Rotational Motion PYQ — JEE Main 2005 · Shift 0 · Q181 with Solution

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Correct answer: C

Interpret the axis carefully

The question asks for the moment of inertia of a uniform semicircular disc of mass $M$ and radius $r$ about a line perpendicular to the plane of the disc through the center.

Here, for a semicircular disc, the natural meaning of “center” in such JEE problems is the center of the full circle from which the semicircle is obtained (the midpoint of the diameter), not the center of mass.

Use area density

For a uniform semicircular disc,

\sigma = \frac{M}{\text{area of semicircle}} = \frac{M}{\frac{1}{2}\pi r^2} = \frac{2M}{\pi r^2}

Take a thin circular ring element

Consider a thin semicircular ring of radius $x$ and thickness $dx$ .

Full circumference of radius $x$ is $2\pi x$
Semicircular arc length is $\pi x$

So its area is

dA = \pi x\,dx

Hence its mass is

dm = \sigma\, dA = \sigma \pi x\,dx

Every point of this ring element is at distance $x$ from the axis, so its contribution to moment of inertia is

dI = x^2\,dm = x^2(\sigma\pi x\,dx)=\sigma\pi x^3dx

Integrate from $0$ to $r$

I = \int_0^r \sigma\pi x^3\,dx

I = \sigma\pi \left[\frac{x^4}{4}\right]_0^r

I = \sigma\pi \frac{r^4}{4}

Now substitute $\sigma = \dfrac{2M}{\pi r^2}$ :

I = \frac{2M}{\pi r^2}\cdot \pi \cdot \frac{r^4}{4}

I = \frac{2M r^2}{4} = \frac{1}{2}Mr^2

Match with the options

I = \frac{1}{2}Mr^2

So the correct option is C.

Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

They agree.

More from Rotational Motion

View written solutionFree

Correct answer: C

Interpret the axis carefully

The question asks for the moment of inertia of a uniform semicircular disc of mass $M$ and radius $r$ about a line perpendicular to the plane of the disc through the center.

Use area density

For a uniform semicircular disc,

\sigma = \frac{M}{\text{area of semicircle}} = \frac{M}{\frac{1}{2}\pi r^2} = \frac{2M}{\pi r^2}

Take a thin circular ring element

Consider a thin semicircular ring of radius $x$ and thickness $dx$ .

Full circumference of radius $x$ is $2\pi x$
Semicircular arc length is $\pi x$

So its area is

dA = \pi x\,dx

Hence its mass is

dm = \sigma\, dA = \sigma \pi x\,dx

Every point of this ring element is at distance $x$ from the axis, so its contribution to moment of inertia is

dI = x^2\,dm = x^2(\sigma\pi x\,dx)=\sigma\pi x^3dx

Integrate from $0$ to $r$

I = \int_0^r \sigma\pi x^3\,dx

I = \sigma\pi \left[\frac{x^4}{4}\right]_0^r

I = \sigma\pi \frac{r^4}{4}

Now substitute $\sigma = \dfrac{2M}{\pi r^2}$ :

I = \frac{2M}{\pi r^2}\cdot \pi \cdot \frac{r^4}{4}

I = \frac{2M r^2}{4} = \frac{1}{2}Mr^2

Match with the options

I = \frac{1}{2}Mr^2

So the correct option is C.

Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

They agree.

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