355412
When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be
1 \(60 \%\)
2 \(500 \%\)
3 \(6 \%\)
4 \(600 \%\)
Explanation:
\(K E_{2}=36 K E_{1}\) Relation between \(K E\) and momentum is, \(p=\sqrt{2 m K E} \Rightarrow p \propto \sqrt{K E}\) \(\dfrac{p_{2}}{p_{1}}=\sqrt{\dfrac{K E_{2}}{K E_{1}}} \Rightarrow \sqrt{\dfrac{36 K E_{2}}{K E_{1}}}=6\) \(\Rightarrow p_{2}=6 p_{1}\) So, \(\dfrac{p_{2}-p_{1}}{p_{1}} \times 100 \%=\dfrac{6 p_{1}-p_{1}}{p_{1}} \times 100=500 \%\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355413
When a man increases his speed by \(2\,m{s^{ - 1}}\), he finds that his kinetic energy is doubled, the original speed of the man is
1 \(2(\sqrt 2 + 1)m{s^{ - 1}}\)
2 \(4.5\;m{s^{ - 1}}\)
3 \(2(\sqrt 2 - 1)m{s^{ - 1}}\)
4 None of these
Explanation:
Man possesses kinetic energy, because of his velocity \((v)\). When \(m\) is mass of man, then \(K=\dfrac{1}{2} m v^{2}\) Let \(v_{1}=v, m_{1}=m_{2}=m\) \(v_{2}=(v+2) m s^{-1}\) Then \(K_{2}=2 K_{1}\) Then, \(\dfrac{K_{1}}{K_{2}}=\dfrac{v_{1}^{2}}{v_{2}^{2}} \Rightarrow \dfrac{K_{1}}{2 K_{1}}=\dfrac{v^{2}}{(v+2)^{2}}\) \(\begin{gathered}\Rightarrow v^{2}-4 v-4=0 \\\Rightarrow v_{1}=\dfrac{4+\sqrt{16+16}}{2} \\v_{1}=2(\sqrt{2}+1) m s^{-1}\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355414
If \(\vec{P}\) and \(\vec{Q}\) are the nonzero positive vectors making angle of \(30^{\circ}\) with each other, then the angle (in degrees) between additive inverse of \(\vec{P}\) with \(\overrightarrow{\mathrm{Q}}\) will be
1 150
2 130
3 120
4 170
Explanation:
If \(\vec P\) is multiplied by negative real number, then we get a vector acting in the opposite direction of \(\vec P\) i.e., \(-\vec P\).
PHXI06:WORK ENERGY AND POWER
355415
An athlete in the olympic games covers a distance of 100 \(m\) in 10 \(s\). His kinetic energy can be estimated to be in the range
1 \(2,000\;J - 5,000\;J\)
2 \(200\;J - 500\;J\)
3 \(20,000\;J - 50,000\;J\)
4 \(2 \times {10^5}J - 3 \times {10^5}J\)
Explanation:
The average speed of the athelete \(\begin{aligned}& v=\dfrac{100}{10}=10 m / s \\& \therefore K . E=\dfrac{1}{2} m v^{2}\end{aligned}\) If mass is \(40 \mathrm{~kg}\) then, \(K . E=\dfrac{1}{2} \times 40 \times(10)^{2}=2000 J\) If mass is \(100 \mathrm{~kg}\) then, \(K . E=\dfrac{1}{2} \times 100 \times(10)^{2}=5000 J\)
PHXI06:WORK ENERGY AND POWER
355416
The sum of the magnitudes of two forces acting at a point is \(8\;N.\) The resultant of these forces is perpendicular to the smaller force and has a magnitude of \(4\;N.\) If the smaller force is of magnitude \(x\), then the value of \(x\) is
355412
When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be
1 \(60 \%\)
2 \(500 \%\)
3 \(6 \%\)
4 \(600 \%\)
Explanation:
\(K E_{2}=36 K E_{1}\) Relation between \(K E\) and momentum is, \(p=\sqrt{2 m K E} \Rightarrow p \propto \sqrt{K E}\) \(\dfrac{p_{2}}{p_{1}}=\sqrt{\dfrac{K E_{2}}{K E_{1}}} \Rightarrow \sqrt{\dfrac{36 K E_{2}}{K E_{1}}}=6\) \(\Rightarrow p_{2}=6 p_{1}\) So, \(\dfrac{p_{2}-p_{1}}{p_{1}} \times 100 \%=\dfrac{6 p_{1}-p_{1}}{p_{1}} \times 100=500 \%\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355413
When a man increases his speed by \(2\,m{s^{ - 1}}\), he finds that his kinetic energy is doubled, the original speed of the man is
1 \(2(\sqrt 2 + 1)m{s^{ - 1}}\)
2 \(4.5\;m{s^{ - 1}}\)
3 \(2(\sqrt 2 - 1)m{s^{ - 1}}\)
4 None of these
Explanation:
