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 \(\mathrm{mg}\) in magnitude
2 constant and greater than \(\mathrm{mg}\) in magnitude
3 variable but always greater than \(\mathrm{mg}\)
4 at first greater than \(\mathrm{mg}\), and later becomes equla to \(\mathrm{mg}\)
Explanation:
At first greater than \(\mathrm{mg}\), and later becomes equal to \(\mathrm{mg}\)
NCERT Examplar- WPE-3
3 RBTS PAPER
162585
A body of mass \(1 \mathrm{~kg}\) is thrown upwards with a velocity \(20 \mathrm{~ms}^{-1}\). It momentarily comes to rest after attaining a height of \(18 \mathrm{~m}\). How much energy is lost due to air friction? \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
1 \(20 \mathrm{~J}\)
2 \(30 \mathrm{~J}\)
3 \(40 \mathrm{~J}\)
4 \(10 \mathrm{~J}\)
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 \(=\mathrm{KE}-\mathrm{PE}\) \( =\frac{1}{2} m v^2-m g h \) \( =\frac{1}{2} \times 1 \times 400-1 \times 18 \times 10 \) \( =200-180=20 \mathrm{~J} \)
NCERT-I-78
3 RBTS PAPER
162586
A ball moving with velocities \(2 \mathbf{~ m s}^{-1}\) collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 , then their velocities (in \(\mathrm{ms}^{-1}\) ) 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 \( m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \) \( \Rightarrow & v_1=\left u_1+\left u_2 \)
Substituting \(\mathrm{u}_1=2 \mathrm{~ms}^{-1}, \mathrm{u}_2=0, \mathrm{~m}_1=\mathrm{m}\) and \(\mathrm{m}_2=2 \mathrm{~m}, \mathrm{e}=0.5\)
We get \(v_1=\frac{m-m}{m+2 m} \times 2\) \( \Rightarrow \mathrm{v}_1=0 \)
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.
Explanation:
Since the balls are in contact in the deformed state, their total energy is in the form of potential energy and kinetic energy, so kinetic energy is less than the total energy. The energy spent against friction is disspated in the form of heat which is not available for doing work.
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 \(\mathrm{mg}\) in magnitude
2 constant and greater than \(\mathrm{mg}\) in magnitude
3 variable but always greater than \(\mathrm{mg}\)
4 at first greater than \(\mathrm{mg}\), and later becomes equla to \(\mathrm{mg}\)
Explanation:
At first greater than \(\mathrm{mg}\), and later becomes equal to \(\mathrm{mg}\)
NCERT Examplar- WPE-3
3 RBTS PAPER
162585
A body of mass \(1 \mathrm{~kg}\) is thrown upwards with a velocity \(20 \mathrm{~ms}^{-1}\). It momentarily comes to rest after attaining a height of \(18 \mathrm{~m}\). How much energy is lost due to air friction? \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
1 \(20 \mathrm{~J}\)
2 \(30 \mathrm{~J}\)
3 \(40 \mathrm{~J}\)
4 \(10 \mathrm{~J}\)
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 \(=\mathrm{KE}-\mathrm{PE}\) \( =\frac{1}{2} m v^2-m g h \) \( =\frac{1}{2} \times 1 \times 400-1 \times 18 \times 10 \) \( =200-180=20 \mathrm{~J} \)
NCERT-I-78
3 RBTS PAPER
162586
A ball moving with velocities \(2 \mathbf{~ m s}^{-1}\) collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 , then their velocities (in \(\mathrm{ms}^{-1}\) ) 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 \( m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \) \( \Rightarrow & v_1=\left u_1+\left u_2 \)
Substituting \(\mathrm{u}_1=2 \mathrm{~ms}^{-1}, \mathrm{u}_2=0, \mathrm{~m}_1=\mathrm{m}\) and \(\mathrm{m}_2=2 \mathrm{~m}, \mathrm{e}=0.5\)
We get \(v_1=\frac{m-m}{m+2 m} \times 2\) \( \Rightarrow \mathrm{v}_1=0 \)
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.
Explanation:
Since the balls are in contact in the deformed state, their total energy is in the form of potential energy and kinetic energy, so kinetic energy is less than the total energy. The energy spent against friction is disspated in the form of heat which is not available for doing work.
