JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2024
- Hard
- Numerical
A uniform thin metal plate of mass \(10 \mathrm{~kg}\) with dimensions is shown. The ratio of \(x\) and \(y\) coordinates of center of mass of plate in \(\frac{n}{9}\). The value of \(n\) is _______.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(15\)
Step-by-step Solution
Detailed explanation
\(\mathrm{m}_1 =\sigma \times 5=10 \mathrm{Kg}\)
\(\mathrm{m}_1 =\sigma \times 5=10 \mathrm{Kg} \quad \mathrm{m}_2=\sigma \times 1=2 \mathrm{Kg}\)
\(\mathrm{m}_3=\sigma \times 6=12 \mathrm{Kg}\)
\(\Rightarrow \mathrm{m}_1 \mathrm{x}_1+\mathrm{m}_2 \mathrm{x}_2=\mathrm{m}_3 \mathrm{x}_3\)
\(10 \mathrm{x}_1+2(1.5)=12(1.5) \Rightarrow \mathrm{x}_1=1.5 \mathrm{~cm}\)
\(\Rightarrow \mathrm{m}_1 \mathrm{y}_1+\mathrm{m}_2 \mathrm{y}_2=\mathrm{m}_3 \mathrm{y}_3\)
\(10 \mathrm{y}_1+2(1.5)=12 \times 1 \Rightarrow \mathrm{y}_1=0.9 \mathrm{~cm}\)
\(\frac{\mathrm{x}_1}{\mathrm{y}_1}=\frac{1.5}{0.9}=\frac{15}{9}\)
\(\mathrm{n}=15\)

\(\mathrm{m}_1 =\sigma \times 5=10 \mathrm{Kg} \quad \mathrm{m}_2=\sigma \times 1=2 \mathrm{Kg}\)
\(\mathrm{m}_3=\sigma \times 6=12 \mathrm{Kg}\)
\(\Rightarrow \mathrm{m}_1 \mathrm{x}_1+\mathrm{m}_2 \mathrm{x}_2=\mathrm{m}_3 \mathrm{x}_3\)
\(10 \mathrm{x}_1+2(1.5)=12(1.5) \Rightarrow \mathrm{x}_1=1.5 \mathrm{~cm}\)
\(\Rightarrow \mathrm{m}_1 \mathrm{y}_1+\mathrm{m}_2 \mathrm{y}_2=\mathrm{m}_3 \mathrm{y}_3\)
\(10 \mathrm{y}_1+2(1.5)=12 \times 1 \Rightarrow \mathrm{y}_1=0.9 \mathrm{~cm}\)
\(\frac{\mathrm{x}_1}{\mathrm{y}_1}=\frac{1.5}{0.9}=\frac{15}{9}\)
\(\mathrm{n}=15\)

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