JEE Physics System Of Particles and Rotational Motion MCQs Set 04

Answer: (d) \( \dfrac{PQ+PR}{3} \)

Explanation:
1. Take P as origin: positions are 0, PQ and PR.
2. \( x_{cm} = \dfrac{0 + PQ + PR}{3} \).

Teacher's Note:
1. Choose the origin at one of the particles to simplify the sum. The masses are equal, so they cancel.

Answer: (d) \( \dfrac{R}{20} \)

Explanation:
1. The hole has radius R/4, so its mass is \( \tfrac{1}{16} \) of the full disc, and its centre is at \( R - \tfrac{R}{4} = \tfrac{3R}{4} \) from the centre.
2. \( x_{cm} = \dfrac{-\tfrac{m}{16}\cdot\tfrac{3R}{4}}{m - \tfrac{m}{16}} = -\dfrac{3R/64}{15/16} = -\dfrac{R}{20} \).
3. The centre of mass is R/20 from the centre, on the side opposite to the hole.

Teacher's Note:
1. Treat the hole as a negative mass. The mass is proportional to the area, i.e. to the square of the radius.

Answer: (b) In inverse ratio of the masses

Explanation:
1. From \( m_1r_1 = m_2r_2 \) about the centre of mass, \( r_1/r_2 = m_2/m_1 \).
2. The distances are in the inverse ratio of the masses, so the centre of mass is closer to the heavier particle.

Teacher's Note:
1. Think of a seesaw: the heavier child sits closer to the pivot. The centre of mass is the balance point.

Answer: (b) May lie within, outside, or on the surface of the body

Explanation:
1. The position depends on the shape and the mass distribution.
2. For a solid sphere it lies inside, for a ring or a bent rod it lies outside the material.

Teacher's Note:
1. Remember the ring and the horseshoe: the centre of mass can be at a point where there is no mass at all.

Answer: (c) The point of intersection of the medians

Explanation:
1. All masses are equal, so the centre of mass is the centroid of the triangle formed by the centres.
2. The centroid is the point where the medians meet.

Teacher's Note:
1. For equal masses the centre of mass is the geometric centroid of the points.

Answer: (a) \( \dfrac{5}{2} \)

Explanation:
1. The three particles (total 6 g) have their centre of mass at the origin, so they behave like a 6 g particle at the origin.
2. \( \vec R = \dfrac{6\times 0 + 4\vec r_4}{10} = (1,2,3) \Rightarrow \vec r_4 = 2.5\,(1,2,3) \).
3. So \( \alpha = \tfrac{5}{2} \).

Teacher's Note:
1. Replace the group of particles by one particle at its centre of mass. This saves a lot of algebra.

Answer: (c) \( \left(\dfrac{a}{2}, \dfrac{a\sqrt3}{6}\right) \)

Explanation:
1. \( x_{cm} = \dfrac{0 + a + a/2}{3} = \dfrac{a}{2} \).
2. \( y_{cm} = \dfrac{0 + 0 + a\sqrt3/2}{3} = \dfrac{a\sqrt3}{6} \).

Teacher's Note:
1. The three points form an equilateral triangle, and the centre of mass is its centroid.

Answer: (b) \( \left(\dfrac{a}{3}, \dfrac{a}{3}\right) \)

Explanation:
1. Centres of the rods: OA at \( (a/2, 0) \), OB at \( (0, a/2) \), AB at \( (a/2, a/2) \), all of equal mass m.
2. \( x_{cm} = \dfrac{m(a/2) + 0 + m(a/2)}{3m} = \dfrac{a}{3} \), and by symmetry \( y_{cm} = \dfrac{a}{3} \).

Teacher's Note:
1. Replace each uniform rod by a point mass at its midpoint. Do not use the centroid of the triangular plate here; the rods are the only mass, and equal masses at the midpoints happen to give the same point.

