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PHXI03:MOTION IN A STRAIGHT LINE

362345 Assertion :
A body falling (in some case), may do so with constant velocity.
Reason :
The body falls freely, when acceleration of a body is equal to acceleration due to gravity.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

When an object falls freely, it experiences only gravitational force acting vertically downward. This downward force causes the object to accelerate, increasing its velocity, which reaches its maximum when the object makes contact with the ground. $\Rightarrow$ Reason is true.
If the downward acceleration due to gravity is counteracted by an upward retarding force, the object falls at a constant velocity. This constant velocity is referred to as the terminal velocity of the object. $\Rightarrow$ Assertion is true.
But they are different facts. So correct option is (2).

PHXI03:MOTION IN A STRAIGHT LINE

362346 With what velocity a ball should be projected vertically so that the distance covered by it in $5^{t h}$ second is twice the distance it covers in its $6^{t h}$ second $(g = 10 m / s^{2})$ ?

1

58. 8 m / s

2

49 m / s

3

65 m / s

4

19.6 m / s

Explanation:

Let $u$ be the initial speed
Given that
$u - 10 (5 - \frac{1}{2}) = 2 [u - 10 (6 - \frac{1}{2})]$
$u - 45 = 2 u - 110$
$u = 65 m s^{- 1}$

PHXI03:MOTION IN A STRAIGHT LINE

362347 A balloon rises from rest on the ground with constant acceleration $1 {m s}^{- 2}$ . A stone is dropped when the balloon has risen to a height of 40 $m$ . Find the time taken by the stone to reach the ground.

1

1 s

2

5 s

3

3 s

4

4 s

Explanation:

$h = \frac{1}{2} a t^{2} \Rightarrow t = \sqrt{\frac{2 h}{a}} = \sqrt{\frac{2 \times 40 \times 8}{10}} = 8 s$
Velocity after $8 s, v = 0 + 1 \times 8 = 8 {m s}^{- 1}$
$40 = - 10 t + \frac{1}{2} 10 t^{2}$
$10 t^{2} - 2 \times 10 t - 2 \times 40 = 0$
$t^{2} - 2 t - 8 = 0$
$t - 4 t + 2 t - 8 = 0$
$t = 4 s$

PHXI03:MOTION IN A STRAIGHT LINE

362348 A ball is dropped into a well in which the water level is at a depth $h$ below the top. If the speed of sound is $c$ , then the time after which the splash is heard will be given by

1

h [\sqrt{\frac{2}{g h}} - \frac{1}{c}]

2

h [\sqrt{\frac{2}{g h}} + \frac{1}{c}]

3

h [\frac{2}{g} - \frac{1}{c}]

4

h [\frac{2}{g} + \frac{1}{c}]

Explanation:

Time of fall $= \sqrt{\frac{2 h}{g}}$
Time taken by the sound to come out $= \frac{h}{c}$
Total time $= \sqrt{\frac{2 h}{g}} + \frac{h}{c} = h [\sqrt{\frac{2}{g h}} + \frac{1}{c}]$

PHXI03:MOTION IN A STRAIGHT LINE

362345 Assertion :
A body falling (in some case), may do so with constant velocity.
Reason :
The body falls freely, when acceleration of a body is equal to acceleration due to gravity.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

When an object falls freely, it experiences only gravitational force acting vertically downward. This downward force causes the object to accelerate, increasing its velocity, which reaches its maximum when the object makes contact with the ground. $\Rightarrow$ Reason is true.
If the downward acceleration due to gravity is counteracted by an upward retarding force, the object falls at a constant velocity. This constant velocity is referred to as the terminal velocity of the object. $\Rightarrow$ Assertion is true.
But they are different facts. So correct option is (2).

PHXI03:MOTION IN A STRAIGHT LINE

362346 With what velocity a ball should be projected vertically so that the distance covered by it in $5^{t h}$ second is twice the distance it covers in its $6^{t h}$ second $(g = 10 m / s^{2})$ ?

