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 Multiple Choice QuestionsMultiple Choice Questions

1.

A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to 8 x 10-4 J by the end of the second revolution after the beginning of the motion?

  • 0.15 m/s2

  • 0.18 m/s2

  • 0.2 m/s2

  • 0.1 m/s2


D.

0.1 m/s2

Given, mass of particle. m = 0.01 kg

Radius of circle along which particle is moving , r = 6.4 cm

Kinetic energy of particle, K.E. = 8 x 10-4 J

rightwards double arrow space 1 half mv squared space equals space 8 space straight x space 10 to the power of negative 4 end exponent space straight J

rightwards double arrow space straight v squared space equals space fraction numerator 16 space straight x space 10 to the power of negative 4 end exponent over denominator 0.01 end fraction
space space space space space space space space space space space equals 16 straight x 10 to the power of negative 2 end exponent space space space space space space space space space space space... space left parenthesis straight i right parenthesis

Given that, KE of particle is equal to 8 x 10-4 J by the end of second revolution after the beginning of the motion of particle.

It means, initial velocity (u) is 0 m/s at this moment.

Now, using the Newton's 3rd equation of motion,

v2 = u2 + 2as

rightwards double arrow space space space straight v squared space equals space 2 straight a subscript straight t straight s

rightwards double arrow space straight v squared space equals space 2 straight a subscript straight t space left parenthesis 4 πr right parenthesis

rightwards double arrow space straight a subscript straight t space equals space fraction numerator straight v squared over denominator 8 πr end fraction
space space space space space space space space space equals space fraction numerator 16 space straight x space 10 to the power of negative 2 end exponent over denominator 8 space straight x space 3.14 space straight x space 6.4 space straight x space 10 to the power of negative 2 end exponent end fraction

space space space space straight a subscript straight t space equals space 0.1 space straight m divided by straight s squared

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2.

A particle has initial velocity (2i+3j) and acceleration (0.31i +0.2j). The magnitude of velocity after 10 s will be

  • 9 square root of 2 units
  • 5 square root of 2 units
  • 5 units

  • 9 Units


B.

5 square root of 2 units

From the equation of motion, we knows,
v=u+ at
v= (2i+3j)
a=(0.3i +0.2j)
v=(2i+3j) +(0.3i +0.2j) x 10
v=5i+5j
straight v equals space 5 square root of 2 space units

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3.

A particle of unit mass undergoes one-dimensional motion such that its velocity according to
V(x) =  βx-2n where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by 

  • -2nβ2 x-2n-1

  • -2nβx-4n-1

  • -2β x-2n+1

  • -2nβ2e-4n+1


B.

-2nβx-4n-1

Given, v = βx-2n
straight a equals space dv over dt equals dx over dt. dv over dx
straight a equals straight v dv over dx equals left parenthesis space βx to the power of negative 2 straight n end exponent right parenthesis left parenthesis negative 2 straight n space βx to the power of negative 2 straight n minus 1 end exponent right parenthesis
straight a equals negative 2 nβ squared straight x to the power of negative 4 straight n minus 1 end exponent

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4.

If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1 s and 2 s is,

  • 3A + 7B

  • fraction numerator 3 over denominator 2 space end fraction A space plus space 7 over 3 B
  • straight A over 2 plus B over 3
  • fraction numerator 3 over denominator 2 space end fraction A space plus space 4 B

B.

fraction numerator 3 over denominator 2 space end fraction A space plus space 7 over 3 B

Velocity of the particles is given as,
v = At + Bt2, where a and B are constants.

rightwards double arrow space dx over dt space equals space At space plus space Bt squared space space open square brackets because space straight v space equals dx over dt close square brackets
rightwards double arrow space dx space equals space left parenthesis At space plus space Bt squared right parenthesis space dt
Integrating both sides, we get
space space space space space space integral subscript straight x subscript 1 end subscript superscript straight x subscript 2 end superscript space d x space equals space integral subscript 1 superscript 2 space left parenthesis A t space plus space B t squared right parenthesis space d t
increment x space equals space x subscript 2 space minus space x subscript 1
space space space space space space equals space A integral subscript 1 superscript 2 t. space d t space plus space B integral subscript 1 superscript 2 t squared. d t
space space space space space space equals space A open square brackets t squared over 2 close square brackets subscript 1 superscript 2 space plus space B space open square brackets t cubed over 3 close square brackets subscript 1 superscript 2
space space space space space space equals space A over 2 left parenthesis 2 squared space minus space 1 squared right parenthesis space plus space B over 3 left parenthesis 2 cubed minus space 1 cubed right parenthesis

Therefore, distance travelled between 1 s and 2 s is,

increment straight x equals straight A over 2 straight x left parenthesis 3 right parenthesis plus straight B over 3 left parenthesis 7 right parenthesis
space space space space space space equals space fraction numerator 3 straight A over denominator 2 end fraction plus fraction numerator 7 straight B over denominator 3 end fraction

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5.

