass of loop₁ = m₁ = 2 π r₁ * d = 2π r d
Mass of loop₂ = m₂ = 2 π r₂ * d = 2π n r d
Moment of Inertia of a circular loop about an axis perpendicular to the plane of the loop and passing through the center of the loop :
I = M R²,
M = Mass, R = radius
MOI of given loops:
I₁ = (2π r d) r² = 2 π d r³
I₂ = (2π n r d) (n r)² = 2 π d n³ r³
Given I₂ = 8 I₁
=> 2π d n³ r³ = 8 * 2 π d r³
=> n³ = 8
=> n = 2.
Circular loops (rings) of the wire:
Radii: r₁ = r , r₂ = n * r
Mass of the wire per unit length = d
Mass of loop₁ = m₁ = 2 π r₁ * d = 2π r d
Mass of loop₂ = m₂ = 2 π r₂ * d = 2π n r d
Moment of Inertia of a circular loop about an axis perpendicular to the plane of the loop and passing through the center of the loop :
I = M R²,
M = Mass, R = radius
MOI of given loops:
I₁ = (2π r d) r² = 2 π d r³
I₂ = (2π n r d) (n r)² = 2 π d n³ r³
Given I₂ = 8 I₁
=> 2π d n³ r³ = 8 * 2 π d r³
=> n³ = 8
=> n = 2.
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