QBManager

2 RBTS PAPER

160778 Rocket engines lift a rocket from the earth surface, because hot gases with high velocity

1 heat up the air which lifts the rocket

2 push against the earth

3 react against the rocket and push it up

4 push against the air

NCERT-I-98

2 RBTS PAPER

160779 The force on a rocket moving with a velocity of $300 m / s$ is $210 N$ . Then the rate of combustion of the fuel will be

1

3.5 kg / \sec

2

2.1 kg / \sec

3

0.7 kg / \sec

4

1.4 kg / \sec

Explanation:

$\begin{aligned} (3) \\ ∴ 210 = 300 \times \frac{dm}{dt} \\ ∴ \frac{dm}{dt} = \frac{210}{300} = 0.7 kg / \sec \end{aligned}$

NCERT-I-95

2 RBTS PAPER

160780 A bullet of mass $200 g$ is fired at a speed of $5 m / s$ . The gun of mass $1 kg$ rebounds backward with a velocity of

1

10 m / s

2

- 1 m / s

3

0.1 m / s

4

0.01 m / s

Explanation:

$m_{1}$ - mass of bullet $= 200 g = 0.2 kg$
$m_{2}$ - mass of gun $= 1 kg$
$V_{1}$ - velocity of bullet $= 5 m / s$
$V_{2}$ - velocity of gun $=$ ?
According to law of conservation of linear momentum,
Linear momentum $=$ Linear momentum before firing after firing
$\begin{aligned} 0 & = m_{1} V_{1} + m_{2} V_{2} \\ V_{2} & = - m_{1} V_{1} / m_{2} \\ = - 0.2 \times 5 \times 1 \\ V_{2} & = - 1 m / s \end{aligned}$
Velocity of the gun $= - 1 m / s$

NCERT-I-98

2 RBTS PAPER

160781 A rigid ball of mass $m$ strikes a rigid wall at $60^{\circ}$ and gets reflected without loss of speed as shown in the figure.
The value of impulse imparted by the wall in the ball will be:

1

mv / 2

2

mv

3

mv / 3

4

2 mv

Explanation:

Impulse = Change in momentum
$= 2 mv \cos 60^{\circ} = mv$

NCERT-I-98

2 RBTS PAPER

160782 An object kept on a smooth inclined plane of an angle $θ$ can be kept stationary relative to the incline by giving a horizontal acceleration to the inclined plane given by:

1

g \cos θ

2

g \tan θ

3

g \sin θ

4 None of these

Explanation:

$g \tan θ$

NCERT-I-102

2 RBTS PAPER

160778 Rocket engines lift a rocket from the earth surface, because hot gases with high velocity

1 heat up the air which lifts the rocket

2 push against the earth

3 react against the rocket and push it up

4 push against the air

NCERT-I-98

2 RBTS PAPER

160779 The force on a rocket moving with a velocity of $300 m / s$ is $210 N$ . Then the rate of combustion of the fuel will be

1

3.5 kg / \sec

2

2.1 kg / \sec

3

0.7 kg / \sec

4

1.4 kg / \sec

Explanation:

$\begin{aligned} (3) \\ ∴ 210 = 300 \times \frac{dm}{dt} \\ ∴ \frac{dm}{dt} = \frac{210}{300} = 0.7 kg / \sec \end{aligned}$

NCERT-I-95

2 RBTS PAPER

160780 A bullet of mass $200 g$ is fired at a speed of $5 m / s$ . The gun of mass $1 kg$ rebounds backward with a velocity of

1

10 m / s

2

- 1 m / s

3

0.1 m / s

4

0.01 m / s

Explanation:

$m_{1}$ - mass of bullet $= 200 g = 0.2 kg$
$m_{2}$ - mass of gun $= 1 kg$
$V_{1}$ - velocity of bullet $= 5 m / s$
$V_{2}$ - velocity of gun $=$ ?
According to law of conservation of linear momentum,
Linear momentum $=$ Linear momentum before firing after firing
$\begin{aligned} 0 & = m_{1} V_{1} + m_{2} V_{2} \\ V_{2} & = - m_{1} V_{1} / m_{2} \\ = - 0.2 \times 5 \times 1 \\ V_{2} & = - 1 m / s \end{aligned}$
Velocity of the gun $= - 1 m / s$

