NEET Test Series from KOTA - 10 Papers In MS WORD
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3 RBTS PAPER
162580
A mass of \(20 \mathrm{~kg}\) moving with a speed of \(10 \mathrm{~m} / \mathrm{s}\) collides with another stationary mass of \(5 \mathrm{~kg}\). As a result of the collision, the two masses stick together. The kinetic energy of the composite mass will be:
1 \(1200 \mathrm{~J}\)
2 \(600 \mathrm{~J}\)
3 \(800 \mathrm{~J}\)
4 \(1000 \mathrm{~J}\)
Explanation:
As there is no external non-conservative force on the system (both masses altogether) the linear momentum should be conserved. Thus we should have \(m_1 v_1+m_2 v_2=\left(m_1+m_2\right) V\) (inelastic collision) \(\mathrm{V}\) is the common velocity after the collision. putting values we get \(20(10)+5(0)=(20+5) V\) or \(V=8 \mathrm{~m} / \mathrm{s}\) so the new KE will be \( K=\frac{1}{2} m V^2=(0.5)(25)(64)=800 J \)
NCERT-I-84
3 RBTS PAPER
162581
Assertion (A): Inelastic collision of two bodies he momentum and energy of each bodws conserved. Reason (R): If two bodies stick to each other yfter colliding, the collision is said t/ be perfectly inelastic.
1 Both \(A\) and \(R\) are true and is the correct explanation of \(A\).
2 Both \(A\) and \(R\) are truebur is not the correct explanation of \(A\).
3 \(A\) is true but \(R\).
4 \(\mathrm{A}\) is false byt
Explanation:
In an elastic collision both the momentum and kinetic energy remains conserved. This rule is not for individual bodies but for the centre of mass of the system of bodies before and after the collision. The collision is perfectly in elastic, if two bodies stick to each other.
NCERT-I-84
3 RBTS PAPER
162582
Initially mas \(1>\) held such that spring is in relaxed condinin. If mass \(m\) is suddenly released, maximury Elopngation in spring will be :
1 \(\frac{m g}{K}\)
2 \(\frac{2 m g}{K}\)
3 \(\frac{m g}{2 K}\)
4 \(\frac{m g}{4 K}\)
Explanation:
Since block is suddenly released so \( \frac{1}{2} k x^2=m g x \) \(x=\frac{2 m g}{K}\).
NCERT-I-82
3 RBTS PAPER
162583
If distance is plotted against \(x\)-axis and kinetic energy against \(y\)-axis for an object moving along \(x\)-axis, then the slope of the graph so obtained is proportional to :
1 Distance
2 Kinetic energy
3 Velocity
4 Acceleration
Explanation:
Acceleration
From work energy theorem \(\Delta \mathrm{K}=\mathrm{W}\), \( \Delta \mathrm{K}=\mathrm{F} . \Delta \mathrm{x} \) \(\frac{\Delta K}{\Delta x}=F\), \( \frac{\Delta K}{\Delta x}=m a \) \(\frac{\Delta K}{\Delta x} \propto a\) \(\tan \theta \propto a\) where slope \(m \propto \tan \theta\}\).
162580
A mass of \(20 \mathrm{~kg}\) moving with a speed of \(10 \mathrm{~m} / \mathrm{s}\) collides with another stationary mass of \(5 \mathrm{~kg}\). As a result of the collision, the two masses stick together. The kinetic energy of the composite mass will be:
1 \(1200 \mathrm{~J}\)
2 \(600 \mathrm{~J}\)
3 \(800 \mathrm{~J}\)
4 \(1000 \mathrm{~J}\)
Explanation:
As there is no external non-conservative force on the system (both masses altogether) the linear momentum should be conserved. Thus we should have \(m_1 v_1+m_2 v_2=\left(m_1+m_2\right) V\) (inelastic collision) \(\mathrm{V}\) is the common velocity after the collision. putting values we get \(20(10)+5(0)=(20+5) V\) or \(V=8 \mathrm{~m} / \mathrm{s}\) so the new KE will be \( K=\frac{1}{2} m V^2=(0.5)(25)(64)=800 J \)
NCERT-I-84
3 RBTS PAPER
162581
Assertion (A): Inelastic collision of two bodies he momentum and energy of each bodws conserved. Reason (R): If two bodies stick to each other yfter colliding, the collision is said t/ be perfectly inelastic.