1

58. 8 m / s

2

49 m / s

3

65 m / s

4

19.6 m / s

Explanation:

Let $u$ be the initial speed
Given that
$u - 10 (5 - \frac{1}{2}) = 2 [u - 10 (6 - \frac{1}{2})]$
$u - 45 = 2 u - 110$
$u = 65 m s^{- 1}$

PHXI03:MOTION IN A STRAIGHT LINE

362347 A balloon rises from rest on the ground with constant acceleration $1 {m s}^{- 2}$ . A stone is dropped when the balloon has risen to a height of 40 $m$ . Find the time taken by the stone to reach the ground.

1

1 s

2

5 s

3

3 s

4

4 s

Explanation:

$h = \frac{1}{2} a t^{2} \Rightarrow t = \sqrt{\frac{2 h}{a}} = \sqrt{\frac{2 \times 40 \times 8}{10}} = 8 s$
Velocity after $8 s, v = 0 + 1 \times 8 = 8 {m s}^{- 1}$
$40 = - 10 t + \frac{1}{2} 10 t^{2}$
$10 t^{2} - 2 \times 10 t - 2 \times 40 = 0$
$t^{2} - 2 t - 8 = 0$
$t - 4 t + 2 t - 8 = 0$
$t = 4 s$

PHXI03:MOTION IN A STRAIGHT LINE

362348 A ball is dropped into a well in which the water level is at a depth $h$ below the top. If the speed of sound is $c$ , then the time after which the splash is heard will be given by

1

h [\sqrt{\frac{2}{g h}} - \frac{1}{c}]

2

h [\sqrt{\frac{2}{g h}} + \frac{1}{c}]

3

h [\frac{2}{g} - \frac{1}{c}]

4

h [\frac{2}{g} + \frac{1}{c}]

Explanation:

Time of fall $= \sqrt{\frac{2 h}{g}}$
Time taken by the sound to come out $= \frac{h}{c}$
Total time $= \sqrt{\frac{2 h}{g}} + \frac{h}{c} = h [\sqrt{\frac{2}{g h}} + \frac{1}{c}]$

PHXI03:MOTION IN A STRAIGHT LINE

362345 Assertion :
A body falling (in some case), may do so with constant velocity.
Reason :
The body falls freely, when acceleration of a body is equal to acceleration due to gravity.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

When an object falls freely, it experiences only gravitational force acting vertically downward. This downward force causes the object to accelerate, increasing its velocity, which reaches its maximum when the object makes contact with the ground. $\Rightarrow$ Reason is true.
If the downward acceleration due to gravity is counteracted by an upward retarding force, the object falls at a constant velocity. This constant velocity is referred to as the terminal velocity of the object. $\Rightarrow$ Assertion is true.
But they are different facts. So correct option is (2).

PHXI03:MOTION IN A STRAIGHT LINE

362346 With what velocity a ball should be projected vertically so that the distance covered by it in $5^{t h}$ second is twice the distance it covers in its $6^{t h}$ second $(g = 10 m / s^{2})$ ?

1

58. 8 m / s

2

49 m / s

3

65 m / s

4

19.6 m / s

Explanation:

Let $u$ be the initial speed
Given that
$u - 10 (5 - \frac{1}{2}) = 2 [u - 10 (6 - \frac{1}{2})]$
$u - 45 = 2 u - 110$
$u = 65 m s^{- 1}$

PHXI03:MOTION IN A STRAIGHT LINE

362347 A balloon rises from rest on the ground with constant acceleration $1 {m s}^{- 2}$ . A stone is dropped when the balloon has risen to a height of 40 $m$ . Find the time taken by the stone to reach the ground.

