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There are two circular iron discs A and B having masses in the ratio 1:2 and diameter in the ratio 2:1. The ratio of their moment of inertia is
- a)4:1
- b)1:3
- c)2:1
- d)8:1
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
There are two circular iron discs A and B having masses in the ratio 1...
Given,
Mass of A=1,
Mass of B=2.
diameter if A=2,
diameter if B=1.
radius (r) of A=d/2=2/2=1.
radius (r) of B=d/2=1/2.
we know ,
moment of inertia of disc=MR2/2.
moment of inertia (I)of A/moment of inertia (I)of B=MR2/2/MR2/2.
(I) of A/(I) of B=1×12/2/2×(1/2)2/2.
=1×1/2/2×(1/4)/2.
=1/2/(1/2)/2.
=1/2/1/4.
=4/2.
=2/1.
Mass of A=1,
Mass of B=2.
diameter if A=2,
diameter if B=1.
radius (r) of A=d/2=2/2=1.
radius (r) of B=d/2=1/2.
we know ,
moment of inertia of disc=MR2/2.
moment of inertia (I)of A/moment of inertia (I)of B=MR2/2/MR2/2.
(I) of A/(I) of B=1×12/2/2×(1/2)2/2.
=1×1/2/2×(1/4)/2.
=1/2/(1/2)/2.
=1/2/1/4.
=4/2.
=2/1.
Most Upvoted Answer
There are two circular iron discs A and B having masses in the ratio 1...
Given:
Mass ratio of A and B = 1:2
Diameter ratio of A and B = 2:1
To find: Ratio of moment of inertia of A and B
Solution:
Let the masses of A and B be m1 and m2 respectively.
And let the diameters of A and B be d1 and d2 respectively.
We know that moment of inertia of a circular disc is given by:
I = (1/2) * m * r^2
where m is the mass of the disc and r is the radius of the disc.
Now, we can write:
m1/m2 = 1/2 (given)
d1/d2 = 2/1 (given)
As we know that the mass is proportional to the volume and volume is proportional to the cube of the diameter (assuming the thickness is the same in both discs), we can write:
m1/m2 = (d1/2)^3 / (d2/2)^3
m1/m2 = (d1^3/d2^3) / 8
m1/m2 = (d1^3/d2^3) * (1/8)
Now, substituting the value of m1/m2 in terms of d1 and d2 in the formula of moment of inertia, we get:
I1/I2 = (1/2) * m1 * (d1/4)^2 / [(1/2) * m2 * (d2/4)^2]
I1/I2 = (d1^2/d2^2) * (m1/m2)
I1/I2 = (d1^2/d2^2) * (d1^3/d2^3) * (1/8)
I1/I2 = (d1^5/d2^5) * (1/8)
I1/I2 = (d1^5/d2^5) * (1/8)
I1/I2 = (1/2) * (d1^2/d2^2)
Therefore, the ratio of moment of inertia of A and B is (d1^2/d2^2) i.e. 2:1 (using the given value of diameter ratio).
Hence, the correct option is (c) 2:1.
Mass ratio of A and B = 1:2
Diameter ratio of A and B = 2:1
To find: Ratio of moment of inertia of A and B
Solution:
Let the masses of A and B be m1 and m2 respectively.
And let the diameters of A and B be d1 and d2 respectively.
We know that moment of inertia of a circular disc is given by:
I = (1/2) * m * r^2
where m is the mass of the disc and r is the radius of the disc.
Now, we can write:
m1/m2 = 1/2 (given)
d1/d2 = 2/1 (given)
As we know that the mass is proportional to the volume and volume is proportional to the cube of the diameter (assuming the thickness is the same in both discs), we can write:
m1/m2 = (d1/2)^3 / (d2/2)^3
m1/m2 = (d1^3/d2^3) / 8
m1/m2 = (d1^3/d2^3) * (1/8)
Now, substituting the value of m1/m2 in terms of d1 and d2 in the formula of moment of inertia, we get:
I1/I2 = (1/2) * m1 * (d1/4)^2 / [(1/2) * m2 * (d2/4)^2]
I1/I2 = (d1^2/d2^2) * (m1/m2)
I1/I2 = (d1^2/d2^2) * (d1^3/d2^3) * (1/8)
I1/I2 = (d1^5/d2^5) * (1/8)
I1/I2 = (d1^5/d2^5) * (1/8)
I1/I2 = (1/2) * (d1^2/d2^2)
Therefore, the ratio of moment of inertia of A and B is (d1^2/d2^2) i.e. 2:1 (using the given value of diameter ratio).
Hence, the correct option is (c) 2:1.
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Community Answer
There are two circular iron discs A and B having masses in the ratio 1...
Given, Mass of A=1, Mass of B=2.diameter if A=2, diameter if B=1.radius (r) of A=d/2=2/2=1.radius (r) of B=d/2=1/2.we know , moment of inertia of disc=MR^2/2.moment of inertia (I)of A/moment of inertia (I)of B=MR^2/2/MR^2/2.(I) of A/(I) of B=1×1^2/2/2×(1/2)^2/2.=1×1/2/2×(1/4)/2.=1/2/(1/2)/2.=1/2/1/4.=4/2.=2/1.
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There are two circular iron discs A and B having masses in the ratio 1:2 and diameter in the ratio 2:1. The ratio of their moment of inertia isa)4:1b)1:3c)2:1d)8:1Correct answer is option 'C'. Can you explain this answer?
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