QBManager

Motion in Plane

143729 A boy runs on a horizontal road with a speed of $4 m / s$ while it is raining. He sees that the rain is making an angle $θ$ with the vertical while running from West to East. However, when he runs from East to West, the angle is $α$ . The rain is pouring down at an angle $45^{\circ}$ with the vertical normal and at a speed of $8 m / s$ as shown in the figure. The ratio $\frac{\tan θ}{\tan α}$ is

1

(1 - \sqrt{2})^{2}

2

(1 + \sqrt{2})^{2}

3

(1 + \sqrt{2})

4

(\sqrt{2} - 1)

Explanation:

B Given, $v = 4 m / s$
The boy is running from East to west, the angle is $α$ and the rain is pouring down at an angle $45^{\circ}$
Then, the relative motion between rain and boy.

$∴ \tan A = \frac{4 \sin 45^{\circ}}{8 - 4 \cos 45^{\circ}}$
$\tan A = \frac{4 (\frac{1}{\sqrt{2}})}{8 - 4 (\frac{1}{\sqrt{2}})} = \frac{1}{(2 \sqrt{2} - 1)}$
$\tan B = \frac{4 \sin (90^{\circ} + 45^{\circ})}{8 - 4 \cos (90^{\circ} + 45^{\circ})} = \frac{1}{(2 \sqrt{2} + 1)}$
From above figure,
$∵ \tan θ = \tan (A + 45^{\circ}) = \frac{\tan A + \tan 45^{\circ}}{1 - \tan A \times \tan 45^{\circ}}$
$= \frac{\frac{1}{2 \sqrt{2} - 1} + 1}{1 - \frac{1}{2 \sqrt{2} - 1} \times 1} = \frac{\frac{1 + 2 \sqrt{2} - 1}{2 \sqrt{2} - 1}}{\frac{2 \sqrt{2} - 1 - 1}{2 \sqrt{2} - 1}} = \frac{\sqrt{2}}{\sqrt{2} - 1}$
and similarly
$∵ \tan α = \tan (B - 45^{\circ}) = \frac{\tan B - \tan 45^{\circ}}{1 + \tan B \tan 45^{\circ}}$
$= \frac{\frac{1}{2 \sqrt{2} + 1} - 1}{1 + \frac{1}{2 \sqrt{2} + 1} \times 1} = \frac{\frac{- 2 \sqrt{2} + 1 - 1}{2 \sqrt{2} + 1}}{\frac{2 \sqrt{2} + 1 + 1}{2 \sqrt{2} + 2}} = \frac{- 2 \sqrt{2}}{2 \sqrt{2} + 2}$
$= \frac{- \sqrt{2}}{\sqrt{2} + 1}$
Hence, $\frac{\tan θ}{\tan α} = \frac{\frac{\sqrt{2}}{\sqrt{2} - 1}}{\frac{- \sqrt{2}}{\sqrt{2} + 1}} = \frac{\sqrt{2}}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{- \sqrt{2}}$
$= \frac{- (\sqrt{2} + 1)}{(\sqrt{2} - 1)} \times \frac{(\sqrt{2} + 1)}{(\sqrt{2} + 1)} = \frac{- (\sqrt{2} + 1)^{2}}{2 - 1}$
$= - (1 + \sqrt{2})^{2} ≃ (1 + \sqrt{2})^{2}$

TS- EAMCET-03.05.2019

Motion in Plane

143730 Assertion: The magnitude of velocity of two boats relative to river is same. Both boats start simultaneously from same point on the bank may reach opposite bank simultaneously moving along different paths.
Reason: For boats to cross the river in same time. The component of their velocity relative to river in direction normal to flow should be same.

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

AIIMS-25.05.2019(E) Shift-2

Motion in Plane

143731 A train of $150 m$ length is going towards North direction at a speed of $10 {ms}^{- 1}$ . A parrot files at a speed of $5 {ms}^{- 1}$ towards South direction parallel to the railway track. The time taken by the parrot to cross the train is equal to-

1

12 s

2

8 s

3

15 s

4

10 s

Explanation:

D Given that, length of train $= 150 m$
Speed of train $= 10 m / \sec$ (North direction)
Speed of parrot $= 5 m / \sec$ (South direction)

Relative velocity of parrot with respect to train
$v = v_{T} + v_{p}$
$v = 10 + 5 = 15 m / \sec$
As we know that
$Speed = distance \times time$
$15 = 150 \times$ time
Time $= \frac{150}{15} = 10 \sec$

BCECE-2017

Motion in Plane

143732 What will be the minimum speed of the roller coaster so that the passenger at the top, when becomes upside down, do not fall out? Consider the acceleration due to gravity as $10 m / s^{2}$ , and the radius of curvature of the roller coaster is $10 m$ .

