Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

 Multiple Choice QuestionsMultiple Choice Questions

1.

A man is at height of 100 m. he sees a car which makes an angle of π6 with man's position. If the car moves to a point where angle is π3 what is the distance moved by it? 

  • 1003m

  • 2003m

  • 2003m

  • 1503m


C.

2003m

In ABCtan 600=100xx=1003             .....(1)In ABDtan 300=100x+yx+y=1003   y  =1003-1003   from (1)              =2003m           


2.

The angular velocity of second hand clock, of a clock is

  • π6rad s-1

  • π60rad s-1

  • π30rad s-1

  • π15rad s-1


C.

π30rad s-1

Angular velocity=θt=2π60rad s-1                              =π30rad s-1Reason :- A second hand makes an angle of 2π i.e 3600 in one minute i.e 60 second


3.

If ar and at represent radial and tangential accelerations, the motion of a particle will be uniformly circular if :

  • ar = 0 and at = 0

  • ar = 0 and at ≠ 0

  • ar ≠ 0 and at = 0

  • ar ≠ 0 and at ≠ 0


C.

ar ≠ 0 and at = 0

If ar = 0 and at = 0, then motion is uniform translatory.

If ar = 0 but at ≠ 0, then motion is accelerated translatory.

If ar ≠ 0 but at = 0, then motion is uniform circular.

If ar ≠ 0 and at ≠ 0, then motion is a non-uniform circular.


4.

An arrow is shot, its range is 200 m and its time of flight is 5 s. If g =10 m/s2, then the horizontal component of velocity of the arrow is :

  • 12.5 m/s

  • 250 m/s

  • 31.25 m/s

  • 40 m/s


D.

40 m/s

R = u2 sin2θg =200mor u2 (2sinθcosθ)g = 200m

Flight time = 2u sinθg = 200m

u cosθ = 40 m/s (Horizontal velocity)


5.

If vectors  P = ai^ + aj^ + 3k^  and  Q = ai^ - 2j^ - k^ are perpendicular to each other, then the positive value of  a  is

  • 3

  • 1

  • 2

  • 0


A.

3

      Vector   P = ai^ + aj^ + 3k^  

and vector  Q = ai^ - 2j^  -  k^

If two vectors are perpendicular to each other, then 

          P · Q = 0

 ai^ + aj^ + 3k^  · ai^ - 2j^ - k^    = 0

⇒   a2 - 2a - 3 = 0

Solving this quadratic equation, we get a = 3 or  - 1.

Therefore positive value of a is 3.


6.

Which of the following velocity-time graphs shows a realistic situation for a body in motion?


B.

Other graph shows more than one velocity of the particle at single instant of time which is not practically possible.


7.

At the uppermost point of a projectile, its velocity and acceleration are at an angle of

  • 0°

  • 90°

  • 45°

  • 180°


B.

90°

At the uppermost point of a projectile, the vertical component of the velocity of projection becomes zero, while the horizontal component remains constant. And the acceleration (due to gravity) is always vertically downwards. Therefore at the uppermost point of a projectile, its velocity and acceleration are at an angle of 90°.


8.

For angles of projection of a projectile at angles (45° - θ) and (45° + θ), the horizontal ranges described by the projectile are in the ratio of

  • 1:1

  • 2:3

  • 1:2

  • 2:1


A.

1:1

For complementary angles of projection, their horizontal ranges will be same.
We know that, horizontal ranges for complementary angles of projection will be same.
The projectiles are projected at angles left parenthesis 45 degree minus straight theta right parenthesis and left parenthesis 45 degree plus straight theta right parenthesis which are complementary to each other i.e., two angles add up to give 90 degree. Hence, horizontal ranges will be equal. Thus, the required ratio is 1:1.

2420 Views

9.

The maximum range of a gun on horizontal terrain is 16 km if g=10m/s2. What must be the muzzle velocity of the shell?

  • 200 m/s

  • 100 m/s

  • 400 m/s

  • 300 m/s


C.

400 m/s

The velocity with which a bullet or shell leaves the muzzle of a gun.
 
We  know that in projection of the particle for maximum range, θ = 45o
 
Now maximum rangeR =μ2sin 2θgR=μ2 sin 90og        sin 90o =1μ =Rg               .....(1)Given :- Rmax = 16 km= 16×103 mg = 10 m/s2Now from equation(1)μ =16×103×10=400 m/s
 
 

10.

A lead shot of  1 mm diameter falls through a long column of glycerine. The variation of its velocity  'v'  with distance covered is represented by


A.

Initially due to the action of gravity, the lead shot will move with increasing velocity for some time. Then due to the viscosity of the glycerine column, the lead shot will attain a constant terminal velocity. As initially. there is some upthrust on the shot due to glycerine the increase of velocity will not be fully linear. So the variation is shown by plot (a).