140351
A spring of spring constant $5 \times 10^{3} \mathrm{~N} / \mathrm{m}$ is stretched by $4 \mathrm{~cm}$ from its outstretched position. Then the work required to stretch it further by $\mathbf{4} \mathbf{~ c m}$ is
140352
A simple pendulum of length ' $L$ ' has mass ' $m$ ' and it oscillates freely with amplitude ' $A$ '. At extreme position, its potential energy is $(\mathrm{g}=$ acceleration due to gravity)
1 $\frac{\mathrm{mgA}}{\mathrm{L}}$
2 $\frac{\mathrm{mgA}^{2}}{\mathrm{~L}}$
3 $\frac{\mathrm{mgA}^{2}}{2 \mathrm{~L}}$
4 $\frac{\mathrm{mgA}}{2 \mathrm{~L}}$
Explanation:
C Given, Length of pendulum $=\mathrm{L}$, Mass $=\mathrm{m}$, Amplitude $=\mathrm{A}$ We know that, Potential Energy (P.E) $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Time period of pendulum $(T)=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ $\text { or } \frac{\mathrm{T}}{2 \pi} =\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ $\because \omega =\frac{2 \pi}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}$ Putting the value of $\omega$ in equation (i), we get - $\therefore \quad \text { P.E } =\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\right)^{2} \mathrm{~A}^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{g}}{\mathrm{L}} \mathrm{A}^{2}=\frac{\mathrm{mgA}^{2}}{2 \mathrm{~L}}$
MHT CET-2017
Oscillations
140353
A simple pendulum is released from $A$ as shown. If $\mathrm{m}$ and $\boldsymbol{l}$ represent the mass of the bob and length of the pendulum, the gain in kinetic energy at $B$ is
1 $\frac{\mathrm{mgl}}{2}$
2 $\frac{\mathrm{mgl}}{\sqrt{2}}$
3 $\frac{\sqrt{3}}{2} \mathrm{mgl}$
4 $\frac{2}{\sqrt{3}} \mathrm{mgl}$
5 $\mathrm{mgl}$
Explanation:
C $\mathrm{h}=l-l \cos 30^{\circ}$ $\mathrm{h}=l\left(1-\cos 30^{\circ}\right)$ Let, $\mathrm{c}$ is the point on datum line. Potential energy at point $\mathrm{B}=\mathrm{mgh}=\operatorname{mg}\left(l-l \cos 30^{\circ}\right)$ $=\operatorname{mg} l\left(1-\frac{\sqrt{3}}{2}\right)$ Potential energy at point $\mathrm{A}=\mathrm{mg} l$ Applying energy conservation- Total energy at point $\mathrm{A}=$ Total energy at point $\mathrm{B}$ $(\mathrm{P} . \mathrm{E})_{\mathrm{A}}+(\mathrm{K} . \mathrm{E})_{\mathrm{A}}=(\mathrm{P} \cdot \mathrm{E})_{\mathrm{B}}+(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}$ $\mathrm{mg} l+0==\mathrm{mg} l\left(1-\frac{\sqrt{3}}{2}\right)+(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}$ $(\mathrm{K} . \mathrm{E})_{\mathrm{B}}=\frac{\sqrt{3}}{2} \mathrm{mg} l$
Kerala CEE 2007
Oscillations
140354
A long spring is stretched by $2 \mathrm{~cm}$ and its potential energy is $U$. If the spring is stretched by $10 \mathrm{~cm}$; its potential energy will be
1 $U / 5$
2 $U / 25$
3 $5 \mathrm{U}$
4 $25 \mathrm{U}$
Explanation:
D Given, $\mathrm{x}_{1}=2 \mathrm{~cm}, \quad \mathrm{x}_{2}=10 \mathrm{~cm}$ We Know that, Elastic Potential energy of a spring $\mathrm{U}=\frac{1}{2} \mathrm{kx}^{2}$ $\mathrm{U} \propto \mathrm{x}^{2}$ So, $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\left(\frac{\mathrm{x}_{2}}{\mathrm{x}_{1}}\right)^{2}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\left(\frac{10}{2}\right)^{2}=(5)^{2}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=25$ $\mathrm{U}_{2}=25 \mathrm{U}_{1}$
