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.

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.


2.

A particle starting from the origin (0,0) moves in a straight line in the (x,y) plane. Its coordinates at a later time are  ( square root of 3 comma space 3). The path of the particle makes with the x -axis an angle of:

  • 30o

  • 45o C

  • 60o C

  • 00


C.

60o C

The slope of the path of the particle gives the measure of angle required.



Draw the situation as shown. OA represents the path of the particle starting from origin O (0,0). Draw a perpendicular from point A to X- axis. Let path pf the particle makes and angle θ with the x -axis, then 

tan θ = slope of line OA


AB over OB space equals space fraction numerator 3 over denominator square root of 3 end fraction space equals space square root of 3
theta space equals space 60 to the power of 0

1512 Views

3.

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

4.

A car runs at aconstant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The average velocity and average speed for each circular lap respectively is

  • 0, 0

  • 0, 10 m/s
  • 10 m/s,  10m/s

  • 10 m/s,  0


B.

0, 10 m/s

In completing a circular lap, car's displacement is zero.
Average velocity is defined as the ratio of displacement to time taken while the average displacement to time taken while the average speed of a particle in a given interval of time is defined as the ratio of distance travelled to the time taken.
On a circular path in completing one turn, the distance travelled is 2 πr while displacement is zero.
Hence, average velocity = fraction numerator displacement over denominator time minus interval end fraction
                             equals 0 over straight t space equals 0

Average space speed space equals space fraction numerator Distance over denominator Time minus interval end fraction
space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 2 πr over denominator straight t end fraction equals space fraction numerator 2 cross times 3.14 cross times 100 over denominator 62.8 end fraction space equals space 10 straight m divided by straight s


519 Views

5.

A car moves from X to Y with a uniform speed vu and returns to Y with a uniform speed vd. The average speed for this round trip is:

  • fraction numerator 2 straight v subscript straight d straight v subscript straight u over denominator straight v subscript straight d plus straight v subscript straight u end fraction
  • square root of straight v subscript straight u straight v subscript straight d end root
  • fraction numerator straight v subscript straight d straight v subscript straight d over denominator straight v subscript straight d plus straight v subscript straight u end fraction
  • fraction numerator straight v subscript straight u space plus straight v subscript straight d over denominator 2 end fraction

A.

fraction numerator 2 straight v subscript straight d straight v subscript straight u over denominator straight v subscript straight d plus straight v subscript straight u end fraction

The average speed of a body in a given time interval is defined as the ratio of distance travelled to the time taken.

Average space speed space equals space fraction numerator Distance space travelled over denominator Time space taken end fraction
Let t1 and t2 be times taken by the car to go from X and Y and then Y to X respectively.

Then, 

straight t subscript 1 space plus straight t subscript 2 space equals space XY over straight v subscript straight u space plus XY over straight v subscript straight d space equals space XY space open parentheses fraction numerator straight v subscript straight u space plus straight v subscript straight d over denominator straight v subscript straight u straight v subscript straight d end fraction close parentheses
Total space distance space travelled
equals space XY space plus space XY space equals space 2 XY
Therefore comma space average space speed space of space the space car space for space this space round space trip space is
straight v subscript av space equals space fraction numerator 2 XY over denominator XY open parentheses begin display style fraction numerator straight v subscript straight u plus straight v subscript straight d over denominator straight v subscript straight u straight v subscript straight d end fraction end style close parentheses end fraction
or
straight v subscript av space equals space fraction numerator 2 straight v subscript straight u straight v subscript straight d over denominator straight v subscript straight u plus straight v subscript straight d end fraction

1176 Views

6. bold A with bold rightwards arrow on top bold space and bold space bold B with bold rightwards arrow on top are two vectors and θ is the angle between them if vertical line bold A with bold rightwards arrow on top bold space straight x bold space bold B with bold rightwards arrow on top bold vertical line bold space equals bold space square root of bold 3 bold space bold left parenthesis bold A with bold rightwards arrow on top bold. bold B with bold rightwards arrow on top bold right parenthesis the value of θ is:
  • 60o

  • 45o

  • 30o

  • 90o


A.