1

20 m / s

2

10 m / s

3

15 m / s

4

25 m / s

Explanation:

B Given that, acceleration due to gravity (g) = $10 m / \sec$
Radius of curvature of roller coaster $= 10 m$
$∵ F_{c} = {ma}_{c}$
$mg = {ma}_{c}$
$g = a_{c}$
$∵ a_{c} = \frac{v^{2}}{r}$
$v^{2} = {ra}_{c} \Rightarrow v = \sqrt{{ra}_{c}}$
$v = \sqrt{10 \times 10} = \sqrt{100}$
$v = 10 m / \sec$

TS EAMCET 04.08.2021

Motion in Plane

143729 A boy runs on a horizontal road with a speed of $4 m / s$ while it is raining. He sees that the rain is making an angle $θ$ with the vertical while running from West to East. However, when he runs from East to West, the angle is $α$ . The rain is pouring down at an angle $45^{\circ}$ with the vertical normal and at a speed of $8 m / s$ as shown in the figure. The ratio $\frac{\tan θ}{\tan α}$ is

1

(1 - \sqrt{2})^{2}

2

(1 + \sqrt{2})^{2}

3

(1 + \sqrt{2})

4

(\sqrt{2} - 1)

Explanation:

B Given, $v = 4 m / s$
The boy is running from East to west, the angle is $α$ and the rain is pouring down at an angle $45^{\circ}$
Then, the relative motion between rain and boy.

$∴ \tan A = \frac{4 \sin 45^{\circ}}{8 - 4 \cos 45^{\circ}}$
$\tan A = \frac{4 (\frac{1}{\sqrt{2}})}{8 - 4 (\frac{1}{\sqrt{2}})} = \frac{1}{(2 \sqrt{2} - 1)}$
$\tan B = \frac{4 \sin (90^{\circ} + 45^{\circ})}{8 - 4 \cos (90^{\circ} + 45^{\circ})} = \frac{1}{(2 \sqrt{2} + 1)}$
From above figure,
$∵ \tan θ = \tan (A + 45^{\circ}) = \frac{\tan A + \tan 45^{\circ}}{1 - \tan A \times \tan 45^{\circ}}$
$= \frac{\frac{1}{2 \sqrt{2} - 1} + 1}{1 - \frac{1}{2 \sqrt{2} - 1} \times 1} = \frac{\frac{1 + 2 \sqrt{2} - 1}{2 \sqrt{2} - 1}}{\frac{2 \sqrt{2} - 1 - 1}{2 \sqrt{2} - 1}} = \frac{\sqrt{2}}{\sqrt{2} - 1}$
and similarly
$∵ \tan α = \tan (B - 45^{\circ}) = \frac{\tan B - \tan 45^{\circ}}{1 + \tan B \tan 45^{\circ}}$
$= \frac{\frac{1}{2 \sqrt{2} + 1} - 1}{1 + \frac{1}{2 \sqrt{2} + 1} \times 1} = \frac{\frac{- 2 \sqrt{2} + 1 - 1}{2 \sqrt{2} + 1}}{\frac{2 \sqrt{2} + 1 + 1}{2 \sqrt{2} + 2}} = \frac{- 2 \sqrt{2}}{2 \sqrt{2} + 2}$
$= \frac{- \sqrt{2}}{\sqrt{2} + 1}$
Hence, $\frac{\tan θ}{\tan α} = \frac{\frac{\sqrt{2}}{\sqrt{2} - 1}}{\frac{- \sqrt{2}}{\sqrt{2} + 1}} = \frac{\sqrt{2}}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{- \sqrt{2}}$
$= \frac{- (\sqrt{2} + 1)}{(\sqrt{2} - 1)} \times \frac{(\sqrt{2} + 1)}{(\sqrt{2} + 1)} = \frac{- (\sqrt{2} + 1)^{2}}{2 - 1}$
$= - (1 + \sqrt{2})^{2} ≃ (1 + \sqrt{2})^{2}$

TS- EAMCET-03.05.2019

Motion in Plane

143730 Assertion: The magnitude of velocity of two boats relative to river is same. Both boats start simultaneously from same point on the bank may reach opposite bank simultaneously moving along different paths.
Reason: For boats to cross the river in same time. The component of their velocity relative to river in direction normal to flow should be same.