140351
A spring of spring constant $5 \times 10^{3} \mathrm{~N} / \mathrm{m}$ is stretched by $4 \mathrm{~cm}$ from its outstretched position. Then the work required to stretch it further by $\mathbf{4} \mathbf{~ c m}$ is
140352
A simple pendulum of length ' $L$ ' has mass ' $m$ ' and it oscillates freely with amplitude ' $A$ '. At extreme position, its potential energy is $(\mathrm{g}=$ acceleration due to gravity)
1 $\frac{\mathrm{mgA}}{\mathrm{L}}$
2 $\frac{\mathrm{mgA}^{2}}{\mathrm{~L}}$
3 $\frac{\mathrm{mgA}^{2}}{2 \mathrm{~L}}$
4 $\frac{\mathrm{mgA}}{2 \mathrm{~L}}$
Explanation:
C Given, Length of pendulum $=\mathrm{L}$, Mass $=\mathrm{m}$, Amplitude $=\mathrm{A}$ We know that, Potential Energy (P.E) $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Time period of pendulum $(T)=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ $\text { or } \frac{\mathrm{T}}{2 \pi} =\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ $\because \omega =\frac{2 \pi}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}$ Putting the value of $\omega$ in equation (i), we get - $\therefore \quad \text { P.E } =\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\right)^{2} \mathrm{~A}^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{g}}{\mathrm{L}} \mathrm{A}^{2}=\frac{\mathrm{mgA}^{2}}{2 \mathrm{~L}}$
MHT CET-2017
Oscillations
140353
A simple pendulum is released from $A$ as shown. If $\mathrm{m}$ and $\boldsymbol{l}$ represent the mass of the bob and length of the pendulum, the gain in kinetic energy at $B$ is
1 $\frac{\mathrm{mgl}}{2}$
2 $\frac{\mathrm{mgl}}{\sqrt{2}}$
3 $\frac{\sqrt{3}}{2} \mathrm{mgl}$
4 $\frac{2}{\sqrt{3}} \mathrm{mgl}$
5 $\mathrm{mgl}$
Explanation:
C $\mathrm{h}=l-l \cos 30^{\circ}$ $\mathrm{h}=l\left(1-\cos 30^{\circ}\right)$ Let, $\mathrm{c}$ is the point on datum line. Potential energy at point $\mathrm{B}=\mathrm{mgh}=\operatorname{mg}\left(l-l \cos 30^{\circ}\right)$ $=\operatorname{mg} l\left(1-\frac{\sqrt{3}}{2}\right)$ Potential energy at point $\mathrm{A}=\mathrm{mg} l$ Applying energy conservation- Total energy at point $\mathrm{A}=$ Total energy at point $\mathrm{B}$ $(\mathrm{P} . \mathrm{E})_{\mathrm{A}}+(\mathrm{K} . \mathrm{E})_{\mathrm{A}}=(\mathrm{P} \cdot \mathrm{E})_{\mathrm{B}}+(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}$ $\mathrm{mg} l+0==\mathrm{mg} l\left(1-\frac{\sqrt{3}}{2}\right)+(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}$ $(\mathrm{K} . \mathrm{E})_{\mathrm{B}}=\frac{\sqrt{3}}{2} \mathrm{mg} l$
Kerala CEE 2007
Oscillations
140354
A long spring is stretched by $2 \mathrm{~cm}$ and its potential energy is $U$. If the spring is stretched by $10 \mathrm{~cm}$; its potential energy will be
1 $U / 5$
2 $U / 25$
3 $5 \mathrm{U}$
4 $25 \mathrm{U}$
Explanation:
D Given, $\mathrm{x}_{1}=2 \mathrm{~cm}, \quad \mathrm{x}_{2}=10 \mathrm{~cm}$ We Know that, Elastic Potential energy of a spring $\mathrm{U}=\frac{1}{2} \mathrm{kx}^{2}$ $\mathrm{U} \propto \mathrm{x}^{2}$ So, $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\left(\frac{\mathrm{x}_{2}}{\mathrm{x}_{1}}\right)^{2}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\left(\frac{10}{2}\right)^{2}=(5)^{2}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=25$ $\mathrm{U}_{2}=25 \mathrm{U}_{1}$
