148974
A solid sphere of mass $1 \mathrm{~kg}$ and radius $10 \mathrm{~cm}$ rolls without slipping on a horizontal surface, with velocity of $10 \mathrm{~cm} / \mathrm{s}$. The total kinetic energy of sphere is
1 $0.007 \mathrm{~J}$
2 $0.05 \mathrm{~J}$
3 $0.01 \mathrm{~J}$
4 $0.07 \mathrm{~J}$
Explanation:
A Given, $\text { Mass }(\mathrm{m})=1 \mathrm{~kg}$ $\text { Radius }(\mathrm{r})=10 \mathrm{~cm}=0.1 \mathrm{~m} .$ $\text { Velocity }(\mathrm{v})=10 \mathrm{~cm} / \mathrm{s}=0.1 \mathrm{~m} / \mathrm{s}$ The momentum of inertia of solid sphere (I) $=\frac{2}{5} \mathrm{mr}^{2}$ We know that, $\omega=\frac{\mathrm{v}}{\mathrm{r}}$ Total kinetic energy of a rolling body without slipping (K.E.) $=$ Rotational kinetic energy + linear kinetic energy $\text { K.E. } =\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{5} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{5} \mathrm{mv}^{2}$ $=\frac{7}{10} \mathrm{mv}^{2}$ $=\frac{7}{10} \times 1 \times(0.1)^{2}$ $= 7 \times 10^{-3}$ $= 0.007 \mathrm{~J}$ Thus, total kinetic energy is $0.007 \mathrm{~J}$.
MHT-CET 2020
Work, Energy and Power
148975
A solid sphere is rolling on a frictionless surface with translational velocity ' $v$ '. It climbs that inclined plane from ' $A$ ' to ' $B$ ' and then moves away from $B$ on the smooth horizontal surface. The value of ' $v$ ' should be
A We know, Potential energy $=$ Translational energy + Rotational energy $\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}$ Moment of inertia of solid sphere - $\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}$ And$ $\omega=\frac{\mathrm{v}}{\mathrm{r}}$ Putting the value of $I$ and $\omega$ in equation (i)- $\therefore \frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\mathrm{mgh}$ $\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=m g h$ $\frac{7}{10} \mathrm{mv}^{2}=\mathrm{mgh}$ $\mathrm{v}^{2}\left(\frac{7}{10}\right)=\mathrm{gh}$ $\mathrm{v}^{2}=\frac{10}{7} \mathrm{gh}$ $\mathrm{v}=\sqrt{\frac{10}{7} \mathrm{gh}}=\left[\frac{10 \mathrm{gh}}{7}\right]^{1 / 2}$ Hence, To climb the inclined surface velocity should be greater than, $\sqrt{\frac{10}{7} \mathrm{gh}}$.
MHT-CET 2020
Work, Energy and Power
148976
A machine which is $70 \%$ efficient raised a 10 kg body through a certain distance and spends $100 \mathrm{~J}$ energy. The body is then released. On reaching the ground, the kinetic energy of the body will be
1 0
2 $70 \mathrm{~J}$
3 $50 \mathrm{~J}$
4 $35 \mathrm{~J}$
Explanation:
B Given, mass (m) $=10 \mathrm{~kg}$ Efficiency of machine $=70 \%$ Actual work done by machine, $\mathrm{W}=70 \% \times 100$ $\mathrm{~W}=70 \text { Joule }$ Now, $\mathrm{W}=\mathrm{mgh}$ $70=10 \times 10 \times \mathrm{h} \quad\left[\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]$ $\frac{70}{100} =\mathrm{h}$ $\mathrm{h} =0.70 \mathrm{~m}$ Then, applying conservation of energy- $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}$ $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}=\mathrm{mgh}$ $=10 \times 10 \times 0.70$ $\mathrm{K} . \mathrm{E}=70$ Joule
AP EAMCET (18.09.2020) Shift-II
Work, Energy and Power
148977
