Two bodies of mass 1 kg have position vectors,bold i with bold hat on top bold space bold plus bold space 2 bold j with bold hat on top bold space bold plus bold k with bold hat on top bold space bold and bold space bold minus 3 bold i with bold hat on top bold space bold minus 2 bold j with bold hat on top bold space bold plus bold k with bold hat on top  respectively.The centre of mass of this system has a position vector

  • negative 2 bold i with hat on top space plus 2 space bold k with hat on top
  • negative 2 bold i with bold hat on top minus bold j with bold hat on top space plus bold k with hat on top
  • 2 bold i with bold hat on top minus bold j with bold hat on top space minus bold k with hat on top
  • 2 bold i with bold hat on top minus bold j with bold hat on top space minus bold k with hat on top

B.

negative 2 bold i with bold hat on top minus bold j with bold hat on top space plus bold k with hat on top

The position vector of centre of mass




Point to know:
The centre of mass changes its position only under the translatory motion. There is no effect of rotatory motion on centre of mass of the body.

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If bold F with bold rightwards arrow on top is the force acting on a particle having position vector straight r with rightwards arrow on top space a n d space tau with rightwards arrow on top be the the torque of this force about the origin, then

  • straight r with rightwards arrow on top. straight tau with rightwards arrow on top space not equal to space 0 space and space bold F with bold rightwards arrow on top space. space straight tau with rightwards arrow on top space equals space 0
  • straight r with rightwards arrow on top. straight tau with rightwards arrow on top space greater than space space 0 space and space bold F with bold rightwards arrow on top space. space straight tau with rightwards arrow on top space less than space 0
  • straight r with rightwards arrow on top. straight tau with rightwards arrow on top space equals space space 0 space and space bold F with bold rightwards arrow on top space. space straight tau with rightwards arrow on top space equals space 0
  • straight r with rightwards arrow on top. straight tau with rightwards arrow on top space equals space space 0 space and space bold F with bold rightwards arrow on top space. space straight tau with rightwards arrow on top space equals space 0

C.

straight r with rightwards arrow on top. straight tau with rightwards arrow on top space equals space space 0 space and space bold F with bold rightwards arrow on top space. space straight tau with rightwards arrow on top space equals space 0

Torque is an axial vector ie, its direction is always perpendicular to the plane containing vectors  

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The distance travelled by a particle starting from rest and moving with an acceleration 4/3 ms-2 in the third second is

  • 6 m

  • 4 m

  • 10/3

  • 10/3


C.

10/3

Distance travelled by  the particle in nth second


Where u is initial speed and a is an acceleration of the particle. 

Here, n = 3, u =0, a = 4/3 m/s2

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A particle moves in the x-y plane according to rule x = a sin ωt and y = a cos ωt. The particle follows

  • an elliptical path

  • a circular path 

  • a parabolic path

  • a parabolic path


B.

a circular path 



This is the equation of a circle, so the particle follows a circular path.
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A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 s is s1 and that covered in the first 20 s is s2, then

  • s2 = 2s1

  • s2 = 3s1

  • s2 = 4 s1

  • s2 = 4 s1


C.

s2 = 4 s1

If the particle is moving in a straight line under the action of a constant force then distance covered s = ut + at2 /2
since the body start from rest u = 0

therefore, s = at2/2

Now,



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