Four squares plates A- B- C and D having densities -1- -2- -3 and -4 are kept as shown in figure- If -1-0-5-2- -2-2-3- -3-44- then location of centre of mass of the system is given by

Name: NEET - 11 - Physics - Waves - Session 10 - W06
Uploaded: 2022-07-04T10:36:42+05:30
Description: Given σ_1, σ_2, σ_3 and σ_4 be the densities of four squares. Given, σ_1=0.5σ_2 σ_2=2σ_3 σ_3=σ_44 Let σ_1=σ (say) ⇒σ_2=2 σ σ_3=σ σ_4=4 σ Density (σ)=mA⇒ m=σ A ...

Given σ_1, σ_2, σ_3 and σ_4 be the densities of four squares. Given, σ_1=0.5σ_2 σ_2=2σ_3 σ_3=σ_44 Let σ_1=σ (say) ⇒σ_2=2 σ σ_3=σ σ_4=4 σ Density (σ)=mA⇒ m=σ A ...

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Byju's Answer

Standard XIII

Physics

COM of Center of Masses

Four squares ...

Question

Four squares plates $A, B, C$ and $D$ having densities $σ_{1}, σ_{2}, σ_{3}$ and $σ_{4}$ are kept as shown in figure. If $σ_{1} = 0.5 σ_{2}; σ_{2} = 2 σ_{3}; σ_{3} = \frac{σ_{4}}{4}$ , then location of centre of mass of the system is given by

(\frac{9 L}{16}, \frac{9 L}{8})

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(\frac{7 L}{8}, \frac{9 L}{8})

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(\frac{11 L}{16}, \frac{9 L}{8})

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(\frac{9 L}{8}, \frac{9 L}{8})

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Solution

The correct option is B $(\frac{7 L}{8}, \frac{9 L}{8})$
Given $σ_{1}, σ_{2}, σ_{3}$ and $σ_{4}$ be the densities of four squares.

Given, $σ_{1} = 0.5 σ_{2}$
$σ_{2} = 2 σ_{3}$
$σ_{3} = \frac{σ_{4}}{4}$
Let $σ_{1} = σ$ (say) $\Rightarrow σ_{2} = 2 σ$
$σ_{3} = σ$
$σ_{4} = 4 σ$
Density $(σ) = \frac{m}{A} \Rightarrow m = σ A$
From figure:
COM of square $A (x_{1}, y_{1}) = (\frac{L}{2}, \frac{L}{2})$
COM of square $B (x_{2}, y_{2}) = (\frac{3 L}{2}, \frac{L}{2})$
COM of square $C (x_{3}, y_{3}) = (\frac{3 L}{2}, \frac{3 L}{2})$
COM of square $D (x_{4}, y_{4}) = (\frac{L}{2}, \frac{3 L}{2})$
Let $(x_{c m}, y_{c m})$ be the co-ordinates of COM of system.
$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$

$= \frac{[σ (\frac{L}{2}) + 2 σ (\frac{3 L}{2}) + σ (\frac{3 L}{2}) + 4 σ (\frac{L}{2})] A}{[σ + 2 σ + σ + 4 σ] A}$

$= \frac{\frac{L}{2} + 3 L + \frac{3 L}{2} + 2 L}{8}$
$= \frac{7 L}{8} m$

$y_{c m} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$= \frac{σ A (\frac{L}{2}) + 2 σ A (\frac{L}{2}) + σ A (\frac{3 L}{2}) + 4 σ A (\frac{3 L}{2})}{8 σ A}$
$= \frac{\frac{L}{2} + L + \frac{3 L}{2} + 6 L}{8}$
$= \frac{9 L}{8} m$
Location of COM $(x_{c m}, y_{c m}) = (\frac{7 L}{8}, \frac{9 L}{8})$

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Similar questions

Q. Four squares plates

A, B, C

and

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having densities

σ_{1}, σ_{2}, σ_{3}

and

σ_{4}

are kept as shown in figure. If

σ_{1} = 0.5 σ_{2}; σ_{2} = 2 σ_{3}; σ_{3} = \frac{σ_{4}}{4}

, then location of centre of mass of the system is given by

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Table-1 Table-2

$(A)$ If $σ_{1} + σ_{2} = 0$	$(P)$ Electric field in region $III$ is towards right.
$(B)$ If $σ_{1} + σ_{2} > 0$	$(Q)$ Electric field in region $I$ is zero.
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$(S)$ None.
$(T)$ Nothing can be said.