A particle is moving on $\mathrm{X}$-axis has potential energy $U=2-20 x+5 x^{2} J$ along $X$-axis. The particle is released at $x=-3$. The maximum value of $x$ will be ( $x$ is in metre and $U$ is in Joules)
1 $5 \mathrm{~m}$
2 $3 \mathrm{~m}$
3 $7 \mathrm{~m}$
4 $8 \mathrm{~m}$
Explanation:
C Given that, $\mathrm{U}=2-20 \mathrm{x}+5 \mathrm{x}^{2} \mathrm{~J}$ We know that, Force $(\mathrm{F})=-\frac{\mathrm{dU}}{\mathrm{dx}}$ $\mathrm{F}=-[-20+10 \mathrm{x}]$ $\mathrm{F}=20-10 \mathrm{x}$ If particle is in equilibrium position then $\mathrm{F}=0$ Putting the value of $F$ in equation (i), we get- $0=20-10 \mathrm{x}$ $20=10 \mathrm{x}$ $\mathrm{x}=2 \mathrm{~m}$ The particle is released from $\mathrm{x}=-3$. So, amplitude from $\mathrm{x}=-3$ and $\mathrm{x}=2$ is $5 \mathrm{~m}$. The maximum value of $\mathrm{x}$ will be- $x=5+2$ $x=7 \mathrm{~m}$
AP EAMCET (18.09.2020) Shift-I
Work, Energy and Power
148978
A particle moves in a straight line with its retardation proportional to its displacement. The loss of its kinetic energy for any displacement $x$ is proportional to
1 $\frac{1}{x}$
2 $x$
3 $x^{2}$
4 $e^{x}$
Explanation:
C Given that, particles move in a straight line with its retardation proportional to its displacement. $\mathrm{a} \propto \mathrm{x}$ $\mathrm{a}=-\mathrm{kx}$ Where, $\mathrm{k}=$ proportionality constant $\mathrm{vdv}=\mathrm{kx} \mathrm{dx}$ Integrating both side, we get- $\int_{v_{i}}^{v_{f}} v d v=-\int_{0}^{x} k x d x$ Where $v_{i}$ and $v_{f}$ are initial and final velocity of particle respectively. $\left(\frac{\mathrm{v}^{2}}{2}\right)_{\mathrm{v}_{\mathrm{i}}}^{\mathrm{v}_{\mathrm{f}}}=-\mathrm{k}\left(\frac{\mathrm{x}^{2}}{2}\right)_{0}^{\mathrm{x}}$ $\frac{\mathrm{v}_{\mathrm{f}}^{2}}{2}-\frac{\mathrm{v}_{\mathrm{i}}^{2}}{2}=-\mathrm{k} \frac{\mathrm{x}^{2}}{2}$ $\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{i}}^{2}=-\frac{\mathrm{mkx}^{2}}{2}$ $(\text { K.E. })_{\text {final }}-(\text { K.E. })_{\text {initial }}=-\frac{1}{2} \mathrm{mkx}^{2}$ Hence, loss in kinetic energy $\propto \mathrm{x}^{2}$
148974
A solid sphere of mass $1 \mathrm{~kg}$ and radius $10 \mathrm{~cm}$ rolls without slipping on a horizontal surface, with velocity of $10 \mathrm{~cm} / \mathrm{s}$. The total kinetic energy of sphere is
1 $0.007 \mathrm{~J}$
2 $0.05 \mathrm{~J}$
3 $0.01 \mathrm{~J}$
4 $0.07 \mathrm{~J}$
Explanation:
A Given, $\text { Mass }(\mathrm{m})=1 \mathrm{~kg}$ $\text { Radius }(\mathrm{r})=10 \mathrm{~cm}=0.1 \mathrm{~m} .$ $\text { Velocity }(\mathrm{v})=10 \mathrm{~cm} / \mathrm{s}=0.1 \mathrm{~m} / \mathrm{s}$ The momentum of inertia of solid sphere (I) $=\frac{2}{5} \mathrm{mr}^{2}$ We know that, $\omega=\frac{\mathrm{v}}{\mathrm{r}}$ Total kinetic energy of a rolling body without slipping (K.E.) $=$ Rotational kinetic energy + linear kinetic energy $\text { K.E. } =\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{5} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{5} \mathrm{mv}^{2}$ $=\frac{7}{10} \mathrm{mv}^{2}$ $=\frac{7}{10} \times 1 \times(0.1)^{2}$ $= 7 \times 10^{-3}$ $= 0.007 \mathrm{~J}$ Thus, total kinetic energy is $0.007 \mathrm{~J}$.
