CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
25
You visited us 25 times! Enjoying our articles? Unlock Full Access!
Question

A semicircular portion of radius r=10 cm is cut from a uniform rectangular plate as shown in the figure. The distance of center of mass C of the remaining plate, from point O is (COM of semicircle is at distance 4r3π from the base)


A
7.75 cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
5 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
17 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4.6 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 7.75 cm
Remaining area = Area of rectangle - Area of semicircular portion,

Area of rectangle A1=2r2
Area of semicircular portion A2=πr22

If r1 and r2 are the positions of centre of mass of the rectangle and the semicircular portion with respect to O.
r1=r2 and r2=4r3π
(left of origin is taken -ve)

Therefore, center of mass of remaining portion
=A1×(r2)A2×(4r3π)A1A2

=⎜ ⎜ ⎜ ⎜2r2×(r2)πr22×(4r3π)2r2πr22⎟ ⎟ ⎟ ⎟
=⎜ ⎜ ⎜2r324r362r2πr22⎟ ⎟ ⎟=2r3(4π)
(Putting r=10 cm (given))
=2×103(4π)=7.75 cm

So, the distance of center of mass C of remaining plate, from point O is 7.75 cm.

flag
Suggest Corrections
thumbs-up
5
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon