Motion in A Plane Test

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Motion in A Plane Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

The vector projection of a $3\hat i + 4 \hat k$ vector on y-axis is

A
5

B
4

C
3

D
Zero

Solution

As the multiple of $\hat j$ in the given vector is zero therefore this vector lies in XZ plane and projection of this vector on y-axis is zero.
Question 2
1 / -0

A force of 5 N acts on a particle along a direction making an angle of 60° with vertical. Its vertical component be

A
10 N

B
3 N

C
4 N

D
2.5 N

Solution

The component of force in vertical direction

= F cos $\theta$ = F cos 60°

= 5 x $\frac{1}{2}$ = 2.5 N
Question 3
1 / -0

The direction of cosines of the $\vec A=2\hat i+4\hat j-5\hat k$ are

A
$\frac{2}{\sqrt{45}}, \frac{4}{\sqrt{45}}$ and $\frac{-5}{\sqrt{45}}$

B
$\frac{1}{\sqrt{45}}, \frac{2}{\sqrt{45}}$ and $\frac{3}{\sqrt{45}}$

C
$\frac{4}{\sqrt{45}}$ , 0 and $\frac{4}{\sqrt{45}}$

D
$\frac{4}{\sqrt{45}}, \frac{2}{\sqrt{45}}$ and $\frac{5}{\sqrt{45}}$

Solution

$\vec A = 2\hat i + 4\hat j - 5\hat k$

$|\vec A| = \sqrt{(2)^2 + (4)^2 + (-5)^2}$

= $\sqrt{45}$

cos $\alpha$ = $\frac{2}{\sqrt{45}}$

cos $\beta$ = $\frac{4}{\sqrt{45}}$

cos $\gamma$ = $\frac{-5}{\sqrt{45}}$
Question 4
1 / -0

How much minimum number of coplanar vectors having different magnitudes can be added to give zero resultant?

A
2

B
3

C
4

D
5

Solution

$\vec F_3 = \vec F_1 + \vec F_2$

There should be minimum three coplanar vectors having different magnitude which should be added to give zero resultant.
Question 5
1 / -0

A hall has the dimensions (10m) × (12m) × (14m). A fly starting at one corner ends up at a diametrically opposite corner. What is the magnitude of its displacement?

A
17 m

B
26 m

C
36 m

D
20 m

Solution

Diagonal of the hall = $\sqrt{l^2 + b^2 + h^2}$

= $\sqrt{10^2 + 12^2 + 14^2}$

= $\sqrt{100 + 144 + 196}$

= $\sqrt{400}$

= 20 m
Question 6
1 / -0

The unit vector along $\hat i + \hat j$ is

A
$\hat k$

B
$\hat i + \hat j$

C
$\frac{\hat i + \hat j}{\sqrt 2}$

D
$\frac{\hat i + \hat j}{2}$

Solution

$\hat R = \frac{\vec R}{|R|}$

= $\frac{\hat i + \hat j}{\sqrt{1^2 + 1^2}}$

= $\frac{1}{\sqrt 2} \hat i + \frac{1}{\sqrt 2} \hat j$
Question 7
1 / -0

Five equal forces of 10 N each are applied at one point and all are lying in one plane. If the angles between them are equal, the resultant force will be

A
Zero

B
10 N

C
20 N

D
$10 \sqrt 2 N$

Solution

If the angle between all forces which are equal and lying in one plane are equal then resultant force will be zero.
Question 8
1 / -0

If a unit vector is represented by $0.5 \hat i + 0.8 \hat j + c \hat k$ , then the value of c is

A
1

B
$\sqrt{0.11}$

C
$\sqrt{0.01}$

D
$\sqrt{0.39}$

Solution

Magnitude of unit vector = 1

$\sqrt{(0.5)^2 + (0.8)^2 + c^2} = 1$

c = $\sqrt{0.11}$
Question 9
1 / -0

A boy walks uniformally along the sides of a rectangular park of size 400 m× 300 m, starting from one corner to the other corner diagonally opposite. Which of the following statement is incorrect?

A
He has travelled a distance of 700 m

B
His displacement is 700 m

C
His displacement is 500 m

D
His velocity is not uniform throughout the walk

Solution

Displacement $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$

AC = $\sqrt{(AB)^2 + (BC)^2}$

= $\sqrt{(400)^2 + (300)^2}$

= 500 m

= AB + BC = 400 + 300

= 700 m
Question 10
1 / -0

A stone is projected horizontally with a 5 m/s from the top of a plane inclined at an angle 45° with the horizontal. How far from the point of projection will the particle strike the plane?

