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 space straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top space equals space straight B with rightwards arrow on top space straight x space straight A with rightwards arrow on top then the angle between A and B isπ
  • π

  • π/3

  • π/2

  • π/4


A.

π

open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses space equals space left parenthesis straight B with rightwards arrow on top space straight x straight A with rightwards arrow on top right parenthesis
rightwards double arrow space open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses minus space left parenthesis straight B with rightwards arrow on top space straight x straight A with rightwards arrow on top right parenthesis space equals space 0
rightwards double arrow space open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses plus space left parenthesis straight B with rightwards arrow on top space straight x straight A with rightwards arrow on top right parenthesis space equals space 0
open square brackets therefore space left parenthesis straight B with rightwards arrow on top space straight x space straight A with rightwards arrow on top right parenthesis space equals space minus space left parenthesis straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top right parenthesis close square brackets
2 space open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses space equals 0
rightwards double arrow space space 2 AB space sin space straight theta space equals space 0
sin space straight theta space equals space 0 space space left square bracket space because space vertical line straight A with rightwards arrow on top vertical line space equals space straight A space not equal to 0 comma space vertical line straight B with rightwards arrow on top vertical line space equals space straight B not equal to space 0 right square bracket
straight theta space equals space 0 space or space straight pi space
163 Views

2.

A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flights in the two cases, then the product of the two time of flights is proportional to

  • R2

  • 1/R2

  • 1/R

  • R


D.

R

straight t subscript 1 straight t subscript 2 space equals space fraction numerator 2 straight u squared sin space 2 straight theta over denominator straight g squared end fraction space equals space fraction numerator 2 straight R over denominator straight g end fraction
251 Views

3.

A projectile can have the same range R for two angles of projection. If T1 and T2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

  • 1/R2

  • 1/R

  • R

  • R2


C.

R

We know in advance that range of projectile is same for complementary angles i.e. for θ and (900 - θ )

straight T subscript 1 space equals space fraction numerator 2 straight u space sin space straight theta over denominator straight g end fraction

straight T subscript 2 space equals space fraction numerator 2 straight u space sin space left parenthesis 90 to the power of 0 minus straight theta right parenthesis over denominator straight g end fraction space equals space fraction numerator 2 space straight u space cos space straight theta over denominator straight g end fraction
and space straight R space equals space fraction numerator straight u squared space sin space 2 straight theta over denominator straight g end fraction
Therefore comma space straight T subscript 1 space straight T subscript 2 space equals space fraction numerator 2 space straight u space sin space straight theta over denominator straight g end fraction space straight x space fraction numerator 2 space straight u space cos space straight theta over denominator straight g end fraction
space equals space fraction numerator 2 straight u squared space left parenthesis 2 space sin space straight theta space cos space straight theta right parenthesis over denominator straight g squared end fraction
space equals space fraction numerator 2 straight u squared space left parenthesis sin space 2 straight theta right parenthesis over denominator straight g squared end fraction
space equals space 2 straight R divided by straight g
space equals space straight T subscript 1 straight T subscript 2 space proportional to space straight R

271 Views

4.

A body of mass m accelerates uniformly from rest to v1 in time t1. The instantaneous power delivered to the body as a function of time t is

  • fraction numerator mv subscript 1 straight t over denominator straight t subscript 1 end fraction
  • fraction numerator mv subscript 1 superscript 2 space straight t over denominator straight t subscript 1 superscript 2 end fraction
  • fraction numerator mv subscript 1 straight t squared over denominator straight t subscript 1 end fraction
  • fraction numerator mv subscript 1 superscript 2 straight t over denominator straight t subscript 1 end fraction

B.

fraction numerator mv subscript 1 superscript 2 space straight t over denominator straight t subscript 1 superscript 2 end fraction

Let the constant acceleration of body of mass m is a.
From equation of motion

v1 = 0 + at1
⇒ a = t2/t= ...... (i)
At an instant t, the velocity v of the body v = 0 + at

straight v space equals space straight v subscript 1 over straight t subscript 1 straight t space space.... space left parenthesis ii right parenthesis
therefore space instantaneous space power
straight p space equals Fv
space equals mav space space left parenthesis because space straight F space equals space ma right parenthesis
equals space straight m open parentheses straight v subscript 1 over straight t subscript 1 close parentheses space straight x open parentheses straight v subscript 1 over straight t subscript 1 space straight x space straight t close parentheses space left square bracket space From space equ space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
space equals space fraction numerator mv subscript 1 superscript 2 straight t over denominator straight t subscript 1 superscript 2 end fraction

515 Views

5.

