In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then BC \over AB is close to :

Center of Mass PYQ — JEE Main 2016 · 10 Apr · Shift 1 · Q45 with Solution

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Correct answer: B

Let the wire consist of two straight uniform segments: $AB$ and $BC$ joined at $B$ .

Since the centre of mass lies vertically below $A$ , its horizontal coordinate relative to $A$ must be zero.

From the figure, take:

$AB$ making $60^\circ$ below the horizontal,
$BC$ making $30^\circ$ below the horizontal to the right.

Let $AB=L_1, \qquad BC=L_2$ and let linear mass density be uniform, say $\lambda$ .

1. Coordinates of midpoints of the two segments

Choose $A$ as origin.

Segment $AB$

Its midpoint is at distance $L_1/2$ from $A$ along the direction of $AB$ . Hence its coordinates are $x_1=\frac{L_1}{2}\cos 60^\circ=\frac{L_1}{4},$ $y_1=-\frac{L_1}{2}\sin 60^\circ=-\frac{\sqrt3 L_1}{4}.$

Mass of segment $AB$ is $m_1=\lambda L_1.$

Segment $BC$

First find coordinates of point $B$ : $x_B=L_1\cos 60^\circ=\frac{L_1}{2},$ $y_B=-L_1\sin 60^\circ=-\frac{\sqrt3 L_1}{2}.$

Midpoint of $BC$ is at distance $L_2/2$ from $B$ along direction $30^\circ$ below the horizontal. So

=\frac{L_1}{2}+\frac{\sqrt3 L_2}{4},$$ $$y_2=y_B-\frac{L_2}{2}\sin 30^\circ =-\frac{\sqrt3 L_1}{2}-\frac{L_2}{4}.$$ Mass of segment $BC$ is $$m_2=\lambda L_2.$$ --- ## 2. Condition for centre of mass to lie vertically below $A$ This means the $x$-coordinate of the combined centre of mass is zero: $$x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}=0.$$ Substitute values: $$\lambda L_1\left(\frac{L_1}{4}\right)+\lambda L_2\left(\frac{L_1}{2}+\frac{\sqrt3 L_2}{4}\right)=0.$$ After cancelling $\lambda$ and multiplying by $4$: $$L_1^2+2L_1L_2+\sqrt3 L_2^2=0.$$ This cannot be zero for positive lengths, so the interpretation of horizontal directions must be that in the figure $AB$ is to the **left** of $A$ while $BC$ is to the right from $B$. So take midpoint of $AB$ at $$x_1=-\frac{L_1}{4}.$$ Then the point $B$ has coordinate $$x_B=-\frac{L_1}{2}.$$ Thus midpoint of $BC$ becomes $$x_2=-\frac{L_1}{2}+\frac{\sqrt3 L_2}{4}.$$ Now apply $x_{cm}=0$: $$\lambda L_1\left(-\frac{L_1}{4}\right)+\lambda L_2\left(-\frac{L_1}{2}+\frac{\sqrt3 L_2}{4}\right)=0.$$ Cancel $\lambda$ and multiply by $4$: $$-L_1^2-2L_1L_2+\sqrt3 L_2^2=0.$$ Rearrange: $$\sqrt3 L_2^2-2L_1L_2-L_1^2=0.$$ Let $$r=\frac{L_2}{L_1}=\frac{BC}{AB}.$$ Then $$\sqrt3 r^2-2r-1=0.$$ --- ## 3. Solve the quadratic $$r=\frac{2\pm\sqrt{4+4\sqrt3}}{2\sqrt3} =\frac{1\pm\sqrt{1+\sqrt3}}{\sqrt3}.$$ Take the positive root: $$r=\frac{1+\sqrt{1+\sqrt3}}{\sqrt3}.$$ Using $\sqrt3\approx1.732$, $$1+\sqrt3\approx2.732, \qquad \sqrt{2.732}\approx1.653.$$ So $$r\approx\frac{1+1.653}{1.732} =\frac{2.653}{1.732} \approx1.53.$$ Thus $$\frac{BC}{AB}\approx1.5.$$ --- ## 4. Compare with options Closest option is: $$\boxed{1.5}$$ which is **Option C**. --- ## 5. Comparison with stored answer Stored correct answer is **B: 1.37**, but the calculation gives **C: 1.5**. So I **disagree** with the stored answer. The stored answer likely has an error unless the figure geometry is different from the standard interpretation.

