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 Multiple Choice QuestionsMultiple Choice Questions

1.

A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals µ. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(=QR), are, respectively close to :


  • 0.2 and 6.5 m

  • 0.2 and 3.5 m

  • 0.29 and 3.5 m

  • 0.29 and 6.5 m


C.

0.29 and 3.5 m

Energy lost over path PQ = μ mg cos θ x 4



Energy lost over path QR = μ mgx

i.e μ mg cos 30°  x 4 = μ mgx  (∴ θ = 30°)

straight x space equals space 2 square root of 3 space equals space 3.45 space straight m
From Q to R energy loss is half of the total energy loss.

i.e μ mgx = mgh/2
μ = 0.29

The values of the coefficient of friction μ and the distance x (=OR) are 0.29 and 3.5 m

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2.

A particle is moving with velocity,straight v space equals space straight k left parenthesis straight y space bold i with hat on top space plus space straight x space bold j with hat on top right parenthesis where K is a constant. The general equation for its path is

  • y = x2 + constant

  • y2 =  x + constant

  • xy = constant

  • y2 = x2 + constant


D.

y2 = x2 + constant

The velocity of the particle,straight v space equals space straight k left parenthesis straight y space bold i with hat on top space plus space straight x space bold j with hat on top right parenthesis 
rightwards double arrow space dx over dt space equals space ky comma space dy over dt space equals space kx
dy over dx space equals space dy over dt straight x dt over dx space equals space kx over ky
space ydy space equals space xdx
straight y squared space equals space straight x squared space space plus space straight C

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3.

A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F (t) = F0e–bt in the x direction. Its speed v(t) is depicted by which of the following curves?


C.

straight F equals space straight F subscript 0 straight e to the power of negative bt end exponent
rightwards double arrow space straight a space equals space straight F over straight m space equals space straight F subscript 0 over straight m straight e to the power of negative bt end exponent
rightwards double arrow space dv over dt space equals straight F subscript 0 over straight m straight e to the power of negative bt end exponent
integral dv space equals space integral subscript 0 superscript straight t straight F over straight m straight e to the power of negative bt end exponent dt
rightwards double arrow space straight v space equals straight F over straight m open square brackets table row cell negative 1 end cell row straight b end table close square brackets open square brackets table row cell straight e to the power of negative bt end exponent end cell end table close square brackets subscript 0 superscript straight t
rightwards double arrow space straight v equals space straight F over mb left square bracket straight e to the power of negative bt end exponent right square bracket
straight v equals space 0 space at space straight t equals space 0
and space straight v rightwards arrow space straight F over mb space as space straight t rightwards arrow infinity

So, velocity increases continuously and attains a maximum value of v= F/mb as t ∞
614 Views

4.

A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be

  • 20 square root of 2m
  • 10 m

  • 10 square root of 2 m
  • 20 m


D.

20 m

Maximum speed with which the boy can throw stone is
straight u space equals space square root of 2 gh end root
space equals space square root of 2 space straight x 10 straight x 10 end root
space equals 10 square root of 2 space straight m divided by straight s
Range is maximum when projectile is thrown at angle of 45o, Thus,
straight R subscript max space equals space fraction numerator straight u squared over denominator straight g space end fraction space equals space fraction numerator left parenthesis 10 square root of 2 right parenthesis squared over denominator 10 end fraction space equals space 20 space straight m

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5.

The figure shows the position –time (x – t) graph of one–dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is

  • 0.4 (Ns)

  • 0.8 Ns

  • 1.6 Ns

  • 0.2 Ns


B.

0.8 Ns

From the graph it is a straight line, so uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph.
(∴ impulse (l) = f x t
⇒ I  = mat
 space equals space fraction numerator straight m space left parenthesis straight v subscript 2 minus straight v subscript 1 right parenthesis over denominator straight t end fraction space straight t space equals space mv subscript 2 space minus space mv subscript 1
Initial velocity, v1 = 2/2 = 1 ms-1
Final velocity v2 = 2/2 = - 1 ms-1
pi  = mv1 = 0.4 N-s
pf  = mv2 = -0.4 N-s
J = pf-pi = -0.4-0.4
 = - 0.8 N-s
|J| = 0.8 N-s

438 Views

6.

An object moving with a speed of 6.25m/s, is decelerated at a rate given by dv/dt =- 2.5√v, where v is the instantaneous speed. The time taken by the object, to come to rest, would be

  • 2 s

  • 4 s

  • 8 s

  • 1 s


A.

2 s

dv over dt space equals space minus space 2.5 square root of straight v
fraction numerator dv over denominator square root of straight v end fraction space equals space minus 2.5 dt
integral subscript 6.25 end subscript superscript 0 straight v to the power of negative 1 divided by 2 end exponent space dv space equals space minus 2.5 integral subscript 0 superscript straight t dt
minus 2.5 left square bracket straight t right square bracket subscript 0 superscript straight t space equals space left square bracket 2 space straight v to the power of 1 divided by 2 end exponent right square bracket subscript 6.25 end subscript superscript 0
straight t space equals 2 straight s
379 Views

7.

A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed v in the x-y plane as shown in the figure:

Which of the following statements is false for the angular momentum → L about the origin?

