NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI05:LAWS OF MOTION
363545
A force \(F_{1}\) of \(500\,N\) is required to push a car of mass \(1000\,kg\) slowly at constant speed on a levelled road. If a force \(F_{2}\) of \(1000\,N\) is applied, then the acceleration of the car will be
1 zero
2 \(1.5\,\,\;m{s^{ - 2}}\)
3 \(1\;\,\,m{s^{ - 2}}\)
4 \(0.5\,\,m{s^{ - 2}}\)
Explanation:
We know that a force of \(500 N\) does not produce any acceleration as it pushes the car slowly at constant speed. Out of \(1000\,N\) force, only \(500\;N\) produces acceleration, thus acceleration produced, \(a = \frac{{{\text{ force }}}}{{{\text{ mass }}}} = \frac{{500}}{{1000}} = 0.5\;m{s^{ - 2}}\)
PHXI05:LAWS OF MOTION
363546
A body of mass \(2\,kg\) has an initial velocity of \(3\;m{s^{ - 1}}\) along \(O E\) and it is subjected to a force of \(4 N\) in a direction perpendicular to \(O E\). The distance of body from \(O\) after \(4 s\) will be.
1 \(12\;\,m\)
2 \(20\;\,m\)
3 \(8\,\,m\)
4 \(48\,\,m\)
Explanation:
The acceleration of the body perpendicular to \(O E\) is a
363547
The position vector of a particle related to time \(t\) is given by \(\vec{r}=\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right) m\).The direction of net force experienced by the particle is
1 positive \(y\)-axis
2 positive \(x\)-axis
3 positive \(z\)-axis
4 in \(x-y\) plane.
Explanation:
Velocity of the particle, \(\vec{v}=\dfrac{d \vec{r}}{d t}=\dfrac{d}{d t}\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right)\) or \(\quad \vec{v}=10 \hat{i}+30 t \hat{j}\) Acceleration of the particle, \(\vec{a}=\dfrac{d \vec{v}}{d t}=30 \hat{j}\) The particle has acceleration only along positive \(y\)-axis. According to Newton's second law, net force must also acts along positive \(y\)-axis.
JEE - 2023
PHXI05:LAWS OF MOTION
363548
A cricket player catches a ball of mass \(120\,g\) moving with \(25\,\;m/s\) speed. If the catching process is completed in \(0.1 s\) then the magnitude of force exerted by the ball on the hand of player will be (in \(S I\) unit)
1 30
2 24
3 12
4 25
Explanation:
Force exerted by the ball on the hand of player, \(F = \frac{{m|v - u|}}{t} = \frac{{120 \times {{10}^{ - 3}} \times 25}}{{0.1}}\) \( = 300 \times {10^{ - 1}} = 30\;N\)
363545
A force \(F_{1}\) of \(500\,N\) is required to push a car of mass \(1000\,kg\) slowly at constant speed on a levelled road. If a force \(F_{2}\) of \(1000\,N\) is applied, then the acceleration of the car will be
1 zero
2 \(1.5\,\,\;m{s^{ - 2}}\)
3 \(1\;\,\,m{s^{ - 2}}\)
4 \(0.5\,\,m{s^{ - 2}}\)
Explanation:
We know that a force of \(500 N\) does not produce any acceleration as it pushes the car slowly at constant speed. Out of \(1000\,N\) force, only \(500\;N\) produces acceleration, thus acceleration produced, \(a = \frac{{{\text{ force }}}}{{{\text{ mass }}}} = \frac{{500}}{{1000}} = 0.5\;m{s^{ - 2}}\)
PHXI05:LAWS OF MOTION
363546
A body of mass \(2\,kg\) has an initial velocity of \(3\;m{s^{ - 1}}\) along \(O E\) and it is subjected to a force of \(4 N\) in a direction perpendicular to \(O E\). The distance of body from \(O\) after \(4 s\) will be.
1 \(12\;\,m\)
2 \(20\;\,m\)
3 \(8\,\,m\)
4 \(48\,\,m\)
Explanation:
The acceleration of the body perpendicular to \(O E\) is a
363547
The position vector of a particle related to time \(t\) is given by \(\vec{r}=\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right) m\).The direction of net force experienced by the particle is
1 positive \(y\)-axis
2 positive \(x\)-axis
3 positive \(z\)-axis
4 in \(x-y\) plane.
Explanation:
Velocity of the particle, \(\vec{v}=\dfrac{d \vec{r}}{d t}=\dfrac{d}{d t}\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right)\) or \(\quad \vec{v}=10 \hat{i}+30 t \hat{j}\) Acceleration of the particle, \(\vec{a}=\dfrac{d \vec{v}}{d t}=30 \hat{j}\) The particle has acceleration only along positive \(y\)-axis. According to Newton's second law, net force must also acts along positive \(y\)-axis.
