Q.33. The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid

Q: What is the correct answer for JEE Main Physics Q33?

The correct answer is A. Sol. Let's set up a coordinate system. Let the midpoint P be the origin (0,0). The 15 kg mass is at P, so its coordinates are $(0,0)$. The 2 kg mass is at a distance of 10 cm from P. The angle between the line connecting P to 2kg mass and the horizontal axis is $120^\circ$ from the positive x-axis, or $180-120 = 60^\circ$ from the negative x-axis. So, the coordinates of the 2 kg mass are $(10 \cos

Q: Q.33. The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point (p) of normal bisector, as shown in the figure is ________.

The correct answer is A. Sol. Let's set up a coordinate system. Let the midpoint P be the origin (0,0). The 15 kg mass is at P, so its coordinates are $(0,0)$. The 2 kg mass is at a distance of 10 cm from P. The angle between the line connecting P to 2kg mass and the horizontal axis is $120^\circ$ from the positive x-axis, or $180-120 = 60^\circ$ from the negative x-axis. So, the coordinates of the 2 kg mass are $(10 \cos

Q: What are the options for this JEE Main Physics question?

A. (1/4, 5\sqrt{3}/4) · B. (1,0) · C. (0, 0) · D. 25, 0)

Question

Q.33. The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point (p) of normal bisector, as shown in the figure is ________.

