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**Objective Questions (Level 2) Single Correct Option**

(c) 5 ms

Applying sine formula in

Î” AOB'

Comparing Eqs. (i) and (ii)

sin Î± - cos Î± tan 37Â°

or 4 sin Î± - 3 cos Î± = 32 cos Î± - 28 sin Î± + 24 sin Î± + 21 cos Î±

or 32 cos Î± + 21 cos Î± + 3 cos Î± = 28 sin Î± - 24 sin Î± + 4 sin Î±

or 56 cos Î± = 8 sin Î±

tan Î± = 7/1

From Eq. (i)

âˆ´

Option (b) is correct.

At the time the body acquires terminal speed its acceleration (a) must be zero.

Thus

- 4v + 8 = 0

v = 2 m/s

Terminal speed = 2 m/s.

Option (b) is correct.

**simultaneously. Then (a) they will collide at (10, 20, 10)m (b) they will collide at (10, 10, 10) m (c) they will never collide (d) None of the above**

âˆ´

âˆ´

âˆ´ Option (c) is correct.**Ques 4: Velocity of the river with respect to ground is given by v _{0}. Width of the river is d. A swimmer swims (with respect to water) perpendicular to the current with acceleration a = 2t (where t is time) starting from rest from the origin O at t = 0. The equation of trajectory of the path followed by the swimmer is**

or âˆ« dv_{y} = âˆ« 2t dt

or v_{y} = t^{2} + C_{1}

At t = 0, v_{y} = 0 (given)

âˆ´ C_{1} = 0

âˆ´ v_{y} = t^{2}

â‡’

or âˆ« dy = âˆ« t^{2} + C_{2}

or

At t = 0, Y = 0 (given)

âˆ´ â€¦(i)

Displacement along x-axis at time t:

x = v_{0}t

or

Substituting above value of t in Eq. (i)

Option (a) is correct.**Ques 5: The relation between time t and displacement x is t = Î±x ^{2} + Î²x where Î± and Î² are constants. The retardation is **

Differentiating w.r.t. time t

or v (Î² + 2Î± Ã— x ) = 1

or v = (Î² + 2Î± Ã— x )

âˆ´

âˆ´ Retardation = 2Î±v

Option (a) is correct.

âˆ´

or

or

As at x = 0 car is at rest C = 0

âˆ´

For v to be maximum.

i.e., a - bx = 0

or x = a/b

Substituting x = a/b in Eq. (i),

âˆ´

Car will come to rest when

x = 2a/b

âˆ´ Distance between two stations = 2a/b

Option (a) is correct.

**Ans:****Sol: **Force = - kx^{2}

â‡’

or

or

Now at x = a, v = 0

âˆ´

Thus

âˆ´ Velocity at x = 0

Option (d) is correct.**Ques 8: A particle is moving mX-Y plane such that vx = 4 + 4t and v _{y} = 4t. If the initial position of the particle is (1, 2). Then the equation of trajectory will be **

(b) y = 2x

(c) x

(d) None of these

or âˆ« dx = âˆ«(4 + 4t) dt

and âˆ« dy = âˆ« 4t dt

or x = 4t + 2t

âˆ´ C

Thus x = 4t + 2t

and y = 2t

At t = 0 x = 1

and at t = 0, y = 2

âˆ´ C

Thus y = 2t

Substituting value of t in Eq. (i) in terms of y using Eq. (ii) we wonâ€™t get relationship as mentioned in option (a), (b) or (c).

âˆ´ Option (d) is correct.

âˆ´ Magnitude of acceleration

Option (a) is correct.**Ques 10: A particle is falling freely under gravity. In first t second it covers distance x _{1} and in the next t second it covers distance x_{2}, then t is given by**

Option (a) is correct.

(b) 2 ms

(d) 4 ms

or

Option (c) is correct.

i.e., t = 7 s

Option (a) is correct.

i.e., sin Î¸ = 3/4

or

Option (c) is correct.

= 22.36 s

Option (a) is correct.

âˆ´

â‡’ t' = 6 s

âˆ´ OB = 50 s

Let OA = t

âˆ´ AB = 56 - t

â‡’ t = 20 s

âˆ´ maximum acceleration

= 1.2 m/s

Option (b) is correct.

âˆ´ OA

i.e., AB

= (4 - 0.2t)

or AB

or AB

For AB to be minimum

0.24t - 4.2 = 0

or t = 17.5 s

âˆ´ (AB)

= 36.75 - 73.5 + 37

(AB)

= 7500 cm

Option (d) is correct.

Acceleration of ball w.r.t. elevator

Final displacement of ball w.r.t. elevator = - 2m

âˆ´

= 2.13 s

Option (a) is correct.

= (15 + 10) m/s

= 25 m/s

Now v

âˆ´ 0

or H = 31.25 m

Maximum height by ball as measured from ground = 31.25 + 2 + 50

= 83.25 m

Option (c) is correct.

Option (d) is correct.

will be maximum, when

ds/dt = 0

i.e., 15 - 15t = 0

i.e., t = 1s

= 12 m

Option (a) is correct.

i.e., displacement of the particles are equal and which is possible when

Area of Î” ACB = Area of Î” A' B' C

[Area OBCA' Î¸O being common]

or Area of Î” APC = Area A' P' C

or AP Ã— PC = A' P' Ã— P' C

or (PC tan Î¸) PC = (P' C tan Î¸) P' C

or PC = P' C'

âˆ´OR = PC + P' C' = 2 PC = 8 s.

Option (c) is correct.

(a) 5 ms-1

(b) 10 ms

(c) 15 ms

(d) data insufficient

Option (b) is correct.

and at t = 4 s common velocity = 8 ms

To find velocity of A at t = 10 s

or

âˆ´ v

Option (d) is correct.

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