NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Work, Energy and Power
149015
A particle with total energy $E$ is moving in a potential energy region u(x). Motion of the particle is restricted to the region when
1 u(x) $ \lt $ E
2 $\mathrm{u}(\mathrm{x})=0$
3 $\mathrm{u}(\mathrm{x}) \leq \mathrm{E}$
4 $u(x)>E$
Explanation:
C Given, A particle with total energy $\mathrm{E}$ is moving in a potential energy region $\mathrm{u}(\mathrm{x})$. We know that, Total Energy $=$ Potential energy + Kinetic energy $\mathrm{E}=\mathrm{u}(\mathrm{x})+\mathrm{KE}(\mathrm{x})$ $\mathrm{KE}(\mathrm{x})=\mathrm{E}-\mathrm{u}(\mathrm{x})$ It means when particle is moving kinetic energy must be positive. Then, $\mathrm{KE}(\mathrm{x}) \geq 0$ $E-u(x)=K E(x) \geq 0$ $E \geq u(x)$ Therefore, $\mathrm{u}(\mathrm{x}) \leq \mathrm{E}$
COMEDK 2013
Work, Energy and Power
149016
A bomb of mass $18 \mathrm{~kg}$ at rest explodes into two pieces of masses $6 \mathrm{~kg}$ and $12 \mathrm{~kg}$. The velocity of $12 \mathrm{~kg}$ mass is $4 \mathrm{~m} / \mathrm{s}$. The kinetic energy of the other mass is
1 $288 \mathrm{~J}$
2 $192 \mathrm{~J}$
3 $96 \mathrm{~J}$
4 $144 \mathrm{~J}$
Explanation:
B Given, Mass of bomb $(\mathrm{m})=18 \mathrm{~kg}$ Velocity of $18 \mathrm{~kg}$ bomb (v) $=0 \mathrm{~m} / \mathrm{s}$ Mass of second bomb $\left(\mathrm{m}_{2}\right)=12 \mathrm{~kg}$ Velocity of second bomb $\left(\mathrm{v}_{2}\right)=4 \mathrm{~m} / \mathrm{sec}$ Mass of first bomb $\left(\mathrm{m}_{1}\right)=6 \mathrm{~kg}$ Velocity of first bomb $\left(\mathrm{v}_{1}\right)=$ ? By using law of conservation of momentum, $\mathrm{mv}= \mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}$ $18 \times 0 =\left(6 \times \mathrm{v}_{1}\right)+(12 \times 4)$ $0 =6 \mathrm{v}_{1}+48$ $6 \mathrm{v}_{1} =-48$ $\mathrm{v}_{1} =-\frac{48}{6}$ $\mathrm{v}_{1} =-8 \mathrm{~m} / \mathrm{sec}$ Kinetic energy of $6 \mathrm{~kg}$ mass (K.E.) $=\frac{1}{2} \mathrm{~m}_{1}\left(\mathrm{v}_{1}\right)^{2}$ $\text { K.E. }=\frac{1}{2} \times 6 \times(-8)^{2}$ $\text { K.E. }=\frac{1}{2} \times 6 \times 8 \times 8$ $\text { K.E. }=192 \text { Joule }$
COMEDK 2015
Work, Energy and Power
149017
Assertion: If collision occurs between two elastic bodies their kinetic energy decreases during the time of collision. Reason: During collision intermolecular space decreases and hence elastic potential energy increases.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A If collision occurs between two elastic bodies then kinetic energy of bodies decreases because, at time of collision intermolecular space of bodies decrease. Intermolecular attraction and repulsion lead to change in potential energy. It means, decreasing the intermolecular space lead to decrease the elastic potential.
