NEET Test Series from KOTA - 10 Papers In MS WORD
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Oscillations
140356
The work done in stretching a spring of force constant $\mathrm{k}$ from length $\boldsymbol{l}_{\mathbf{1}}$ to $\boldsymbol{l}_{\mathbf{2}}$ is
B Let, a spring of force constant $\mathrm{k}$ is stretched from length $l_{1}$ to $l_{2}$. Energy stored in a stretched spring, $\mathrm{U}_{1}=\frac{1}{2} \mathrm{k} \times l_{1}^{2}$ $\mathrm{U}_{2}=\frac{1}{2} \mathrm{k} \times l_{2}^{2}$ $\therefore$ Workdone is stored in stretching the spring, $\mathrm{W}=\mathrm{U}_{2}-\mathrm{U}_{1}$ $\mathrm{~W}=\frac{1}{2} \mathrm{k} \times l_{2}^{2}-\frac{1}{2} \mathrm{k} \times l_{1}^{2}$ $\mathrm{~W}=\frac{1}{2} \mathrm{k}\left(l_{2}^{2}-l_{1}^{2}\right)$
CG PET- 2004
Oscillations
140357
What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass? (Assume the collision to be head- on elastic collision)
1 $50.0 \%$
2 $66.6 \%$
3 $55.5 \%$
4 $33.3 \%$
Explanation:
C We know that, for elastic collision $\mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{u}-0$ $\mathrm{u}=\mathrm{v}_{2}-\mathrm{v}_{1}$ From momentum conservation- $\mathrm{mu}=\mathrm{mv}_{1}+5 \mathrm{mv}_{2}$ $\mathrm{mu}=\mathrm{m}\left(\mathrm{v}_{1}+5 \mathrm{v}_{2}\right)$ $\mathrm{u}=\mathrm{v}_{1}+5 \mathrm{v}_{2}$ Adding equation (i) \& equation (ii), we get- $u+u=v_{2}-v_{1}+v_{1}+5 v_{2}$ $2 u=6 v_{2}$ $u=3 v_{2}$ $v_{2}=\frac{u}{3}$ Now, initial kinetic energy of first mass - $\mathrm{k}=\frac{1}{2} \mathrm{mu}^{2}$ And final kinetic energy of second mass $\mathrm{k}^{\prime}=\frac{1}{2} \times 5 \mathrm{~m} \times\left(\frac{\mathrm{u}}{3}\right)^{2}$ $\mathrm{k}=\frac{1}{2} \times 5 \mathrm{~m} \times \frac{\mathrm{u}^{2}}{9}$ $\mathrm{k}^{\prime}=\frac{1}{2} \mathrm{mu}^{2} \times \frac{5}{9}$ $\mathrm{k}^{\prime}=\mathrm{k} \times \frac{5}{9}=\frac{5}{9} \mathrm{k}$ Percentage energy transferred $=\frac{\frac{5}{9} \mathrm{k}}{\mathrm{k}} \times 100=55.55 \%$
JEE Main-27.06.2022
Oscillations
140358
The displacement of a simple harmonic motion of amplitude $6 \mathrm{~cm}$ when its kinetic energy is equal to its potential energy is
1 $2 \sqrt{2} \mathrm{~cm}$
2 $2 \mathrm{~cm}$
3 $3 \sqrt{2} \mathrm{~cm}$
4 $\frac{3}{\sqrt{2}} \mathrm{~cm}$
Explanation:
C Given, $\mathrm{A}=6 \mathrm{~cm}$ Lets displacement be $\mathrm{x} \mathrm{cm}$. When kinetic energy is equal to potential energy. then $\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ $\mathrm{~A}^{2}-\mathrm{x}^{2}=\mathrm{x}^{2}$ $2 \mathrm{x}^{2}=\mathrm{A}^{2}$ $2 \mathrm{x}^{2}=6^{2}$ $\mathrm{x}^{2}=18$ $\mathrm{x}=3 \sqrt{2} \mathrm{~cm}$
AP EAMCET-25.08.2021
Oscillations