A
$5\sqrt 2 m$

B
$11\sqrt 2 m$

C
$12\sqrt 2 m$

D
$15\sqrt 2 m$

Solution

Range = $\frac{2 u^2}{g \;sin \theta} = 5\sqrt2 m$
Question 11
1 / -0

Two balls are projected at an angle θ and (90°−θ) to the horizontal with the same speed. The ratio of their maximum vertical heights is:

A
1:1

B
$tan \theta:1$

C
$1:tan\theta$

D
$tan^2\theta : 1$

Solution

$\frac{H_1}{H_2} = \frac{\frac{u^2sin^2 \theta}{2g}}{\frac{u^2sin^2(90^o - \theta)}{2g}}$

= $tan^2 \theta$
Question 12
1 / -0

The vector that must be added to the vector $\hat i - 3 \hat j + 2 \hat k$ and $3\hat i - 6 \hat j + 7 \hat k$ so that the resultant vector is a unit vector along the y-axis, is

A
$4\hat i - 2 \hat j + 5 \hat k$

B
$-4\hat i - 2 \hat j + 5 \hat k$

C
$3\hat i - 4 \hat j + 5 \hat k$

D
Null vector

Solution

Unit vector along y axis = $\hat j$ , so the required vector

= $\hat j$ - $[(i - 3\hat j + 2 \hat k) + (3 \hat i + 6 \hat j - 7 \hat k)]$

= $-4\hat i - 2 \hat j + 5 \hat k$
Question 13
1 / -0

A particle crossing the origin of co-ordinates at time t = 0, moves in the xy-plane with a constant acceleration a in the y-direction. If its equation of motion is y = $bx^2$ (b is a constant), its velocity component in the x-direction is

A
$\sqrt{\frac{2b}{a}}$

B
$\sqrt{\frac{a}{2b}}$

C
$\sqrt{\frac{a}{b}}$

D
$\sqrt{\frac{b}{a}}$

Solution

y = $bx^2.$ Differentiating w.r.t to t on both $\frac{dy}{dx}$ = $2bx$ $\frac{dx}{dt}$ ⇒ $v_y = 2bxv_x$

Again differentiating w.r.t to t on both sides we get

$\frac{dv_y}{dt} = 2bv_x \frac{dx}{dt} + 2bx \frac{dv_x}{dt}$ = $2bv_x^2 + 0$

[ $\frac{dv_x}{dt} = 0$ , because the particle had constant acceleration along y - direction]

Now, $\frac{dv_y}{dt} = a = 2bv_x^2;$ $v^2_x = \frac{a}{2b}$

⇒ $v_x = \sqrt{\frac{a}{2b}}$
Question 14
1 / -0

A plane flying horizontally at a height of 1500 m with a velocity of 200 $ms^{-1}$ passes directly overhead on antiaircraft gun. Then the angle with the horizontal at which the gun should be fired from the shell with a muzzle velocity of 400 $ms^{-1}$ to hit the plane, is

A
90°

B
60°

C
30°

D
45°

Solution

Horizontal distance covered should be same for the time of collision.

400 cos $\theta$ = 200

or cos $\theta$ = $\frac{1}{2}$ or

$\theta$ = 60°
Question 15
1 / -0

The equation of trajectory of projectile is given where x and y are in meter. The maximum range of the projectile is y = $\frac{x}{\sqrt 3} - \frac{gx^2}{20}$

A
$\frac{8}{3}m$

B
$\frac{4}{3}m$

C
$\frac{3}{4}m$

D
$\frac{3}{8}m$

Solution

Comparing the given equation with the equation of trajectory of a projectile,

y = x tan $\theta$ - $\frac{gx^2}{2u^2cos^2 \theta}$ , we get, tan $\theta$ = $\frac{1}{\sqrt 3}$

⇒ $\theta$ = 30° and $2u^2cos^2 \theta = 20$

⇒ $u^2 = \frac{20}{2cos^2 \theta} = \frac{40}{3}$

Now,

$R_{max} = \frac{u^2}{g} = \frac{40}{3 \times 10} = \frac{4}{3} m$
Question 16
1 / -0

Assertion: In projectile motion, the angle between the instantaneous velocity and acceleration at the highest point is 180°.

Reason: At the highest point, velocity of projectile will be in horizontal direction only.

(A) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(B) If both assertion and reason are true but reason is not the correct explanation of the assertion.

(C) If assertion is true but reason is false.

(D) If the assertion and reason both are false.

(E) If assertion is false but reason is true.