MHT-CET 2020
Work, Energy and Power
148975
A solid sphere is rolling on a frictionless surface with translational velocity ' $v$ '. It climbs that inclined plane from ' $A$ ' to ' $B$ ' and then moves away from $B$ on the smooth horizontal surface. The value of ' $v$ ' should be
A We know, Potential energy $=$ Translational energy + Rotational energy $\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}$ Moment of inertia of solid sphere - $\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}$ And$ $\omega=\frac{\mathrm{v}}{\mathrm{r}}$ Putting the value of $I$ and $\omega$ in equation (i)- $\therefore \frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\mathrm{mgh}$ $\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=m g h$ $\frac{7}{10} \mathrm{mv}^{2}=\mathrm{mgh}$ $\mathrm{v}^{2}\left(\frac{7}{10}\right)=\mathrm{gh}$ $\mathrm{v}^{2}=\frac{10}{7} \mathrm{gh}$ $\mathrm{v}=\sqrt{\frac{10}{7} \mathrm{gh}}=\left[\frac{10 \mathrm{gh}}{7}\right]^{1 / 2}$ Hence, To climb the inclined surface velocity should be greater than, $\sqrt{\frac{10}{7} \mathrm{gh}}$.
MHT-CET 2020
Work, Energy and Power
148976
A machine which is $70 \%$ efficient raised a 10 kg body through a certain distance and spends $100 \mathrm{~J}$ energy. The body is then released. On reaching the ground, the kinetic energy of the body will be
1 0
2 $70 \mathrm{~J}$
3 $50 \mathrm{~J}$
4 $35 \mathrm{~J}$
Explanation:
B Given, mass (m) $=10 \mathrm{~kg}$ Efficiency of machine $=70 \%$ Actual work done by machine, $\mathrm{W}=70 \% \times 100$ $\mathrm{~W}=70 \text { Joule }$ Now, $\mathrm{W}=\mathrm{mgh}$ $70=10 \times 10 \times \mathrm{h} \quad\left[\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]$ $\frac{70}{100} =\mathrm{h}$ $\mathrm{h} =0.70 \mathrm{~m}$ Then, applying conservation of energy- $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}$ $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}=\mathrm{mgh}$ $=10 \times 10 \times 0.70$ $\mathrm{K} . \mathrm{E}=70$ Joule
AP EAMCET (18.09.2020) Shift-II
Work, Energy and Power
148977
A particle is moving on $\mathrm{X}$-axis has potential energy $U=2-20 x+5 x^{2} J$ along $X$-axis. The particle is released at $x=-3$. The maximum value of $x$ will be ( $x$ is in metre and $U$ is in Joules)
1 $5 \mathrm{~m}$
2 $3 \mathrm{~m}$
3 $7 \mathrm{~m}$
4 $8 \mathrm{~m}$
Explanation:
C Given that, $\mathrm{U}=2-20 \mathrm{x}+5 \mathrm{x}^{2} \mathrm{~J}$ We know that, Force $(\mathrm{F})=-\frac{\mathrm{dU}}{\mathrm{dx}}$ $\mathrm{F}=-[-20+10 \mathrm{x}]$ $\mathrm{F}=20-10 \mathrm{x}$ If particle is in equilibrium position then $\mathrm{F}=0$ Putting the value of $F$ in equation (i), we get- $0=20-10 \mathrm{x}$ $20=10 \mathrm{x}$ $\mathrm{x}=2 \mathrm{~m}$ The particle is released from $\mathrm{x}=-3$. So, amplitude from $\mathrm{x}=-3$ and $\mathrm{x}=2$ is $5 \mathrm{~m}$. The maximum value of $\mathrm{x}$ will be- $x=5+2$ $x=7 \mathrm{~m}$
AP EAMCET (18.09.2020) Shift-I
Work, Energy and Power
148978
A particle moves in a straight line with its retardation proportional to its displacement. The loss of its kinetic energy for any displacement $x$ is proportional to
1 $\frac{1}{x}$
2 $x$
3 $x^{2}$
4 $e^{x}$
Explanation:
C Given that, particles move in a straight line with its retardation proportional to its displacement. $\mathrm{a} \propto \mathrm{x}$ $\mathrm{a}=-\mathrm{kx}$ Where, $\mathrm{k}=$ proportionality constant $\mathrm{vdv}=\mathrm{kx} \mathrm{dx}$ Integrating both side, we get- $\int_{v_{i}}^{v_{f}} v d v=-\int_{0}^{x} k x d x$ Where $v_{i}$ and $v_{f}$ are initial and final velocity of particle respectively. $\left(\frac{\mathrm{v}^{2}}{2}\right)_{\mathrm{v}_{\mathrm{i}}}^{\mathrm{v}_{\mathrm{f}}}=-\mathrm{k}\left(\frac{\mathrm{x}^{2}}{2}\right)_{0}^{\mathrm{x}}$ $\frac{\mathrm{v}_{\mathrm{f}}^{2}}{2}-\frac{\mathrm{v}_{\mathrm{i}}^{2}}{2}=-\mathrm{k} \frac{\mathrm{x}^{2}}{2}$ $\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{i}}^{2}=-\frac{\mathrm{mkx}^{2}}{2}$ $(\text { K.E. })_{\text {final }}-(\text { K.E. })_{\text {initial }}=-\frac{1}{2} \mathrm{mkx}^{2}$ Hence, loss in kinetic energy $\propto \mathrm{x}^{2}$
148974
A solid sphere of mass $1 \mathrm{~kg}$ and radius $10 \mathrm{~cm}$ rolls without slipping on a horizontal surface, with velocity of $10 \mathrm{~cm} / \mathrm{s}$. The total kinetic energy of sphere is
1 $0.007 \mathrm{~J}$
2 $0.05 \mathrm{~J}$
3 $0.01 \mathrm{~J}$
4 $0.07 \mathrm{~J}$
Explanation:
A Given, $\text { Mass }(\mathrm{m})=1 \mathrm{~kg}$ $\text { Radius }(\mathrm{r})=10 \mathrm{~cm}=0.1 \mathrm{~m} .$ $\text { Velocity }(\mathrm{v})=10 \mathrm{~cm} / \mathrm{s}=0.1 \mathrm{~m} / \mathrm{s}$ The momentum of inertia of solid sphere (I) $=\frac{2}{5} \mathrm{mr}^{2}$ We know that, $\omega=\frac{\mathrm{v}}{\mathrm{r}}$ Total kinetic energy of a rolling body without slipping (K.E.) $=$ Rotational kinetic energy + linear kinetic energy $\text { K.E. } =\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{5} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{5} \mathrm{mv}^{2}$ $=\frac{7}{10} \mathrm{mv}^{2}$ $=\frac{7}{10} \times 1 \times(0.1)^{2}$ $= 7 \times 10^{-3}$ $= 0.007 \mathrm{~J}$ Thus, total kinetic energy is $0.007 \mathrm{~J}$.
MHT-CET 2020
Work, Energy and Power
148975
A solid sphere is rolling on a frictionless surface with translational velocity ' $v$ '. It climbs that inclined plane from ' $A$ ' to ' $B$ ' and then moves away from $B$ on the smooth horizontal surface. The value of ' $v$ ' should be
A We know, Potential energy $=$ Translational energy + Rotational energy $\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}$ Moment of inertia of solid sphere - $\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}$ And$ $\omega=\frac{\mathrm{v}}{\mathrm{r}}$ Putting the value of $I$ and $\omega$ in equation (i)- $\therefore \frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\mathrm{mgh}$ $\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=m g h$ $\frac{7}{10} \mathrm{mv}^{2}=\mathrm{mgh}$ $\mathrm{v}^{2}\left(\frac{7}{10}\right)=\mathrm{gh}$ $\mathrm{v}^{2}=\frac{10}{7} \mathrm{gh}$ $\mathrm{v}=\sqrt{\frac{10}{7} \mathrm{gh}}=\left[\frac{10 \mathrm{gh}}{7}\right]^{1 / 2}$ Hence, To climb the inclined surface velocity should be greater than, $\sqrt{\frac{10}{7} \mathrm{gh}}$.