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

AIIMS-25.05.2019(E) Shift-2

Motion in Plane

143731 A train of $150 m$ length is going towards North direction at a speed of $10 {ms}^{- 1}$ . A parrot files at a speed of $5 {ms}^{- 1}$ towards South direction parallel to the railway track. The time taken by the parrot to cross the train is equal to-

1

12 s

2

8 s

3

15 s

4

10 s

Explanation:

D Given that, length of train $= 150 m$
Speed of train $= 10 m / \sec$ (North direction)
Speed of parrot $= 5 m / \sec$ (South direction)

Relative velocity of parrot with respect to train
$v = v_{T} + v_{p}$
$v = 10 + 5 = 15 m / \sec$
As we know that
$Speed = distance \times time$
$15 = 150 \times$ time
Time $= \frac{150}{15} = 10 \sec$

BCECE-2017

Motion in Plane

143732 What will be the minimum speed of the roller coaster so that the passenger at the top, when becomes upside down, do not fall out? Consider the acceleration due to gravity as $10 m / s^{2}$ , and the radius of curvature of the roller coaster is $10 m$ .

1

20 m / s

2

10 m / s

3

15 m / s

4

25 m / s

Explanation:

B Given that, acceleration due to gravity (g) = $10 m / \sec$
Radius of curvature of roller coaster $= 10 m$
$∵ F_{c} = {ma}_{c}$
$mg = {ma}_{c}$
$g = a_{c}$
$∵ a_{c} = \frac{v^{2}}{r}$
$v^{2} = {ra}_{c} \Rightarrow v = \sqrt{{ra}_{c}}$
$v = \sqrt{10 \times 10} = \sqrt{100}$
$v = 10 m / \sec$

TS EAMCET 04.08.2021

Motion in Plane

143729 A boy runs on a horizontal road with a speed of $4 m / s$ while it is raining. He sees that the rain is making an angle $θ$ with the vertical while running from West to East. However, when he runs from East to West, the angle is $α$ . The rain is pouring down at an angle $45^{\circ}$ with the vertical normal and at a speed of $8 m / s$ as shown in the figure. The ratio $\frac{\tan θ}{\tan α}$ is

1

(1 - \sqrt{2})^{2}

2

(1 + \sqrt{2})^{2}

3

(1 + \sqrt{2})

4

(\sqrt{2} - 1)

Explanation:

B Given, $v = 4 m / s$
The boy is running from East to west, the angle is $α$ and the rain is pouring down at an angle $45^{\circ}$
Then, the relative motion between rain and boy.

$∴ \tan A = \frac{4 \sin 45^{\circ}}{8 - 4 \cos 45^{\circ}}$
$\tan A = \frac{4 (\frac{1}{\sqrt{2}})}{8 - 4 (\frac{1}{\sqrt{2}})} = \frac{1}{(2 \sqrt{2} - 1)}$
$\tan B = \frac{4 \sin (90^{\circ} + 45^{\circ})}{8 - 4 \cos (90^{\circ} + 45^{\circ})} = \frac{1}{(2 \sqrt{2} + 1)}$
From above figure,
$∵ \tan θ = \tan (A + 45^{\circ}) = \frac{\tan A + \tan 45^{\circ}}{1 - \tan A \times \tan 45^{\circ}}$
$= \frac{\frac{1}{2 \sqrt{2} - 1} + 1}{1 - \frac{1}{2 \sqrt{2} - 1} \times 1} = \frac{\frac{1 + 2 \sqrt{2} - 1}{2 \sqrt{2} - 1}}{\frac{2 \sqrt{2} - 1 - 1}{2 \sqrt{2} - 1}} = \frac{\sqrt{2}}{\sqrt{2} - 1}$
and similarly
$∵ \tan α = \tan (B - 45^{\circ}) = \frac{\tan B - \tan 45^{\circ}}{1 + \tan B \tan 45^{\circ}}$
$= \frac{\frac{1}{2 \sqrt{2} + 1} - 1}{1 + \frac{1}{2 \sqrt{2} + 1} \times 1} = \frac{\frac{- 2 \sqrt{2} + 1 - 1}{2 \sqrt{2} + 1}}{\frac{2 \sqrt{2} + 1 + 1}{2 \sqrt{2} + 2}} = \frac{- 2 \sqrt{2}}{2 \sqrt{2} + 2}$
$= \frac{- \sqrt{2}}{\sqrt{2} + 1}$
Hence, $\frac{\tan θ}{\tan α} = \frac{\frac{\sqrt{2}}{\sqrt{2} - 1}}{\frac{- \sqrt{2}}{\sqrt{2} + 1}} = \frac{\sqrt{2}}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{- \sqrt{2}}$
$= \frac{- (\sqrt{2} + 1)}{(\sqrt{2} - 1)} \times \frac{(\sqrt{2} + 1)}{(\sqrt{2} + 1)} = \frac{- (\sqrt{2} + 1)^{2}}{2 - 1}$
$= - (1 + \sqrt{2})^{2} ≃ (1 + \sqrt{2})^{2}$