140359
A spring is stretched by $5 \mathrm{~cm}$ by a force $10 \mathrm{~N}$. The time period of the oscillations when a mass of $2 \mathbf{~ k g}$ is suspended by it is
1 $0.0628 \mathrm{~s}$
2 $6.28 \mathrm{~s}$
3 $3.14 \mathrm{~s}$
4 $0.628 \mathrm{~s}$
Explanation:
D Given, Stretched $(\mathrm{x})=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Force $(\mathrm{F})=10 \mathrm{~N}$. $\operatorname{Mass}(\mathrm{M})=2 \mathrm{Kg}$ Now, $\quad \mathrm{F}=\mathrm{K} \mathrm{x}$ $10=5 \times 10^{-2} \times \mathrm{K}$ $\mathrm{K}=\frac{1000}{5}$ $\mathrm{~K}=200 \mathrm{~N} / \mathrm{m}$ Now for spring-mass system undergoing SH.M, $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{2}{200}}$ $\mathrm{~T}=\frac{2 \pi}{10}=0.628 \mathrm{sec}$
140356
The work done in stretching a spring of force constant $\mathrm{k}$ from length $\boldsymbol{l}_{\mathbf{1}}$ to $\boldsymbol{l}_{\mathbf{2}}$ is
B Let, a spring of force constant $\mathrm{k}$ is stretched from length $l_{1}$ to $l_{2}$. Energy stored in a stretched spring, $\mathrm{U}_{1}=\frac{1}{2} \mathrm{k} \times l_{1}^{2}$ $\mathrm{U}_{2}=\frac{1}{2} \mathrm{k} \times l_{2}^{2}$ $\therefore$ Workdone is stored in stretching the spring, $\mathrm{W}=\mathrm{U}_{2}-\mathrm{U}_{1}$ $\mathrm{~W}=\frac{1}{2} \mathrm{k} \times l_{2}^{2}-\frac{1}{2} \mathrm{k} \times l_{1}^{2}$ $\mathrm{~W}=\frac{1}{2} \mathrm{k}\left(l_{2}^{2}-l_{1}^{2}\right)$
CG PET- 2004
Oscillations
140357
What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass? (Assume the collision to be head- on elastic collision)
1 $50.0 \%$
2 $66.6 \%$
3 $55.5 \%$
4 $33.3 \%$
Explanation:
C We know that, for elastic collision $\mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{u}-0$ $\mathrm{u}=\mathrm{v}_{2}-\mathrm{v}_{1}$ From momentum conservation- $\mathrm{mu}=\mathrm{mv}_{1}+5 \mathrm{mv}_{2}$ $\mathrm{mu}=\mathrm{m}\left(\mathrm{v}_{1}+5 \mathrm{v}_{2}\right)$ $\mathrm{u}=\mathrm{v}_{1}+5 \mathrm{v}_{2}$ Adding equation (i) \& equation (ii), we get- $u+u=v_{2}-v_{1}+v_{1}+5 v_{2}$ $2 u=6 v_{2}$ $u=3 v_{2}$ $v_{2}=\frac{u}{3}$ Now, initial kinetic energy of first mass - $\mathrm{k}=\frac{1}{2} \mathrm{mu}^{2}$ And final kinetic energy of second mass $\mathrm{k}^{\prime}=\frac{1}{2} \times 5 \mathrm{~m} \times\left(\frac{\mathrm{u}}{3}\right)^{2}$ $\mathrm{k}=\frac{1}{2} \times 5 \mathrm{~m} \times \frac{\mathrm{u}^{2}}{9}$ $\mathrm{k}^{\prime}=\frac{1}{2} \mathrm{mu}^{2} \times \frac{5}{9}$ $\mathrm{k}^{\prime}=\mathrm{k} \times \frac{5}{9}=\frac{5}{9} \mathrm{k}$ Percentage energy transferred $=\frac{\frac{5}{9} \mathrm{k}}{\mathrm{k}} \times 100=55.55 \%$
JEE Main-27.06.2022
Oscillations
140358