MHT-CET 2020
Work, Energy and Power
148976
A machine which is $70 \%$ efficient raised a 10 kg body through a certain distance and spends $100 \mathrm{~J}$ energy. The body is then released. On reaching the ground, the kinetic energy of the body will be
1 0
2 $70 \mathrm{~J}$
3 $50 \mathrm{~J}$
4 $35 \mathrm{~J}$
Explanation:
B Given, mass (m) $=10 \mathrm{~kg}$ Efficiency of machine $=70 \%$ Actual work done by machine, $\mathrm{W}=70 \% \times 100$ $\mathrm{~W}=70 \text { Joule }$ Now, $\mathrm{W}=\mathrm{mgh}$ $70=10 \times 10 \times \mathrm{h} \quad\left[\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]$ $\frac{70}{100} =\mathrm{h}$ $\mathrm{h} =0.70 \mathrm{~m}$ Then, applying conservation of energy- $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}$ $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}=\mathrm{mgh}$ $=10 \times 10 \times 0.70$ $\mathrm{K} . \mathrm{E}=70$ Joule
AP EAMCET (18.09.2020) Shift-II
Work, Energy and Power
148977
A particle is moving on $\mathrm{X}$-axis has potential energy $U=2-20 x+5 x^{2} J$ along $X$-axis. The particle is released at $x=-3$. The maximum value of $x$ will be ( $x$ is in metre and $U$ is in Joules)
1 $5 \mathrm{~m}$
2 $3 \mathrm{~m}$
3 $7 \mathrm{~m}$
4 $8 \mathrm{~m}$
Explanation:
C Given that, $\mathrm{U}=2-20 \mathrm{x}+5 \mathrm{x}^{2} \mathrm{~J}$ We know that, Force $(\mathrm{F})=-\frac{\mathrm{dU}}{\mathrm{dx}}$ $\mathrm{F}=-[-20+10 \mathrm{x}]$ $\mathrm{F}=20-10 \mathrm{x}$ If particle is in equilibrium position then $\mathrm{F}=0$ Putting the value of $F$ in equation (i), we get- $0=20-10 \mathrm{x}$ $20=10 \mathrm{x}$ $\mathrm{x}=2 \mathrm{~m}$ The particle is released from $\mathrm{x}=-3$. So, amplitude from $\mathrm{x}=-3$ and $\mathrm{x}=2$ is $5 \mathrm{~m}$. The maximum value of $\mathrm{x}$ will be- $x=5+2$ $x=7 \mathrm{~m}$
AP EAMCET (18.09.2020) Shift-I
Work, Energy and Power
148978
A particle moves in a straight line with its retardation proportional to its displacement. The loss of its kinetic energy for any displacement $x$ is proportional to
1 $\frac{1}{x}$
2 $x$
3 $x^{2}$
4 $e^{x}$
Explanation:
C Given that, particles move in a straight line with its retardation proportional to its displacement. $\mathrm{a} \propto \mathrm{x}$ $\mathrm{a}=-\mathrm{kx}$ Where, $\mathrm{k}=$ proportionality constant $\mathrm{vdv}=\mathrm{kx} \mathrm{dx}$ Integrating both side, we get- $\int_{v_{i}}^{v_{f}} v d v=-\int_{0}^{x} k x d x$ Where $v_{i}$ and $v_{f}$ are initial and final velocity of particle respectively. $\left(\frac{\mathrm{v}^{2}}{2}\right)_{\mathrm{v}_{\mathrm{i}}}^{\mathrm{v}_{\mathrm{f}}}=-\mathrm{k}\left(\frac{\mathrm{x}^{2}}{2}\right)_{0}^{\mathrm{x}}$ $\frac{\mathrm{v}_{\mathrm{f}}^{2}}{2}-\frac{\mathrm{v}_{\mathrm{i}}^{2}}{2}=-\mathrm{k} \frac{\mathrm{x}^{2}}{2}$ $\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{i}}^{2}=-\frac{\mathrm{mkx}^{2}}{2}$ $(\text { K.E. })_{\text {final }}-(\text { K.E. })_{\text {initial }}=-\frac{1}{2} \mathrm{mkx}^{2}$ Hence, loss in kinetic energy $\propto \mathrm{x}^{2}$
148974
A solid sphere of mass $1 \mathrm{~kg}$ and radius $10 \mathrm{~cm}$ rolls without slipping on a horizontal surface, with velocity of $10 \mathrm{~cm} / \mathrm{s}$. The total kinetic energy of sphere is
1 $0.007 \mathrm{~J}$
2 $0.05 \mathrm{~J}$
3 $0.01 \mathrm{~J}$
4 $0.07 \mathrm{~J}$
Explanation:
A Given, $\text { Mass }(\mathrm{m})=1 \mathrm{~kg}$ $\text { Radius }(\mathrm{r})=10 \mathrm{~cm}=0.1 \mathrm{~m} .$ $\text { Velocity }(\mathrm{v})=10 \mathrm{~cm} / \mathrm{s}=0.1 \mathrm{~m} / \mathrm{s}$ The momentum of inertia of solid sphere (I) $=\frac{2}{5} \mathrm{mr}^{2}$ We know that, $\omega=\frac{\mathrm{v}}{\mathrm{r}}$ Total kinetic energy of a rolling body without slipping (K.E.) $=$ Rotational kinetic energy + linear kinetic energy $\text { K.E. } =\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{5} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{5} \mathrm{mv}^{2}$ $=\frac{7}{10} \mathrm{mv}^{2}$ $=\frac{7}{10} \times 1 \times(0.1)^{2}$ $= 7 \times 10^{-3}$ $= 0.007 \mathrm{~J}$ Thus, total kinetic energy is $0.007 \mathrm{~J}$.
