NEET Test Series from KOTA - 10 Papers In MS WORD
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Work, Energy and Power
148961
The energy required to accelerate a car from $10 \mathrm{~ms}^{-1}$ to $20 \mathrm{~ms}^{-1}$ is times the energy required to accelerate the car from rest to 10 $\mathrm{ms}^{-1}$
1 Equal
2 4 times
3 2 times
4 3 times
Explanation:
D Given, Energy required accelerating from $10 \mathrm{~m} / \mathrm{s}$ to $20 \mathrm{~m} / \mathrm{s}-$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m}(\Delta \mathrm{v})^{2}=\frac{1}{2} \mathrm{~m}\left(20^{2}-10^{2}\right)$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m}(400-100)$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m} \times 300$ $\mathrm{E}_{1} =150 \mathrm{~m}$ And energy required accelerating from rest to $10 \mathrm{~m} / \mathrm{s}$ - $\mathrm{E}_{2} =\frac{1}{2} \mathrm{~m}\left(10^{2}-0^{2}\right)$ $\mathrm{E}_{2} =\frac{1}{2} \mathrm{~m} \times 100$ $\mathrm{E}_{2} =50 \mathrm{~m}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} =\frac{150 \mathrm{~m}}{50 \mathrm{~m}}$ $\mathrm{E}_{1} =3 \mathrm{E}_{2}$
AP EAMCET (Medical)-07.10.2020
Work, Energy and Power
148962
The kinetic energy of an electron which is accelerated through a potential difference of $100 \mathrm{~V}$ will be
D Given, Charge of electron $(\mathrm{e})=1.6 \times 10^{-19} \mathrm{C}$ Potential difference $(\Delta \mathrm{V})=100 \mathrm{~V}$ We know, Energy acquired by electron when it is acceleration through a potential difference $(100 \mathrm{~V})$ So, kinetic energy, $\mathrm{KE}=\mathrm{E} =\mathrm{e} \times \Delta \mathrm{V}$ $=1.6 \times 10^{-19} \times 100$ $=1.6 \times 10^{-17} \mathrm{~J}$
AP EAMCET-24.09.2020
Work, Energy and Power
148963
A tangential force $F$ acts along the rim of a ring of radius $R$ and displaces the ring through an angle $\theta$. The work done by the force is :
1 $\mathrm{FR}^{2} \theta$
2 $\frac{\mathrm{FR}}{\theta}$
3 $\frac{\text { FR } \theta}{2}$
4 $\operatorname{FR} \theta$
Explanation:
D Given, $\text { Displacement }=\theta$ $\text { Arc length }=\mathrm{x}$ Then, $\quad \mathrm{x}=\mathrm{R} \theta$ We know, $\text { Work done }=\text { F.x }$ $=F(R \theta)$ Hence, work done by the force is FR $\theta$.
AP EAMCET-24.09.2020
Work, Energy and Power
148965
Two object of masses $m_{1}$ and $m_{2}$ posses equal kinetic energies. If $p_{1}$ and $p_{2}$ are their respective momentum, then $p_{1}: p_{2}$ is
1 $\mathrm{m}_{1}: \mathrm{m}_{2}$
2 $\mathrm{m}_{2}: \mathrm{m}_{1}$
3 $\sqrt{\mathrm{m}_{1}}: \sqrt{\mathrm{m}_{2}}$
4 $\mathrm{m}_{1}^{2}: \mathrm{m}_{2}^{2}$
Explanation:
C We know, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} =\frac{1}{2} \mathrm{mv}^{2} \times \frac{\mathrm{m}}{\mathrm{m}}=\frac{(\mathrm{mv})^{2}}{2 \mathrm{~m}}$ $\mathrm{KE} =\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ For $\mathrm{p}_{1}$ $\mathrm{KE}=\frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}_{1}}$ $\mathrm{p}_{1}=\sqrt{2 \mathrm{~m}_{1} \mathrm{KE}}$ For $\mathrm{p}_{2}$, $\mathrm{KE}=\frac{\mathrm{p}_{2}^{2}}{2 \mathrm{~m}_{2}}$ $\mathrm{p}_{2}=\sqrt{2 \mathrm{~m}_{2} \mathrm{KE}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\sqrt{\frac{2 \mathrm{~m}_{1} \mathrm{KE}}{2 \mathrm{~m}_{2} \mathrm{KE}}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}}$ Thus the momentum ratio, $\mathrm{p}_{1}: \mathrm{p}_{2}=\sqrt{\mathrm{m}_{1}}: \sqrt{\mathrm{m}_{2}}$