  • bold L space equals space minus fraction numerator mv over denominator square root of 2 end fraction space straight R space straight k with hat on top comma space when space the space paticle space is space moving space from space straight A space to space straight B
  • straight L space equals space mv space open parentheses fraction numerator straight R over denominator square root of 2 end fraction space plus straight a close parentheses space bold k with bold hat on top comma space when space the space particle space is space moving space from space straight B space to space straight C
  • straight L space equals space mv space open parentheses fraction numerator straight R over denominator square root of 2 end fraction space minus straight a close parentheses space straight k with hat on top comma space when space the space particle space is space moving space from space straight B space to space straight C
  • bold L space equals space minus fraction numerator mv over denominator square root of 2 end fraction space straight R space straight k with hat on top comma space when space the space paticle space is space moving space from space straight D space to space straight A

B.

straight L space equals space mv space open parentheses fraction numerator straight R over denominator square root of 2 end fraction space plus straight a close parentheses space bold k with bold hat on top comma space when space the space particle space is space moving space from space straight B space to space straight C

For a particle of mass, m is moving along the side of a square a. Such that
Angular momentum L about the origin

equals space straight L space equals space straight r space straight x space straight p space equals space straight r subscript straight p space sin space straight theta space straight n with hat on top space or space straight L space equals space straight r space left parenthesis straight p right parenthesis straight n with hat on top
When space straight a space particle space is space moving space from space straight D space to space straight A
straight L space equals space fraction numerator straight R over denominator square root of 2 end fraction space mv space left parenthesis negative straight k with hat on top right parenthesis
straight A space particle space is space moving space from space straight A space to space straight B
straight L space equals space fraction numerator straight R over denominator square root of 2 end fraction space mv space left parenthesis negative straight k with hat on top right parenthesis
and space it space moves space from space straight C space to space straight D
space straight L space equals space open parentheses fraction numerator straight R over denominator square root of 2 end fraction plus straight a close parentheses mv space left parenthesis straight k with hat on top right parenthesis
For space straight B space to space straight C comma space we space have space
straight L space space equals space open parentheses fraction numerator straight R over denominator square root of 2 end fraction plus straight a close parentheses mv space left parenthesis straight k with hat on top right parenthesis


1517 Views

8.

A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to:

  • turn left

  • turn right

  • go straight

  • turn left and right alternately


A.

turn left

As, the wheel rolls forward the radius of the wheel, decreases along AB hence for the same number of rotations it moves less distance along AB, hence it turns left.

763 Views

9.

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

  • square root of GM over straight R end root
  • square root of 2 square root of 2 GM over straight R end root
  • square root of GM over straight R left parenthesis 1 plus 2 square root of 2 right parenthesis end root
  • 1 half square root of GM over straight R left parenthesis 1 plus 2 square root of 2 right parenthesis end root

D.

1 half square root of GM over straight R left parenthesis 1 plus 2 square root of 2 right parenthesis end root

Net force acting on any one particle M,
 =fraction numerator GM squared over denominator left parenthesis 2 straight R right parenthesis squared end fraction space plus space fraction numerator GM squared over denominator left parenthesis straight R square root of 2 right parenthesis squared end fraction cos space 45 to the power of straight o space plus space fraction numerator GM squared over denominator left parenthesis straight R square root of 2 right parenthesis squared end fraction Cos space 45 to the power of straight o
space equals space GM squared over straight R squared open parentheses 1 fourth plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses


This force will equal to centripetal force
So, Mv squared over straight R space equals space GM squared over straight R squared open parentheses 1 fourth plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses
straight v space equals space square root of fraction numerator GM over denominator 4 straight R end fraction left parenthesis 1 plus 2 square root of 2 right parenthesis end root space equals space 1 half square root of GM over straight R left parenthesis 2 square root of 2 plus 1 right parenthesis end root

658 Views

10.

A projectile is given an initial velocity ofopen parentheses straight i with hat on top space plus 2 straight j with hat on top close parentheses straight m divided by straight s  where is along the ground and bold j with bold hat on top is along the vertical. If g = 10 m/s2, the equation of its trajectory is: 

  • y = x-5x2

  • y = 2x-5x2

  • 4y = 2x- 5x2

  • 4y = 2x-25x2


B.

y = 2x-5x2

Initial velocity, space straight V space equals space left parenthesis straight i with hat on top plus 2 straight j with hat on top right parenthesis space straight m divided by straight s
Magnitude of velocity,
straight v equals square root of left parenthesis 1 right parenthesis squared space plus left parenthesis 2 right parenthesis squared end root space equals space square root of 5 space straight m divided by straight s end root
the equation of trajectory of the projectile
straight y space equals space straight x space tan space straight theta space minus space fraction numerator gx squared over denominator 2 straight u squared end fraction left parenthesis 1 plus space tan squared straight theta right parenthesis
left square bracket because space tan space straight theta space equals space straight y over straight x space equals 2 over 1 space equals 2 right square bracket
straight y space equals space straight x.2 space minus space fraction numerator 10 left parenthesis straight x right parenthesis squared over denominator 2 left parenthesis square root of 5 right parenthesis squared end fraction left square bracket 1 plus left parenthesis 2 right parenthesis squared right square bracket
equals space 2 straight x minus fraction numerator 10 left parenthesis straight x squared right parenthesis over denominator 2 space straight x 5 end fraction left parenthesis 1 plus 4 right parenthesis space equals space 2 straight x minus 5 straight x squared

197 Views