MHT-CET 2020
Work, Energy and Power
148975
A solid sphere is rolling on a frictionless surface with translational velocity ' $v$ '. It climbs that inclined plane from ' $A$ ' to ' $B$ ' and then moves away from $B$ on the smooth horizontal surface. The value of ' $v$ ' should be
A We know, Potential energy $=$ Translational energy + Rotational energy $\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}$ Moment of inertia of solid sphere - $\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}$ And$ $\omega=\frac{\mathrm{v}}{\mathrm{r}}$ Putting the value of $I$ and $\omega$ in equation (i)- $\therefore \frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\mathrm{mgh}$ $\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=m g h$ $\frac{7}{10} \mathrm{mv}^{2}=\mathrm{mgh}$ $\mathrm{v}^{2}\left(\frac{7}{10}\right)=\mathrm{gh}$ $\mathrm{v}^{2}=\frac{10}{7} \mathrm{gh}$ $\mathrm{v}=\sqrt{\frac{10}{7} \mathrm{gh}}=\left[\frac{10 \mathrm{gh}}{7}\right]^{1 / 2}$ Hence, To climb the inclined surface velocity should be greater than, $\sqrt{\frac{10}{7} \mathrm{gh}}$.
MHT-CET 2020
Work, Energy and Power
148976
A machine which is $70 \%$ efficient raised a 10 kg body through a certain distance and spends $100 \mathrm{~J}$ energy. The body is then released. On reaching the ground, the kinetic energy of the body will be
1 0
2 $70 \mathrm{~J}$
3 $50 \mathrm{~J}$
4 $35 \mathrm{~J}$
Explanation:
B Given, mass (m) $=10 \mathrm{~kg}$ Efficiency of machine $=70 \%$ Actual work done by machine, $\mathrm{W}=70 \% \times 100$ $\mathrm{~W}=70 \text { Joule }$ Now, $\mathrm{W}=\mathrm{mgh}$ $70=10 \times 10 \times \mathrm{h} \quad\left[\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]$ $\frac{70}{100} =\mathrm{h}$ $\mathrm{h} =0.70 \mathrm{~m}$ Then, applying conservation of energy- $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}$ $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}=\mathrm{mgh}$ $=10 \times 10 \times 0.70$ $\mathrm{K} . \mathrm{E}=70$ Joule
AP EAMCET (18.09.2020) Shift-II
Work, Energy and Power
148977
A particle is moving on $\mathrm{X}$-axis has potential energy $U=2-20 x+5 x^{2} J$ along $X$-axis. The particle is released at $x=-3$. The maximum value of $x$ will be ( $x$ is in metre and $U$ is in Joules)
1 $5 \mathrm{~m}$
2 $3 \mathrm{~m}$
3 $7 \mathrm{~m}$
4 $8 \mathrm{~m}$
Explanation:
C Given that, $\mathrm{U}=2-20 \mathrm{x}+5 \mathrm{x}^{2} \mathrm{~J}$ We know that, Force $(\mathrm{F})=-\frac{\mathrm{dU}}{\mathrm{dx}}$ $\mathrm{F}=-[-20+10 \mathrm{x}]$ $\mathrm{F}=20-10 \mathrm{x}$ If particle is in equilibrium position then $\mathrm{F}=0$ Putting the value of $F$ in equation (i), we get- $0=20-10 \mathrm{x}$ $20=10 \mathrm{x}$ $\mathrm{x}=2 \mathrm{~m}$ The particle is released from $\mathrm{x}=-3$. So, amplitude from $\mathrm{x}=-3$ and $\mathrm{x}=2$ is $5 \mathrm{~m}$. The maximum value of $\mathrm{x}$ will be- $x=5+2$ $x=7 \mathrm{~m}$
AP EAMCET (18.09.2020) Shift-I
Work, Energy and Power
148978
A particle moves in a straight line with its retardation proportional to its displacement. The loss of its kinetic energy for any displacement $x$ is proportional to
1 $\frac{1}{x}$
2 $x$
3 $x^{2}$