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1. Coordinates of midpoints of the two segments

Choose

A

as origin.

Segment

AB

Its midpoint is at distance

L_1/2

from

A

along the direction of

AB

. Hence its coordinates are

x_1=\frac{L_1}{2}\cos 60^\circ=\frac{L_1}{4},

y_1=-\frac{L_1}{2}\sin 60^\circ=-\frac{\sqrt3 L_1}{4}.

Mass of segment

AB

m_1=\lambda L_1.

Segment

BC

First find coordinates of point

B

x_B=L_1\cos 60^\circ=\frac{L_1}{2},

y_B=-L_1\sin 60^\circ=-\frac{\sqrt3 L_1}{2}.

Midpoint of

BC

is at distance

L_2/2

from

B

along direction

30^\circ

below the horizontal. So

=\frac{L_1}{2}+\frac{\sqrt3 L_2}{4},$$ $$y_2=y_B-\frac{L_2}{2}\sin 30^\circ =-\frac{\sqrt3 L_1}{2}-\frac{L_2}{4}.$$ Mass of segment $BC$ is $$m_2=\lambda L_2.$$ --- ## 2. Condition for centre of mass to lie vertically below $A$ This means the $x$-coordinate of the combined centre of mass is zero: $$x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}=0.$$ Substitute values: $$\lambda L_1\left(\frac{L_1}{4}\right)+\lambda L_2\left(\frac{L_1}{2}+\frac{\sqrt3 L_2}{4}\right)=0.$$ After cancelling $\lambda$ and multiplying by $4$: $$L_1^2+2L_1L_2+\sqrt3 L_2^2=0.$$ This cannot be zero for positive lengths, so the interpretation of horizontal directions must be that in the figure $AB$ is to the **left** of $A$ while $BC$ is to the right from $B$. So take midpoint of $AB$ at $$x_1=-\frac{L_1}{4}.$$ Then the point $B$ has coordinate $$x_B=-\frac{L_1}{2}.$$ Thus midpoint of $BC$ becomes $$x_2=-\frac{L_1}{2}+\frac{\sqrt3 L_2}{4}.$$ Now apply $x_{cm}=0$: $$\lambda L_1\left(-\frac{L_1}{4}\right)+\lambda L_2\left(-\frac{L_1}{2}+\frac{\sqrt3 L_2}{4}\right)=0.$$ Cancel $\lambda$ and multiply by $4$: $$-L_1^2-2L_1L_2+\sqrt3 L_2^2=0.$$ Rearrange: $$\sqrt3 L_2^2-2L_1L_2-L_1^2=0.$$ Let $$r=\frac{L_2}{L_1}=\frac{BC}{AB}.$$ Then $$\sqrt3 r^2-2r-1=0.$$ --- ## 3. Solve the quadratic $$r=\frac{2\pm\sqrt{4+4\sqrt3}}{2\sqrt3} =\frac{1\pm\sqrt{1+\sqrt3}}{\sqrt3}.$$ Take the positive root: $$r=\frac{1+\sqrt{1+\sqrt3}}{\sqrt3}.$$ Using $\sqrt3\approx1.732$, $$1+\sqrt3\approx2.732, \qquad \sqrt{2.732}\approx1.653.$$ So $$r\approx\frac{1+1.653}{1.732} =\frac{2.653}{1.732} \approx1.53.$$ Thus $$\frac{BC}{AB}\approx1.5.$$ --- ## 4. Compare with options Closest option is: $$\boxed{1.5}$$ which is **Option C**. --- ## 5. Comparison with stored answer Stored correct answer is **B: 1.37**, but the calculation gives **C: 1.5**. So I **disagree** with the stored answer. The stored answer likely has an error unless the figure geometry is different from the standard interpretation.

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