The relation between time t and distance x is t=ax2 +bx where a and b are constants. The acceleration is

  • −2abv2

  • 2bv3

  • −2av3

  • 2av2


C.

−2av3

 if t = ax2+bx
then, differentiate both side with respect to time (t)
 
dtdt = 2axdxdt +bdxdt1 = 2axv +bvv.(2ax + b) = 1(2ax + b) = 1/v
 
Again differentiate both side with respect to time (t)
2adxdt = -v2.dvdt2av = - v-2.accelerationacceleration = -2av3
Hence, retardation= -2av3

6.

A particle is moving eastwards with a velocity of 5 m/s in 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is

  • fraction numerator 1 over denominator square root of 2 end fraction straight m divided by straight s squared spacetowards north-east
  • 1 half straight m divided by straight s squared spacetowards north.
  • zero

  • fraction numerator 1 over denominator square root of 2 end fraction straight m divided by straight s squared spacetowards north-west

D.

fraction numerator 1 over denominator square root of 2 end fraction straight m divided by straight s squared spacetowards north-west


straight a with rightwards arrow on top space equals space fraction numerator stack straight V subscript straight f with rightwards arrow on top minus stack straight V subscript straight i with rightwards arrow on top over denominator straight t end fraction space
space equals space fraction numerator 5 straight j with hat on top minus 5 straight i with hat on top over denominator 10 end fraction space equals 1 half space left parenthesis straight j with hat on top minus straight i with hat on top right parenthesis
therefore straight a space equals space fraction numerator 1 over denominator square root of 2 end fraction ms to the power of negative 2 end exponent space towards space north space west
199 Views

7.

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place in a plane. It follows that

  • its velocity is constant

  • its acceleration is constant

  • its kinetic energy is constant

  • it moves in a straight line.


C.

its kinetic energy is constant

When a force of constant magnitude acts on the velocity of particle perpendicularly, then there is no change in the kinetic energy of the particle. Hence, kinetic energy remains constant.

188 Views

8.

A car starting from rest accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate f/2 to come to rest. If the total distance traversed is 15 S, then

  • S=ft

  • S= ft2/72

  • S = 1/2 ft2

  • S = 1/4 ft2


B.

S= ft2/72



straight S space equals space fraction numerator ft subscript 1 superscript 2 over denominator 2 end fraction
straight v subscript straight o space equals square root of 2 Sf end root
During space retardation
straight S subscript 2 space equals space 2 straight S
During space constant space velocity
15 straight S space minus 3 straight S space equals space 12 space straight S space equals straight v subscript straight o straight t
rightwards double arrow space straight S space equals ft squared over 72
275 Views

9.

A parachutist after bailing outfalls 50 m without friction. When a parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bail out?

  • 91 m

  • 182 m

  • 293 m

  • 111 m


C.

293 m

straight s equals space 50 space plus open parentheses fraction numerator 3 squared minus left parenthesis 2 straight x 10 straight x 50 right parenthesis over denominator 2 left parenthesis negative 2 right parenthesis end fraction close parentheses
space equals space 293 space straight m
142 Views

10.

Which of the following statements is false for a particle moving in a circle with a constant angular speed?

  • The velocity vector is tangent to the circle.

  • The acceleration vector is tangent to the circle.

  • The acceleration vector points to the centre of the circle.

  • The velocity and acceleration vectors are perpendicular to each other.


B.

The acceleration vector is tangent to the circle.

For a particle moving in a circle with constant angular speed, the velocity vector is always tangent to the circle and the acceleration vector always points towards the centre of the circle or is always along the radius of the circle. Since, the tangential vector is perpendicular to radial vector, therefore, velocity vector will be perpendicular to the acceleration vector. But in no case acceleration vector is tangent to the circle.
165 Views