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 Multiple Choice QuestionsMultiple Choice Questions

1.

An earthquake generates both transverse (S) and longitudinal (P) sound waves in the earth. The speed of S waves is about 4.5 km/s and that of P waves is about 8.0 km/s. A seismograph records P and S waves from an earthquake. The first P wave arrives 4.0 min before the first S wave. The epicentre of the earthquake is located at a distance about

  • 25 km

  • 250 km

  • 2500 km

  • 5000 km


C.

2500 km

Velocity of the S wave is v1 = 4.5 km/s

The velocity of the P waves is v2 = 8.0 km/s

Let the time taken by the S and P waves to reach the seismograph be t1 and  t2

It is given that

    t1 - t2 = 4 min

               = 4 × 60 sec

    t1 - t2 = 240 sec              .....(i)

Let the distance of the epicentre (km) be S. Then 

    S = v1t1 = v2t2

⇒  4.5 × t1 - 8t2 = 0

⇒              t14.58t1            ....(ii)

Using (i) and (ii)

      t1 - t2 = 240

⇒   t1 1 - 4.58 = 240

⇒     t1240 × 83.5

⇒          = 548.5 s

∴      S = v1t1 

           = 4.5 × 548.5

       S = 2468.6 

       S ≈ 2500 km


2.

S.I unit of velocity is 

  • ms

  • m sec-1

  • m hr-1

  • m/hr


B.

m sec-1

Velocity is a physical quantity, both magnitude and sirection are needed to define it. 

  S = Vt

S = distance

V = Velocity

t = time

The SI unit of distance is Meter (m)

The SI unit of time is second ( sec)

So the unit of velocity in SI unit is meter per sec i.e  m/s.


3.

Speed in kilometre per hour in S.I unit is represented as

  • KMPH

  • Kmhr-1

  • Kmh-1

  • kilometre/hour


B.

Kmhr-1

Kilometres per hour is a unit of measurement, which measures speed or velocity. The unit  symbol is km/h  or km h-1.

By definition, an object travelling at a speed of 1 km/h in a hour moves 1 kilometre.


4.

A wheel rotates with constant acceleration of 2.0 radian/sec2 .If the wheel starts from rest the number of revolution it makes in the first ten second, will be approximately

  • 32

  • 24

  • 16

  • 8


C.

16

Angular acceleration= 2 rad/s2

Time interval t= 10sec

Angular displacement θ = ?

Now by θ =ωt+12αt2             ....(i)It is an equation of angular variables of motion corresponding to the linear variable based equations ss=ut+12αt2(i)  θ = 0+12αt2θ =αt22=2×1002=100n=θ2π=1002×3.14=15.92 16


5.

A ball is thrown vertically upwards. Which of the following  plots represents the speed-time graph of the ball during its flight if the air resistance is not ignored?

  •  


D.

During the upward motion, the speed of the body decreases and will be zero at the highest point (since the gravitational force acting downward), afterward, the body starts downward motion and its speed increases.


6.

A car covers 2/5 of a certain distance with speed v1 and rest 3/5 part with velocity v2. The average speed of the car is :

  • 12v1v2

  • 5v1v23v1+2v2

  • 2v1v2v1+v2

  • v1+v22


B.

5v1v23v1+2v2

Let total distance covered = s

   t12/5sv1 and t23/5sv2

Average speed = total distance total time 

                   v = st1+ t2

or                v = s2s5v1+3s5v2

Hence,         v = 5v1v22v2+3v1


7.

Apparent frequency of sound of engine is changing in the ratio 5/3 under the condition that engine is first aprroaching and then receding away from the observer. If velocity of sound is 340 m/s, then velocity of engine is

  • 340 m/s

  • 170 m/s

  • 85 m/s

  • 310 m/s


C.

85 m/s

When source is moving towards stationary observer, its apparent frequency

                η1 = ηνν - vs

where, ν velocity of sound 

         νs velocity of source

When source is moving away from observer, its apparent frequency

             η2 = ηνν + vs

         η1η2 = νν - νs × ν + νsv

             53 = ν + νsν - νs

By solving, 

             νs = ν4 = 3404

                 = 85 m/s 


8.

The radius of earth is 6400 km and g = 10 m/s2. In order that a body of 5 kg weight is zero at the equator, the angular speed is

  • 1800rad/sec

  • 11600rad/sec

  • 1400rad/sec

  • 180rad/sec


A.

1800rad/sec

Given:- radius of earth=6400km=6400×103, g=10m/s2

g'=0, m=0(weight is Zero at equator)

g=Rω2  ω2=gR

Angular speedω=gR=106400×103=1800rad/sec


9.

A body starting from rest moves along a straight line with a constant acceleration. The variation of speed (v) with distance ( s ) is represented by the graph


B.

For a body moving with constant acceleration α

      v = u + αt

Since the body starts from rest

      u = 0

∴     v = αt

which is a straight line passing through the origin. Hence the graph is (b)


10.

A particle starts from rest and has an acceleration  of 2 m/s2 for  10 sec. After that, it travels for 30 sec with constant speed and then undergoes a retardation of 4 m/s2  and comes back to rest. The total distance covered by the particle is

  • 650 m

  • 750 m

  • 700 m

  • 800 m


B.

750 m

Initial velocity  ( u ) = 0

Acceleration (α1 ) = 2 m/s2 

time during acceleration (t1) = 10 sec

Time during constant velocity (t2 ) = 30 sec

and retardation ( α2 ) = - 4 m/s2 

( negative sign due to retardation )

Distance covered by the particle during acceleration

    s1 = ut112 α1t1 2

         = ( 0 × 10 ) + 12 × 2 × (10)2

   s1 = 100 m                 ......(i)

And velocity of the particle a the end of acceleration 

  v = u  + 12 α1t1 2   

      = 0 + ( 2 × 10 ) 

   v = 20 m/s

Therefore distance covered by the particle during constant velocity

  s2 = v × t2 

      = 20 × 30⇒

  s2 = 600 m                    .....(ii)

Relation for the distance covered by the particle during retardation ( s) is

   v2 = u2 + 2α2s3

⇒   ( 0 )2 = ( 20 )2 + 2 × ( - 4 ) × s3

⇒    400 = 8 s3

⇒   s34008

⇒   s3 = 50 m

Therefore total distance covered by the particle 

  s = s1  + s2 + s3

    = 100 +600 + 50

 s = 750 m