JEE - 2023
PHXI05:LAWS OF MOTION
363548
A cricket player catches a ball of mass \(120\,g\) moving with \(25\,\;m/s\) speed. If the catching process is completed in \(0.1 s\) then the magnitude of force exerted by the ball on the hand of player will be (in \(S I\) unit)
1 30
2 24
3 12
4 25
Explanation:
Force exerted by the ball on the hand of player, \(F = \frac{{m|v - u|}}{t} = \frac{{120 \times {{10}^{ - 3}} \times 25}}{{0.1}}\) \( = 300 \times {10^{ - 1}} = 30\;N\)
363545
A force \(F_{1}\) of \(500\,N\) is required to push a car of mass \(1000\,kg\) slowly at constant speed on a levelled road. If a force \(F_{2}\) of \(1000\,N\) is applied, then the acceleration of the car will be
1 zero
2 \(1.5\,\,\;m{s^{ - 2}}\)
3 \(1\;\,\,m{s^{ - 2}}\)
4 \(0.5\,\,m{s^{ - 2}}\)
Explanation:
We know that a force of \(500 N\) does not produce any acceleration as it pushes the car slowly at constant speed. Out of \(1000\,N\) force, only \(500\;N\) produces acceleration, thus acceleration produced, \(a = \frac{{{\text{ force }}}}{{{\text{ mass }}}} = \frac{{500}}{{1000}} = 0.5\;m{s^{ - 2}}\)
PHXI05:LAWS OF MOTION
363546
A body of mass \(2\,kg\) has an initial velocity of \(3\;m{s^{ - 1}}\) along \(O E\) and it is subjected to a force of \(4 N\) in a direction perpendicular to \(O E\). The distance of body from \(O\) after \(4 s\) will be.
1 \(12\;\,m\)
2 \(20\;\,m\)
3 \(8\,\,m\)
4 \(48\,\,m\)
Explanation:
The acceleration of the body perpendicular to \(O E\) is a
363547
The position vector of a particle related to time \(t\) is given by \(\vec{r}=\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right) m\).The direction of net force experienced by the particle is
1 positive \(y\)-axis
2 positive \(x\)-axis
3 positive \(z\)-axis
4 in \(x-y\) plane.
Explanation:
Velocity of the particle, \(\vec{v}=\dfrac{d \vec{r}}{d t}=\dfrac{d}{d t}\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right)\) or \(\quad \vec{v}=10 \hat{i}+30 t \hat{j}\) Acceleration of the particle, \(\vec{a}=\dfrac{d \vec{v}}{d t}=30 \hat{j}\) The particle has acceleration only along positive \(y\)-axis. According to Newton's second law, net force must also acts along positive \(y\)-axis.
JEE - 2023
PHXI05:LAWS OF MOTION
363548
A cricket player catches a ball of mass \(120\,g\) moving with \(25\,\;m/s\) speed. If the catching process is completed in \(0.1 s\) then the magnitude of force exerted by the ball on the hand of player will be (in \(S I\) unit)
1 30
2 24
3 12
4 25
Explanation:
Force exerted by the ball on the hand of player, \(F = \frac{{m|v - u|}}{t} = \frac{{120 \times {{10}^{ - 3}} \times 25}}{{0.1}}\) \( = 300 \times {10^{ - 1}} = 30\;N\)
363545
A force \(F_{1}\) of \(500\,N\) is required to push a car of mass \(1000\,kg\) slowly at constant speed on a levelled road. If a force \(F_{2}\) of \(1000\,N\) is applied, then the acceleration of the car will be
1 zero
2 \(1.5\,\,\;m{s^{ - 2}}\)
3 \(1\;\,\,m{s^{ - 2}}\)
4 \(0.5\,\,m{s^{ - 2}}\)
Explanation:
We know that a force of \(500 N\) does not produce any acceleration as it pushes the car slowly at constant speed. Out of \(1000\,N\) force, only \(500\;N\) produces acceleration, thus acceleration produced, \(a = \frac{{{\text{ force }}}}{{{\text{ mass }}}} = \frac{{500}}{{1000}} = 0.5\;m{s^{ - 2}}\)
PHXI05:LAWS OF MOTION
363546
A body of mass \(2\,kg\) has an initial velocity of \(3\;m{s^{ - 1}}\) along \(O E\) and it is subjected to a force of \(4 N\) in a direction perpendicular to \(O E\). The distance of body from \(O\) after \(4 s\) will be.
1 \(12\;\,m\)
2 \(20\;\,m\)
3 \(8\,\,m\)
4 \(48\,\,m\)
Explanation:
The acceleration of the body perpendicular to \(O E\) is a
363547
The position vector of a particle related to time \(t\) is given by \(\vec{r}=\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right) m\).The direction of net force experienced by the particle is
1 positive \(y\)-axis
2 positive \(x\)-axis
3 positive \(z\)-axis
4 in \(x-y\) plane.
Explanation:
Velocity of the particle, \(\vec{v}=\dfrac{d \vec{r}}{d t}=\dfrac{d}{d t}\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right)\) or \(\quad \vec{v}=10 \hat{i}+30 t \hat{j}\) Acceleration of the particle, \(\vec{a}=\dfrac{d \vec{v}}{d t}=30 \hat{j}\) The particle has acceleration only along positive \(y\)-axis. According to Newton's second law, net force must also acts along positive \(y\)-axis.
JEE - 2023
PHXI05:LAWS OF MOTION
363548
A cricket player catches a ball of mass \(120\,g\) moving with \(25\,\;m/s\) speed. If the catching process is completed in \(0.1 s\) then the magnitude of force exerted by the ball on the hand of player will be (in \(S I\) unit)
1 30
2 24
3 12
4 25
Explanation:
Force exerted by the ball on the hand of player, \(F = \frac{{m|v - u|}}{t} = \frac{{120 \times {{10}^{ - 3}} \times 25}}{{0.1}}\) \( = 300 \times {10^{ - 1}} = 30\;N\)