60o

bold A with bold rightwards arrow on top bold space bold x bold space bold B with bold rightwards arrow on top bold space equals space AB space sin space space and space stack bold A bold. with bold rightwards arrow on top bold B with bold rightwards arrow on top bold space equals space AB space cos space straight theta.
Given comma space vertical line bold A with bold rightwards arrow on top bold space bold x bold space bold B with bold rightwards arrow on top vertical line space equals space square root of 3 open parentheses stack bold A bold. with bold rightwards arrow on top bold B with bold rightwards arrow on top bold space close parentheses
rightwards double arrow space AB space sin space space straight theta space equals square root of 3 space AB space cos space space straight theta
tan space equals space square root of 3
space straight theta space equals space 60 to the power of 0
838 Views

7.

The vectors bold A with bold rightwards arrow on top space and space bold B with bold rightwards arrow on top are such that a:
open vertical bar straight A with rightwards arrow on top plus straight B with rightwards arrow on top close vertical bar space equals space open vertical bar straight A with rightwards arrow on top space minus space straight B with rightwards arrow on top close vertical bar
The  angle between the two vectors is

  • 90 degree
  • 60 degree
  • 75 degree
  • 45 degree

A.

90 degree

As we have given
        open vertical bar straight A with rightwards arrow on top plus straight B with rightwards arrow on top close vertical bar space equals open vertical bar straight A with rightwards arrow on top minus straight B with rightwards arrow on top close vertical bar
or space space space square root of straight A squared plus straight B squared plus 2 AB space Cosθ end root
space equals square root of straight A squared plus straight B squared minus 2 AB space Cos space straight theta end root
where straight theta is the angle between straight A with rightwards arrow on top space and space straight B with rightwards arrow on top
Squaring both sides, we have
straight A squared plus straight B squared plus 2 AB space Cosθ space equals space straight A squared plus straight B squared minus 2 AB space cosθ
or space space space space space space space space space space space space space space space space space space 4 AB space cosθ space equals space 0
As space space space space space space space space space space space space space space space space space AB not equal to 0
therefore space space space space space space space space space space cosθ space equals space 0 space equals space cos space 90 degree
therefore space space space space space space space space space space space space space space space space space space space space space space space straight theta space equals space 90 degree space space space space space space space space space space space space

Hence, angle between stack straight A space with rightwards arrow on top and space straight B with rightwards arrow on top space is space 90 degree

358 Views

8.

The position x of a particle with respect to time t along x- axis is given by x = 9t2 -t3 where x is in meter and t in second. What will be the position of this particle when it achieves maximum speed along the +x direction? 

  • 32 m

  • 54 m

  • 81 m 

  • 24 m


B.

54 m

At the instant when speed is maximum, its acceleration is zero.

Given, the position x of particle with respect to time t along x- axis

x = 9t2-t3 ... (i)
differentiating Eq. (i) with respect to time, we get speed, ie,
straight v space equals space dx over dt space equals space straight d over dt left parenthesis 9 straight t squared minus straight t cubed space right parenthesis
straight v space equals space 18 space straight t space minus space 3 straight t squared space... space left parenthesis ii right parenthesis

Again differentiating Eq. (ii) with respect to time, we get acceleration, ie,

straight a space equals space dv over dt space equals straight d over dt left parenthesis 9 straight t squared minus straight t cubed right parenthesis
straight a space equals space 18 minus 6 straight t
Now, when speed of particle is maximum, its acceleration is zero, ie,

a= 0
18-6t = 0 or t = 3s
Putting in eq (i) We, obtain the position of a particle at that time. 

x = 9 (3)2 - (3)3 = 9 (9) -27 
= 81-27 = 54 m

1607 Views

9.

A particle moving along x- axis has acceleration f, at time t, given f = fo (1-t/T), where fo and T are constants. The particle at t =0 and the instant when f = 0, the particle's velocity (vx) is:

  • foT

  • foT2/2

  • foT2

  • foT/2


D.

foT/2

1137 Views

10.

The distance travelled by a particle starting from rest and moving with an acceleration 4/3 ms-2 in the third second is

  • 6 m

  • 4 m

  • 10/3

  • 19/3


C.

10/3

Distance travelled by  the particle in nth second

straight S subscript straight n to the power of th end subscript space equals space straight u space plus 1 half space straight a space left parenthesis 2 straight n minus 1 right parenthesis
Where u is initial speed and a is an acceleration of the particle. 

Here, n = 3, u =0, a = 4/3 m/s2

straight S subscript 3 rd end subscript space space equals space 0 space plus space 1 half space straight x 4 over 3 space straight x space left parenthesis 2 space straight x space 3 minus 1 right parenthesis
space equals space 4 over 6 space straight x space 5 space equals space 10 over 3 space straight m

521 Views