1

1 s

2

5 s

3

3 s

4

4 s

Explanation:

$h = \frac{1}{2} a t^{2} \Rightarrow t = \sqrt{\frac{2 h}{a}} = \sqrt{\frac{2 \times 40 \times 8}{10}} = 8 s$
Velocity after $8 s, v = 0 + 1 \times 8 = 8 {m s}^{- 1}$
$40 = - 10 t + \frac{1}{2} 10 t^{2}$
$10 t^{2} - 2 \times 10 t - 2 \times 40 = 0$
$t^{2} - 2 t - 8 = 0$
$t - 4 t + 2 t - 8 = 0$
$t = 4 s$

PHXI03:MOTION IN A STRAIGHT LINE

362348 A ball is dropped into a well in which the water level is at a depth $h$ below the top. If the speed of sound is $c$ , then the time after which the splash is heard will be given by

1

h [\sqrt{\frac{2}{g h}} - \frac{1}{c}]

2

h [\sqrt{\frac{2}{g h}} + \frac{1}{c}]

3

h [\frac{2}{g} - \frac{1}{c}]

4

h [\frac{2}{g} + \frac{1}{c}]

Explanation:

Time of fall $= \sqrt{\frac{2 h}{g}}$
Time taken by the sound to come out $= \frac{h}{c}$
Total time $= \sqrt{\frac{2 h}{g}} + \frac{h}{c} = h [\sqrt{\frac{2}{g h}} + \frac{1}{c}]$

PHXI03:MOTION IN A STRAIGHT LINE

362345 Assertion :
A body falling (in some case), may do so with constant velocity.
Reason :
The body falls freely, when acceleration of a body is equal to acceleration due to gravity.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

When an object falls freely, it experiences only gravitational force acting vertically downward. This downward force causes the object to accelerate, increasing its velocity, which reaches its maximum when the object makes contact with the ground. $\Rightarrow$ Reason is true.
If the downward acceleration due to gravity is counteracted by an upward retarding force, the object falls at a constant velocity. This constant velocity is referred to as the terminal velocity of the object. $\Rightarrow$ Assertion is true.
But they are different facts. So correct option is (2).

PHXI03:MOTION IN A STRAIGHT LINE

362346 With what velocity a ball should be projected vertically so that the distance covered by it in $5^{t h}$ second is twice the distance it covers in its $6^{t h}$ second $(g = 10 m / s^{2})$ ?

1

58. 8 m / s

2

49 m / s

3

65 m / s

4

19.6 m / s

Explanation:

Let $u$ be the initial speed
Given that
$u - 10 (5 - \frac{1}{2}) = 2 [u - 10 (6 - \frac{1}{2})]$
$u - 45 = 2 u - 110$
$u = 65 m s^{- 1}$

PHXI03:MOTION IN A STRAIGHT LINE

362347 A balloon rises from rest on the ground with constant acceleration $1 {m s}^{- 2}$ . A stone is dropped when the balloon has risen to a height of 40 $m$ . Find the time taken by the stone to reach the ground.

1

1 s

2

5 s

3

3 s

4

4 s

Explanation:

$h = \frac{1}{2} a t^{2} \Rightarrow t = \sqrt{\frac{2 h}{a}} = \sqrt{\frac{2 \times 40 \times 8}{10}} = 8 s$
Velocity after $8 s, v = 0 + 1 \times 8 = 8 {m s}^{- 1}$
$40 = - 10 t + \frac{1}{2} 10 t^{2}$
$10 t^{2} - 2 \times 10 t - 2 \times 40 = 0$
$t^{2} - 2 t - 8 = 0$
$t - 4 t + 2 t - 8 = 0$
$t = 4 s$

PHXI03:MOTION IN A STRAIGHT LINE

362348 A ball is dropped into a well in which the water level is at a depth $h$ below the top. If the speed of sound is $c$ , then the time after which the splash is heard will be given by

1

h [\sqrt{\frac{2}{g h}} - \frac{1}{c}]

2

h [\sqrt{\frac{2}{g h}} + \frac{1}{c}]

3

h [\frac{2}{g} - \frac{1}{c}]

4

h [\frac{2}{g} + \frac{1}{c}]

Explanation:

Time of fall $= \sqrt{\frac{2 h}{g}}$
Time taken by the sound to come out $= \frac{h}{c}$
Total time $= \sqrt{\frac{2 h}{g}} + \frac{h}{c} = h [\sqrt{\frac{2}{g h}} + \frac{1}{c}]$