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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 \(\mathrm{mg}\) in magnitude
2 constant and greater than \(\mathrm{mg}\) in magnitude
3 variable but always greater than \(\mathrm{mg}\)
4 at first greater than \(\mathrm{mg}\), and later becomes equla to \(\mathrm{mg}\)
Explanation:
At first greater than \(\mathrm{mg}\), and later becomes equal to \(\mathrm{mg}\)
NCERT Examplar- WPE-3
3 RBTS PAPER
162585
A body of mass \(1 \mathrm{~kg}\) is thrown upwards with a velocity \(20 \mathrm{~ms}^{-1}\). It momentarily comes to rest after attaining a height of \(18 \mathrm{~m}\). How much energy is lost due to air friction? \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
1 \(20 \mathrm{~J}\)
2 \(30 \mathrm{~J}\)
3 \(40 \mathrm{~J}\)
4 \(10 \mathrm{~J}\)
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 \(=\mathrm{KE}-\mathrm{PE}\) \( =\frac{1}{2} m v^2-m g h \) \( =\frac{1}{2} \times 1 \times 400-1 \times 18 \times 10 \) \( =200-180=20 \mathrm{~J} \)
NCERT-I-78
3 RBTS PAPER
162586
A ball moving with velocities \(2 \mathbf{~ m s}^{-1}\) collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 , then their velocities (in \(\mathrm{ms}^{-1}\) ) 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 \( m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \) \( \Rightarrow & v_1=\left u_1+\left u_2 \)
Substituting \(\mathrm{u}_1=2 \mathrm{~ms}^{-1}, \mathrm{u}_2=0, \mathrm{~m}_1=\mathrm{m}\) and \(\mathrm{m}_2=2 \mathrm{~m}, \mathrm{e}=0.5\)
We get \(v_1=\frac{m-m}{m+2 m} \times 2\) \( \Rightarrow \mathrm{v}_1=0 \)
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.
Explanation:
Since the balls are in contact in the deformed state, their total energy is in the form of potential energy and kinetic energy, so kinetic energy is less than the total energy. The energy spent against friction is disspated in the form of heat which is not available for doing work.
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 \(\mathrm{mg}\) in magnitude
2 constant and greater than \(\mathrm{mg}\) in magnitude
3 variable but always greater than \(\mathrm{mg}\)
4 at first greater than \(\mathrm{mg}\), and later becomes equla to \(\mathrm{mg}\)
Explanation:
At first greater than \(\mathrm{mg}\), and later becomes equal to \(\mathrm{mg}\)
NCERT Examplar- WPE-3
3 RBTS PAPER
162585
A body of mass \(1 \mathrm{~kg}\) is thrown upwards with a velocity \(20 \mathrm{~ms}^{-1}\). It momentarily comes to rest after attaining a height of \(18 \mathrm{~m}\). How much energy is lost due to air friction? \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
1 \(20 \mathrm{~J}\)
2 \(30 \mathrm{~J}\)
3 \(40 \mathrm{~J}\)
4 \(10 \mathrm{~J}\)
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 \(=\mathrm{KE}-\mathrm{PE}\) \( =\frac{1}{2} m v^2-m g h \) \( =\frac{1}{2} \times 1 \times 400-1 \times 18 \times 10 \) \( =200-180=20 \mathrm{~J} \)
NCERT-I-78
3 RBTS PAPER
162586
A ball moving with velocities \(2 \mathbf{~ m s}^{-1}\) collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 , then their velocities (in \(\mathrm{ms}^{-1}\) ) 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 \( m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \) \( \Rightarrow & v_1=\left u_1+\left u_2 \)
Substituting \(\mathrm{u}_1=2 \mathrm{~ms}^{-1}, \mathrm{u}_2=0, \mathrm{~m}_1=\mathrm{m}\) and \(\mathrm{m}_2=2 \mathrm{~m}, \mathrm{e}=0.5\)
We get \(v_1=\frac{m-m}{m+2 m} \times 2\) \( \Rightarrow \mathrm{v}_1=0 \)
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.
Explanation:
Since the balls are in contact in the deformed state, their total energy is in the form of potential energy and kinetic energy, so kinetic energy is less than the total energy. The energy spent against friction is disspated in the form of heat which is not available for doing work.