The displacement of a simple harmonic motion of amplitude $6 \mathrm{~cm}$ when its kinetic energy is equal to its potential energy is
1 $2 \sqrt{2} \mathrm{~cm}$
2 $2 \mathrm{~cm}$
3 $3 \sqrt{2} \mathrm{~cm}$
4 $\frac{3}{\sqrt{2}} \mathrm{~cm}$
Explanation:
C Given, $\mathrm{A}=6 \mathrm{~cm}$ Lets displacement be $\mathrm{x} \mathrm{cm}$. When kinetic energy is equal to potential energy. then $\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ $\mathrm{~A}^{2}-\mathrm{x}^{2}=\mathrm{x}^{2}$ $2 \mathrm{x}^{2}=\mathrm{A}^{2}$ $2 \mathrm{x}^{2}=6^{2}$ $\mathrm{x}^{2}=18$ $\mathrm{x}=3 \sqrt{2} \mathrm{~cm}$
AP EAMCET-25.08.2021
Oscillations
140359
A spring is stretched by $5 \mathrm{~cm}$ by a force $10 \mathrm{~N}$. The time period of the oscillations when a mass of $2 \mathbf{~ k g}$ is suspended by it is
1 $0.0628 \mathrm{~s}$
2 $6.28 \mathrm{~s}$
3 $3.14 \mathrm{~s}$
4 $0.628 \mathrm{~s}$
Explanation:
D Given, Stretched $(\mathrm{x})=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Force $(\mathrm{F})=10 \mathrm{~N}$. $\operatorname{Mass}(\mathrm{M})=2 \mathrm{Kg}$ Now, $\quad \mathrm{F}=\mathrm{K} \mathrm{x}$ $10=5 \times 10^{-2} \times \mathrm{K}$ $\mathrm{K}=\frac{1000}{5}$ $\mathrm{~K}=200 \mathrm{~N} / \mathrm{m}$ Now for spring-mass system undergoing SH.M, $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{2}{200}}$ $\mathrm{~T}=\frac{2 \pi}{10}=0.628 \mathrm{sec}$
140356
The work done in stretching a spring of force constant $\mathrm{k}$ from length $\boldsymbol{l}_{\mathbf{1}}$ to $\boldsymbol{l}_{\mathbf{2}}$ is
B Let, a spring of force constant $\mathrm{k}$ is stretched from length $l_{1}$ to $l_{2}$. Energy stored in a stretched spring, $\mathrm{U}_{1}=\frac{1}{2} \mathrm{k} \times l_{1}^{2}$ $\mathrm{U}_{2}=\frac{1}{2} \mathrm{k} \times l_{2}^{2}$ $\therefore$ Workdone is stored in stretching the spring, $\mathrm{W}=\mathrm{U}_{2}-\mathrm{U}_{1}$ $\mathrm{~W}=\frac{1}{2} \mathrm{k} \times l_{2}^{2}-\frac{1}{2} \mathrm{k} \times l_{1}^{2}$ $\mathrm{~W}=\frac{1}{2} \mathrm{k}\left(l_{2}^{2}-l_{1}^{2}\right)$
CG PET- 2004
Oscillations
140357
What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass? (Assume the collision to be head- on elastic collision)
1 $50.0 \%$
2 $66.6 \%$
3 $55.5 \%$
4 $33.3 \%$
Explanation:
C We know that, for elastic collision $\mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{u}-0$ $\mathrm{u}=\mathrm{v}_{2}-\mathrm{v}_{1}$ From momentum conservation- $\mathrm{mu}=\mathrm{mv}_{1}+5 \mathrm{mv}_{2}$ $\mathrm{mu}=\mathrm{m}\left(\mathrm{v}_{1}+5 \mathrm{v}_{2}\right)$ $\mathrm{u}=\mathrm{v}_{1}+5 \mathrm{v}_{2}$ Adding equation (i) \& equation (ii), we get- $u+u=v_{2}-v_{1}+v_{1}+5 v_{2}$ $2 u=6 v_{2}$ $u=3 v_{2}$ $v_{2}=\frac{u}{3}$ Now, initial kinetic energy of first mass - $\mathrm{k}=\frac{1}{2} \mathrm{mu}^{2}$ And final kinetic energy of second mass $\mathrm{k}^{\prime}=\frac{1}{2} \times 5 \mathrm{~m} \times\left(\frac{\mathrm{u}}{3}\right)^{2}$ $\mathrm{k}=\frac{1}{2} \times 5 \mathrm{~m} \times \frac{\mathrm{u}^{2}}{9}$ $\mathrm{k}^{\prime}=\frac{1}{2} \mathrm{mu}^{2} \times \frac{5}{9}$ $\mathrm{k}^{\prime}=\mathrm{k} \times \frac{5}{9}=\frac{5}{9} \mathrm{k}$ Percentage energy transferred $=\frac{\frac{5}{9} \mathrm{k}}{\mathrm{k}} \times 100=55.55 \%$
JEE Main-27.06.2022
Oscillations
140358
The displacement of a simple harmonic motion of amplitude $6 \mathrm{~cm}$ when its kinetic energy is equal to its potential energy is
1 $2 \sqrt{2} \mathrm{~cm}$
2 $2 \mathrm{~cm}$
3 $3 \sqrt{2} \mathrm{~cm}$
4 $\frac{3}{\sqrt{2}} \mathrm{~cm}$
Explanation:
C Given, $\mathrm{A}=6 \mathrm{~cm}$ Lets displacement be $\mathrm{x} \mathrm{cm}$. When kinetic energy is equal to potential energy. then $\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ $\mathrm{~A}^{2}-\mathrm{x}^{2}=\mathrm{x}^{2}$ $2 \mathrm{x}^{2}=\mathrm{A}^{2}$ $2 \mathrm{x}^{2}=6^{2}$ $\mathrm{x}^{2}=18$ $\mathrm{x}=3 \sqrt{2} \mathrm{~cm}$
AP EAMCET-25.08.2021
Oscillations
140359
A spring is stretched by $5 \mathrm{~cm}$ by a force $10 \mathrm{~N}$. The time period of the oscillations when a mass of $2 \mathbf{~ k g}$ is suspended by it is
1 $0.0628 \mathrm{~s}$
2 $6.28 \mathrm{~s}$
3 $3.14 \mathrm{~s}$
4 $0.628 \mathrm{~s}$
Explanation:
D Given, Stretched $(\mathrm{x})=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Force $(\mathrm{F})=10 \mathrm{~N}$. $\operatorname{Mass}(\mathrm{M})=2 \mathrm{Kg}$ Now, $\quad \mathrm{F}=\mathrm{K} \mathrm{x}$ $10=5 \times 10^{-2} \times \mathrm{K}$ $\mathrm{K}=\frac{1000}{5}$ $\mathrm{~K}=200 \mathrm{~N} / \mathrm{m}$ Now for spring-mass system undergoing SH.M, $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{2}{200}}$ $\mathrm{~T}=\frac{2 \pi}{10}=0.628 \mathrm{sec}$
140356
The work done in stretching a spring of force constant $\mathrm{k}$ from length $\boldsymbol{l}_{\mathbf{1}}$ to $\boldsymbol{l}_{\mathbf{2}}$ is
B Let, a spring of force constant $\mathrm{k}$ is stretched from length $l_{1}$ to $l_{2}$. Energy stored in a stretched spring, $\mathrm{U}_{1}=\frac{1}{2} \mathrm{k} \times l_{1}^{2}$ $\mathrm{U}_{2}=\frac{1}{2} \mathrm{k} \times l_{2}^{2}$ $\therefore$ Workdone is stored in stretching the spring, $\mathrm{W}=\mathrm{U}_{2}-\mathrm{U}_{1}$ $\mathrm{~W}=\frac{1}{2} \mathrm{k} \times l_{2}^{2}-\frac{1}{2} \mathrm{k} \times l_{1}^{2}$ $\mathrm{~W}=\frac{1}{2} \mathrm{k}\left(l_{2}^{2}-l_{1}^{2}\right)$