The velocity of a projectile at the initial point A is (2 i + 3 j) m/s. Its velocity (in m/s) at point B is,

  • -2 i - 3 j

  • -2 i + 3 j

  • i - 3 j

  • i + 3 j


C.

i - 3 j

From the fig,, the X component remain unchanged, while the Y-component is reverse. Then, the velocity at point B is (2 i - j) m/s.

2149 Views

6.

A balloon with mass m is descending down with an acceleration a (where a <g). How much mass should be removed from it so that it starts moving up with an acceleration a?

  • fraction numerator 2 ma over denominator straight g plus straight a end fraction
  • fraction numerator 2 ma over denominator straight g minus straight a end fraction
  • fraction numerator ma over denominator straight g plus straight a end fraction
  • fraction numerator ma over denominator straight g minus straight a end fraction

A.

fraction numerator 2 ma over denominator straight g plus straight a end fraction

When the balloon is descending down with acceleration a,


So, mg - B = mx A             ... (i)
where B is the buoyant force.

We assume here that while removing same mass, the volume of balloon and hence buoyant force will not change.

Let, us assume the new mass of the balloon is m'.

So, mass removed is (m-m')

Therefore,

B-m'g = m'x a         ... (ii)

On solving equations (i) and (ii), we have

mg - B = m x a

B - m'g = m' x a

mg - m'g = ma +  m'a

(mg - ma) = m' (g+a) = m (g-a) = m' (g+a)

That is,

straight m apostrophe space equals space fraction numerator straight m space left parenthesis straight g minus straight a right parenthesis over denominator straight g plus straight a end fraction
That is, mass removed is m-m'

equals space straight m space open square brackets fraction numerator 1 minus left parenthesis straight g minus straight a right parenthesis over denominator left parenthesis straight g plus straight a right parenthesis end fraction close square brackets
equals space straight m open square brackets fraction numerator left parenthesis straight g plus straight a right parenthesis minus left parenthesis straight g minus straight a right parenthesis over denominator left parenthesis straight g plus straight a right parenthesis end fraction close square brackets
equals space straight m open square brackets fraction numerator straight g plus straight a minus straight g plus straight a over denominator straight g plus straight a end fraction close square brackets
equals space increment straight m space equals space fraction numerator 2 ma over denominator straight g plus straight a end fraction

900 Views

7.

What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

  • square root of 2 gR end root
  • square root of 3 gR end root
  • square root of 5 gR end root
  • square root of gR

C.

square root of 5 gR end root

The question is illustrated in the figure below,

Let, the tension at point A be TA.

Using Newton's second law, we have

straight T subscript straight A space minus space mg space equals space mv subscript straight c squared over straight R
Energy at point A = 1 half m v subscript o squared space space space space space space space space space... space left parenthesis i right parenthesis
Energy at point C is,

1 half m v subscript c squared space plus space m g space x space 2 R space space space space space space space space space space... space left parenthesis i i right parenthesis
At point C, using Newton's second law,

straight T subscript straight c space plus space mg space equals space mv subscript straight c squared over straight R
In order to complete a loop, Tc greater-than or slanted equal to space 0
So,
 mg space equals space mv subscript straight c squared over straight R
rightwards double arrow space straight v subscript straight c space equals space square root of gR space space space space space space space space... space left parenthesis iii right parenthesis

From equation (i) and (ii)

Using the principle of conservation of energy,

1 half m v subscript o squared space equals space 1 half m v subscript c squared space plus space 2 space m g R
rightwards double arrow space 1 half m v subscript o squared italic space italic equals italic space italic 1 over italic 2 m g R italic space italic plus italic space italic 2 m g R italic space x italic space italic 2
italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space open square brackets v subscript c italic space italic equals italic space square root of g R end root close square brackets
italic rightwards double arrow italic space v subscript o to the power of italic 2 italic space italic equals italic space g R italic space italic plus italic space italic 4 g R
italic rightwards double arrow italic space italic space v subscript o italic space italic equals italic space square root of italic 5 g R end root
5996 Views

8.

A particle  moves so that its position vector is given by r = cos ωt space straight x with hat on top space plus space sin space ωt space straight y with hat on top where straight omega is a cosntant. which of the following is true?

  • velocity and acceleration both are parallel to r

  • velocity is perpendicular to r and acceleration is directed towards the origin

  • velocity is perpendicular to r and acceleration is directed away from the origin

  • velocity and acceleration both are perpendicular to r


B.

velocity is perpendicular to r and acceleration is directed towards the origin

Position vector of the article is given by,
straight r space equals space cos space ωt space stack straight x space with hat on top space plus space sin space ωt space straight y with hat on top
where straight omega is a constant.
Velocity of the particle is,

straight v space equals space dr over dt space equals space straight d over dt open parentheses cos space straight omega space straight t space straight x with hat on top space plus space sin space ωt space straight y with hat on top close parentheses
straight v space equals space left parenthesis negative sinωt right parenthesis straight omega stack straight x space with hat on top space plus space left parenthesis cos space ωt right parenthesis straight omega straight y with hat on top

space space space equals space minus space straight omega space left parenthesis space sin space ωt space straight x with hat on top space minus space cos space ωt space straight y with hat on top right parenthesis