Man possesses kinetic energy, because of his velocity \((v)\). When \(m\) is mass of man, then \(K=\dfrac{1}{2} m v^{2}\) Let \(v_{1}=v, m_{1}=m_{2}=m\) \(v_{2}=(v+2) m s^{-1}\) Then \(K_{2}=2 K_{1}\) Then, \(\dfrac{K_{1}}{K_{2}}=\dfrac{v_{1}^{2}}{v_{2}^{2}} \Rightarrow \dfrac{K_{1}}{2 K_{1}}=\dfrac{v^{2}}{(v+2)^{2}}\) \(\begin{gathered}\Rightarrow v^{2}-4 v-4=0 \\\Rightarrow v_{1}=\dfrac{4+\sqrt{16+16}}{2} \\v_{1}=2(\sqrt{2}+1) m s^{-1}\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355414
If \(\vec{P}\) and \(\vec{Q}\) are the nonzero positive vectors making angle of \(30^{\circ}\) with each other, then the angle (in degrees) between additive inverse of \(\vec{P}\) with \(\overrightarrow{\mathrm{Q}}\) will be
1 150
2 130
3 120
4 170
Explanation:
If \(\vec P\) is multiplied by negative real number, then we get a vector acting in the opposite direction of \(\vec P\) i.e., \(-\vec P\).
PHXI06:WORK ENERGY AND POWER
355415
An athlete in the olympic games covers a distance of 100 \(m\) in 10 \(s\). His kinetic energy can be estimated to be in the range
1 \(2,000\;J - 5,000\;J\)
2 \(200\;J - 500\;J\)
3 \(20,000\;J - 50,000\;J\)
4 \(2 \times {10^5}J - 3 \times {10^5}J\)
Explanation:
The average speed of the athelete \(\begin{aligned}& v=\dfrac{100}{10}=10 m / s \\& \therefore K . E=\dfrac{1}{2} m v^{2}\end{aligned}\) If mass is \(40 \mathrm{~kg}\) then, \(K . E=\dfrac{1}{2} \times 40 \times(10)^{2}=2000 J\) If mass is \(100 \mathrm{~kg}\) then, \(K . E=\dfrac{1}{2} \times 100 \times(10)^{2}=5000 J\)
PHXI06:WORK ENERGY AND POWER
355416
The sum of the magnitudes of two forces acting at a point is \(8\;N.\) The resultant of these forces is perpendicular to the smaller force and has a magnitude of \(4\;N.\) If the smaller force is of magnitude \(x\), then the value of \(x\) is
355412
When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be
1 \(60 \%\)
2 \(500 \%\)
3 \(6 \%\)
4 \(600 \%\)
Explanation:
\(K E_{2}=36 K E_{1}\) Relation between \(K E\) and momentum is, \(p=\sqrt{2 m K E} \Rightarrow p \propto \sqrt{K E}\) \(\dfrac{p_{2}}{p_{1}}=\sqrt{\dfrac{K E_{2}}{K E_{1}}} \Rightarrow \sqrt{\dfrac{36 K E_{2}}{K E_{1}}}=6\) \(\Rightarrow p_{2}=6 p_{1}\) So, \(\dfrac{p_{2}-p_{1}}{p_{1}} \times 100 \%=\dfrac{6 p_{1}-p_{1}}{p_{1}} \times 100=500 \%\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355413
When a man increases his speed by \(2\,m{s^{ - 1}}\), he finds that his kinetic energy is doubled, the original speed of the man is
1 \(2(\sqrt 2 + 1)m{s^{ - 1}}\)
2 \(4.5\;m{s^{ - 1}}\)
3 \(2(\sqrt 2 - 1)m{s^{ - 1}}\)
4 None of these
Explanation:
Man possesses kinetic energy, because of his velocity \((v)\). When \(m\) is mass of man, then \(K=\dfrac{1}{2} m v^{2}\) Let \(v_{1}=v, m_{1}=m_{2}=m\) \(v_{2}=(v+2) m s^{-1}\) Then \(K_{2}=2 K_{1}\) Then, \(\dfrac{K_{1}}{K_{2}}=\dfrac{v_{1}^{2}}{v_{2}^{2}} \Rightarrow \dfrac{K_{1}}{2 K_{1}}=\dfrac{v^{2}}{(v+2)^{2}}\) \(\begin{gathered}\Rightarrow v^{2}-4 v-4=0 \\\Rightarrow v_{1}=\dfrac{4+\sqrt{16+16}}{2} \\v_{1}=2(\sqrt{2}+1) m s^{-1}\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355414
If \(\vec{P}\) and \(\vec{Q}\) are the nonzero positive vectors making angle of \(30^{\circ}\) with each other, then the angle (in degrees) between additive inverse of \(\vec{P}\) with \(\overrightarrow{\mathrm{Q}}\) will be
1 150
2 130
3 120
4 170
Explanation:
If \(\vec P\) is multiplied by negative real number, then we get a vector acting in the opposite direction of \(\vec P\) i.e., \(-\vec P\).
