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Rotational Motion question

2020 · 9 Jan · Shift 1 · Q49
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Rotational Motion question

2020 · 9 Jan · Shift 1 · Q49

JEE MainPhysicsRotational MotionMCQ+4 / −1
JEE Main 2020 (Online) 9th January Morning Slot Physics - Rotational Motion Question 152 English Three solid spheres each of mass m and diameter d are stuck together such that the lines connecting the centres form an equilateral triangle of side of length d. The ratio I0/IA of moment of inertia I0 of the system about an axis passing the centroid and about center of any of the spheres IA and perpendicular to the plane of the triangle is :
  1. A
    1323{{13} \over {23}}2313​
  2. B
    2313{{23} \over {13}}1323​
  3. C
    1513{{15} \over {13}}1315​
  4. D
    1315{{13} \over {15}}1513​
View written solutionFree

Correct answer: A

  1. Given data
  • Three identical solid spheres
  • Mass of each sphere =m= m=m
  • Diameter of each sphere =d= d=d so radius of each sphere is r=d2r = \frac d2r=2d​
  • Their centres form an equilateral triangle of side ddd.

We need the ratio IOIA\frac{I_O}{I_A}IA​IO​​ where:

  • IOI_OIO​ = moment of inertia of the system about an axis through the centroid of the triangle and perpendicular to its plane.
  • IAI_AIA​ = moment of inertia of the system about an axis through the centre of any one sphere and perpendicular to the plane.

  1. Moment of inertia of one solid sphere about its own centre

For a solid sphere, Icm=25mr2=25m(d2)2=md210I_{\text{cm}} = \frac{2}{5}mr^2 = \frac{2}{5}m\left(\frac d2\right)^2 = \frac{md^2}{10}Icm​=52​mr2=52​m(2d​)2=10md2​


  1. Distance of each sphere centre from centroid of the equilateral triangle

For an equilateral triangle of side ddd, distance from centroid to each vertex is R=d3R = \frac{d}{\sqrt{3}}R=3​d​

So for each sphere, using parallel axis theorem about the centroidal axis, Ione about O=Icm+mR2I_{\text{one about }O} = I_{\text{cm}} + mR^2Ione about O​=Icm​+mR2 =md210+m(d3)2= \frac{md^2}{10} + m\left(\frac{d}{\sqrt{3}}\right)^2=10md2​+m(3​d​)2 =md210+md23= \frac{md^2}{10} + \frac{md^2}{3}=10md2​+3md2​

For three spheres, IO=3(md210+md23)I_O = 3\left(\frac{md^2}{10} + \frac{md^2}{3}\right)IO​=3(10md2​+3md2​) =3md2(110+13)= 3md^2\left(\frac{1}{10} + \frac{1}{3}\right)=3md2(101​+31​) =3md2(1330)= 3md^2\left(\frac{13}{30}\right)=3md2(3013​) =1310md2= \frac{13}{10}md^2=1013​md2


  1. Moment of inertia about axis through centre of one sphere

Let the axis pass through the centre of sphere AAA and be perpendicular to the plane.

  • For sphere AAA itself: IA(1)=md210I_A^{(1)} = \frac{md^2}{10}IA(1)​=10md2​

  • The other two sphere centres are each at distance ddd from this axis, since side of triangle is ddd.

For each of those spheres, IA(2)=Icm+md2=md210+md2=1110md2I_A^{(2)} = I_{\text{cm}} + md^2 = \frac{md^2}{10} + md^2 = \frac{11}{10}md^2IA(2)​=Icm​+md2=10md2​+md2=1011​md2

Thus total IA=md210+2(1110md2)I_A = \frac{md^2}{10} + 2\left(\frac{11}{10}md^2\right)IA​=10md2​+2(1011​md2) =md210+2210md2= \frac{md^2}{10} + \frac{22}{10}md^2=10md2​+1022​md2 =2310md2= \frac{23}{10}md^2=1023​md2


  1. Required ratio

IOIA=1310md22310md2=1323\frac{I_O}{I_A} = \frac{\frac{13}{10}md^2}{\frac{23}{10}md^2} = \frac{13}{23}IA​IO​​=1023​md21013​md2​=2313​


  1. Option check
  • A: 1323\frac{13}{23}2313​ ✅
  • B: 2313\frac{23}{13}1323​ ❌
  • C: 1513\frac{15}{13}1315​ ❌
  • D: 1315\frac{13}{15}1513​ ❌

So the correct option is A.

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