TS- EAMCET-03.05.2019

Motion in Plane

143730 Assertion: The magnitude of velocity of two boats relative to river is same. Both boats start simultaneously from same point on the bank may reach opposite bank simultaneously moving along different paths.
Reason: For boats to cross the river in same time. The component of their velocity relative to river in direction normal to flow should be same.

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

AIIMS-25.05.2019(E) Shift-2

Motion in Plane

143731 A train of $150 m$ length is going towards North direction at a speed of $10 {ms}^{- 1}$ . A parrot files at a speed of $5 {ms}^{- 1}$ towards South direction parallel to the railway track. The time taken by the parrot to cross the train is equal to-

1

12 s

2

8 s

3

15 s

4

10 s

Explanation:

D Given that, length of train $= 150 m$
Speed of train $= 10 m / \sec$ (North direction)
Speed of parrot $= 5 m / \sec$ (South direction)

Relative velocity of parrot with respect to train
$v = v_{T} + v_{p}$
$v = 10 + 5 = 15 m / \sec$
As we know that
$Speed = distance \times time$
$15 = 150 \times$ time
Time $= \frac{150}{15} = 10 \sec$

BCECE-2017

Motion in Plane

143732 What will be the minimum speed of the roller coaster so that the passenger at the top, when becomes upside down, do not fall out? Consider the acceleration due to gravity as $10 m / s^{2}$ , and the radius of curvature of the roller coaster is $10 m$ .

1

20 m / s

2

10 m / s

3

15 m / s

4

25 m / s

Explanation:

B Given that, acceleration due to gravity (g) = $10 m / \sec$
Radius of curvature of roller coaster $= 10 m$
$∵ F_{c} = {ma}_{c}$
$mg = {ma}_{c}$
$g = a_{c}$
$∵ a_{c} = \frac{v^{2}}{r}$
$v^{2} = {ra}_{c} \Rightarrow v = \sqrt{{ra}_{c}}$
$v = \sqrt{10 \times 10} = \sqrt{100}$
$v = 10 m / \sec$

TS EAMCET 04.08.2021

Motion in Plane

143729 A boy runs on a horizontal road with a speed of $4 m / s$ while it is raining. He sees that the rain is making an angle $θ$ with the vertical while running from West to East. However, when he runs from East to West, the angle is $α$ . The rain is pouring down at an angle $45^{\circ}$ with the vertical normal and at a speed of $8 m / s$ as shown in the figure. The ratio $\frac{\tan θ}{\tan α}$ is

1

(1 - \sqrt{2})^{2}

2

(1 + \sqrt{2})^{2}

3

(1 + \sqrt{2})

4

(\sqrt{2} - 1)

Explanation:

B Given, $v = 4 m / s$
The boy is running from East to west, the angle is $α$ and the rain is pouring down at an angle $45^{\circ}$
Then, the relative motion between rain and boy.