A
A

B
B

C
C

D
D

Solution

At the highest point, vertical component of velocity becomes zero so there will be only horizontal velocity and it is perpendicular to the acceleration due to gravity.
Question 17
1 / -0

If a particle moves in a circle describing equal angles in equal times, its velocity vector

A
Remains constant

B
Changes in magnitude

C
Changes in direction

D
Changes both in magnitude and direction

Solution

It is always directed in the direction of tangent to circle.
Question 18
1 / -0

Assertion: The trajectory of projectile is quadratic in y and linear in x.

Reason: y component of trajectory is independent of x-component.

(A) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(B) If both assertion and reason are true but reason is not the correct explanation of the assertion.

(C) If assertion is true but reason is false.

(D) If the assertion and reason both are false.

A
A

B
B

C
C

D
D

Solution

y = x tan $\theta$ - $\frac{gx^2}{2u^2cos^2\theta}$

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Motion in A Plane Test - 11

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Motion in A Plane Test - 11

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Question 1

5

4

3

Zero

Solution

Question 2

10 N

3 N

4 N

2.5 N

Solution

Question 3

2√45,4√45\frac{2}{\sqrt{45}}, \frac{4}{\sqrt{45}} and −5√45\frac{-5}{\sqrt{45}}

1√45,2√45\frac{1}{\sqrt{45}}, \frac{2}{\sqrt{45}} and 3√45\frac{3}{\sqrt{45}}

4√45\frac{4}{\sqrt{45}}, 0 and 4√45\frac{4}{\sqrt{45}}

4√45,2√45\frac{4}{\sqrt{45}}, \frac{2}{\sqrt{45}} and 5√45\frac{5}{\sqrt{45}}

Solution

Question 4

2

3

4

5

Solution

Question 5

17 m

26 m

36 m

20 m

Solution

Question 6

ˆk\hat k

ˆi+ˆj\hat i + \hat j

ˆi+ˆj√2\frac{\hat i + \hat j}{\sqrt 2}

ˆi+ˆj2\frac{\hat i + \hat j}{2}

Solution

Question 7

Zero

10 N

20 N

10√2N10 \sqrt 2 N

Solution

Question 8

1

√0.11\sqrt{0.11}

√0.01\sqrt{0.01}

√0.39\sqrt{0.39}

Solution

Question 9

He has travelled a distance of 700 m

His displacement is 700 m

His displacement is 500 m

His velocity is not uniform throughout the walk

Solution

Question 10

5√2m5\sqrt 2 m

11√2m11\sqrt 2 m

12√2m12\sqrt 2 m

15√2m15\sqrt 2 m

Solution

Question 11

1:1

tanθ:1tan \theta:1

1:tanθ1:tan\theta

tan2θ:1tan^2\theta : 1

Solution

Question 12

4ˆi−2ˆj+5ˆk4\hat i - 2 \hat j + 5 \hat k

−4ˆi−2ˆj+5ˆk-4\hat i - 2 \hat j + 5 \hat k

3ˆi−4ˆj+5ˆk3\hat i - 4 \hat j + 5 \hat k

Null vector

Solution

$\frac{2}{\sqrt{45}}, \frac{4}{\sqrt{45}}$ and $\frac{-5}{\sqrt{45}}$

$\frac{1}{\sqrt{45}}, \frac{2}{\sqrt{45}}$ and $\frac{3}{\sqrt{45}}$

$\frac{4}{\sqrt{45}}$ , 0 and $\frac{4}{\sqrt{45}}$

$\frac{4}{\sqrt{45}}, \frac{2}{\sqrt{45}}$ and $\frac{5}{\sqrt{45}}$

$\hat k$

$\hat i + \hat j$

$\frac{\hat i + \hat j}{\sqrt 2}$

$\frac{\hat i + \hat j}{2}$

$10 \sqrt 2 N$

$\sqrt{0.11}$

$\sqrt{0.01}$

$\sqrt{0.39}$

$5\sqrt 2 m$

$11\sqrt 2 m$

$12\sqrt 2 m$

$15\sqrt 2 m$

$tan \theta:1$

$1:tan\theta$

$tan^2\theta : 1$

$4\hat i - 2 \hat j + 5 \hat k$

$-4\hat i - 2 \hat j + 5 \hat k$

$3\hat i - 4 \hat j + 5 \hat k$

$\sqrt{\frac{2b}{a}}$

$\sqrt{\frac{a}{2b}}$

$\sqrt{\frac{a}{b}}$

$\sqrt{\frac{b}{a}}$

$\frac{8}{3}m$

$\frac{4}{3}m$

$\frac{3}{4}m$

$\frac{3}{8}m$