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 Multiple Choice QuestionsMultiple Choice Questions

101.

A man is at a distance of  6 m from a bus. The bus begins to move with a constant acceleration of 3ms.In order to catch the bus, the minimum speed with which the man should run towards the bus is

  • 2 m s-1

  • 4 m s-1

  • 6 m s-1

  • 8 m s-1


C.

6 m s-1

If the man did not run, the bus would be at a distance s1 at time t given by

   s1 = 6 + 12α t2

       = 6 + 12 × 3 × t2

   s1 = 6 + 32t2

If v is the speed of man, he would cover a distance s2 = vt in time t.

To catch the bus,

           s1 = s2

     6 + 32 t2 = v t

⇒     t2 - 2 v3t + 4 = 0

which gives

         t = 2 v6 ±  4 v29 - 16 12  

Now t will be real if  4 v29 - 16 is positive or zero.

Minimum v corresponds to 4 v216 - 16 = 0 which gives v = 6 m s-1        


102.

Two identical charged spheres suspended froma  common point by two massless strings of lengths l, are initially at a distance d (d<< l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velcoity v. Then, v varies as a function of the distance x between the sphere, as

  • vproportional tox

  • straight v space proportional to space straight x to the power of begin inline style bevelled fraction numerator negative 1 over denominator 2 end fraction end style end exponent
  • vproportional tox-1

  • vproportional tox-1


B.

straight v space proportional to space straight x to the power of begin inline style bevelled fraction numerator negative 1 over denominator 2 end fraction end style end exponent

Two identical charged spheres suspended from a common point by two massless strings of length L.

In increment space ABC,
tan space theta space equals space fraction numerator F over denominator m g end fraction

rightwards double arrow space straight F over mg space equals space tan space straight theta space... space left parenthesis straight i right parenthesis
The charge begins to leak from both the sphere at a constant rate. As a result, the spheres approach each other with velocity v.

Therefore, equation (i) can be rewritten as,
fraction numerator Kq squared over denominator straight X squared mg end fraction space equals space fraction numerator X divided by 2 over denominator square root of l squared minus begin display style x squared over 4 end style end root end fraction

rightwards double arrow space fraction numerator Kq squared over denominator straight X squared mg end fraction space equals space fraction numerator straight x over denominator 2 straight l end fraction
rightwards double arrow space straight q squared space proportional to space straight x cubed
rightwards double arrow space straight q space proportional to space straight x to the power of begin inline style bevelled 3 over 2 end style end exponent
rightwards double arrow space dq over dt space proportional to space fraction numerator straight d space left parenthesis straight x to the power of bevelled 3 over 2 end exponent right parenthesis over denominator dx end fraction. space dx over dt
rightwards double arrow space dq over dt space proportional to space straight x to the power of begin inline style bevelled 1 half end style end exponent
rightwards double arrow space straight v space proportional to space 1 over straight x to the power of bevelled 1 half end exponent
rightwards double arrow space straight v space proportional to space straight x to the power of negative 1 divided by 2 end exponent

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103.

A body is at rest at x = 0 and t = 0, it starts moving in the positive X-direction with a constant acceleration. At the same instant, another body passes through x = O moving in the positive X-direction with a constant speed. The position of the first body is given by x1 (t) after time t and that of second body by x2 (t) after the same time interval. Which of the following graphs correctly describes (x− x2) as a function of time t ?


C.

We have, distance travelled by particle 1 increases with time as                  x1 = 12 at2Distance travelled by particle 2 is proportional to t                  x2 = vt    x1 - x2 = 12 at2 - vt Graph of x1 - x2 vs t is parabola.At             t = 0 , x1 - x2 = 0Also slope   = xt = at - vFor t < va , the slope is negativeFor t = va , the slope is zeroFor t > va , the slope is positiveHence, option (c) is correct.


104.

The x and y coordinates of the particle at any time are x = 5t – 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2 s is

  • 0

  • 5 m/s2

  • -4 m/s2

  • -4 m/s2


C.