AIIMS-2011
Work, Energy and Power
149018
Assertion: $n$ small balls each of mass $m$ colliding elastically each second on surface with velocity $u$. The force experienced by the surface is $2 \mathrm{mnu}$. Reason: On elastic collision, the ball rebounds with the same velocity.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A In elastic collision, kinetic energy remains conserved. Therefore, the ball rebounds with the same velocity. Given, Mass $=\mathrm{m} \mathrm{kg}$, velocity $=\mathrm{u} \mathrm{m} / \mathrm{s}$ We know, In elastic collision change in momentum $\Delta \mathrm{p} =\mathrm{mu}-(-\mathrm{mu})$ $=2 \mathrm{mu}$ For $\mathrm{n}$ balls, $\Delta \mathrm{p}=\mathrm{n}(2 \mathrm{mu})=2 \mathrm{n} \mathrm{mu}$ According to Newton's second law - $\mathrm{F}=\frac{\Delta \mathrm{p}}{\mathrm{t}}=\frac{2 \mathrm{nmu}}{1}$ $\mathrm{~F}=2 \mathrm{nmu}$
149015
A particle with total energy $E$ is moving in a potential energy region u(x). Motion of the particle is restricted to the region when
1 u(x) $ \lt $ E
2 $\mathrm{u}(\mathrm{x})=0$
3 $\mathrm{u}(\mathrm{x}) \leq \mathrm{E}$
4 $u(x)>E$
Explanation:
C Given, A particle with total energy $\mathrm{E}$ is moving in a potential energy region $\mathrm{u}(\mathrm{x})$. We know that, Total Energy $=$ Potential energy + Kinetic energy $\mathrm{E}=\mathrm{u}(\mathrm{x})+\mathrm{KE}(\mathrm{x})$ $\mathrm{KE}(\mathrm{x})=\mathrm{E}-\mathrm{u}(\mathrm{x})$ It means when particle is moving kinetic energy must be positive. Then, $\mathrm{KE}(\mathrm{x}) \geq 0$ $E-u(x)=K E(x) \geq 0$ $E \geq u(x)$ Therefore, $\mathrm{u}(\mathrm{x}) \leq \mathrm{E}$
COMEDK 2013
Work, Energy and Power
149016
A bomb of mass $18 \mathrm{~kg}$ at rest explodes into two pieces of masses $6 \mathrm{~kg}$ and $12 \mathrm{~kg}$. The velocity of $12 \mathrm{~kg}$ mass is $4 \mathrm{~m} / \mathrm{s}$. The kinetic energy of the other mass is
1 $288 \mathrm{~J}$
2 $192 \mathrm{~J}$
3 $96 \mathrm{~J}$
4 $144 \mathrm{~J}$
Explanation:
B Given, Mass of bomb $(\mathrm{m})=18 \mathrm{~kg}$ Velocity of $18 \mathrm{~kg}$ bomb (v) $=0 \mathrm{~m} / \mathrm{s}$ Mass of second bomb $\left(\mathrm{m}_{2}\right)=12 \mathrm{~kg}$ Velocity of second bomb $\left(\mathrm{v}_{2}\right)=4 \mathrm{~m} / \mathrm{sec}$ Mass of first bomb $\left(\mathrm{m}_{1}\right)=6 \mathrm{~kg}$ Velocity of first bomb $\left(\mathrm{v}_{1}\right)=$ ? By using law of conservation of momentum, $\mathrm{mv}= \mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}$ $18 \times 0 =\left(6 \times \mathrm{v}_{1}\right)+(12 \times 4)$ $0 =6 \mathrm{v}_{1}+48$ $6 \mathrm{v}_{1} =-48$ $\mathrm{v}_{1} =-\frac{48}{6}$ $\mathrm{v}_{1} =-8 \mathrm{~m} / \mathrm{sec}$ Kinetic energy of $6 \mathrm{~kg}$ mass (K.E.) $=\frac{1}{2} \mathrm{~m}_{1}\left(\mathrm{v}_{1}\right)^{2}$ $\text { K.E. }=\frac{1}{2} \times 6 \times(-8)^{2}$ $\text { K.E. }=\frac{1}{2} \times 6 \times 8 \times 8$ $\text { K.E. }=192 \text { Joule }$
COMEDK 2015
Work, Energy and Power
149017
Assertion: If collision occurs between two elastic bodies their kinetic energy decreases during the time of collision. Reason: During collision intermolecular space decreases and hence elastic potential energy increases.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A If collision occurs between two elastic bodies then kinetic energy of bodies decreases because, at time of collision intermolecular space of bodies decrease. Intermolecular attraction and repulsion lead to change in potential energy. It means, decreasing the intermolecular space lead to decrease the elastic potential.
