Consider a thin uniform shell as shown. Let I₁, I₂, I₃ and I₄ be the moments of inertia of the object about axes AA′, BB′, CC′ and DD′ respectively. I₁ + I₂ + I₃ + I₄ = nI₁/2. Then n − 5 = ...
Physics · Mixed Chapter
JEE Mains

Question

Consider a thin uniform shell as shown. Let I₁, I₂, I₃ and I₄ be the moments of inertia of the object about axes AA′, BB′, CC′ and DD′ respectively. I₁ + I₂ + I₃ + I₄ = nI₁/2. Then n − 5 = ...

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: For a thin uniform hemispherical shell of radius R and mass M, the moment of inertia about either diameter through the sphere centre is (2/3)MR². Thus I₃=I₄=(2/3)MR². Step 2: Its centre of mass is R/2 above the sphere centre. Shift the horizontal diameter axis upward by R to AA′: I₁=I₃+MR²−2R(MR/2)=I₃=(2/3)MR². Step 3: Shift the vertical diameter axis a distance R right to BB′: I₂=I₄+MR²=(5/3)MR². Step 4: I₁+I₂+I₃+I₄=(2+5+2+2)MR²/3=(11/3)MR²=(11/2)I₁. Hence n=11 and n−5=6.