JEE Main 2022 · 28 Jul · Shift 1 · Physics Q60

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PhysicsInteger+4 / -1Question 60 of 90

Four identical discs each of mass 'M' and diameter 'a' are arranged in a small plane as shown in figure. If the moment of inertia of the system about OO' is $\dfrac x4Ma^2$ . Then, the value of $x$ will be ______.

Four touching discs in a diamond: top and bottom discs centred on the vertical axis OO', two discs either side of the axis touching it.

Your answer

Answer and worked solution

Official answer: 3

Disc of radius $\frac a2$ about a diameter: $\frac{Ma^2}{16}$ .

Axial discs (top and bottom): $\frac{Ma^2}{16}$ each. Side discs, by parallel axes: $\frac{Ma^2}{16}+\frac{Ma^2}4$ each.

Total $\frac{12}{16}Ma^2=\frac34Ma^2$ , so $x=3$ .

Answer: 3

Topics:Parallel and perpendicular axis theorems Moment of inertia

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JEE Main 2022 · 28 Jul · Shift 1, Physics question 60