$∴ \tan A = \frac{4 \sin 45^{\circ}}{8 - 4 \cos 45^{\circ}}$
$\tan A = \frac{4 (\frac{1}{\sqrt{2}})}{8 - 4 (\frac{1}{\sqrt{2}})} = \frac{1}{(2 \sqrt{2} - 1)}$
$\tan B = \frac{4 \sin (90^{\circ} + 45^{\circ})}{8 - 4 \cos (90^{\circ} + 45^{\circ})} = \frac{1}{(2 \sqrt{2} + 1)}$
From above figure,
$∵ \tan θ = \tan (A + 45^{\circ}) = \frac{\tan A + \tan 45^{\circ}}{1 - \tan A \times \tan 45^{\circ}}$
$= \frac{\frac{1}{2 \sqrt{2} - 1} + 1}{1 - \frac{1}{2 \sqrt{2} - 1} \times 1} = \frac{\frac{1 + 2 \sqrt{2} - 1}{2 \sqrt{2} - 1}}{\frac{2 \sqrt{2} - 1 - 1}{2 \sqrt{2} - 1}} = \frac{\sqrt{2}}{\sqrt{2} - 1}$
and similarly
$∵ \tan α = \tan (B - 45^{\circ}) = \frac{\tan B - \tan 45^{\circ}}{1 + \tan B \tan 45^{\circ}}$
$= \frac{\frac{1}{2 \sqrt{2} + 1} - 1}{1 + \frac{1}{2 \sqrt{2} + 1} \times 1} = \frac{\frac{- 2 \sqrt{2} + 1 - 1}{2 \sqrt{2} + 1}}{\frac{2 \sqrt{2} + 1 + 1}{2 \sqrt{2} + 2}} = \frac{- 2 \sqrt{2}}{2 \sqrt{2} + 2}$
$= \frac{- \sqrt{2}}{\sqrt{2} + 1}$
Hence, $\frac{\tan θ}{\tan α} = \frac{\frac{\sqrt{2}}{\sqrt{2} - 1}}{\frac{- \sqrt{2}}{\sqrt{2} + 1}} = \frac{\sqrt{2}}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{- \sqrt{2}}$
$= \frac{- (\sqrt{2} + 1)}{(\sqrt{2} - 1)} \times \frac{(\sqrt{2} + 1)}{(\sqrt{2} + 1)} = \frac{- (\sqrt{2} + 1)^{2}}{2 - 1}$
$= - (1 + \sqrt{2})^{2} ≃ (1 + \sqrt{2})^{2}$

TS- EAMCET-03.05.2019

Motion in Plane

143730 Assertion: The magnitude of velocity of two boats relative to river is same. Both boats start simultaneously from same point on the bank may reach opposite bank simultaneously moving along different paths.
Reason: For boats to cross the river in same time. The component of their velocity relative to river in direction normal to flow should be same.

1 If both assertion and reason are true and reason is the correct explanation of assertion.

2 If both assertion and reason are true but reason is not the correct explanation of assertion.

3 If assertion is true but reason is false.

4 If both assertion and reason are false.

AIIMS-25.05.2019(E) Shift-2

Motion in Plane

143731 A train of $150 m$ length is going towards North direction at a speed of $10 {ms}^{- 1}$ . A parrot files at a speed of $5 {ms}^{- 1}$ towards South direction parallel to the railway track. The time taken by the parrot to cross the train is equal to-

1

12 s

2

8 s

3

15 s

4

10 s

Explanation:

D Given that, length of train $= 150 m$
Speed of train $= 10 m / \sec$ (North direction)
Speed of parrot $= 5 m / \sec$ (South direction)

Relative velocity of parrot with respect to train
$v = v_{T} + v_{p}$
$v = 10 + 5 = 15 m / \sec$
As we know that
$Speed = distance \times time$
$15 = 150 \times$ time
Time $= \frac{150}{15} = 10 \sec$

BCECE-2017

Motion in Plane

143732 What will be the minimum speed of the roller coaster so that the passenger at the top, when becomes upside down, do not fall out? Consider the acceleration due to gravity as $10 m / s^{2}$ , and the radius of curvature of the roller coaster is $10 m$ .

1

20 m / s

2

10 m / s

3

15 m / s

4

25 m / s

Explanation:

B Given that, acceleration due to gravity (g) = $10 m / \sec$
Radius of curvature of roller coaster $= 10 m$
$∵ F_{c} = {ma}_{c}$
$mg = {ma}_{c}$
$g = a_{c}$
$∵ a_{c} = \frac{v^{2}}{r}$
$v^{2} = {ra}_{c} \Rightarrow v = \sqrt{{ra}_{c}}$
$v = \sqrt{10 \times 10} = \sqrt{100}$
$v = 10 m / \sec$

TS EAMCET 04.08.2021