QBManager

3 RBTS PAPER

162584 A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is

1 constant and equal to

mg

in magnitude

2 constant and greater than

mg

in magnitude

3 variable but always greater than

mg

4 at first greater than

mg

, and later becomes equla to

mg

Explanation:

At first greater than $mg$ , and later becomes equal to $mg$

NCERT Examplar- WPE-3

3 RBTS PAPER

162585 A body of mass $1 kg$ is thrown upwards with a velocity $20 {ms}^{- 1}$ . It momentarily comes to rest after attaining a height of $18 m$ . How much energy is lost due to air friction? $(g = 10 {ms}^{- 2})$

1

20 J

2

30 J

3

40 J

4

10 J

Explanation:

The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy.
Initially body possesses only kinetic energy and after attaining a height the kinetic energy is zero.
Therefore, loss of energy $= KE - PE$
$= \frac{1}{2} m v^{2} - m g h$
$= \frac{1}{2} \times 1 \times 400 - 1 \times 18 \times 10$
$= 200 - 180 = 20 J$

NCERT-I-78

3 RBTS PAPER

162586 A ball moving with velocities $2 {m s}^{- 1}$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 , then their velocities (in ${ms}^{- 1}$ ) after collision will be

1 0,1

2 1,1

3

1, 0.5

4 0,2

Explanation:

If two bodies collide head on with coefficient of restitution
$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$

From the law of conservation of linear momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$Misplaced &$

Substituting $u_{1} = 2 {ms}^{- 1}, u_{2} = 0, m_{1} = m$ and $m_{2} = 2 m, e = 0.5$

We get $v_{1} = \frac{m - m}{m + 2 m} \times 2$
$\Rightarrow v_{1} = 0$

Similarly, $Missing or unrecognized delimiter for \left$
$Missing or unrecognized delimiter for \left$

NCERT-I-84

3 RBTS PAPER

162587 Assertion (A): In an elastic collision of two billiard balls, the total KE is conserved during the short time of collision of the balls (i.e., when they are in contact).
Reason (R): Energy spent against friction does not follow the law of conservation of energy.

1 Both

A

and

R

are true and

R

is the correct explanation of

A

.

2 Both

A

and

R

are true but

R

is not the correct explanation of

A

.

3

A

is true but

R

is false.

4 Both

A

and

R

are false.

NCERT-I-84

3 RBTS PAPER

162584 A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is

1 constant and equal to

mg

in magnitude

2 constant and greater than

mg

in magnitude

3 variable but always greater than

mg

4 at first greater than

mg

, and later becomes equla to

mg

Explanation:

At first greater than $mg$ , and later becomes equal to $mg$

NCERT Examplar- WPE-3

3 RBTS PAPER

162585 A body of mass $1 kg$ is thrown upwards with a velocity $20 {ms}^{- 1}$ . It momentarily comes to rest after attaining a height of $18 m$ . How much energy is lost due to air friction? $(g = 10 {ms}^{- 2})$

1

20 J

2

30 J

3

40 J

4

10 J

Explanation:

The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy.
Initially body possesses only kinetic energy and after attaining a height the kinetic energy is zero.
Therefore, loss of energy $= KE - PE$
$= \frac{1}{2} m v^{2} - m g h$
$= \frac{1}{2} \times 1 \times 400 - 1 \times 18 \times 10$
$= 200 - 180 = 20 J$

NCERT-I-78

3 RBTS PAPER

162586 A ball moving with velocities $2 {m s}^{- 1}$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 , then their velocities (in ${ms}^{- 1}$ ) after collision will be

1 0,1

2 1,1

3

1, 0.5

4 0,2

Explanation:

If two bodies collide head on with coefficient of restitution
$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$

From the law of conservation of linear momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$Misplaced &$

Substituting $u_{1} = 2 {ms}^{- 1}, u_{2} = 0, m_{1} = m$ and $m_{2} = 2 m, e = 0.5$

We get $v_{1} = \frac{m - m}{m + 2 m} \times 2$
$\Rightarrow v_{1} = 0$

Similarly, $Missing or unrecognized delimiter for \left$
$Missing or unrecognized delimiter for \left$

NCERT-I-84

3 RBTS PAPER

162587 Assertion (A): In an elastic collision of two billiard balls, the total KE is conserved during the short time of collision of the balls (i.e., when they are in contact).
Reason (R): Energy spent against friction does not follow the law of conservation of energy.

1 Both

A

and

R

are true and

R

is the correct explanation of

A

.

2 Both

A

and

R

are true but

R

is not the correct explanation of

A

.

3

A

is true but

R

is false.

4 Both

A

and

R

are false.

