Four solid spheres each of diameter 5 ~cm and mass 0.5 ~kg ar placed with their centres at the corners of a square of side 4 ~cm . The moment of inertia of the system about the diagonal of the square is N × 10^ -4

Correct answer: 9 aligned r & =(d)/(2)= 5 2 ~cm & = 5 2 ×10^ -2 ~m m & =0.5 ~kg a & =4 ~cm & =4 ×10^ -2 ~m I_ X X & =I_1+I_2+I_3+I_4 aligned aligned =& [(2)/(5)m r²+m( a 2 )^2]+(2)/(5)m r² & +[(2)/(5)m r²+m( a 2 )^2]+(2)/(5)m r² aligned Substituting the values, we get aligned & I_ X X & =9 ×10^ -4 Kgm^ -2 & N & =9 aligned Answer is 9. Analysis of Question (i) Question is simple. (ii) Only theorem of parallel axes is to be used properly.

JEE Advanced 2011Paper 1

Exam: JEE Advanced
Year: 2011
Paper: Paper 1
Subject: Physics
Chapter: Rotational Motion

JEE Advanced 2011 · Paper 1 · Physics · Rotational Motion

Four solid spheres each of diameter 5 ~cm and mass 0.5 ~kg ar...

Four solid spheres each of diameter

\sqrt{5} \mathrm{~cm}

and mass

0.5 \mathrm{~kg}

ar placed with their centres at the corners of a square of side

4 \mathrm{~cm}

. The moment of inertia of the system about the diagonal of the square is

N \times 10^{-4} \mathrm{~kg}-\mathrm{m}^2

, then

N

Four solid spheres each of diameter 5 ~cm and mass 0.5 ~kg ar placed with their centres at the corners of a square of

Correct answer

Step-by-step solution

\begin{aligned}r & =\frac{d}{2}=\frac{\sqrt{5}}{2} \mathrm{~cm}\\ & =\frac{\sqrt{5}}{2} \times10^{-2} \mathrm{~m}\\ m & =0.5 \mathrm{~kg}\\ a & =4 \mathrm{~cm}\\ & =4 \times10^{-2} \mathrm{~m}\\ I_{X X} & =I_1+I_2+I_3+I_4 \end{aligned}

<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_advanced%252FwyYDFv4zwC7Wj6gaWYMemyJb5bu5w2S0rDm7FTzCbw4.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2FwyYDFv4zwC7Wj6gaWYMemyJb5bu5w2S0rDm7FTzCbw4.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2FwyYDFv4zwC7Wj6gaWYMemyJb5bu5w2S0rDm7FTzCbw4.original.fullsize.png?alt=media">

\begin{aligned} =& {\left[\frac{2}{5}m r²+m\left(\frac{a}{\sqrt{2}}\right)^2\right]+\frac{2}{5}m r² } \\ & +\left[\frac{2}{5}m r²+m\left(\frac{a}{\sqrt{2}}\right)^2\right]+\frac{2}{5}m r² \end{aligned}

Substituting the values, we get

\begin{aligned}& I_{X X} & =9 \times10^{-4} \mathrm{Kgm}^{-2}\\ \therefore \quad & N & =9 \end{aligned}

Answer is 9. Analysis of Question (i) Question is simple. (ii) Only theorem of parallel axes is to be used properly.

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