140351
A spring of spring constant $5 \times 10^{3} \mathrm{~N} / \mathrm{m}$ is stretched by $4 \mathrm{~cm}$ from its outstretched position. Then the work required to stretch it further by $\mathbf{4} \mathbf{~ c m}$ is
140352
A simple pendulum of length ' $L$ ' has mass ' $m$ ' and it oscillates freely with amplitude ' $A$ '. At extreme position, its potential energy is $(\mathrm{g}=$ acceleration due to gravity)
1 $\frac{\mathrm{mgA}}{\mathrm{L}}$
2 $\frac{\mathrm{mgA}^{2}}{\mathrm{~L}}$
3 $\frac{\mathrm{mgA}^{2}}{2 \mathrm{~L}}$
4 $\frac{\mathrm{mgA}}{2 \mathrm{~L}}$
Explanation:
C Given, Length of pendulum $=\mathrm{L}$, Mass $=\mathrm{m}$, Amplitude $=\mathrm{A}$ We know that, Potential Energy (P.E) $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Time period of pendulum $(T)=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ $\text { or } \frac{\mathrm{T}}{2 \pi} =\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ $\because \omega =\frac{2 \pi}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}$ Putting the value of $\omega$ in equation (i), we get - $\therefore \quad \text { P.E } =\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\right)^{2} \mathrm{~A}^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{g}}{\mathrm{L}} \mathrm{A}^{2}=\frac{\mathrm{mgA}^{2}}{2 \mathrm{~L}}$
MHT CET-2017
Oscillations
140353
A simple pendulum is released from $A$ as shown. If $\mathrm{m}$ and $\boldsymbol{l}$ represent the mass of the bob and length of the pendulum, the gain in kinetic energy at $B$ is
1 $\frac{\mathrm{mgl}}{2}$
2 $\frac{\mathrm{mgl}}{\sqrt{2}}$
3 $\frac{\sqrt{3}}{2} \mathrm{mgl}$
4 $\frac{2}{\sqrt{3}} \mathrm{mgl}$
5 $\mathrm{mgl}$
Explanation:
C $\mathrm{h}=l-l \cos 30^{\circ}$ $\mathrm{h}=l\left(1-\cos 30^{\circ}\right)$ Let, $\mathrm{c}$ is the point on datum line. Potential energy at point $\mathrm{B}=\mathrm{mgh}=\operatorname{mg}\left(l-l \cos 30^{\circ}\right)$ $=\operatorname{mg} l\left(1-\frac{\sqrt{3}}{2}\right)$ Potential energy at point $\mathrm{A}=\mathrm{mg} l$ Applying energy conservation- Total energy at point $\mathrm{A}=$ Total energy at point $\mathrm{B}$ $(\mathrm{P} . \mathrm{E})_{\mathrm{A}}+(\mathrm{K} . \mathrm{E})_{\mathrm{A}}=(\mathrm{P} \cdot \mathrm{E})_{\mathrm{B}}+(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}$ $\mathrm{mg} l+0==\mathrm{mg} l\left(1-\frac{\sqrt{3}}{2}\right)+(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}$ $(\mathrm{K} . \mathrm{E})_{\mathrm{B}}=\frac{\sqrt{3}}{2} \mathrm{mg} l$
Kerala CEE 2007
Oscillations
140354
A long spring is stretched by $2 \mathrm{~cm}$ and its potential energy is $U$. If the spring is stretched by $10 \mathrm{~cm}$; its potential energy will be
1 $U / 5$
2 $U / 25$
3 $5 \mathrm{U}$
4 $25 \mathrm{U}$
Explanation:
D Given, $\mathrm{x}_{1}=2 \mathrm{~cm}, \quad \mathrm{x}_{2}=10 \mathrm{~cm}$ We Know that, Elastic Potential energy of a spring $\mathrm{U}=\frac{1}{2} \mathrm{kx}^{2}$ $\mathrm{U} \propto \mathrm{x}^{2}$ So, $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\left(\frac{\mathrm{x}_{2}}{\mathrm{x}_{1}}\right)^{2}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\left(\frac{10}{2}\right)^{2}=(5)^{2}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=25$ $\mathrm{U}_{2}=25 \mathrm{U}_{1}$