NCERT-I-98

2 RBTS PAPER

160781 A rigid ball of mass $m$ strikes a rigid wall at $60^{\circ}$ and gets reflected without loss of speed as shown in the figure.
The value of impulse imparted by the wall in the ball will be:

1

mv / 2

2

mv

3

mv / 3

4

2 mv

Explanation:

Impulse = Change in momentum
$= 2 mv \cos 60^{\circ} = mv$

NCERT-I-98

2 RBTS PAPER

160782 An object kept on a smooth inclined plane of an angle $θ$ can be kept stationary relative to the incline by giving a horizontal acceleration to the inclined plane given by:

1

g \cos θ

2

g \tan θ

3

g \sin θ

4 None of these

Explanation:

$g \tan θ$

NCERT-I-102

2 RBTS PAPER

160778 Rocket engines lift a rocket from the earth surface, because hot gases with high velocity

1 heat up the air which lifts the rocket

2 push against the earth

3 react against the rocket and push it up

4 push against the air

NCERT-I-98

2 RBTS PAPER

160779 The force on a rocket moving with a velocity of $300 m / s$ is $210 N$ . Then the rate of combustion of the fuel will be

1

3.5 kg / \sec

2

2.1 kg / \sec

3

0.7 kg / \sec

4

1.4 kg / \sec

Explanation:

$\begin{aligned} (3) \\ ∴ 210 = 300 \times \frac{dm}{dt} \\ ∴ \frac{dm}{dt} = \frac{210}{300} = 0.7 kg / \sec \end{aligned}$

NCERT-I-95

2 RBTS PAPER

160780 A bullet of mass $200 g$ is fired at a speed of $5 m / s$ . The gun of mass $1 kg$ rebounds backward with a velocity of

1

10 m / s

2

- 1 m / s

3

0.1 m / s

4

0.01 m / s

Explanation:

$m_{1}$ - mass of bullet $= 200 g = 0.2 kg$
$m_{2}$ - mass of gun $= 1 kg$
$V_{1}$ - velocity of bullet $= 5 m / s$
$V_{2}$ - velocity of gun $=$ ?
According to law of conservation of linear momentum,
Linear momentum $=$ Linear momentum before firing after firing
$\begin{aligned} 0 & = m_{1} V_{1} + m_{2} V_{2} \\ V_{2} & = - m_{1} V_{1} / m_{2} \\ = - 0.2 \times 5 \times 1 \\ V_{2} & = - 1 m / s \end{aligned}$
Velocity of the gun $= - 1 m / s$

NCERT-I-98

2 RBTS PAPER

160781 A rigid ball of mass $m$ strikes a rigid wall at $60^{\circ}$ and gets reflected without loss of speed as shown in the figure.
The value of impulse imparted by the wall in the ball will be:

1

mv / 2

2

mv

3

mv / 3

4

2 mv

Explanation:

Impulse = Change in momentum
$= 2 mv \cos 60^{\circ} = mv$

NCERT-I-98

2 RBTS PAPER

160782 An object kept on a smooth inclined plane of an angle $θ$ can be kept stationary relative to the incline by giving a horizontal acceleration to the inclined plane given by:

1

g \cos θ

2

g \tan θ

3

g \sin θ

4 None of these

Explanation:

$g \tan θ$

NCERT-I-102

2 RBTS PAPER

160778 Rocket engines lift a rocket from the earth surface, because hot gases with high velocity

1 heat up the air which lifts the rocket

2 push against the earth

3 react against the rocket and push it up

4 push against the air

NCERT-I-98

2 RBTS PAPER

160779 The force on a rocket moving with a velocity of $300 m / s$ is $210 N$ . Then the rate of combustion of the fuel will be

1

3.5 kg / \sec

2

2.1 kg / \sec

3

0.7 kg / \sec

4

1.4 kg / \sec

Explanation:

$\begin{aligned} (3) \\ ∴ 210 = 300 \times \frac{dm}{dt} \\ ∴ \frac{dm}{dt} = \frac{210}{300} = 0.7 kg / \sec \end{aligned}$