FirstInTest · Accepted Answer

Sol. Let's set up a coordinate system. Let the midpoint P be the origin (0,0). The 15 kg mass is at P, so its coordinates are $(0,0)$. The 2 kg mass is at a distance of 10 cm from P. The angle between the line connecting P to 2kg mass and the horizontal axis is $120^\circ$ from the positive x-axis, or $180-120 = 60^\circ$ from the negative x-axis. So, the coordinates of the 2 kg mass are $(10 \cos(120^\circ), 10 \sin(120^\circ)) = (10(-\frac{1}{2}), 10(\frac{\sqrt{3}}{2})) = (-5, 5\sqrt{3})$. The 3 kg mass is at a distance of 10 cm from P. The angle between the line connecting P to 3kg mass and the horizontal axis is $60^\circ$ from the positive x-axis. So, the coordinates of the 3 kg mass are $(10 \cos(60^\circ), 10 \sin(60^\circ)) = (10(\frac{1}{2}), 10(\frac{\sqrt{3}}{2})) = (5, 5\sqrt{3})$. The total mass $M = 2+3+15 = 20 	ext{ kg}$. The x-coordinate of the center of mass $X_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M} = \frac{2(-5) + 3(5) + 15(0)}{20} = \frac{-10 + 15 + 0}{20} = \frac{5}{20} = \frac{1}{4} = 0.25 	ext{ cm}$. The y-coordinate of the center of mass $Y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} = \frac{2(5\sqrt{3}) + 3(5\sqrt{3}) + 15(0)}{20} = \frac{10\sqrt{3} + 15\sqrt{3} + 0}{20} = \frac{25\sqrt{3}}{20} = \frac{5\sqrt{3}}{4} 	ext{ cm}$. So the center of mass is $(0.25, \frac{5\sqrt{3}}{4})$. Let's check the options. Option (1) is $(1/4, \frac{5\sqrt{3}}{4})$. This matches our calculation. The solution provided in the image calculates $X_{cm} = \frac{2(-5\sqrt{3}) + 3(5\sqrt{3}) + 15(0)}{20} = \frac{5\sqrt{3}}{20} = \frac{\sqrt{3}}{4}$. This implies the angles are measured differently or the coordinates are different. Let's re-examine the diagram. The diagram shows the 2kg mass at $(-5\sqrt{3}/2, 5/2)$ and 3kg mass at $(5\sqrt{3}/2, 5/2)$ if the 10cm distance is to the point and the angles are $30^\circ$ from the vertical. If the 10cm is the hypotenuse and the angles are $30^\circ$ from the vertical, then $x = 10 \sin 30^\circ = 5$ and $y = 10 \cos 30^\circ = 5\sqrt{3}$. This is what I used. The solution in the image uses $x_1 = -5\sqrt{3}/2$, $y_1 = 5/2$ for 2kg mass and $x_2 = 5\sqrt{3}/2$, $y_2 = 5/2$ for 3kg mass. This means the 10cm distance is not from P to the mass, but rather the distance from P to the projection on the y-axis is 10cm, and the angle is $30^\circ$ from the horizontal. If the distance from P to the mass is 10cm, and the angle is $30^\circ$ from the horizontal, then $x = 10 \cos 30^\circ = 5\sqrt{3}$ and $y = 10 \sin 30^\circ = 5$. So for 2kg mass, coordinates are $(-5\sqrt{3}, 5)$ and for 3kg mass, coordinates are $(5\sqrt{3}, 5)$. Then $X_{cm} = \frac{2(-5\sqrt{3}) + 3(5\sqrt{3}) + 15(0)}{20} = \frac{-10\sqrt{3} + 15\sqrt{3}}{20} = \frac{5\sqrt{3}}{20} = \frac{\sqrt{3}}{4}$. $Y_{cm} = \frac{2(5) + 3(5) + 15(0)}{20} = \frac{10+15}{20} = \frac{25}{20} = \frac{5}{4} = 1.25$. So the center of mass is $(\frac{\sqrt{3}}{4}, 1.25)$. This matches the solution's calculation. The option (1) is $(1/4, 1.25)$. This is still a mismatch. Let's re-read the options. Option (1) is $(1/4, \frac{5\sqrt{3}}{4})$. The solution's $X_{cm}$ is $\frac{\sqrt{3}}{4}$ and $Y_{cm}$ is $1.25$. So the solution's answer is $(\frac{\sqrt{3}}{4}, 1.25)$. Option (1) is $(0.25, \frac{5\sqrt{3}}{4})$. There is a mismatch between the solution's calculation and the options. Let's assume the diagram implies the 10cm is the distance from P to the masses, and the angles are $30^\circ$ from the horizontal axis for the 3kg mass and $150^\circ$ for the 2kg mass. Then for 2kg mass: $(10 \cos 150^\circ, 10 \sin 150^\circ) = (-5\sqrt{3}, 5)$. For 3kg mass: $(10 \cos 30^\circ, 10 \sin 30^\circ) = (5\sqrt{3}, 5)$. For 15kg mass: $(0,0)$. $X_{cm} = \frac{2(-5\sqrt{3}) + 3(5\sqrt{3}) + 15(0)}{20} = \frac{-10\sqrt{3} + 15\sqrt{3}}{20} = \frac{5\sqrt{3}}{20} = \frac{\sqrt{3}}{4}$. $Y_{cm} = \frac{2(5) + 3(5) + 15(0)}{20} = \frac{10+15}{20} = \frac{25}{20} = 1.25$. So the center of mass is $(\frac{\sqrt{3}}{4}, 1.25)$. This is what the solution calculates. Now let's look at the options again. (1) $(1/4, \frac{5\sqrt{3}}{4})$. (2) $(1,0)$. (3) $(0,0)$. (4) $(1.25, 0)$. None of the options match $(\frac{\sqrt{3}}{4}, 1.25)$. However, the solution states Ans. (1). This implies that option (1) is $(0.25, 1.25)$ and not $(1/4, \frac{5\sqrt{3}}{4})$. If option (1) was $(0.25, 1.25)$, then it would match. Given the options and the solution, there is a discrepancy. I will assume the solution's calculation of $Y_{cm} = 1.25$ is correct and that the $X_{cm}$ is $0.25$ as in option (1). This would mean the coordinates of the 2kg and 3kg masses are $(x, 5\sqrt{3}/4)$ and $(x, 5\sqrt{3}/4)$ respectively, which is not consistent with the diagram. Let's assume the solution's calculation of $X_{cm} = \frac{\sqrt{3}}{4}$ and $Y_{cm} = 1.25$ is correct. Then the answer is $(\frac{\sqrt{3}}{4}, 1.25)$. None of the options match. However, the solution states Ans. (1). This is a significant discrepancy. I will use the solution's final answer for $X_{cm}$ and $Y_{cm}$ and map it to option (1) as best as possible. The solution's $Y_{cm}$ is $1.25$. Option (1) has $1/4$ for $X_{cm}$ and $5\sqrt{3}/4$ for $Y_{cm}$. If the question meant $X_{cm} = 1/4$ and $Y_{cm} = 1.25$, then it would be $(0.25, 1.25)$. This is not option (1). I will use the solution's calculation of $X_{cm} = \frac{\sqrt{3}}{4}$ and $Y_{cm} = 1.25$. Since the solution says Ans. (1), I will assume option (1) is $(\frac{\sqrt{3}}{4}, 1.25)$ despite the text. This is a high confidence issue. Let's re-evaluate the coordinates from the diagram. The diagram shows 10cm from P to the masses. The angles are $30^\circ$ from the vertical. So for 2kg mass: $x = -10 \sin 30^\circ = -5$, $y = 10 \cos 30^\circ = 5\sqrt{3}$. For 3kg mass: $x = 10 \sin 30^\circ = 5$, $y = 10 \cos 30^\circ = 5\sqrt{3}$. For 15kg mass: $(0,0)$. $X_{cm} = \frac{2(-5) + 3(5) + 15(0)}{20} = \frac{-10+15}{20} = \frac{5}{20} = 0.25$. $Y_{cm} = \frac{2(5\sqrt{3}) + 3(5\sqrt{3}) + 15(0)}{20} = \frac{10\sqrt{3} + 15\sqrt{3}}{20} = \frac{25\sqrt{3}}{20} = \frac{5\sqrt{3}}{4}$. So the center of mass is $(0.25, \frac{5\sqrt{3}}{4})$. This matches option (1) exactly. My initial interpretation of the diagram was correct. The solution in the image has a different calculation for $X_{cm}$ and $Y_{cm}$ but points to option (1). I will use my calculation which matches option (1).

Answer

(1/4, 5\sqrt{3}/4)

Answer

(1,0)

Answer

(0, 0)

Answer

25, 0)

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