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Two circular discs A and B have equal masses and uniform thickness but have densities r1 and r2 such that r1 > r2. their moment of inertia is (a) I1 > I2 (b) I1 >> I2 (c) I1 < I2 (d) I1 = I2

Two circular discs A and B have equal masses and uniform thickness but have densities r1 and r2 such that r1 > r2. their moment of inertia is
(a) I1 > I2 (b) I1 >> I2
(c) I1 < I2 (d) I1 = I2

Grade:12

3 Answers

Arun
25750 Points
3 years ago
disc-A and disc-B has same mass. Density of disc-A is ρ1 and density of disc-B is ρ2 .
 
Since it is given that ρ1> ρ2 , then volume of disc-B is greater than volume of disc-A. 
But both disc has same thickness. Hence cross section area of disc-B is more or radius of disc-B is more.
 
For same mass of two circular discs, moment of inertia is greater for the disc that has greater radius. 
 
Hence disc-B has greater moment of inertia compare to disc-A
Vikas TU
14149 Points
3 years ago
mass = volume × density ⇒M=(πR2t)d…(1)
where t is thickness
I=MR^2/2 …(2)
From (1) and (2), we get :
I=M^2/2πdt⇒I∝1/d
⇒IA
Vikas TU
14149 Points
3 years ago
Ia is less than Ib …....................................................................................................

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