1 Both \(A\) and \(R\) are true and is the correct explanation of \(A\).
2 Both \(A\) and \(R\) are truebur is not the correct explanation of \(A\).
3 \(A\) is true but \(R\).
4 \(\mathrm{A}\) is false byt
Explanation:
In an elastic collision both the momentum and kinetic energy remains conserved. This rule is not for individual bodies but for the centre of mass of the system of bodies before and after the collision. The collision is perfectly in elastic, if two bodies stick to each other.
NCERT-I-84
3 RBTS PAPER
162582
Initially mas \(1>\) held such that spring is in relaxed condinin. If mass \(m\) is suddenly released, maximury Elopngation in spring will be :
1 \(\frac{m g}{K}\)
2 \(\frac{2 m g}{K}\)
3 \(\frac{m g}{2 K}\)
4 \(\frac{m g}{4 K}\)
Explanation:
Since block is suddenly released so \( \frac{1}{2} k x^2=m g x \) \(x=\frac{2 m g}{K}\).
NCERT-I-82
3 RBTS PAPER
162583
If distance is plotted against \(x\)-axis and kinetic energy against \(y\)-axis for an object moving along \(x\)-axis, then the slope of the graph so obtained is proportional to :
1 Distance
2 Kinetic energy
3 Velocity
4 Acceleration
Explanation:
Acceleration
From work energy theorem \(\Delta \mathrm{K}=\mathrm{W}\), \( \Delta \mathrm{K}=\mathrm{F} . \Delta \mathrm{x} \) \(\frac{\Delta K}{\Delta x}=F\), \( \frac{\Delta K}{\Delta x}=m a \) \(\frac{\Delta K}{\Delta x} \propto a\) \(\tan \theta \propto a\) where slope \(m \propto \tan \theta\}\).
162580
A mass of \(20 \mathrm{~kg}\) moving with a speed of \(10 \mathrm{~m} / \mathrm{s}\) collides with another stationary mass of \(5 \mathrm{~kg}\). As a result of the collision, the two masses stick together. The kinetic energy of the composite mass will be:
1 \(1200 \mathrm{~J}\)
2 \(600 \mathrm{~J}\)
3 \(800 \mathrm{~J}\)
4 \(1000 \mathrm{~J}\)
Explanation:
As there is no external non-conservative force on the system (both masses altogether) the linear momentum should be conserved. Thus we should have \(m_1 v_1+m_2 v_2=\left(m_1+m_2\right) V\) (inelastic collision) \(\mathrm{V}\) is the common velocity after the collision. putting values we get \(20(10)+5(0)=(20+5) V\) or \(V=8 \mathrm{~m} / \mathrm{s}\) so the new KE will be \( K=\frac{1}{2} m V^2=(0.5)(25)(64)=800 J \)
NCERT-I-84
3 RBTS PAPER
162581
Assertion (A): Inelastic collision of two bodies he momentum and energy of each bodws conserved. Reason (R): If two bodies stick to each other yfter colliding, the collision is said t/ be perfectly inelastic.
1 Both \(A\) and \(R\) are true and is the correct explanation of \(A\).
2 Both \(A\) and \(R\) are truebur is not the correct explanation of \(A\).
3 \(A\) is true but \(R\).
4 \(\mathrm{A}\) is false byt
Explanation:
In an elastic collision both the momentum and kinetic energy remains conserved. This rule is not for individual bodies but for the centre of mass of the system of bodies before and after the collision. The collision is perfectly in elastic, if two bodies stick to each other.