AIIMS-2011
Work, Energy and Power
149018
Assertion: $n$ small balls each of mass $m$ colliding elastically each second on surface with velocity $u$. The force experienced by the surface is $2 \mathrm{mnu}$. Reason: On elastic collision, the ball rebounds with the same velocity.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A In elastic collision, kinetic energy remains conserved. Therefore, the ball rebounds with the same velocity. Given, Mass $=\mathrm{m} \mathrm{kg}$, velocity $=\mathrm{u} \mathrm{m} / \mathrm{s}$ We know, In elastic collision change in momentum $\Delta \mathrm{p} =\mathrm{mu}-(-\mathrm{mu})$ $=2 \mathrm{mu}$ For $\mathrm{n}$ balls, $\Delta \mathrm{p}=\mathrm{n}(2 \mathrm{mu})=2 \mathrm{n} \mathrm{mu}$ According to Newton's second law - $\mathrm{F}=\frac{\Delta \mathrm{p}}{\mathrm{t}}=\frac{2 \mathrm{nmu}}{1}$ $\mathrm{~F}=2 \mathrm{nmu}$
149015
A particle with total energy $E$ is moving in a potential energy region u(x). Motion of the particle is restricted to the region when
1 u(x) $ \lt $ E
2 $\mathrm{u}(\mathrm{x})=0$
3 $\mathrm{u}(\mathrm{x}) \leq \mathrm{E}$
4 $u(x)>E$
Explanation:
C Given, A particle with total energy $\mathrm{E}$ is moving in a potential energy region $\mathrm{u}(\mathrm{x})$. We know that, Total Energy $=$ Potential energy + Kinetic energy $\mathrm{E}=\mathrm{u}(\mathrm{x})+\mathrm{KE}(\mathrm{x})$ $\mathrm{KE}(\mathrm{x})=\mathrm{E}-\mathrm{u}(\mathrm{x})$ It means when particle is moving kinetic energy must be positive. Then, $\mathrm{KE}(\mathrm{x}) \geq 0$ $E-u(x)=K E(x) \geq 0$ $E \geq u(x)$ Therefore, $\mathrm{u}(\mathrm{x}) \leq \mathrm{E}$
COMEDK 2013
Work, Energy and Power
149016
A bomb of mass $18 \mathrm{~kg}$ at rest explodes into two pieces of masses $6 \mathrm{~kg}$ and $12 \mathrm{~kg}$. The velocity of $12 \mathrm{~kg}$ mass is $4 \mathrm{~m} / \mathrm{s}$. The kinetic energy of the other mass is
1 $288 \mathrm{~J}$
2 $192 \mathrm{~J}$
3 $96 \mathrm{~J}$
4 $144 \mathrm{~J}$
Explanation:
B Given, Mass of bomb $(\mathrm{m})=18 \mathrm{~kg}$ Velocity of $18 \mathrm{~kg}$ bomb (v) $=0 \mathrm{~m} / \mathrm{s}$ Mass of second bomb $\left(\mathrm{m}_{2}\right)=12 \mathrm{~kg}$ Velocity of second bomb $\left(\mathrm{v}_{2}\right)=4 \mathrm{~m} / \mathrm{sec}$ Mass of first bomb $\left(\mathrm{m}_{1}\right)=6 \mathrm{~kg}$ Velocity of first bomb $\left(\mathrm{v}_{1}\right)=$ ? By using law of conservation of momentum, $\mathrm{mv}= \mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}$ $18 \times 0 =\left(6 \times \mathrm{v}_{1}\right)+(12 \times 4)$ $0 =6 \mathrm{v}_{1}+48$ $6 \mathrm{v}_{1} =-48$ $\mathrm{v}_{1} =-\frac{48}{6}$ $\mathrm{v}_{1} =-8 \mathrm{~m} / \mathrm{sec}$ Kinetic energy of $6 \mathrm{~kg}$ mass (K.E.) $=\frac{1}{2} \mathrm{~m}_{1}\left(\mathrm{v}_{1}\right)^{2}$ $\text { K.E. }=\frac{1}{2} \times 6 \times(-8)^{2}$ $\text { K.E. }=\frac{1}{2} \times 6 \times 8 \times 8$ $\text { K.E. }=192 \text { Joule }$
COMEDK 2015
Work, Energy and Power
149017
Assertion: If collision occurs between two elastic bodies their kinetic energy decreases during the time of collision. Reason: During collision intermolecular space decreases and hence elastic potential energy increases.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A If collision occurs between two elastic bodies then kinetic energy of bodies decreases because, at time of collision intermolecular space of bodies decrease. Intermolecular attraction and repulsion lead to change in potential energy. It means, decreasing the intermolecular space lead to decrease the elastic potential.