-4 m/s2

x = 5t – 2t2    y = 10t

dx over dt space equals space 5 minus 4 straight t space space space space space space space dy over dt space equals space 10
straight v subscript straight x space equals space 5 minus 4 straight t space space space space space space space space space straight V subscript straight y space equals space 10
dv over dt straight x space equals space minus 4 space space space space space space space space dv over dt straight y space equals space 10
straight a subscript straight x space equals space minus space 4 space space space space space space space space space space space straight a subscript straight y space equals space 0

Acceleration of particle at t = 2 s is = –4 m/s2

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105.

The displacement x of a particle varies with time t as x = ae-αt + beβt where a, b, α and β are positive constants. The velocity of the particle will

  • decrease with time

  • be independent of α and β

  • drop to zero when α = β

  • increase with time


D.

increase with time

Given:- 

        x = a e-αt + beβt

We know that

   velocity v = dxdt

              v  =  d dt ae-αt + beβt 

                  = -aα e-αt + bβ eβt

              v = -a αe-αt +  eβt

               v = A + B

where A = - aα e-αt , B = bβ e-βt

The value ot term A = -aα e-αt increases and of term B = bβ e-βt increases with time. As a result velocity goes on increasing with time. 


106.

A monkey of mass 30 kg climbs on a massless rope whose breaking strength is 450 N. The rope will break if the monkey (Take g = 10 m/s2 )

  • climbs up with a uniform speed of 5 m/s

  • climbs down with an acceleration 4 m/s2

  • climbs up with an acceleration 6 m/s2

  • climbs down with a uniform speed of 5 m/s2


C.

climbs up with an acceleration 6 m/s2

To move up with an acceleration 'a', the monkey will push the rope downward with a force

               F = ma = 30 a

Now,  Tmax = mg + ma = 30 × 10 + 30 a

⇒        450 = 300 + 30 a

⇒            a = 5 m/s2

Hence, the rope will break if the monkey climbs up with an acceleration more than 5 m/s2.


107.

If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is,

  • 90o

  • 45o

  • 180o

  • 180o


A.

90o

There are two vectors P and Q.

It is given that,

open vertical bar straight P plus straight Q close vertical bar space equals open vertical bar P minus Q close vertical bar
Let, angle between P and Q is straight ϕ,
straight P squared plus straight Q squared space plus space 2 PQ space cosϕ space equals space straight P squared space plus space straight Q squared space minus space 2 space PQ space cos space straight ϕ
rightwards double arrow space 4 PQ space cos space straight ϕ space equals space 0

rightwards double arrow space cos space straight ϕ space equals space 0 space left square bracket because space straight P comma space straight Q space not equal to space 0 right square bracket

rightwards double arrow space straight ϕ space equals space straight pi over 2 equals space 90 to the power of straight o

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108.

A body is whirled in a horizontal circle of radius 25 cm. It has an angular velocity of 13 rad/s. What is its linear velocity at any point on circular path?

  • 2 m/s

  • 3 m/s

  • 3.25 m/s

  • 4.25 m/s


C.

3.25 m/s

Given:- r = 25 cm

          r = 0.25  m

  ω = 13 rad/s

   Linear speed  = Radius × angular speed

            V = rω

          = 0.25 × 13

        3.25 m/s


109.

If A + B = C and that C is perpendicular to A. What is the angle between A and B, if A = B

  • π4 rad

  • π2 rad

  • 3 π4 rad

  • π rad


C.

3 π4 rad

Given:-

      A + B = C

Since, B = C - A

Also

   C  A

    B2 = 2 A2

⇒   B = 2 A

Now

    A2 + B2 + 2A B cosθ = C2 = A2

∴          cos θ = - 12

This gives

       θ = 3π4 rad


110.

A ball is dropped from the top of a building 100 m high. At the same instant another ball is thrown upwards with a velocity of 40 m/s from the bottom of the building. The two balls will meet after

  • 3 s

  • 2 s

  • 2.5 s

  • 5 s


C.

2.5 s

Let balls meet after t s. The distance travelled by the ball coming down is 

          s112 gt2

Distance travelled by the other ball

        s2 = 40 t - 12 gt2

         s1 + s2 = 100 m

∴        12 gt2 + 40 t - 12gt2 = 100m

                  t = 10040

∴                 t = 2.5 s