148961
The energy required to accelerate a car from $10 \mathrm{~ms}^{-1}$ to $20 \mathrm{~ms}^{-1}$ is times the energy required to accelerate the car from rest to 10 $\mathrm{ms}^{-1}$
1 Equal
2 4 times
3 2 times
4 3 times
Explanation:
D Given, Energy required accelerating from $10 \mathrm{~m} / \mathrm{s}$ to $20 \mathrm{~m} / \mathrm{s}-$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m}(\Delta \mathrm{v})^{2}=\frac{1}{2} \mathrm{~m}\left(20^{2}-10^{2}\right)$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m}(400-100)$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m} \times 300$ $\mathrm{E}_{1} =150 \mathrm{~m}$ And energy required accelerating from rest to $10 \mathrm{~m} / \mathrm{s}$ - $\mathrm{E}_{2} =\frac{1}{2} \mathrm{~m}\left(10^{2}-0^{2}\right)$ $\mathrm{E}_{2} =\frac{1}{2} \mathrm{~m} \times 100$ $\mathrm{E}_{2} =50 \mathrm{~m}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} =\frac{150 \mathrm{~m}}{50 \mathrm{~m}}$ $\mathrm{E}_{1} =3 \mathrm{E}_{2}$
AP EAMCET (Medical)-07.10.2020
Work, Energy and Power
148962
The kinetic energy of an electron which is accelerated through a potential difference of $100 \mathrm{~V}$ will be
D Given, Charge of electron $(\mathrm{e})=1.6 \times 10^{-19} \mathrm{C}$ Potential difference $(\Delta \mathrm{V})=100 \mathrm{~V}$ We know, Energy acquired by electron when it is acceleration through a potential difference $(100 \mathrm{~V})$ So, kinetic energy, $\mathrm{KE}=\mathrm{E} =\mathrm{e} \times \Delta \mathrm{V}$ $=1.6 \times 10^{-19} \times 100$ $=1.6 \times 10^{-17} \mathrm{~J}$
AP EAMCET-24.09.2020
Work, Energy and Power
148963
A tangential force $F$ acts along the rim of a ring of radius $R$ and displaces the ring through an angle $\theta$. The work done by the force is :
1 $\mathrm{FR}^{2} \theta$
2 $\frac{\mathrm{FR}}{\theta}$
3 $\frac{\text { FR } \theta}{2}$
4 $\operatorname{FR} \theta$
Explanation:
D Given, $\text { Displacement }=\theta$ $\text { Arc length }=\mathrm{x}$ Then, $\quad \mathrm{x}=\mathrm{R} \theta$ We know, $\text { Work done }=\text { F.x }$ $=F(R \theta)$ Hence, work done by the force is FR $\theta$.
AP EAMCET-24.09.2020
Work, Energy and Power
148965
Two object of masses $m_{1}$ and $m_{2}$ posses equal kinetic energies. If $p_{1}$ and $p_{2}$ are their respective momentum, then $p_{1}: p_{2}$ is
1 $\mathrm{m}_{1}: \mathrm{m}_{2}$
2 $\mathrm{m}_{2}: \mathrm{m}_{1}$
3 $\sqrt{\mathrm{m}_{1}}: \sqrt{\mathrm{m}_{2}}$
4 $\mathrm{m}_{1}^{2}: \mathrm{m}_{2}^{2}$
Explanation:
C We know, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} =\frac{1}{2} \mathrm{mv}^{2} \times \frac{\mathrm{m}}{\mathrm{m}}=\frac{(\mathrm{mv})^{2}}{2 \mathrm{~m}}$ $\mathrm{KE} =\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ For $\mathrm{p}_{1}$ $\mathrm{KE}=\frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}_{1}}$ $\mathrm{p}_{1}=\sqrt{2 \mathrm{~m}_{1} \mathrm{KE}}$ For $\mathrm{p}_{2}$, $\mathrm{KE}=\frac{\mathrm{p}_{2}^{2}}{2 \mathrm{~m}_{2}}$ $\mathrm{p}_{2}=\sqrt{2 \mathrm{~m}_{2} \mathrm{KE}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\sqrt{\frac{2 \mathrm{~m}_{1} \mathrm{KE}}{2 \mathrm{~m}_{2} \mathrm{KE}}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}}$ Thus the momentum ratio, $\mathrm{p}_{1}: \mathrm{p}_{2}=\sqrt{\mathrm{m}_{1}}: \sqrt{\mathrm{m}_{2}}$