PHXI06:WORK ENERGY AND POWER
355415
An athlete in the olympic games covers a distance of 100 \(m\) in 10 \(s\). His kinetic energy can be estimated to be in the range
1 \(2,000\;J - 5,000\;J\)
2 \(200\;J - 500\;J\)
3 \(20,000\;J - 50,000\;J\)
4 \(2 \times {10^5}J - 3 \times {10^5}J\)
Explanation:
The average speed of the athelete \(\begin{aligned}& v=\dfrac{100}{10}=10 m / s \\& \therefore K . E=\dfrac{1}{2} m v^{2}\end{aligned}\) If mass is \(40 \mathrm{~kg}\) then, \(K . E=\dfrac{1}{2} \times 40 \times(10)^{2}=2000 J\) If mass is \(100 \mathrm{~kg}\) then, \(K . E=\dfrac{1}{2} \times 100 \times(10)^{2}=5000 J\)
PHXI06:WORK ENERGY AND POWER
355416
The sum of the magnitudes of two forces acting at a point is \(8\;N.\) The resultant of these forces is perpendicular to the smaller force and has a magnitude of \(4\;N.\) If the smaller force is of magnitude \(x\), then the value of \(x\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI06:WORK ENERGY AND POWER
355412
When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be
1 \(60 \%\)
2 \(500 \%\)
3 \(6 \%\)
4 \(600 \%\)
Explanation:
\(K E_{2}=36 K E_{1}\) Relation between \(K E\) and momentum is, \(p=\sqrt{2 m K E} \Rightarrow p \propto \sqrt{K E}\) \(\dfrac{p_{2}}{p_{1}}=\sqrt{\dfrac{K E_{2}}{K E_{1}}} \Rightarrow \sqrt{\dfrac{36 K E_{2}}{K E_{1}}}=6\) \(\Rightarrow p_{2}=6 p_{1}\) So, \(\dfrac{p_{2}-p_{1}}{p_{1}} \times 100 \%=\dfrac{6 p_{1}-p_{1}}{p_{1}} \times 100=500 \%\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355413
When a man increases his speed by \(2\,m{s^{ - 1}}\), he finds that his kinetic energy is doubled, the original speed of the man is
1 \(2(\sqrt 2 + 1)m{s^{ - 1}}\)
2 \(4.5\;m{s^{ - 1}}\)
3 \(2(\sqrt 2 - 1)m{s^{ - 1}}\)
4 None of these
Explanation:
Man possesses kinetic energy, because of his velocity \((v)\). When \(m\) is mass of man, then \(K=\dfrac{1}{2} m v^{2}\) Let \(v_{1}=v, m_{1}=m_{2}=m\) \(v_{2}=(v+2) m s^{-1}\) Then \(K_{2}=2 K_{1}\) Then, \(\dfrac{K_{1}}{K_{2}}=\dfrac{v_{1}^{2}}{v_{2}^{2}} \Rightarrow \dfrac{K_{1}}{2 K_{1}}=\dfrac{v^{2}}{(v+2)^{2}}\) \(\begin{gathered}\Rightarrow v^{2}-4 v-4=0 \\\Rightarrow v_{1}=\dfrac{4+\sqrt{16+16}}{2} \\v_{1}=2(\sqrt{2}+1) m s^{-1}\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355414
If \(\vec{P}\) and \(\vec{Q}\) are the nonzero positive vectors making angle of \(30^{\circ}\) with each other, then the angle (in degrees) between additive inverse of \(\vec{P}\) with \(\overrightarrow{\mathrm{Q}}\) will be
1 150
2 130
3 120
4 170
Explanation:
If \(\vec P\) is multiplied by negative real number, then we get a vector acting in the opposite direction of \(\vec P\) i.e., \(-\vec P\).
