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Consider the following statements
Statement 1: CM of a uniform semicircular disc of radius R=2Rπ from the center.
Statement 2: CM of a uniform semicircular ring of radius R=4R3π from the center.
Statement 3: CM of a solid hemisphere of radius R=4R3π from the center.
Statement 4: CM of a hemisphere shell of radius R=R2 from the center.
Which statements are correct:
A) 1, 2, 4
B) 1, 3, 4
C) 4 only
D) 1, 2 only

Answer
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Hint: Calculate the CM for each of the mentioned masses from the method of integrating the mass and finding the center of mass. Take an infinitesimal element correspondingly and proceed accordingly.

Complete step by step solution:
For CM of a uniform semicircular ring, let us assume that the ring is kept in the y>0 region. Let us consider dm at an angle θ from the positive x-axis, which subtends an angle dθ at the origin. The length of this dm is dl=Rdθ.
Length of a semicircular ring, l=πR
If mass for length l=M, then mass for length dl=Mldl.
dm=MlRdθ
Therefore, CMx=1M0πRcosθMπRRdθ
CMx=0
And, CMy=1M0πRsinθMπRRdθ
CMy=2Rπ

For CM of a semicircular disc, let us assume that the disc is kept in the y>0 region. Let us consider a ring on the disc of radius r, of thickness dr and mass dm.
Mass per unit area of the disc:
MA=M(πR22)=2MπR2
Area of the considered ring:
Ar=πrdr
Therefore, dm=MAAr=2MπR2πrdr=2MR2rdr
Hence, CMy=1M0Mydm=1M0R2rπ2MR2rdr
CMy=4πR20Rr2dr=4R3π

For CM of solid Hemisphere, let us assume that the hemisphere is kept in the y>0 region. Let us assume a disc contained inside the hemisphere at a height y, of thickness dy and mass dm, with radius r.
Mass per unit volume of the hemisphere:
MV=M(23πR3)=3M2πR3
Let the volume of the disc considered be dV.
dV=πr2dy
By applying Pythagoras theorem:
y2+r2=R2 r2=R2y2
Therefore, dV=π(R2y2)dy
dm=MVdV=3M2πR2π(R2y2)dy
CMy=1M0Mydm=1M0Ry3M2R3(R2y2)dy
CMy=32R30R(yR2y3)dy
CMy=3R8

For the CM of a hemispherical shell, let us assume that the hemispherical shell is kept in the region y>0. Let us consider a ring contained at the periphery of the shell at height y, radius r and subtends and dθ at the origin, θ is the angle at which the ring is considered from the x-axis. Therefore, the thickness of the ring is Rdθ. Let the mass of this ring be dm .
Let dA be the surface area of the ring.
dA=2πrRdθ
From trigonometry, it can be easily proved that r=Rcosθ and y=Rsinθ .
dA=2πRcosθRdθ=2πR2cosθdθ
Mass per unit area of the hemispherical shell:
MA=M2πR2
dm=M2πR22πR2cosθdθ=Mcosθdθ
CMy=1M0Mydm=1M0π2RsinθMcosθdθ
CMy=R20π2sin(2θ)dθ
CMy=R2
Therefore, only statement 4 is correct.

Hence, option (C) is correct.

Note: Do the integration very carefully as many errors can be done while doing integration. Don’t forget to write constants while doing integration. Take proper care of the limits of the integration. Do the manipulation(s) carefully, if required.
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