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 Multiple Choice QuestionsMultiple Choice Questions

11.

A police jeep is chasing with velocity of 45 km/h. A thief in another jeep moving with velocity 153 km/h. Police fires a bullet with muzzle velocity of 180 m/s. The velocity it will strike the car of the thief is

  • 150 m/s

  • 27 m/s

  • 450 m/s

  • 250 m/s


A.

150 m/s

Relative velocity is the vector difference between the velocities of two bodies, the velocity of a body with respect to another regarded as being at rest.

Here, we calculate the relative velocity of the jeep:

 153 -45 =108 km/h

km/hour =100060×60=518= 108 ×518 =30 m/s

Therefore, striking speed = relative speed of a bullet with respect to the thief's car:

 180 - 30 = 150 m/s 


12.

A stone is thrown with an initial speed of 4.9m/s from a bridge in vertically upward direction.It falls down in water after 2sec. The height of bridge is

  • 24.7m

  • 19.8 m

  • 9.8 m

  • 4.9 m


C.

9.8 m

Here :- Initial velocity u=-4.9m/s (-ve is due to vertically upward motion)

Total time t = 2sec

The height of the bridge is given by

s =ut+12at2

Here we consider a=gacceleration due to gravity

s=-4.9×2+12×9.8×22 =-9.8+19.6 =9.8m


13.

A ball of mass 10 g moving with acceleration of 20 m/s2 is hit by a force which acts on it for 0.1 sec. The impulsive force is

  • 1.2Ns

  • 0.3Ns

  • 0.1Ns

  • 0.5Ns


B.

0.3Ns

Here:- Mass of the ball m= 10 g ; time dt =0.1 sec ;

acceleration of the ball = 20m/s2

Force F =ma =0.15×20 =3N

 

Impulsive forces are forces that act over short times producing rapid changes in motion

 Impulsive force = F×dt = 3×0.1 =0.3 Ns


14.

A person is standing in an elevator. In which situation he finds his weight less?

  • When the elevator moves upward with constant acceleration

  • when the elevator moves downward with constant acceleration

  • when the elevator moves upward with uniform velocity

  • When the elevator moves downward with uniform velocity


B.

when the elevator moves downward with constant acceleration

When elevator moves downward with constant acceleration. 

     α = g.

Then person standing in an elevator find himself weightless.

(g' with respect to the lift is   g - α

  


15.

A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by
        straight x equals 40 straight t space plus 12 straight t minus straight t cubed
How long would the particle travel before coming to rest?

  • 24m

  • 40m

  • 56m

  • 56m


C.

56m

Velocity is rate of change of distance or displacement.
Distance travelled by the particle is
                      straight x equals space 40 space plus 12 straight t minus straight t cubed
We know that, velocity is rate of change of distance i.e., straight v equals dx over dt.
therefore space space space straight v space equals space straight d over dt left parenthesis 40 plus 12 straight t minus straight t cubed right parenthesis
space space space space space space equals space 0 plus 12 minus 3 straight t squared

but final velocity v = 0
therefore space space 12 minus 3 straight t squared space equals space 0
or space space space space space space space space straight t squared space equals space 12 over 3 space equals 4
or space space space space space space straight t space equals space 2 straight s
Hence, distance travelled by the particle before coming to rest is given by
                   straight x equals 40 plus 12 left parenthesis 2 right parenthesis minus left parenthesis 2 right parenthesis cubed
space space equals 40 plus 24 minus 8 space equals space 64 minus 8
space space equals space 56 space straight m

            

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16.

" Parsec " is the  unit of

  • time

  • distance

  • frequency

  • angular acceleration


A.

time

Parasec (pc) is an astronomical unit of length equal to the distance at which baseline of one astronomical unit subtends at an angle of one sec of arc.

1 parsec = 3.085677  × 1016 m = 3.26 light years


17.

Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimension of permittivity is

  • [ M L-2 T2 A ]

  • [ M-1 L-3 T4 A2 ]

  • [ M L T-2 A ]

  • [ M L2 T-1 A2 ]


B.

[ M-1 L-3 T4 A2 ]

Force of attraction between two charges is given by

     F = 14πεo q1q2r2

        = A2 L2M L T-2 A2

       ( where 4π is dimensionless )

    F = [ M-1 L-3 T4 A2 ]


18.

A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the relation straight s equals 1 third straight t squared, where t is in s. Work done by the force in 2 s is:

  • 5 over 19 straight J
  • 3 over 8 straight J
  • 8 over 3 straight J
  • 8 over 3 straight J

C.

8 over 3 straight J

If a constant force is applied on the object causing a displacement in it, then it is said that work has been done on the body to displace it.
  Work done by the force  = Force x Displacement
or          W = F x s                            ...(i)
But from Newton's 2nd law, we have
  Force  =  Mass x Acceleration
i.e.,       F = ma                                ...(ii)
Hence, from Eqs. (i) and (ii), we get
straight W space equals space mas space equals space straight m open parentheses fraction numerator straight d squared straight s over denominator dt squared end fraction close parentheses straight s space space space space space space... left parenthesis iii right parenthesis space space space open parentheses because space straight a space equals space fraction numerator straight d squared straight s over denominator dt squared end fraction close parentheses
Now comma space we space have comma space space straight s space equals space 1 third straight t squared
therefore space space space space space space space space space fraction numerator straight d squared straight s over denominator dt squared end fraction space equals space straight d over dt open square brackets straight d over dt open parentheses 1 third straight t squared close parentheses close square brackets
space space space space space space space space space space space space space space space space space space space equals straight d over dt cross times open parentheses 2 over 3 straight t close parentheses
space space space space space space space space space space space space space space space space space space equals 2 over 3 dt over dt space space space space space space
space space space space space space space space space space space space space space space space space space equals 2 over 3 space space space space space space space
space space space space space space space space space space space space space space space
Hence, Eq (ii) becomes
                straight W space equals 2 over 3 ms space equals space 2 over 3 straight m cross times 1 third straight t squared
                   equals 2 over 9 mt squared
We have given
                   straight m equals 3 space kg comma space straight t space equals space 2 straight s
therefore space space space space straight w space equals space 2 over 9 cross times 3 cross times left parenthesis 2 right parenthesis squared space equals space 8 over 3 straight J
 

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19.

A police jeep is chasing with a velocity of 45 km/h, a thief in another jeep moving with velocity 153 km/h. Police fire a bullet with muzzle velocity of 180 m/s. The velocity with which it will strike the car of the thief is :

  • 150 m/s

  • 27 m/s

  • 450 m/s

  • 250 m/s


A.

150 m/s

The relative velocity cf thief's jeep with respect to the police jeep

= 153 - 45 = 108 km/h

108 × 518 = 30 m/s

Therefore, striking speed = relative speed of bullet with respect to the thief's car

= 180 - 30 = 150 m/s


20.

When a ball is thrown up vertically with velocity  vo, it reaches a maximum height of h. If one wishes triple the maximum height then the ball should be thrown with velocity

  • 3 vo

  • 3vo

  • 9vo

  • 3/2 vo


A.

3 vo

According to the Kinematic equation

    v2 = u2 + 2 αs         

 ∴   u2 ∝ s  or  vo 2 ∝ s      ( since  u = vo )     ....(i)

For given condition

     u'2 ∝ 3 s                                      .....(ii)

From eqn. (i) and (ii)

      u'2vo 2 = 3

⇒   u' = 3vo