CG PET- 2004
Oscillations
140357
What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass? (Assume the collision to be head- on elastic collision)
1 $50.0 \%$
2 $66.6 \%$
3 $55.5 \%$
4 $33.3 \%$
Explanation:
C We know that, for elastic collision $\mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{u}-0$ $\mathrm{u}=\mathrm{v}_{2}-\mathrm{v}_{1}$ From momentum conservation- $\mathrm{mu}=\mathrm{mv}_{1}+5 \mathrm{mv}_{2}$ $\mathrm{mu}=\mathrm{m}\left(\mathrm{v}_{1}+5 \mathrm{v}_{2}\right)$ $\mathrm{u}=\mathrm{v}_{1}+5 \mathrm{v}_{2}$ Adding equation (i) \& equation (ii), we get- $u+u=v_{2}-v_{1}+v_{1}+5 v_{2}$ $2 u=6 v_{2}$ $u=3 v_{2}$ $v_{2}=\frac{u}{3}$ Now, initial kinetic energy of first mass - $\mathrm{k}=\frac{1}{2} \mathrm{mu}^{2}$ And final kinetic energy of second mass $\mathrm{k}^{\prime}=\frac{1}{2} \times 5 \mathrm{~m} \times\left(\frac{\mathrm{u}}{3}\right)^{2}$ $\mathrm{k}=\frac{1}{2} \times 5 \mathrm{~m} \times \frac{\mathrm{u}^{2}}{9}$ $\mathrm{k}^{\prime}=\frac{1}{2} \mathrm{mu}^{2} \times \frac{5}{9}$ $\mathrm{k}^{\prime}=\mathrm{k} \times \frac{5}{9}=\frac{5}{9} \mathrm{k}$ Percentage energy transferred $=\frac{\frac{5}{9} \mathrm{k}}{\mathrm{k}} \times 100=55.55 \%$
JEE Main-27.06.2022
Oscillations
140358
The displacement of a simple harmonic motion of amplitude $6 \mathrm{~cm}$ when its kinetic energy is equal to its potential energy is
1 $2 \sqrt{2} \mathrm{~cm}$
2 $2 \mathrm{~cm}$
3 $3 \sqrt{2} \mathrm{~cm}$
4 $\frac{3}{\sqrt{2}} \mathrm{~cm}$
Explanation:
C Given, $\mathrm{A}=6 \mathrm{~cm}$ Lets displacement be $\mathrm{x} \mathrm{cm}$. When kinetic energy is equal to potential energy. then $\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ $\mathrm{~A}^{2}-\mathrm{x}^{2}=\mathrm{x}^{2}$ $2 \mathrm{x}^{2}=\mathrm{A}^{2}$ $2 \mathrm{x}^{2}=6^{2}$ $\mathrm{x}^{2}=18$ $\mathrm{x}=3 \sqrt{2} \mathrm{~cm}$
AP EAMCET-25.08.2021
Oscillations
140359
A spring is stretched by $5 \mathrm{~cm}$ by a force $10 \mathrm{~N}$. The time period of the oscillations when a mass of $2 \mathbf{~ k g}$ is suspended by it is
1 $0.0628 \mathrm{~s}$
2 $6.28 \mathrm{~s}$
3 $3.14 \mathrm{~s}$
4 $0.628 \mathrm{~s}$
Explanation:
D Given, Stretched $(\mathrm{x})=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Force $(\mathrm{F})=10 \mathrm{~N}$. $\operatorname{Mass}(\mathrm{M})=2 \mathrm{Kg}$ Now, $\quad \mathrm{F}=\mathrm{K} \mathrm{x}$ $10=5 \times 10^{-2} \times \mathrm{K}$ $\mathrm{K}=\frac{1000}{5}$ $\mathrm{~K}=200 \mathrm{~N} / \mathrm{m}$ Now for spring-mass system undergoing SH.M, $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{2}{200}}$ $\mathrm{~T}=\frac{2 \pi}{10}=0.628 \mathrm{sec}$