NCERT-I-84

3 RBTS PAPER

162584 A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is

1 constant and equal to

mg

in magnitude

2 constant and greater than

mg

in magnitude

3 variable but always greater than

mg

4 at first greater than

mg

, and later becomes equla to

mg

Explanation:

At first greater than $mg$ , and later becomes equal to $mg$

NCERT Examplar- WPE-3

3 RBTS PAPER

162585 A body of mass $1 kg$ is thrown upwards with a velocity $20 {ms}^{- 1}$ . It momentarily comes to rest after attaining a height of $18 m$ . How much energy is lost due to air friction? $(g = 10 {ms}^{- 2})$

1

20 J

2

30 J

3

40 J

4

10 J

Explanation:

The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy.
Initially body possesses only kinetic energy and after attaining a height the kinetic energy is zero.
Therefore, loss of energy $= KE - PE$
$= \frac{1}{2} m v^{2} - m g h$
$= \frac{1}{2} \times 1 \times 400 - 1 \times 18 \times 10$
$= 200 - 180 = 20 J$

NCERT-I-78

3 RBTS PAPER

162586 A ball moving with velocities $2 {m s}^{- 1}$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 , then their velocities (in ${ms}^{- 1}$ ) after collision will be

1 0,1

2 1,1

3

1, 0.5

4 0,2

Explanation:

If two bodies collide head on with coefficient of restitution
$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$

From the law of conservation of linear momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$Misplaced &$

Substituting $u_{1} = 2 {ms}^{- 1}, u_{2} = 0, m_{1} = m$ and $m_{2} = 2 m, e = 0.5$

We get $v_{1} = \frac{m - m}{m + 2 m} \times 2$
$\Rightarrow v_{1} = 0$

Similarly, $Missing or unrecognized delimiter for \left$
$Missing or unrecognized delimiter for \left$

NCERT-I-84

3 RBTS PAPER

162587 Assertion (A): In an elastic collision of two billiard balls, the total KE is conserved during the short time of collision of the balls (i.e., when they are in contact).
Reason (R): Energy spent against friction does not follow the law of conservation of energy.

1 Both

A

and

R

are true and

R

is the correct explanation of

A

.

2 Both

A

and

R

are true but

R

is not the correct explanation of

A

.

3

A

is true but

R

is false.

4 Both

A

and

R

are false.

NCERT-I-84

3 RBTS PAPER

162584 A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is

1 constant and equal to

mg

in magnitude

2 constant and greater than

mg

in magnitude

3 variable but always greater than

mg

4 at first greater than

mg

, and later becomes equla to

mg

Explanation:

At first greater than $mg$ , and later becomes equal to $mg$

NCERT Examplar- WPE-3

3 RBTS PAPER

162585 A body of mass $1 kg$ is thrown upwards with a velocity $20 {ms}^{- 1}$ . It momentarily comes to rest after attaining a height of $18 m$ . How much energy is lost due to air friction? $(g = 10 {ms}^{- 2})$

1

20 J

2

30 J

3

40 J

4

10 J

Explanation:

The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy.
Initially body possesses only kinetic energy and after attaining a height the kinetic energy is zero.
Therefore, loss of energy $= KE - PE$
$= \frac{1}{2} m v^{2} - m g h$
$= \frac{1}{2} \times 1 \times 400 - 1 \times 18 \times 10$
$= 200 - 180 = 20 J$

NCERT-I-78

3 RBTS PAPER

162586 A ball moving with velocities $2 {m s}^{- 1}$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 , then their velocities (in ${ms}^{- 1}$ ) after collision will be

1 0,1

2 1,1

3

1, 0.5

4 0,2

Explanation:

If two bodies collide head on with coefficient of restitution
$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$

From the law of conservation of linear momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$Misplaced &$

Substituting $u_{1} = 2 {ms}^{- 1}, u_{2} = 0, m_{1} = m$ and $m_{2} = 2 m, e = 0.5$

We get $v_{1} = \frac{m - m}{m + 2 m} \times 2$
$\Rightarrow v_{1} = 0$

Similarly, $Missing or unrecognized delimiter for \left$
$Missing or unrecognized delimiter for \left$

NCERT-I-84

3 RBTS PAPER

162587 Assertion (A): In an elastic collision of two billiard balls, the total KE is conserved during the short time of collision of the balls (i.e., when they are in contact).
Reason (R): Energy spent against friction does not follow the law of conservation of energy.

1 Both

A

and

R

are true and

R

is the correct explanation of

A

.

2 Both

A

and

R

are true but

R

is not the correct explanation of

A

.

3

A

is true but

R

is false.

4 Both

A

and

R

are false.

NCERT-I-84