148961
The energy required to accelerate a car from $10 \mathrm{~ms}^{-1}$ to $20 \mathrm{~ms}^{-1}$ is times the energy required to accelerate the car from rest to 10 $\mathrm{ms}^{-1}$
1 Equal
2 4 times
3 2 times
4 3 times
Explanation:
D Given, Energy required accelerating from $10 \mathrm{~m} / \mathrm{s}$ to $20 \mathrm{~m} / \mathrm{s}-$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m}(\Delta \mathrm{v})^{2}=\frac{1}{2} \mathrm{~m}\left(20^{2}-10^{2}\right)$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m}(400-100)$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m} \times 300$ $\mathrm{E}_{1} =150 \mathrm{~m}$ And energy required accelerating from rest to $10 \mathrm{~m} / \mathrm{s}$ - $\mathrm{E}_{2} =\frac{1}{2} \mathrm{~m}\left(10^{2}-0^{2}\right)$ $\mathrm{E}_{2} =\frac{1}{2} \mathrm{~m} \times 100$ $\mathrm{E}_{2} =50 \mathrm{~m}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} =\frac{150 \mathrm{~m}}{50 \mathrm{~m}}$ $\mathrm{E}_{1} =3 \mathrm{E}_{2}$
AP EAMCET (Medical)-07.10.2020
Work, Energy and Power
148962
The kinetic energy of an electron which is accelerated through a potential difference of $100 \mathrm{~V}$ will be
D Given, Charge of electron $(\mathrm{e})=1.6 \times 10^{-19} \mathrm{C}$ Potential difference $(\Delta \mathrm{V})=100 \mathrm{~V}$ We know, Energy acquired by electron when it is acceleration through a potential difference $(100 \mathrm{~V})$ So, kinetic energy, $\mathrm{KE}=\mathrm{E} =\mathrm{e} \times \Delta \mathrm{V}$ $=1.6 \times 10^{-19} \times 100$ $=1.6 \times 10^{-17} \mathrm{~J}$
AP EAMCET-24.09.2020
Work, Energy and Power
148963
A tangential force $F$ acts along the rim of a ring of radius $R$ and displaces the ring through an angle $\theta$. The work done by the force is :
1 $\mathrm{FR}^{2} \theta$
2 $\frac{\mathrm{FR}}{\theta}$
3 $\frac{\text { FR } \theta}{2}$
4 $\operatorname{FR} \theta$
Explanation:
D Given, $\text { Displacement }=\theta$ $\text { Arc length }=\mathrm{x}$ Then, $\quad \mathrm{x}=\mathrm{R} \theta$ We know, $\text { Work done }=\text { F.x }$ $=F(R \theta)$ Hence, work done by the force is FR $\theta$.
AP EAMCET-24.09.2020
Work, Energy and Power
148965
Two object of masses $m_{1}$ and $m_{2}$ posses equal kinetic energies. If $p_{1}$ and $p_{2}$ are their respective momentum, then $p_{1}: p_{2}$ is
1 $\mathrm{m}_{1}: \mathrm{m}_{2}$
2 $\mathrm{m}_{2}: \mathrm{m}_{1}$
3 $\sqrt{\mathrm{m}_{1}}: \sqrt{\mathrm{m}_{2}}$
4 $\mathrm{m}_{1}^{2}: \mathrm{m}_{2}^{2}$
Explanation:
C We know, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} =\frac{1}{2} \mathrm{mv}^{2} \times \frac{\mathrm{m}}{\mathrm{m}}=\frac{(\mathrm{mv})^{2}}{2 \mathrm{~m}}$ $\mathrm{KE} =\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ For $\mathrm{p}_{1}$ $\mathrm{KE}=\frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}_{1}}$ $\mathrm{p}_{1}=\sqrt{2 \mathrm{~m}_{1} \mathrm{KE}}$ For $\mathrm{p}_{2}$, $\mathrm{KE}=\frac{\mathrm{p}_{2}^{2}}{2 \mathrm{~m}_{2}}$ $\mathrm{p}_{2}=\sqrt{2 \mathrm{~m}_{2} \mathrm{KE}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\sqrt{\frac{2 \mathrm{~m}_{1} \mathrm{KE}}{2 \mathrm{~m}_{2} \mathrm{KE}}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}}$ Thus the momentum ratio, $\mathrm{p}_{1}: \mathrm{p}_{2}=\sqrt{\mathrm{m}_{1}}: \sqrt{\mathrm{m}_{2}}$
