Question

a thick uniform wire is bent into the shape of the letter U .which point indicates location of center of mass of the wire.a is the midpoint of line joining of two parallel sides of u shaped wire

Answer 1

Question 1.

Answer: (C) B.

Point D cannot be the Center of Mass, since it must be symmetrically between the two parallel sides (the system is symmetrical about a line parallel to and midway between the parallel arms of the U-wire).

Point C cannot be the Center of Mass, since it implies that the masses of the parallel arms are zero, which isn't true, since the wire is uniform. (to imagine this, try removing the two arms)

Point A cannot be the Center of Mass since it implies that the mass of the base is zero, which also isn't true. (to imagine, try removing the base)

Point B is correct, since it lies on the line parallel to and midway between the two parallel arms, and also since it shifts from the CoM of the parallel arms towards the base, which it should, since the base has a finite mass.

Question 2.

Answer: (C).

Use the concept of negative mass.

Initially, a disc of radius 4R has (let) mass M, and (let) areal density $\sigma$. So:

$\sigma(\pi(4R)^2) = M$

$\sigma(\pi(R)^2) = M' = \frac{M}{16}$

Meaning, a disc of radius R of the same material shall have mass: $\frac{M}{16}$.

Now, when we remove this smaller disc from the system, we are essentially adding a disc of negative mass to the system.

The center of mass of each single disc is at the geometric center of that disc. So, to calculate center of mass's location, we shall observe, from the center of the disc of radius 4R:

$(x_{cm}, y_{cm}) = (\frac{M(0) + (\frac{-M}{16})(0) + (\frac{-M}{16})(3R)}{M - \frac{M}{16} - \frac{M}{16}}, \frac{M(0) + (\frac{-M}{16})(0) + (\frac{-M}{16})(3R)}{M - \frac{M}{16} - \frac{M}{16}})$

(kindly refer to the figure when reading the above expression)

$(x_{cm}, y_{cm}) = (\frac{\frac{-3MR}{16}}{\frac{14M}{16}}, \frac{\frac{-3MR}{16}}{\frac{14M}{16}}) = (\frac{-3R}{14}, \frac{-3R}{14})$

Question 3.

Answer: (B) 1, 3, 2. (deduced, not determined)

Comparing the 1st and 2nd plates, we can easily see that the more mass removed from the left-hand-side of the y-axis, that is, from the side of the negative x-axis, the more the center of mass shall shift to the right. So, 2 > 1.

Comparing the 1st and 3rd plates, we must see that the removal of the same mass from one point of a lesser-massed system shall cause a greater shift of the center of mass of that system, that if the mass is removed from the corresponding point of a greater-massed system.

When we apply the concept of negative mass, we have:

$x_{cm} = \frac{M_{o}x_{o} - M_{r}x_{r}}{M_{o} - M_{r}}$

In the above expression, if $M_{o}$ is lesser, and for our current case, $x_{o}$ is zero, we see that the shift is more. So, 3 > 1. Note that they CANNOT be equal.

A comparison between 2nd and 3rd plates shall require a mathematical approach, but so far, we've established:

1 has the least shift, since 1 < 2 and 1 < 3.

There's only one option that permits this, that is, B.

Answer 2

Answer is B

You cn slve it by logic also..

OR

consider the three wires to be indvdual rods.

now answer is B

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