NCERT-I-84
3 RBTS PAPER
162582
Initially mas \(1>\) held such that spring is in relaxed condinin. If mass \(m\) is suddenly released, maximury Elopngation in spring will be :
1 \(\frac{m g}{K}\)
2 \(\frac{2 m g}{K}\)
3 \(\frac{m g}{2 K}\)
4 \(\frac{m g}{4 K}\)
Explanation:
Since block is suddenly released so \( \frac{1}{2} k x^2=m g x \) \(x=\frac{2 m g}{K}\).
NCERT-I-82
3 RBTS PAPER
162583
If distance is plotted against \(x\)-axis and kinetic energy against \(y\)-axis for an object moving along \(x\)-axis, then the slope of the graph so obtained is proportional to :
1 Distance
2 Kinetic energy
3 Velocity
4 Acceleration
Explanation:
Acceleration
From work energy theorem \(\Delta \mathrm{K}=\mathrm{W}\), \( \Delta \mathrm{K}=\mathrm{F} . \Delta \mathrm{x} \) \(\frac{\Delta K}{\Delta x}=F\), \( \frac{\Delta K}{\Delta x}=m a \) \(\frac{\Delta K}{\Delta x} \propto a\) \(\tan \theta \propto a\) where slope \(m \propto \tan \theta\}\).
162580
A mass of \(20 \mathrm{~kg}\) moving with a speed of \(10 \mathrm{~m} / \mathrm{s}\) collides with another stationary mass of \(5 \mathrm{~kg}\). As a result of the collision, the two masses stick together. The kinetic energy of the composite mass will be:
1 \(1200 \mathrm{~J}\)
2 \(600 \mathrm{~J}\)
3 \(800 \mathrm{~J}\)
4 \(1000 \mathrm{~J}\)
Explanation:
As there is no external non-conservative force on the system (both masses altogether) the linear momentum should be conserved. Thus we should have \(m_1 v_1+m_2 v_2=\left(m_1+m_2\right) V\) (inelastic collision) \(\mathrm{V}\) is the common velocity after the collision. putting values we get \(20(10)+5(0)=(20+5) V\) or \(V=8 \mathrm{~m} / \mathrm{s}\) so the new KE will be \( K=\frac{1}{2} m V^2=(0.5)(25)(64)=800 J \)
NCERT-I-84
3 RBTS PAPER
162581
Assertion (A): Inelastic collision of two bodies he momentum and energy of each bodws conserved. Reason (R): If two bodies stick to each other yfter colliding, the collision is said t/ be perfectly inelastic.
1 Both \(A\) and \(R\) are true and is the correct explanation of \(A\).
2 Both \(A\) and \(R\) are truebur is not the correct explanation of \(A\).
3 \(A\) is true but \(R\).
4 \(\mathrm{A}\) is false byt
Explanation:
In an elastic collision both the momentum and kinetic energy remains conserved. This rule is not for individual bodies but for the centre of mass of the system of bodies before and after the collision. The collision is perfectly in elastic, if two bodies stick to each other.
NCERT-I-84
3 RBTS PAPER
162582
Initially mas \(1>\) held such that spring is in relaxed condinin. If mass \(m\) is suddenly released, maximury Elopngation in spring will be :
1 \(\frac{m g}{K}\)
2 \(\frac{2 m g}{K}\)
3 \(\frac{m g}{2 K}\)
4 \(\frac{m g}{4 K}\)
Explanation:
Since block is suddenly released so \( \frac{1}{2} k x^2=m g x \) \(x=\frac{2 m g}{K}\).
NCERT-I-82
3 RBTS PAPER
162583
If distance is plotted against \(x\)-axis and kinetic energy against \(y\)-axis for an object moving along \(x\)-axis, then the slope of the graph so obtained is proportional to :
1 Distance
2 Kinetic energy
3 Velocity
4 Acceleration
Explanation:
Acceleration
From work energy theorem \(\Delta \mathrm{K}=\mathrm{W}\), \( \Delta \mathrm{K}=\mathrm{F} . \Delta \mathrm{x} \) \(\frac{\Delta K}{\Delta x}=F\), \( \frac{\Delta K}{\Delta x}=m a \) \(\frac{\Delta K}{\Delta x} \propto a\) \(\tan \theta \propto a\) where slope \(m \propto \tan \theta\}\).