140351
A spring of spring constant $5 \times 10^{3} \mathrm{~N} / \mathrm{m}$ is stretched by $4 \mathrm{~cm}$ from its outstretched position. Then the work required to stretch it further by $\mathbf{4} \mathbf{~ c m}$ is
140352
A simple pendulum of length ' $L$ ' has mass ' $m$ ' and it oscillates freely with amplitude ' $A$ '. At extreme position, its potential energy is $(\mathrm{g}=$ acceleration due to gravity)
1 $\frac{\mathrm{mgA}}{\mathrm{L}}$
2 $\frac{\mathrm{mgA}^{2}}{\mathrm{~L}}$
3 $\frac{\mathrm{mgA}^{2}}{2 \mathrm{~L}}$
4 $\frac{\mathrm{mgA}}{2 \mathrm{~L}}$
Explanation:
C Given, Length of pendulum $=\mathrm{L}$, Mass $=\mathrm{m}$, Amplitude $=\mathrm{A}$ We know that, Potential Energy (P.E) $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ Time period of pendulum $(T)=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ $\text { or } \frac{\mathrm{T}}{2 \pi} =\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ $\because \omega =\frac{2 \pi}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}$ Putting the value of $\omega$ in equation (i), we get - $\therefore \quad \text { P.E } =\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\right)^{2} \mathrm{~A}^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{g}}{\mathrm{L}} \mathrm{A}^{2}=\frac{\mathrm{mgA}^{2}}{2 \mathrm{~L}}$
MHT CET-2017
Oscillations
140353
A simple pendulum is released from $A$ as shown. If $\mathrm{m}$ and $\boldsymbol{l}$ represent the mass of the bob and length of the pendulum, the gain in kinetic energy at $B$ is
1 $\frac{\mathrm{mgl}}{2}$
2 $\frac{\mathrm{mgl}}{\sqrt{2}}$
3 $\frac{\sqrt{3}}{2} \mathrm{mgl}$
4 $\frac{2}{\sqrt{3}} \mathrm{mgl}$
5 $\mathrm{mgl}$
Explanation:
C $\mathrm{h}=l-l \cos 30^{\circ}$ $\mathrm{h}=l\left(1-\cos 30^{\circ}\right)$ Let, $\mathrm{c}$ is the point on datum line. Potential energy at point $\mathrm{B}=\mathrm{mgh}=\operatorname{mg}\left(l-l \cos 30^{\circ}\right)$ $=\operatorname{mg} l\left(1-\frac{\sqrt{3}}{2}\right)$ Potential energy at point $\mathrm{A}=\mathrm{mg} l$ Applying energy conservation- Total energy at point $\mathrm{A}=$ Total energy at point $\mathrm{B}$ $(\mathrm{P} . \mathrm{E})_{\mathrm{A}}+(\mathrm{K} . \mathrm{E})_{\mathrm{A}}=(\mathrm{P} \cdot \mathrm{E})_{\mathrm{B}}+(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}$ $\mathrm{mg} l+0==\mathrm{mg} l\left(1-\frac{\sqrt{3}}{2}\right)+(\mathrm{K} \cdot \mathrm{E})_{\mathrm{B}}$ $(\mathrm{K} . \mathrm{E})_{\mathrm{B}}=\frac{\sqrt{3}}{2} \mathrm{mg} l$
Kerala CEE 2007
Oscillations
140354
A long spring is stretched by $2 \mathrm{~cm}$ and its potential energy is $U$. If the spring is stretched by $10 \mathrm{~cm}$; its potential energy will be
1 $U / 5$
2 $U / 25$
3 $5 \mathrm{U}$
4 $25 \mathrm{U}$
Explanation:
D Given, $\mathrm{x}_{1}=2 \mathrm{~cm}, \quad \mathrm{x}_{2}=10 \mathrm{~cm}$ We Know that, Elastic Potential energy of a spring $\mathrm{U}=\frac{1}{2} \mathrm{kx}^{2}$ $\mathrm{U} \propto \mathrm{x}^{2}$ So, $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\left(\frac{\mathrm{x}_{2}}{\mathrm{x}_{1}}\right)^{2}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\left(\frac{10}{2}\right)^{2}=(5)^{2}$ $\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=25$ $\mathrm{U}_{2}=25 \mathrm{U}_{1}$
