Tag: turning effects of forces

Quizzes (7) Questions (109)

Questions Related to turning effects of forces

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

If linear density of a rod of length 3 m varies as  $\lambda=2+x$, then the position of the centre of gravity of the rod is 

  1. $\dfrac{7}{3}m$

  2. $\dfrac{12}{7}m$

  3. $\dfrac{10}{7}m$

  4. $\dfrac{9}{7}m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The center of mass x_cm is calculated by the integral of (x * lambda * dx) / integral of (lambda * dx) from 0 to L. With lambda = 2 + x and L = 3, the numerator is integral from 0 to 3 of (2x + x^2) dx = [x^2 + x^3/3] from 0 to 3 = 9 + 9 = 18. The denominator is integral from 0 to 3 of (2 + x) dx = [2x + x^2/2] from 0 to 3 = 6 + 4.5 = 10.5. Thus, x_cm = 18 / 10.5 = 180 / 105 = 12 / 7.

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

A bullet of mass 0.01$\mathrm { kg }$ and traveling at a speed of 500$\mathrm { m } / \mathrm { sec }$ strikes a block of which suspended by a string of length 5$\mathrm { m }$ . The centre of gravity of the block is found to vertical distance of 0.1$\mathrm { m }$ . What is the speed of the bullet after it emerges from the block?

  1. 359$\mathrm { m } / \mathrm { s }$

  2. 220$\mathrm { m } / \mathrm { s }$

  3. 204$\mathrm { m } / \mathrm { s }$

  4. 284$\mathrm { m } / \mathrm { s }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By energy conservation of the suspended block we can say 

initial $KE=$ final potential energy
$\dfrac{1}{2}m{v^2} = mgH$
$\dfrac{1}{2}m{v^2} = gH$
$v = \sqrt {2gH} $
$v = \sqrt {2 \times 10 \times 0.1}  = 1.41\,m/s$
now$,$ by momentum conservation 
${P _{bullet}} = {P _{block}} + {P _{bullet}}$
$0.01 \times 500 = 1 \times 1.41 + 0.01 \times {v _f}$
$5 - 1.41 = 0.01 \times {v _f}$
$0.01{v _f} = 3.59$
${v _f} = 359\,m/s$
Hence,
option $(A)$ is correct answer.

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

A metallic rod falls under gravity with its ends pointing east and west. Then

  1. an emf is induced in it as it cuts the magnetic lines of force

  2. no emf is induced at all

  3. two emf of equal nut opposite signs are induced, giving no emf is

  4. its acceleration is equal to the product of g and the radius of the rod.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When rod falls due to earth magnetic field an emf is induced 

Hence,
option $A$ is correct answer.

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

The apparent weight of a person of mass m in an elevator is 2mg. The elevator is moving

  1. up with an acceleration of $\frac{g}{2}$

  2. up with an acceleration of g

  3. up with an acceleration of 2g

  4. down with an acceleration of g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The apparent weight is given by N = m(g + a) for upward acceleration. If N = 2mg, then m(g + a) = 2mg, which simplifies to g + a = 2g, so a = g. The elevator is moving upward with an acceleration of g.

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

A body of mass 5 kg initially moving with speed 10 m/s along x-axis in gravity free space explodes and breaks into three pieces of masses 1 kg, 1 kg and 3 kg. the two pieces of equal masses fly off with the same speed 20 m/s along y-axis and z-axis respectively. what is the velocity of heavier fragment?

  1. $ ( \frac {10}{3} \hat i - \frac {20}{3} \hat j - \frac {40}{3} \hat k ) m/s $

  2. $ ( \frac {50}{3} \hat i - \frac {20}{3} \hat j - \frac {20}{3} \hat k ) m/s $

  3. $ ( \frac {20}{3} \hat i - \frac {20}{3} \hat j - \frac {20}{3} \hat k ) m/s $

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

AU the particles of a system are situated at a distance r from the origin. The distance of the centre of mass of the system from the origin is :

  1. = r

  2. $\leq \, r$

  3. $> r$

  4. $\leq 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The largest distance between the origin and center of mass would be 'r' when there is only one particle.
If there are more than one particle, the center of mass would be inside the circle of radius 'r' centered at the origin.
Hence option B is correct.

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

The position vector of three particles of masses $m _1\, =\,1kg,\, m _2\, =\, 2\, kg$ and $m _3\, =\, 3\, kg$ are $\vec{r} _1\, =\, (\hat{i}\, +\, 4\hat{j}\, +\, \hat{k})\, m,\, \vec{r} _2\, =\, (\hat{i}\, +\, \hat{j}\, +\, \hat{k}) m$ and $\vec{r} _3\, =\, (2\hat{i}\, -\, \hat{j}\, -\, 2\hat{k})$ m respectively. Find the position vector of their center of mass.

  1. $\displaystyle \frac {1}{2}\, (\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$

  2. $\displaystyle \frac {1}{2}\, (\hat{i}\, +\, 3\hat{j}\, -\, \hat{k})\, m$

  3. $\displaystyle \frac {1}{2}\, (\hat{i}\, +\, \hat{j}\, -\, 3\hat{k})\, m$

  4. $\displaystyle \frac {1}{2}\, (3\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The position vector of COM of the.three particles will be given by
$\vec{r} _{COM}\, =\, \displaystyle \frac {m _1\vec{r} _1\, +\, m _2\vec{r} _2\, +\, m _3\vec{r} _3}{m _1\, +\, m _2\, +\, m _3}$
Substituting the values, we get
$\vec{r} _{COM}\, =\, \displaystyle \frac {(1) (\hat{i}\, +\, 4\hat{j}\, +\, \hat{k})\, +\, (2) (\hat{i}\, +\, \hat{j}\, +\, \hat{k})\, +\, (3) (2\hat{i}\, -\, \hat{j}\, -\, 2\hat{k})}{1+2+3}\, =\, \displaystyle \frac {1}{2}\, (3\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$.
Hence, the position vector of their center of mass is $\, \displaystyle \frac {1}{2}\, (3\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$.

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

At which point is the centre of gravity situated in: A circular lamina.

  1. At the centre of radius.

  2. At the centre of semi circular lamina.

  3. At the centre of circular lamina.

  4. can not say
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Centre of gravity means a point from which the weight of a body or system may be considered to act. In uniform gravity it is the same as the centre of mass. For regular bodies centre of gravity lies at the centre of the body. Hence this will be at the centre of the circular lamina.

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

What is the position of centre of gravity of a rectangular lamina?

  1. At the mid point of longer side

  2. At the mid point of shorter side

  3. At the point of intersection of its diagonals

  4. At one of the corners
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Centre of gravity means a point from which the weight of a body or system may be considered to act. In uniform gravity it is the same as the centre of mass. Hence for a regular shaped bodies it will have at the centre of that body. Hence for a rectangle it is nothing but at the point of intersection of diagonals.