AIIMS-2011
Work, Energy and Power
149018
Assertion: $n$ small balls each of mass $m$ colliding elastically each second on surface with velocity $u$. The force experienced by the surface is $2 \mathrm{mnu}$. Reason: On elastic collision, the ball rebounds with the same velocity.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A In elastic collision, kinetic energy remains conserved. Therefore, the ball rebounds with the same velocity. Given, Mass $=\mathrm{m} \mathrm{kg}$, velocity $=\mathrm{u} \mathrm{m} / \mathrm{s}$ We know, In elastic collision change in momentum $\Delta \mathrm{p} =\mathrm{mu}-(-\mathrm{mu})$ $=2 \mathrm{mu}$ For $\mathrm{n}$ balls, $\Delta \mathrm{p}=\mathrm{n}(2 \mathrm{mu})=2 \mathrm{n} \mathrm{mu}$ According to Newton's second law - $\mathrm{F}=\frac{\Delta \mathrm{p}}{\mathrm{t}}=\frac{2 \mathrm{nmu}}{1}$ $\mathrm{~F}=2 \mathrm{nmu}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Work, Energy and Power
149015
A particle with total energy $E$ is moving in a potential energy region u(x). Motion of the particle is restricted to the region when
1 u(x) $ \lt $ E
2 $\mathrm{u}(\mathrm{x})=0$
3 $\mathrm{u}(\mathrm{x}) \leq \mathrm{E}$
4 $u(x)>E$
Explanation:
C Given, A particle with total energy $\mathrm{E}$ is moving in a potential energy region $\mathrm{u}(\mathrm{x})$. We know that, Total Energy $=$ Potential energy + Kinetic energy $\mathrm{E}=\mathrm{u}(\mathrm{x})+\mathrm{KE}(\mathrm{x})$ $\mathrm{KE}(\mathrm{x})=\mathrm{E}-\mathrm{u}(\mathrm{x})$ It means when particle is moving kinetic energy must be positive. Then, $\mathrm{KE}(\mathrm{x}) \geq 0$ $E-u(x)=K E(x) \geq 0$ $E \geq u(x)$ Therefore, $\mathrm{u}(\mathrm{x}) \leq \mathrm{E}$
COMEDK 2013
Work, Energy and Power
149016
A bomb of mass $18 \mathrm{~kg}$ at rest explodes into two pieces of masses $6 \mathrm{~kg}$ and $12 \mathrm{~kg}$. The velocity of $12 \mathrm{~kg}$ mass is $4 \mathrm{~m} / \mathrm{s}$. The kinetic energy of the other mass is
1 $288 \mathrm{~J}$
2 $192 \mathrm{~J}$
3 $96 \mathrm{~J}$
4 $144 \mathrm{~J}$
Explanation:
B Given, Mass of bomb $(\mathrm{m})=18 \mathrm{~kg}$ Velocity of $18 \mathrm{~kg}$ bomb (v) $=0 \mathrm{~m} / \mathrm{s}$ Mass of second bomb $\left(\mathrm{m}_{2}\right)=12 \mathrm{~kg}$ Velocity of second bomb $\left(\mathrm{v}_{2}\right)=4 \mathrm{~m} / \mathrm{sec}$ Mass of first bomb $\left(\mathrm{m}_{1}\right)=6 \mathrm{~kg}$ Velocity of first bomb $\left(\mathrm{v}_{1}\right)=$ ? By using law of conservation of momentum, $\mathrm{mv}= \mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}$ $18 \times 0 =\left(6 \times \mathrm{v}_{1}\right)+(12 \times 4)$ $0 =6 \mathrm{v}_{1}+48$ $6 \mathrm{v}_{1} =-48$ $\mathrm{v}_{1} =-\frac{48}{6}$ $\mathrm{v}_{1} =-8 \mathrm{~m} / \mathrm{sec}$ Kinetic energy of $6 \mathrm{~kg}$ mass (K.E.) $=\frac{1}{2} \mathrm{~m}_{1}\left(\mathrm{v}_{1}\right)^{2}$ $\text { K.E. }=\frac{1}{2} \times 6 \times(-8)^{2}$ $\text { K.E. }=\frac{1}{2} \times 6 \times 8 \times 8$ $\text { K.E. }=192 \text { Joule }$
COMEDK 2015
Work, Energy and Power
149017
Assertion: If collision occurs between two elastic bodies their kinetic energy decreases during the time of collision. Reason: During collision intermolecular space decreases and hence elastic potential energy increases.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A If collision occurs between two elastic bodies then kinetic energy of bodies decreases because, at time of collision intermolecular space of bodies decrease. Intermolecular attraction and repulsion lead to change in potential energy. It means, decreasing the intermolecular space lead to decrease the elastic potential.
AIIMS-2011
Work, Energy and Power
149018
Assertion: $n$ small balls each of mass $m$ colliding elastically each second on surface with velocity $u$. The force experienced by the surface is $2 \mathrm{mnu}$. Reason: On elastic collision, the ball rebounds with the same velocity.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A In elastic collision, kinetic energy remains conserved. Therefore, the ball rebounds with the same velocity. Given, Mass $=\mathrm{m} \mathrm{kg}$, velocity $=\mathrm{u} \mathrm{m} / \mathrm{s}$ We know, In elastic collision change in momentum $\Delta \mathrm{p} =\mathrm{mu}-(-\mathrm{mu})$ $=2 \mathrm{mu}$ For $\mathrm{n}$ balls, $\Delta \mathrm{p}=\mathrm{n}(2 \mathrm{mu})=2 \mathrm{n} \mathrm{mu}$ According to Newton's second law - $\mathrm{F}=\frac{\Delta \mathrm{p}}{\mathrm{t}}=\frac{2 \mathrm{nmu}}{1}$ $\mathrm{~F}=2 \mathrm{nmu}$