Answer: (d) \( \dfrac{L(3A + 2BL)}{3(2A + BL)} \)

Explanation:
1. \( x_{cm} = \dfrac{\int_0^L x(A+Bx)dx}{\int_0^L (A+Bx)dx} = \dfrac{\tfrac{AL^2}{2} + \tfrac{BL^3}{3}}{AL + \tfrac{BL^2}{2}} \).
2. Simplifying, \( x_{cm} = \dfrac{L(3A + 2BL)}{3(2A + BL)} \).

Teacher's Note:
1. Check the limits: B = 0 gives L/2 (uniform rod) and A = 0 gives 2L/3 (density proportional to x).

Answer: (c) \( \dfrac{R}{6} \)

Explanation:
1. The hole has mass m/4 and its centre is at R/2 from the centre.
2. \( y_{cm} = \dfrac{-\tfrac{m}{4}\cdot\tfrac{R}{2}}{m - \tfrac{m}{4}} = -\dfrac{R/8}{3/4} = -\dfrac{R}{6} \).
3. The distance is R/6.

Teacher's Note:
1. Subtract the hole as a negative mass and place it at its own centre.

Answer: (b) \( \left(2\hat i - 2\hat j + \tfrac{2}{3}\hat k\right) \) m

Explanation:
1. \( \vec r_c = \dfrac{1(4,2,-3) + 2(1,-4,2) + 3(2,-2,1)}{6} \).
2. Numerator \( = (4+2+6,\ 2-8-6,\ -3+4+3) = (12, -12, 4) \).
3. \( \vec r_c = (2, -2, \tfrac{2}{3}) \) m.

Teacher's Note:
1. Work component by component. The denominator is the total mass, 6 kg.

Answer: (c) \( \left(\dfrac{2r}{\pi}, \dfrac{2r}{\pi}\right) \)

Explanation:
1. For an arc of half-angle \( \theta \) (radians), the centre of mass is at \( \dfrac{r\sin\theta}{\theta} \) from the centre along the axis of symmetry.
2. Here \( \theta = \tfrac{\pi}{4} \): distance \( = \dfrac{r\sin(\pi/4)}{\pi/4} = \dfrac{2\sqrt2\,r}{\pi} \).
3. Resolving along x and y at 45°: \( x_c = y_c = \dfrac{2r}{\pi} \).

Teacher's Note:
1. A quarter ring is a symmetric arc about the 45° line. Resolve the distance along both axes.

Answer: (c) \( \dfrac{2r}{\pi} \)

Explanation:
1. Half-angle \( \theta = \tfrac{\pi}{2} \), so \( y_c = \dfrac{r\sin(\pi/2)}{\pi/2} = \dfrac{2r}{\pi} \).

Teacher's Note:
1. Semicircular wire gives 2r/π, whereas a semicircular plate gives 4r/3π. Do not mix them up.

Answer: (d) \( \left(\dfrac{4r}{3\pi}, \dfrac{4r}{3\pi}\right) \)

Explanation:
1. For a sector of half-angle \( \theta \), the centre of mass is at \( \dfrac{2r\sin\theta}{3\theta} \) from the centre.
2. With \( \theta = \tfrac{\pi}{4} \): \( \dfrac{2r\sin(\pi/4)}{3\pi/4} = \dfrac{4\sqrt2\,r}{3\pi} \).
3. Resolving at 45°, \( x_c = y_c = \dfrac{4r}{3\pi} \).

Teacher's Note:
1. Same method as for the ring, but with the factor 2/3 for a solid sector.

Answer: (a) \( \dfrac{4R}{3\pi} \)

Explanation:
1. Using \( \dfrac{2R\sin\theta}{3\theta} \) with \( \theta = \tfrac{\pi}{2} \): \( \dfrac{2R}{3\pi/2} = \dfrac{4R}{3\pi} \).

Teacher's Note:
1. Learn the four standard results: 4R/3π (plate), 2R/π (wire), 3R/8 (solid hemisphere), R/2 (hollow hemisphere).

Answer: (d) \( \dfrac{1}{\pi} \) m

Explanation:
1. For a semicircular ring, \( y_c = \dfrac{2r}{\pi} = \dfrac{2\times 0.5}{\pi} = \dfrac{1}{\pi} \) m.