148961
The energy required to accelerate a car from $10 \mathrm{~ms}^{-1}$ to $20 \mathrm{~ms}^{-1}$ is times the energy required to accelerate the car from rest to 10 $\mathrm{ms}^{-1}$
1 Equal
2 4 times
3 2 times
4 3 times
Explanation:
D Given, Energy required accelerating from $10 \mathrm{~m} / \mathrm{s}$ to $20 \mathrm{~m} / \mathrm{s}-$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m}(\Delta \mathrm{v})^{2}=\frac{1}{2} \mathrm{~m}\left(20^{2}-10^{2}\right)$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m}(400-100)$ $\mathrm{E}_{1} =\frac{1}{2} \mathrm{~m} \times 300$ $\mathrm{E}_{1} =150 \mathrm{~m}$ And energy required accelerating from rest to $10 \mathrm{~m} / \mathrm{s}$ - $\mathrm{E}_{2} =\frac{1}{2} \mathrm{~m}\left(10^{2}-0^{2}\right)$ $\mathrm{E}_{2} =\frac{1}{2} \mathrm{~m} \times 100$ $\mathrm{E}_{2} =50 \mathrm{~m}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} =\frac{150 \mathrm{~m}}{50 \mathrm{~m}}$ $\mathrm{E}_{1} =3 \mathrm{E}_{2}$
AP EAMCET (Medical)-07.10.2020
Work, Energy and Power
148962
The kinetic energy of an electron which is accelerated through a potential difference of $100 \mathrm{~V}$ will be
D Given, Charge of electron $(\mathrm{e})=1.6 \times 10^{-19} \mathrm{C}$ Potential difference $(\Delta \mathrm{V})=100 \mathrm{~V}$ We know, Energy acquired by electron when it is acceleration through a potential difference $(100 \mathrm{~V})$ So, kinetic energy, $\mathrm{KE}=\mathrm{E} =\mathrm{e} \times \Delta \mathrm{V}$ $=1.6 \times 10^{-19} \times 100$ $=1.6 \times 10^{-17} \mathrm{~J}$
AP EAMCET-24.09.2020
Work, Energy and Power
148963
A tangential force $F$ acts along the rim of a ring of radius $R$ and displaces the ring through an angle $\theta$. The work done by the force is :
1 $\mathrm{FR}^{2} \theta$
2 $\frac{\mathrm{FR}}{\theta}$
3 $\frac{\text { FR } \theta}{2}$
4 $\operatorname{FR} \theta$
Explanation:
D Given, $\text { Displacement }=\theta$ $\text { Arc length }=\mathrm{x}$ Then, $\quad \mathrm{x}=\mathrm{R} \theta$ We know, $\text { Work done }=\text { F.x }$ $=F(R \theta)$ Hence, work done by the force is FR $\theta$.
AP EAMCET-24.09.2020
Work, Energy and Power
148965
Two object of masses $m_{1}$ and $m_{2}$ posses equal kinetic energies. If $p_{1}$ and $p_{2}$ are their respective momentum, then $p_{1}: p_{2}$ is
1 $\mathrm{m}_{1}: \mathrm{m}_{2}$
2 $\mathrm{m}_{2}: \mathrm{m}_{1}$
3 $\sqrt{\mathrm{m}_{1}}: \sqrt{\mathrm{m}_{2}}$
4 $\mathrm{m}_{1}^{2}: \mathrm{m}_{2}^{2}$
Explanation:
C We know, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} =\frac{1}{2} \mathrm{mv}^{2} \times \frac{\mathrm{m}}{\mathrm{m}}=\frac{(\mathrm{mv})^{2}}{2 \mathrm{~m}}$ $\mathrm{KE} =\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ For $\mathrm{p}_{1}$ $\mathrm{KE}=\frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}_{1}}$ $\mathrm{p}_{1}=\sqrt{2 \mathrm{~m}_{1} \mathrm{KE}}$ For $\mathrm{p}_{2}$, $\mathrm{KE}=\frac{\mathrm{p}_{2}^{2}}{2 \mathrm{~m}_{2}}$ $\mathrm{p}_{2}=\sqrt{2 \mathrm{~m}_{2} \mathrm{KE}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\sqrt{\frac{2 \mathrm{~m}_{1} \mathrm{KE}}{2 \mathrm{~m}_{2} \mathrm{KE}}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}}$ Thus the momentum ratio, $\mathrm{p}_{1}: \mathrm{p}_{2}=\sqrt{\mathrm{m}_{1}}: \sqrt{\mathrm{m}_{2}}$