4 $e^{x}$
Explanation:
C Given that, particles move in a straight line with its retardation proportional to its displacement. $\mathrm{a} \propto \mathrm{x}$ $\mathrm{a}=-\mathrm{kx}$ Where, $\mathrm{k}=$ proportionality constant $\mathrm{vdv}=\mathrm{kx} \mathrm{dx}$ Integrating both side, we get- $\int_{v_{i}}^{v_{f}} v d v=-\int_{0}^{x} k x d x$ Where $v_{i}$ and $v_{f}$ are initial and final velocity of particle respectively. $\left(\frac{\mathrm{v}^{2}}{2}\right)_{\mathrm{v}_{\mathrm{i}}}^{\mathrm{v}_{\mathrm{f}}}=-\mathrm{k}\left(\frac{\mathrm{x}^{2}}{2}\right)_{0}^{\mathrm{x}}$ $\frac{\mathrm{v}_{\mathrm{f}}^{2}}{2}-\frac{\mathrm{v}_{\mathrm{i}}^{2}}{2}=-\mathrm{k} \frac{\mathrm{x}^{2}}{2}$ $\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{i}}^{2}=-\frac{\mathrm{mkx}^{2}}{2}$ $(\text { K.E. })_{\text {final }}-(\text { K.E. })_{\text {initial }}=-\frac{1}{2} \mathrm{mkx}^{2}$ Hence, loss in kinetic energy $\propto \mathrm{x}^{2}$
148974
A solid sphere of mass $1 \mathrm{~kg}$ and radius $10 \mathrm{~cm}$ rolls without slipping on a horizontal surface, with velocity of $10 \mathrm{~cm} / \mathrm{s}$. The total kinetic energy of sphere is
1 $0.007 \mathrm{~J}$
2 $0.05 \mathrm{~J}$
3 $0.01 \mathrm{~J}$
4 $0.07 \mathrm{~J}$
Explanation:
A Given, $\text { Mass }(\mathrm{m})=1 \mathrm{~kg}$ $\text { Radius }(\mathrm{r})=10 \mathrm{~cm}=0.1 \mathrm{~m} .$ $\text { Velocity }(\mathrm{v})=10 \mathrm{~cm} / \mathrm{s}=0.1 \mathrm{~m} / \mathrm{s}$ The momentum of inertia of solid sphere (I) $=\frac{2}{5} \mathrm{mr}^{2}$ We know that, $\omega=\frac{\mathrm{v}}{\mathrm{r}}$ Total kinetic energy of a rolling body without slipping (K.E.) $=$ Rotational kinetic energy + linear kinetic energy $\text { K.E. } =\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{5} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}$ $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{5} \mathrm{mv}^{2}$ $=\frac{7}{10} \mathrm{mv}^{2}$ $=\frac{7}{10} \times 1 \times(0.1)^{2}$ $= 7 \times 10^{-3}$ $= 0.007 \mathrm{~J}$ Thus, total kinetic energy is $0.007 \mathrm{~J}$.
MHT-CET 2020
Work, Energy and Power
148975
A solid sphere is rolling on a frictionless surface with translational velocity ' $v$ '. It climbs that inclined plane from ' $A$ ' to ' $B$ ' and then moves away from $B$ on the smooth horizontal surface. The value of ' $v$ ' should be
A We know, Potential energy $=$ Translational energy + Rotational energy $\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}$ Moment of inertia of solid sphere - $\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}$ And$ $\omega=\frac{\mathrm{v}}{\mathrm{r}}$ Putting the value of $I$ and $\omega$ in equation (i)- $\therefore \frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\mathrm{mgh}$ $\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=m g h$ $\frac{7}{10} \mathrm{mv}^{2}=\mathrm{mgh}$ $\mathrm{v}^{2}\left(\frac{7}{10}\right)=\mathrm{gh}$ $\mathrm{v}^{2}=\frac{10}{7} \mathrm{gh}$ $\mathrm{v}=\sqrt{\frac{10}{7} \mathrm{gh}}=\left[\frac{10 \mathrm{gh}}{7}\right]^{1 / 2}$ Hence, To climb the inclined surface velocity should be greater than, $\sqrt{\frac{10}{7} \mathrm{gh}}$.
