The document Previous year questions (2016-20) - Motion Notes | EduRev is a part of the JEE Course Physics Class 11.

All you need of JEE at this link: JEE

**Q 1. Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, T _{1} and T_{2}. The temperature of the hot reservoir of the first engine is T_{1} and the temperature of the cold reservoir of the second engine is T_{2}. T is temperature of the sink of first engine which is also the source for the second engine. How is T related to T_{1} and T_{2}, if both the engines perform equal amount of work? [2020]** (2)

Ans:

Given that, two ideal Carnot engines in cascade between temperatures T

We know,

work done by first engine = work done by second engine

Here,

So,

We have

Then**Q 2. A particle is moving along the x-axis, with its coordinate as a function of time given by x(t) =10+ 8t - 3t ^{2} . Another particle is moving along the y-axis with its coordinate as a function of time given by y(t) = 5 - 8t^{3}. At t=1 s, the speed of the second particle as measured in the frame of the first particle is given as √v. Then v (in m/s) is ______. [2020]** 580

Ans:

Given that

x(t) =10+ 8t - 3t

So,

At t = 1 s, we have

Relative velocity of second particle with respect to first particle is

(1) is perpendicular to is directed away from the origin

(2) both are perpendicular to

(3) both are parallel to

(4) is perpendicular to is directed towards the origin

Ans:

Given that

So,

and

Since, is perpendicular to

And had opposite direction to or antiparallel to that is, directed toward origin.

Ans:

Given that

u = 0 m/s, H = 100 m, d = 19 m, t = 1/2 s

Let time taken to tavel (H – h) distance be t.

So,

..(1)

Now, time taken to hit the ground is

..(2)

From Eqs. (1) and (2), we get

⇒ a = 8 m/s

**Q 5. A particle starts from the origin at t = 0 with an initial velocity of and moves in the x-y plane with a constant acceleration The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D m. The value of D is(1) 32(2) 50(3) 60(4) 40Ans:** (3)

Given that

From equation of motion in two dimensions, we have

⇒ t

⇒ t = 4 s

So,

Ans:

x

.. (1)

Velocity is given by

⇒ vx = at + b ..(2)

Acceleration is given by

Differentiate Eq. (2) with respect to t, we get

So,

Therefore, n = 3

(1)

(2)

(3)

(4)

Let time taken by car A is t'

v

Put this value in Eq. (1), we get

(1) 11/5

(2) 5/2

(3) 3/2

(4) 25/11

Ans:

Let passenger train and freight train be A and B and v

Relative velocity of A with respect to B is (same direction)

Time taken by the train A to cross the train B

Train moving in opposite direction

Therefore, required ratio is**Q 9. A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5 s? (JEE Main 2019)(1) 10 m(2) 6 m(3) 3 m(4) 9 mAns:** (4)

As we know that, area under v-t curve gives displacement.

therefore,

(1) y = x

(2) y

(3) y

(4) xy = constant

Ans:

Thus,

Now,

Integrating both the sides, we get

y

x = a cos ωt

y = a sin ωt

and

z = aωt

The speed of the particle is (JEE Main 2019)

(1)

(2) aω

(3)

(4) 2aω

Ans:

Velocity coordinates of particle is obtained by differentiating the position coordinate with respect to time, Thus,

Hence, velocity of particle is

(1)

(2)

(3)

(4)

Ans:

Side of a cube is a

**Q 13. Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is (JEE Main 2019)(1) 1 : 16(2) 1 : 2(3) 1 : 4(4) 1 : 8Ans: **(1)

Area covered = πR

A ∝R

Thus, range

Therefore, required ratio is**Q 14. Two vectors have equal magnitudes. The magnitude of **

** is n times the magnitude of The angle between is (JEE Main 2019)(1) (2) (3) (4) Ans:** (1)

We know that

According to the question

Squaring both the sides, we get

(1) 2.5 m

(2) 2.8 m

(3) 10.3 m

(4) 5.1 m

Ans:

At t = 1:

The radius of curvature of the trajectory is

= 2.77 m = 2.8 m

(1) 15m

(2) 20√2m

(3) 5m

(4) 10√2m

Ans:

**(1)**

**(2)**

**(3)**

**(4)**

**Ans: **B**Solution:** In graph ‘2’ initial slope is zero which is not possible, since initial velocity is non zero in all other three graphs.

(1) 75 m

(2) 160 m

(3) 150 m

(4) 100 m

Ans:

u = 40 Km/hr

u = 100/9 m/s

V

a = −1.54 m/s

Now

s = 160m

(3) 112.5 m and 15 s

(4) 112.5 m and 22.5 s

Ans:

s

= 225 + 112.5

= 337.5 m

s

s

⟹ s

For catching up ⟹ s

30t = 1/2 × 3 × t

20 = t

(1)

**(2) **

**(3) **

**(4)Ans:** C

Solving equation (i) and (ii) we get

Ans:

(1)

(2) 15 s

(3)

(4)

Ans:

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

129 videos|243 docs|151 tests

### NCERT Solutions - Motion in a Plane

- Doc | 22 pages
### Types of Vectors and Scalars

- Doc | 3 pages
### Test: Motion In Plane - Introduction To Vectors

- Test | 10 ques | 10 min
### NCERT Textbook - Motion in a Plane

- Doc | 24 pages
### Projectile Motion

- Video | 06:46 min

- Test: Motion In Two Dimensions
- Test | 5 ques | 10 min