MCQMediumJEE 2025Centre of Mass

JEE Physics 2025 Question with Solution

Consider a circular disc of radius 20cm20 \, \text{cm} with center located at the origin. A circular hole of radius 5cm5 \, \text{cm} is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of the center of mass of the residual or remaining disc from the origin will be:

  • A

    2.0cm2.0 \, \text{cm}

  • B

    0.5cm0.5 \, \text{cm}

  • C

    1.5cm1.5 \, \text{cm}

  • D

    1.0cm1.0 \, \text{cm}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A uniform circular disc has radius R=20cmR = 20 \, \text{cm} and its center is at the origin. A circular hole of radius r=5cmr = 5 \, \text{cm} is removed such that the edge of the hole touches the edge of the disc.

Find: The distance of the center of mass of the remaining disc from the origin.

Treat the removed hole as a negative-mass region. Since the disc is uniform, mass is proportional to area.

The center of the hole is at a distance

Rr=205=15cmR - r = 20 - 5 = 15 \, \text{cm}

from the origin along the +x+x-axis.

Using the center of mass relation for composite bodies,

Xcm=A1X1+A2X2A1+A2X_{\text{cm}} = \frac{A_1 X_1 + A_2 X_2}{A_1 + A_2}

For the remaining disc, the removed area is taken as negative:

Xcm=A1x1+(A2)x2A1A2X_{\text{cm}} = \frac{A_1 x_1 + (-A_2)x_2}{A_1 - A_2}

where

A1=πR2=π(20)2=400πcm2A_1 = \pi R^2 = \pi (20)^2 = 400\pi \, \text{cm}^2 A2=πr2=π(5)2=25πcm2A_2 = \pi r^2 = \pi (5)^2 = 25\pi \, \text{cm}^2 x1=0,x2=15cmx_1 = 0, \qquad x_2 = 15 \, \text{cm}

Substituting,

Xcm=(400π)(0)+(25π)(15)400π25πX_{\text{cm}} = \frac{(400\pi)(0) + (-25\pi)(15)}{400\pi - 25\pi} Xcm=375π375π=1cmX_{\text{cm}} = \frac{-375\pi}{375\pi} = -1 \, \text{cm}

The negative sign shows that the center of mass shifts towards the side opposite to the hole. Therefore, its distance from the origin is 1.0cm1.0 \, \text{cm}.

The correct option is D.

Area Comparison Method

Given: Original disc radius 20cm20 \, \text{cm}, removed disc radius 5cm5 \, \text{cm}.

Find: Distance of the center of mass of the remaining part from the origin.

Because the smaller disc touches the boundary of the larger disc, the center of the removed disc is at

15cm15 \, \text{cm}

from the origin.

For a uniform lamina, area can be used in place of mass. The original disc has area 400π400\pi, while the removed disc has area 25π25\pi. Thus the removed part is

25π400π=116\frac{25\pi}{400\pi} = \frac{1}{16}

of the full disc.

So the shift of the center of mass from the origin is

(116)(15)1116=15/1615/16=1cm\frac{\left(\frac{1}{16}\right)(15)}{1 - \frac{1}{16}} = \frac{15/16}{15/16} = 1 \, \text{cm}

The direction is opposite to the removed hole, but the required distance is 1cm1 \, \text{cm}.

Hence, the correct option is D.

Common mistakes

  • Taking the center of the hole at 20cm20 \, \text{cm} from the origin is incorrect. Since the hole touches the outer edge internally, its center is at Rr=15cmR-r = 15 \, \text{cm}, not at the radius of the big disc.

  • Using the removed disc with positive mass or positive area is wrong. The cut-out portion must be treated as a negative mass/area contribution in the center of mass formula.

  • Reporting 1cm-1 \, \text{cm} as the final answer is a sign error in interpretation. The question asks for the distance from the origin, so the answer must be the positive magnitude 1.0cm1.0 \, \text{cm}.

Practice more Centre of Mass questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Related questions