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
Explanation:
Hot gases with ingh velocity react against the rocket and(pondt up.
NCERT-I-98
2 RBTS PAPER
160779
The force on a rocket moving with a velocity of $300 \mathrm{~m} / \mathrm{s}$ is $210 \mathrm{~N}$. Then the rate of combustion of the fuel will be
160780
A bullet of mass $200 \mathrm{~g}$ is fired at a speed of $5 \mathrm{~m} / \mathrm{s}$. The gun of mass $1 \mathrm{~kg}$ rebounds backward with a velocity of
1 $10 \mathrm{~m} / \mathrm{s}$
2 $-1 \mathrm{~m} / \mathrm{s}$
3 $0.1 \mathrm{~m} / \mathrm{s}$
4 $0.01 \mathrm{~m} / \mathrm{s}$
Explanation:
$\mathrm{m}_1$ - mass of bullet $=200 \mathrm{~g}=0.2 \mathrm{~kg}$ $m_2$ - mass of gun $=1 \mathrm{~kg}$ $\mathrm{V}_1$ - velocity of bullet $=5 \mathrm{~m} / \mathrm{s}$ $\mathrm{V}_2$ - velocity of gun $=$ ? According to law of conservation of linear momentum, Linear momentum $=$ Linear momentum before firing after firing $\begin{aligned} 0 & =\mathrm{m}_1 \mathrm{~V}_1+\mathrm{m}_2 \mathrm{~V}_2 \\ \mathrm{~V}_2 & =-\mathrm{m}_1 \mathrm{~V}_1 / \mathrm{m}_2 \\ & =-0.2 \times 5 \times 1 \\ \mathrm{~V}_2 & =-1 \mathrm{~m} / \mathrm{s} \end{aligned}$ Velocity of the gun $=-1 \mathrm{~m} / \mathrm{s}$
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 $\mathrm{mv} / 2$
2 $\mathrm{mv}$
3 $\mathrm{mv} / 3$
4 $2 \mathrm{mv}$
Explanation:
Impulse = Change in momentum $=2 \mathrm{mv} \cos 60^{\circ}=\mathrm{mv}$
NCERT-I-98
2 RBTS PAPER
160782
An object kept on a smooth inclined plane of an angle $\theta$ can be kept stationary relative to the incline by giving a horizontal acceleration to the inclined plane given by:
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
Explanation:
Hot gases with ingh velocity react against the rocket and(pondt up.
NCERT-I-98
2 RBTS PAPER
160779
The force on a rocket moving with a velocity of $300 \mathrm{~m} / \mathrm{s}$ is $210 \mathrm{~N}$. Then the rate of combustion of the fuel will be
160780
A bullet of mass $200 \mathrm{~g}$ is fired at a speed of $5 \mathrm{~m} / \mathrm{s}$. The gun of mass $1 \mathrm{~kg}$ rebounds backward with a velocity of
1 $10 \mathrm{~m} / \mathrm{s}$
2 $-1 \mathrm{~m} / \mathrm{s}$
3 $0.1 \mathrm{~m} / \mathrm{s}$
4 $0.01 \mathrm{~m} / \mathrm{s}$
Explanation:
$\mathrm{m}_1$ - mass of bullet $=200 \mathrm{~g}=0.2 \mathrm{~kg}$ $m_2$ - mass of gun $=1 \mathrm{~kg}$ $\mathrm{V}_1$ - velocity of bullet $=5 \mathrm{~m} / \mathrm{s}$ $\mathrm{V}_2$ - velocity of gun $=$ ? According to law of conservation of linear momentum, Linear momentum $=$ Linear momentum before firing after firing $\begin{aligned} 0 & =\mathrm{m}_1 \mathrm{~V}_1+\mathrm{m}_2 \mathrm{~V}_2 \\ \mathrm{~V}_2 & =-\mathrm{m}_1 \mathrm{~V}_1 / \mathrm{m}_2 \\ & =-0.2 \times 5 \times 1 \\ \mathrm{~V}_2 & =-1 \mathrm{~m} / \mathrm{s} \end{aligned}$ Velocity of the gun $=-1 \mathrm{~m} / \mathrm{s}$
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 $\mathrm{mv} / 2$
2 $\mathrm{mv}$
3 $\mathrm{mv} / 3$
4 $2 \mathrm{mv}$
Explanation:
Impulse = Change in momentum $=2 \mathrm{mv} \cos 60^{\circ}=\mathrm{mv}$
NCERT-I-98
2 RBTS PAPER
160782
An object kept on a smooth inclined plane of an angle $\theta$ can be kept stationary relative to the incline by giving a horizontal acceleration to the inclined plane given by:
NEET Test Series from KOTA - 10 Papers In MS WORD
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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
Explanation:
Hot gases with ingh velocity react against the rocket and(pondt up.