Acceleration space of space the space particle comma
straight a space equals space dv over dt
space space equals straight d over dt open square brackets negative straight omega space sin space ωt space straight x with hat on top space plus straight omega space cos space ωt space straight y with hat on top close square brackets
space equals space minus straight omega squared space cos space ωt space straight x with hat on top space minus space straight omega squared space sin space ωt space straight y with hat on top
space equals space minus straight omega squared space open parentheses cos space ωt space straight x with hat on top space space minus sin space ωt space straight y with hat on top space close parentheses
straight a space equals space minus straight omega squared space straight r space equals straight omega squared space left parenthesis negative straight r right parenthesis space
The particle is at a point P. That is, its position vector is directed as shown below:

Therefore, acceleration is directed towards -r, that is towards "O"

We have

straight v with rightwards harpoon with barb upwards on top. r with rightwards harpoon with barb upwards on top space equals space minus straight omega space left parenthesis sinω space straight t space straight x with hat on top space minus space cos space ωt space straight y with hat on top right parenthesis.
space space space space space space space space space space space space space space space space space space space left parenthesis cos space ωt space straight x with hat on top space plus space sin space ωt space straight y with hat on top right parenthesis 

       equals space minus straight omega space left square bracket space sin space straight omega. space cosωt space plus
space space space space space space space space space space space space 0 plus 0 space minus space sin space ωt. space cos space ωt right square bracket 

equals space minus straight omega space left parenthesis 0 right parenthesis space equals space 0

rightwards double arrow space straight v perpendicular space straight r
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9.

A body of mass 1 kg begins to move under the action of a time dependent force F = left parenthesis 2 straight t space straight i with hat on top space plus space 3 straight t squared space straight j with hat on top right parenthesis N, where straight i with hat on top space a n d space j with hat on top are units vectors along X and Y axis. What power will be developed by the force at the time (t)?

  • (2t2 + 4t4) W

  • (2t3 + 3 t4) W

  • (2t3 + 3t5) W

  • (2t + 3t3)W


C.

(2t3 + 3t5) W

A body of mass 1 kg begins to move under the action of time dependent force,
F = (2t straight i with hat on top+3t2 straight j with hat on top) N,
where stack straight i space with hat on top a n d space j with hat on top are unit vectors along X and Y axis.
F = ma
rightwards double arrow space straight a space equals space straight F over straight m
rightwards double arrow space straight a space equals space fraction numerator left parenthesis 2 straight t space straight i with hat on top space plus space 3 straight t squared space straight j with hat on top right parenthesis over denominator 1 end fraction space space left square bracket space straight m space equals 1 space kg right square bracket
rightwards double arrow space straight a space equals space left parenthesis 2 straight t space straight i with hat on top space plus space 3 straight t squared space straight j with hat on top right parenthesis space straight m divided by straight s squared

Acceleration, a = dv over dt
rightwards double arrow space dv space equals space straight a. space dt space space space space space space... space left parenthesis straight i right parenthesis
Integrating both sides, we get

integral d v space equals space integral a. space d t
space space space space space space space space space equals space integral left parenthesis 2 t space i with hat on top space plus space 3 t squared space j with hat on top right parenthesis space d t
v space equals space t squared space space i with hat on top space plus straight t cubed space straight j with hat on top
Power developed by the force at the time t will be given as,

P = F.v = (2t straight i with hat on top + 3t2 straight j with hat on top).(straight t squared space straight i with hat on top space plus space straight t cubed space straight j with hat on top)
   = (2t. t2 + 3t2.t3)

P = (2t3 + 3t5) W

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10.

A projectile is fired from the surface of the earth with a velocity of 5 m/s and angle straight theta with the horizontal. Another projectile fired from another planet with a velocity of 3 m/s at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is,

  • 3.5

  • 5.9

  • 16.3

  • 110.8


A.

3.5

If the trajectory is same for both the projectiles, their maximum height will be the same.

That is,

(Hmax)1 = (Hmax)2

i.e., 
space space space space space space space space fraction numerator straight u subscript 1 squared space sin squared straight theta over denominator 2 straight g subscript 1 end fraction space equals space fraction numerator straight u subscript 2 squared space sin squared straight theta over denominator 2 straight g subscript 2 end fraction
rightwards double arrow space straight u subscript 1 squared over straight u subscript 2 squared space equals space straight g subscript 1 over straight g subscript 2
rightwards double arrow space fraction numerator left parenthesis 5 right parenthesis squared over denominator left parenthesis 3 right parenthesis squared end fraction space equals space fraction numerator 9.8 over denominator g subscript 2 end fraction
rightwards double arrow space g subscript 2 italic space italic equals italic space fraction numerator italic 9 italic. italic 8 italic space x italic space italic 9 over denominator italic 25 end fraction
italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic equals italic space italic 3 italic. italic 5 italic space m italic divided by s to the power of italic 2

1750 Views