MHT-CET 2020
Work, Energy and Power
148976
A machine which is $70 \%$ efficient raised a 10 kg body through a certain distance and spends $100 \mathrm{~J}$ energy. The body is then released. On reaching the ground, the kinetic energy of the body will be
1 0
2 $70 \mathrm{~J}$
3 $50 \mathrm{~J}$
4 $35 \mathrm{~J}$
Explanation:
B Given, mass (m) $=10 \mathrm{~kg}$ Efficiency of machine $=70 \%$ Actual work done by machine, $\mathrm{W}=70 \% \times 100$ $\mathrm{~W}=70 \text { Joule }$ Now, $\mathrm{W}=\mathrm{mgh}$ $70=10 \times 10 \times \mathrm{h} \quad\left[\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]$ $\frac{70}{100} =\mathrm{h}$ $\mathrm{h} =0.70 \mathrm{~m}$ Then, applying conservation of energy- $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}$ $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}=\mathrm{mgh}$ $=10 \times 10 \times 0.70$ $\mathrm{K} . \mathrm{E}=70$ Joule
AP EAMCET (18.09.2020) Shift-II
Work, Energy and Power
148977
A particle is moving on $\mathrm{X}$-axis has potential energy $U=2-20 x+5 x^{2} J$ along $X$-axis. The particle is released at $x=-3$. The maximum value of $x$ will be ( $x$ is in metre and $U$ is in Joules)
1 $5 \mathrm{~m}$
2 $3 \mathrm{~m}$
3 $7 \mathrm{~m}$
4 $8 \mathrm{~m}$
Explanation:
C Given that, $\mathrm{U}=2-20 \mathrm{x}+5 \mathrm{x}^{2} \mathrm{~J}$ We know that, Force $(\mathrm{F})=-\frac{\mathrm{dU}}{\mathrm{dx}}$ $\mathrm{F}=-[-20+10 \mathrm{x}]$ $\mathrm{F}=20-10 \mathrm{x}$ If particle is in equilibrium position then $\mathrm{F}=0$ Putting the value of $F$ in equation (i), we get- $0=20-10 \mathrm{x}$ $20=10 \mathrm{x}$ $\mathrm{x}=2 \mathrm{~m}$ The particle is released from $\mathrm{x}=-3$. So, amplitude from $\mathrm{x}=-3$ and $\mathrm{x}=2$ is $5 \mathrm{~m}$. The maximum value of $\mathrm{x}$ will be- $x=5+2$ $x=7 \mathrm{~m}$
AP EAMCET (18.09.2020) Shift-I
Work, Energy and Power
148978
A particle moves in a straight line with its retardation proportional to its displacement. The loss of its kinetic energy for any displacement $x$ is proportional to
1 $\frac{1}{x}$
2 $x$
3 $x^{2}$
4 $e^{x}$
Explanation:
C Given that, particles move in a straight line with its retardation proportional to its displacement. $\mathrm{a} \propto \mathrm{x}$ $\mathrm{a}=-\mathrm{kx}$ Where, $\mathrm{k}=$ proportionality constant $\mathrm{vdv}=\mathrm{kx} \mathrm{dx}$ Integrating both side, we get- $\int_{v_{i}}^{v_{f}} v d v=-\int_{0}^{x} k x d x$ Where $v_{i}$ and $v_{f}$ are initial and final velocity of particle respectively. $\left(\frac{\mathrm{v}^{2}}{2}\right)_{\mathrm{v}_{\mathrm{i}}}^{\mathrm{v}_{\mathrm{f}}}=-\mathrm{k}\left(\frac{\mathrm{x}^{2}}{2}\right)_{0}^{\mathrm{x}}$ $\frac{\mathrm{v}_{\mathrm{f}}^{2}}{2}-\frac{\mathrm{v}_{\mathrm{i}}^{2}}{2}=-\mathrm{k} \frac{\mathrm{x}^{2}}{2}$ $\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{i}}^{2}=-\frac{\mathrm{mkx}^{2}}{2}$ $(\text { K.E. })_{\text {final }}-(\text { K.E. })_{\text {initial }}=-\frac{1}{2} \mathrm{mkx}^{2}$ Hence, loss in kinetic energy $\propto \mathrm{x}^{2}$