NCERT-I-98
2 RBTS PAPER
160779
The force on a rocket moving with a velocity of $300 \mathrm{~m} / \mathrm{s}$ is $210 \mathrm{~N}$. Then the rate of combustion of the fuel will be
160780
A bullet of mass $200 \mathrm{~g}$ is fired at a speed of $5 \mathrm{~m} / \mathrm{s}$. The gun of mass $1 \mathrm{~kg}$ rebounds backward with a velocity of
1 $10 \mathrm{~m} / \mathrm{s}$
2 $-1 \mathrm{~m} / \mathrm{s}$
3 $0.1 \mathrm{~m} / \mathrm{s}$
4 $0.01 \mathrm{~m} / \mathrm{s}$
Explanation:
$\mathrm{m}_1$ - mass of bullet $=200 \mathrm{~g}=0.2 \mathrm{~kg}$ $m_2$ - mass of gun $=1 \mathrm{~kg}$ $\mathrm{V}_1$ - velocity of bullet $=5 \mathrm{~m} / \mathrm{s}$ $\mathrm{V}_2$ - velocity of gun $=$ ? According to law of conservation of linear momentum, Linear momentum $=$ Linear momentum before firing after firing $\begin{aligned} 0 & =\mathrm{m}_1 \mathrm{~V}_1+\mathrm{m}_2 \mathrm{~V}_2 \\ \mathrm{~V}_2 & =-\mathrm{m}_1 \mathrm{~V}_1 / \mathrm{m}_2 \\ & =-0.2 \times 5 \times 1 \\ \mathrm{~V}_2 & =-1 \mathrm{~m} / \mathrm{s} \end{aligned}$ Velocity of the gun $=-1 \mathrm{~m} / \mathrm{s}$
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 $\mathrm{mv} / 2$
2 $\mathrm{mv}$
3 $\mathrm{mv} / 3$
4 $2 \mathrm{mv}$
Explanation:
Impulse = Change in momentum $=2 \mathrm{mv} \cos 60^{\circ}=\mathrm{mv}$
NCERT-I-98
2 RBTS PAPER
160782
An object kept on a smooth inclined plane of an angle $\theta$ can be kept stationary relative to the incline by giving a horizontal acceleration to the inclined plane given by:
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
Explanation:
Hot gases with ingh velocity react against the rocket and(pondt up.