NCERT-I-95

2 RBTS PAPER

160780 A bullet of mass $200 g$ is fired at a speed of $5 m / s$ . The gun of mass $1 kg$ rebounds backward with a velocity of

1

10 m / s

2

- 1 m / s

3

0.1 m / s

4

0.01 m / s

Explanation:

$m_{1}$ - mass of bullet $= 200 g = 0.2 kg$
$m_{2}$ - mass of gun $= 1 kg$
$V_{1}$ - velocity of bullet $= 5 m / s$
$V_{2}$ - velocity of gun $=$ ?
According to law of conservation of linear momentum,
Linear momentum $=$ Linear momentum before firing after firing
$\begin{aligned} 0 & = m_{1} V_{1} + m_{2} V_{2} \\ V_{2} & = - m_{1} V_{1} / m_{2} \\ = - 0.2 \times 5 \times 1 \\ V_{2} & = - 1 m / s \end{aligned}$
Velocity of the gun $= - 1 m / s$

NCERT-I-98

2 RBTS PAPER

160781 A rigid ball of mass $m$ strikes a rigid wall at $60^{\circ}$ and gets reflected without loss of speed as shown in the figure.
The value of impulse imparted by the wall in the ball will be:

1

mv / 2

2

mv

3

mv / 3

4

2 mv

Explanation:

Impulse = Change in momentum
$= 2 mv \cos 60^{\circ} = mv$

NCERT-I-98

2 RBTS PAPER

160782 An object kept on a smooth inclined plane of an angle $θ$ can be kept stationary relative to the incline by giving a horizontal acceleration to the inclined plane given by:

1

g \cos θ

2

g \tan θ

3

g \sin θ

4 None of these

Explanation:

$g \tan θ$

NCERT-I-102

2 RBTS PAPER

160778 Rocket engines lift a rocket from the earth surface, because hot gases with high velocity

1 heat up the air which lifts the rocket

2 push against the earth

3 react against the rocket and push it up

4 push against the air

NCERT-I-98

2 RBTS PAPER

160779 The force on a rocket moving with a velocity of $300 m / s$ is $210 N$ . Then the rate of combustion of the fuel will be

1

3.5 kg / \sec

2

2.1 kg / \sec

3

0.7 kg / \sec

4

1.4 kg / \sec

Explanation:

$\begin{aligned} (3) \\ ∴ 210 = 300 \times \frac{dm}{dt} \\ ∴ \frac{dm}{dt} = \frac{210}{300} = 0.7 kg / \sec \end{aligned}$

NCERT-I-95

2 RBTS PAPER

160780 A bullet of mass $200 g$ is fired at a speed of $5 m / s$ . The gun of mass $1 kg$ rebounds backward with a velocity of

1

10 m / s

2

- 1 m / s

3

0.1 m / s

4

0.01 m / s

Explanation:

$m_{1}$ - mass of bullet $= 200 g = 0.2 kg$
$m_{2}$ - mass of gun $= 1 kg$
$V_{1}$ - velocity of bullet $= 5 m / s$
$V_{2}$ - velocity of gun $=$ ?
According to law of conservation of linear momentum,
Linear momentum $=$ Linear momentum before firing after firing
$\begin{aligned} 0 & = m_{1} V_{1} + m_{2} V_{2} \\ V_{2} & = - m_{1} V_{1} / m_{2} \\ = - 0.2 \times 5 \times 1 \\ V_{2} & = - 1 m / s \end{aligned}$
Velocity of the gun $= - 1 m / s$

NCERT-I-98

2 RBTS PAPER

160781 A rigid ball of mass $m$ strikes a rigid wall at $60^{\circ}$ and gets reflected without loss of speed as shown in the figure.
The value of impulse imparted by the wall in the ball will be:

1

mv / 2

2

mv

3

mv / 3

4

2 mv

Explanation:

Impulse = Change in momentum
$= 2 mv \cos 60^{\circ} = mv$

NCERT-I-98

2 RBTS PAPER

160782 An object kept on a smooth inclined plane of an angle $θ$ can be kept stationary relative to the incline by giving a horizontal acceleration to the inclined plane given by:

1

g \cos θ

2

g \tan θ

3

g \sin θ

4 None of these

Explanation:

$g \tan θ$

NCERT-I-102

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