Teacher's Note:
1. Use 2r/π for a thin wire (ring). The radius is 0.5 m, so the answer is 1/π m.

Answer: (d) 6 cm

Explanation:
1. For a solid hemisphere, \( y_c = \dfrac{3R}{8} = \dfrac{3\times 16}{8} = 6 \) cm from the flat base.

Teacher's Note:
1. A solid hemisphere has 3R/8; a hollow hemispherical shell has R/2.

Answer: (b) \( \dfrac{3L}{4} \)

Explanation:
1. \( x_{cm} = \dfrac{\int_0^L x\cdot x^2dx}{\int_0^L x^2dx} = \dfrac{L^4/4}{L^3/3} = \dfrac{3L}{4} \).

Teacher's Note:
1. Mass per unit length is proportional to the density because the cross-section is constant.

Answer: (b) 9 cm

Explanation:
1. R = 28 cm and r = 21 cm. Hole mass fraction \( = (21/28)^2 = \tfrac{9}{16} \); hole centre is 28 - 21 = 7 cm from the centre.
2. \( x_{cm} = -\dfrac{\tfrac{9}{16}\times 7}{1 - \tfrac{9}{16}} = -\dfrac{63/16}{7/16} = -9 \) cm.

Teacher's Note:
1. The area, and so the mass, scales with the square of the diameter. Remove it as a negative mass at the hole's centre.

Answer: (c) \( \dfrac{r}{14} \)

Explanation:
1. The removed sphere has radius r/2, so its mass is \( \tfrac{1}{8} \) of the total. The distance is largest when the hole touches the surface, centre at r/2.
2. \( d = \dfrac{\tfrac{m}{8}\cdot\tfrac{r}{2}}{m - \tfrac{m}{8}} = \dfrac{r/16}{7/8} = \dfrac{r}{14} \).

Teacher's Note:
1. Mass scales as the cube of the radius. The farthest shift occurs when the hole is as far out as possible.

Answer: (c) 1.24 Å

Explanation:
1. With H at the origin and Cl at 1.27 Å: \( x_{cm} = \dfrac{35.5\times 1.27}{36.5} = 1.235 \) Å \( \approx 1.24 \) Å.

Teacher's Note:
1. The centre of mass is very close to the heavy chlorine nucleus. This is why the molecule rotates almost about Cl.

Answer: (d) \( \dfrac{12}{7} \) m

Explanation:
1. \( \int_0^3 x(2+x)dx = [x^2 + x^3/3]_0^3 = 9 + 9 = 18 \).
2. \( \int_0^3 (2+x)dx = [2x + x^2/2]_0^3 = 6 + 4.5 = 10.5 \).
3. \( x_{cm} = \dfrac{18}{10.5} = \dfrac{12}{7} \) m.

Teacher's Note:
1. The centre of mass shifts towards the denser end, so it must be greater than 1.5 m.

Answer: (a) 17.7 cm

Explanation:
1. Each half has length 0.5 m; with the bend at the origin, their centres are at (0.25, 0) and (0, 0.25).
2. \( (x_{cm}, y_{cm}) = (0.125, 0.125) \) m, so the distance \( = 0.125\sqrt2 = 0.177 \) m = 17.7 cm.

Teacher's Note:
1. The original centre is the bend point. Each half is replaced by a point mass at its own midpoint.

Answer: (c) Nearer to M

Explanation:
1. The centre of mass divides the line in the inverse ratio of the masses, so it lies closer to the heavier mass M.
2. It is not exactly at M, only very near it.

Teacher's Note:
1. Be careful with option (d): it is nearly, but not exactly, at M.

Answer: (c) Internal forces on the particles

Explanation:
1. \( \vec R_{cm} = \dfrac{\sum m_i\vec r_i}{M} \) depends on masses and positions.
2. Internal forces cancel in pairs and cannot change the motion of the centre of mass.

Teacher's Note:
1. Only external forces affect the centre of mass.