PHXI06:WORK ENERGY AND POWER
355415
An athlete in the olympic games covers a distance of 100 \(m\) in 10 \(s\). His kinetic energy can be estimated to be in the range
1 \(2,000\;J - 5,000\;J\)
2 \(200\;J - 500\;J\)
3 \(20,000\;J - 50,000\;J\)
4 \(2 \times {10^5}J - 3 \times {10^5}J\)
Explanation:
The average speed of the athelete \(\begin{aligned}& v=\dfrac{100}{10}=10 m / s \\& \therefore K . E=\dfrac{1}{2} m v^{2}\end{aligned}\) If mass is \(40 \mathrm{~kg}\) then, \(K . E=\dfrac{1}{2} \times 40 \times(10)^{2}=2000 J\) If mass is \(100 \mathrm{~kg}\) then, \(K . E=\dfrac{1}{2} \times 100 \times(10)^{2}=5000 J\)
PHXI06:WORK ENERGY AND POWER
355416
The sum of the magnitudes of two forces acting at a point is \(8\;N.\) The resultant of these forces is perpendicular to the smaller force and has a magnitude of \(4\;N.\) If the smaller force is of magnitude \(x\), then the value of \(x\) is
355412
When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be
1 \(60 \%\)
2 \(500 \%\)
3 \(6 \%\)
4 \(600 \%\)
Explanation:
\(K E_{2}=36 K E_{1}\) Relation between \(K E\) and momentum is, \(p=\sqrt{2 m K E} \Rightarrow p \propto \sqrt{K E}\) \(\dfrac{p_{2}}{p_{1}}=\sqrt{\dfrac{K E_{2}}{K E_{1}}} \Rightarrow \sqrt{\dfrac{36 K E_{2}}{K E_{1}}}=6\) \(\Rightarrow p_{2}=6 p_{1}\) So, \(\dfrac{p_{2}-p_{1}}{p_{1}} \times 100 \%=\dfrac{6 p_{1}-p_{1}}{p_{1}} \times 100=500 \%\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355413
When a man increases his speed by \(2\,m{s^{ - 1}}\), he finds that his kinetic energy is doubled, the original speed of the man is
1 \(2(\sqrt 2 + 1)m{s^{ - 1}}\)
2 \(4.5\;m{s^{ - 1}}\)
3 \(2(\sqrt 2 - 1)m{s^{ - 1}}\)
4 None of these
Explanation:
Man possesses kinetic energy, because of his velocity \((v)\). When \(m\) is mass of man, then \(K=\dfrac{1}{2} m v^{2}\) Let \(v_{1}=v, m_{1}=m_{2}=m\) \(v_{2}=(v+2) m s^{-1}\) Then \(K_{2}=2 K_{1}\) Then, \(\dfrac{K_{1}}{K_{2}}=\dfrac{v_{1}^{2}}{v_{2}^{2}} \Rightarrow \dfrac{K_{1}}{2 K_{1}}=\dfrac{v^{2}}{(v+2)^{2}}\) \(\begin{gathered}\Rightarrow v^{2}-4 v-4=0 \\\Rightarrow v_{1}=\dfrac{4+\sqrt{16+16}}{2} \\v_{1}=2(\sqrt{2}+1) m s^{-1}\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355414
If \(\vec{P}\) and \(\vec{Q}\) are the nonzero positive vectors making angle of \(30^{\circ}\) with each other, then the angle (in degrees) between additive inverse of \(\vec{P}\) with \(\overrightarrow{\mathrm{Q}}\) will be
1 150
2 130
3 120
4 170
Explanation:
If \(\vec P\) is multiplied by negative real number, then we get a vector acting in the opposite direction of \(\vec P\) i.e., \(-\vec P\).
PHXI06:WORK ENERGY AND POWER
355415
An athlete in the olympic games covers a distance of 100 \(m\) in 10 \(s\). His kinetic energy can be estimated to be in the range
1 \(2,000\;J - 5,000\;J\)
2 \(200\;J - 500\;J\)
3 \(20,000\;J - 50,000\;J\)
4 \(2 \times {10^5}J - 3 \times {10^5}J\)
Explanation:
The average speed of the athelete \(\begin{aligned}& v=\dfrac{100}{10}=10 m / s \\& \therefore K . E=\dfrac{1}{2} m v^{2}\end{aligned}\) If mass is \(40 \mathrm{~kg}\) then, \(K . E=\dfrac{1}{2} \times 40 \times(10)^{2}=2000 J\) If mass is \(100 \mathrm{~kg}\) then, \(K . E=\dfrac{1}{2} \times 100 \times(10)^{2}=5000 J\)
PHXI06:WORK ENERGY AND POWER
355416
The sum of the magnitudes of two forces acting at a point is \(8\;N.\) The resultant of these forces is perpendicular to the smaller force and has a magnitude of \(4\;N.\) If the smaller force is of magnitude \(x\), then the value of \(x\) is