NCERT-I-98
2 RBTS PAPER
160779
The force on a rocket moving with a velocity of $300 \mathrm{~m} / \mathrm{s}$ is $210 \mathrm{~N}$. Then the rate of combustion of the fuel will be
160780
A bullet of mass $200 \mathrm{~g}$ is fired at a speed of $5 \mathrm{~m} / \mathrm{s}$. The gun of mass $1 \mathrm{~kg}$ rebounds backward with a velocity of
1 $10 \mathrm{~m} / \mathrm{s}$
2 $-1 \mathrm{~m} / \mathrm{s}$
3 $0.1 \mathrm{~m} / \mathrm{s}$
4 $0.01 \mathrm{~m} / \mathrm{s}$
Explanation:
$\mathrm{m}_1$ - mass of bullet $=200 \mathrm{~g}=0.2 \mathrm{~kg}$ $m_2$ - mass of gun $=1 \mathrm{~kg}$ $\mathrm{V}_1$ - velocity of bullet $=5 \mathrm{~m} / \mathrm{s}$ $\mathrm{V}_2$ - velocity of gun $=$ ? According to law of conservation of linear momentum, Linear momentum $=$ Linear momentum before firing after firing $\begin{aligned} 0 & =\mathrm{m}_1 \mathrm{~V}_1+\mathrm{m}_2 \mathrm{~V}_2 \\ \mathrm{~V}_2 & =-\mathrm{m}_1 \mathrm{~V}_1 / \mathrm{m}_2 \\ & =-0.2 \times 5 \times 1 \\ \mathrm{~V}_2 & =-1 \mathrm{~m} / \mathrm{s} \end{aligned}$ Velocity of the gun $=-1 \mathrm{~m} / \mathrm{s}$
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 $\mathrm{mv} / 2$
2 $\mathrm{mv}$
3 $\mathrm{mv} / 3$
4 $2 \mathrm{mv}$
Explanation:
Impulse = Change in momentum $=2 \mathrm{mv} \cos 60^{\circ}=\mathrm{mv}$
NCERT-I-98
2 RBTS PAPER
160782
An object kept on a smooth inclined plane of an angle $\theta$ can be kept stationary relative to the incline by giving a horizontal acceleration to the inclined plane given by:
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
Explanation:
Hot gases with ingh velocity react against the rocket and(pondt up.
NCERT-I-98
2 RBTS PAPER
160779
The force on a rocket moving with a velocity of $300 \mathrm{~m} / \mathrm{s}$ is $210 \mathrm{~N}$. Then the rate of combustion of the fuel will be
160780
A bullet of mass $200 \mathrm{~g}$ is fired at a speed of $5 \mathrm{~m} / \mathrm{s}$. The gun of mass $1 \mathrm{~kg}$ rebounds backward with a velocity of
1 $10 \mathrm{~m} / \mathrm{s}$
2 $-1 \mathrm{~m} / \mathrm{s}$
3 $0.1 \mathrm{~m} / \mathrm{s}$
4 $0.01 \mathrm{~m} / \mathrm{s}$
Explanation:
$\mathrm{m}_1$ - mass of bullet $=200 \mathrm{~g}=0.2 \mathrm{~kg}$ $m_2$ - mass of gun $=1 \mathrm{~kg}$ $\mathrm{V}_1$ - velocity of bullet $=5 \mathrm{~m} / \mathrm{s}$ $\mathrm{V}_2$ - velocity of gun $=$ ? According to law of conservation of linear momentum, Linear momentum $=$ Linear momentum before firing after firing $\begin{aligned} 0 & =\mathrm{m}_1 \mathrm{~V}_1+\mathrm{m}_2 \mathrm{~V}_2 \\ \mathrm{~V}_2 & =-\mathrm{m}_1 \mathrm{~V}_1 / \mathrm{m}_2 \\ & =-0.2 \times 5 \times 1 \\ \mathrm{~V}_2 & =-1 \mathrm{~m} / \mathrm{s} \end{aligned}$ Velocity of the gun $=-1 \mathrm{~m} / \mathrm{s}$
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 $\mathrm{mv} / 2$
2 $\mathrm{mv}$
3 $\mathrm{mv} / 3$
4 $2 \mathrm{mv}$
Explanation:
Impulse = Change in momentum $=2 \mathrm{mv} \cos 60^{\circ}=\mathrm{mv}$
NCERT-I-98
2 RBTS PAPER
160782
An object kept on a smooth inclined plane of an angle $\theta$ can be kept stationary relative to the incline by giving a horizontal acceleration to the inclined plane given by:
