138724
The velocity with which a projectile must be fired so that it escapes earth's gravitation does not depend on:
1 mass of the earth
2 mass of the projectile
3 radius of the projectile's orbit
4 gravitational constant
Explanation:
B The kinetic energy given to the body should be equal to potential energy for body to escape i.e. Potential energy $=$ kinetic energy $\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}}=\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Where $\mathrm{m}=$ mass of projectile, $\mathrm{M}_{\mathrm{e}}=$ mass of earth, $\mathrm{G}=$ gravitational constant $\mathrm{R}_{\mathrm{e}}=$ Radius of earth $\therefore \quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From the above formula it is clear that escape velocity is independent of the mass of the projectile.
UPSEE 2001
Gravitation
138725
The escape velocity from the earth is $\mathbf{1 1 . 2}$ $\mathrm{km} / \mathrm{sec}$. The escape velocity from a planet having twice the radius and the same mean density as the earth, is:
1 $11.2 \mathrm{~km} / \mathrm{sec}$
2 $22.4 \mathrm{~km} / \mathrm{sec}$
3 $15.00 \mathrm{~km} / \mathrm{sec}$
4 $5.8 \mathrm{~km} / \mathrm{sec}$
Explanation:
B We know that, $\text { Escape velocity }\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Where, $\quad \mathrm{G}=$ gravitational constant $\mathrm{M}_{\mathrm{e}}=\text { mass of earth }$ $\mathrm{R}_{\mathrm{e}}=\text { radius of earth }$ For planet $\quad \mathrm{R}_{\mathrm{P}}=2 \mathrm{R}_{\mathrm{e}}$ Same density of planet and earth, $\rho_{\mathrm{P}}=\rho_{\mathrm{e}}=\rho$ $\therefore$ Escape veloctiy of earth - \(\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}_{\mathrm{e}}} \times \frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^3 \rho} \quad\left(\begin{array}{rl}\mathrm{M}_{\mathrm{e}} & =\mathrm{V} \times \rho \\ & =\frac{4}{3} \pi \mathrm{R}^3 \cdot \rho\end{array}\right) \ldots\) Escape velocity for planet - $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{G}}{2 \mathrm{R}_{\mathrm{e}}} \times \frac{4}{3} \pi\left(2 \mathrm{R}_{\mathrm{e}}\right)^{3} \rho}$ Dividing eq. (i) by eq. (ii) we get $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{P}}} =\frac{1}{2}$ $\mathrm{v}_{\mathrm{P}} =2 \mathrm{v}_{\mathrm{e}}$ $=2 \times 11.2=22.4 \mathrm{~km} / \mathrm{s}$
MP PMT 1994
Gravitation
138726
If $v_{0}$ be the orbital velocity of a satellite in a circular orbit close to the earth's surface and $v_{e}$ is the escape velocity from the earth, then relation between the two is
C Given that, Orbital velocity of satellite $=\mathrm{v}_{\mathrm{o}}$ Escape velocity form the earth $=v_{e}$ As we know that, $\mathrm{v}_{\mathrm{o}}=\sqrt{\mathrm{gR}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}$ Now, ratio between orbital velocity and escape velocity $\frac{\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{\mathrm{gR}}{2 \mathrm{gR}}}=\frac{1}{\sqrt{2}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{2} \mathrm{v}_{\mathrm{o}}$
AIIMS-2002
Gravitation
138728
Knowing that the mass of the moon is $1 / 81$ times that of earth and its radius is $1 / 4$ the radius of earth. If the escape velocity at the surface of the earth is $11.2 \mathrm{~km} / \mathrm{sec}$, then the value of escape velocity at the surface of the moon is
1 $2.5 \mathrm{~km} / \mathrm{sec}$
2 $0.14 \mathrm{~km} / \mathrm{sec}$
3 $5 \mathrm{~km} / \mathrm{sec}$
4 $8 \mathrm{~km} / \mathrm{sec}$
Explanation:
A Given that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=11.2 \mathrm{~km} / \mathrm{s}$ Mass of moon $(\mathrm{M})=\frac{\mathrm{M}}{81}$ Radius of moon $(\mathrm{R})=\frac{\mathrm{R}}{4}$ So, escape velocity at the surface of the moon is - $\mathrm{v}_{\mathrm{e}}^{\prime}=\sqrt{\frac{2 \mathrm{GM}^{\prime}}{\mathrm{R}^{\prime}}}=\sqrt{\frac{2 \mathrm{GM}}{81(\mathrm{R} / 4)}}=\frac{2}{9} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}^{\prime}=\frac{2}{9} \times 11.2 \mathrm{~km} / \mathrm{s}=2.488 \approx 2.5 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{e}}^{\prime}=2.5 \mathrm{~km} / \mathrm{s}$
138724
The velocity with which a projectile must be fired so that it escapes earth's gravitation does not depend on:
1 mass of the earth
2 mass of the projectile
3 radius of the projectile's orbit
4 gravitational constant
Explanation:
B The kinetic energy given to the body should be equal to potential energy for body to escape i.e. Potential energy $=$ kinetic energy $\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}}=\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Where $\mathrm{m}=$ mass of projectile, $\mathrm{M}_{\mathrm{e}}=$ mass of earth, $\mathrm{G}=$ gravitational constant $\mathrm{R}_{\mathrm{e}}=$ Radius of earth $\therefore \quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From the above formula it is clear that escape velocity is independent of the mass of the projectile.
UPSEE 2001
Gravitation
138725
The escape velocity from the earth is $\mathbf{1 1 . 2}$ $\mathrm{km} / \mathrm{sec}$. The escape velocity from a planet having twice the radius and the same mean density as the earth, is:
1 $11.2 \mathrm{~km} / \mathrm{sec}$
2 $22.4 \mathrm{~km} / \mathrm{sec}$
3 $15.00 \mathrm{~km} / \mathrm{sec}$
4 $5.8 \mathrm{~km} / \mathrm{sec}$
Explanation:
B We know that, $\text { Escape velocity }\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Where, $\quad \mathrm{G}=$ gravitational constant $\mathrm{M}_{\mathrm{e}}=\text { mass of earth }$ $\mathrm{R}_{\mathrm{e}}=\text { radius of earth }$ For planet $\quad \mathrm{R}_{\mathrm{P}}=2 \mathrm{R}_{\mathrm{e}}$ Same density of planet and earth, $\rho_{\mathrm{P}}=\rho_{\mathrm{e}}=\rho$ $\therefore$ Escape veloctiy of earth - \(\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}_{\mathrm{e}}} \times \frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^3 \rho} \quad\left(\begin{array}{rl}\mathrm{M}_{\mathrm{e}} & =\mathrm{V} \times \rho \\ & =\frac{4}{3} \pi \mathrm{R}^3 \cdot \rho\end{array}\right) \ldots\) Escape velocity for planet - $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{G}}{2 \mathrm{R}_{\mathrm{e}}} \times \frac{4}{3} \pi\left(2 \mathrm{R}_{\mathrm{e}}\right)^{3} \rho}$ Dividing eq. (i) by eq. (ii) we get $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{P}}} =\frac{1}{2}$ $\mathrm{v}_{\mathrm{P}} =2 \mathrm{v}_{\mathrm{e}}$ $=2 \times 11.2=22.4 \mathrm{~km} / \mathrm{s}$
MP PMT 1994
Gravitation
138726
If $v_{0}$ be the orbital velocity of a satellite in a circular orbit close to the earth's surface and $v_{e}$ is the escape velocity from the earth, then relation between the two is
C Given that, Orbital velocity of satellite $=\mathrm{v}_{\mathrm{o}}$ Escape velocity form the earth $=v_{e}$ As we know that, $\mathrm{v}_{\mathrm{o}}=\sqrt{\mathrm{gR}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}$ Now, ratio between orbital velocity and escape velocity $\frac{\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{\mathrm{gR}}{2 \mathrm{gR}}}=\frac{1}{\sqrt{2}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{2} \mathrm{v}_{\mathrm{o}}$
AIIMS-2002
Gravitation
138728
Knowing that the mass of the moon is $1 / 81$ times that of earth and its radius is $1 / 4$ the radius of earth. If the escape velocity at the surface of the earth is $11.2 \mathrm{~km} / \mathrm{sec}$, then the value of escape velocity at the surface of the moon is
1 $2.5 \mathrm{~km} / \mathrm{sec}$
2 $0.14 \mathrm{~km} / \mathrm{sec}$
3 $5 \mathrm{~km} / \mathrm{sec}$
4 $8 \mathrm{~km} / \mathrm{sec}$
Explanation:
A Given that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=11.2 \mathrm{~km} / \mathrm{s}$ Mass of moon $(\mathrm{M})=\frac{\mathrm{M}}{81}$ Radius of moon $(\mathrm{R})=\frac{\mathrm{R}}{4}$ So, escape velocity at the surface of the moon is - $\mathrm{v}_{\mathrm{e}}^{\prime}=\sqrt{\frac{2 \mathrm{GM}^{\prime}}{\mathrm{R}^{\prime}}}=\sqrt{\frac{2 \mathrm{GM}}{81(\mathrm{R} / 4)}}=\frac{2}{9} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}^{\prime}=\frac{2}{9} \times 11.2 \mathrm{~km} / \mathrm{s}=2.488 \approx 2.5 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{e}}^{\prime}=2.5 \mathrm{~km} / \mathrm{s}$
138724
The velocity with which a projectile must be fired so that it escapes earth's gravitation does not depend on:
1 mass of the earth
2 mass of the projectile
3 radius of the projectile's orbit
4 gravitational constant
Explanation:
B The kinetic energy given to the body should be equal to potential energy for body to escape i.e. Potential energy $=$ kinetic energy $\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}}=\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Where $\mathrm{m}=$ mass of projectile, $\mathrm{M}_{\mathrm{e}}=$ mass of earth, $\mathrm{G}=$ gravitational constant $\mathrm{R}_{\mathrm{e}}=$ Radius of earth $\therefore \quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From the above formula it is clear that escape velocity is independent of the mass of the projectile.
UPSEE 2001
Gravitation
138725
The escape velocity from the earth is $\mathbf{1 1 . 2}$ $\mathrm{km} / \mathrm{sec}$. The escape velocity from a planet having twice the radius and the same mean density as the earth, is:
1 $11.2 \mathrm{~km} / \mathrm{sec}$
2 $22.4 \mathrm{~km} / \mathrm{sec}$
3 $15.00 \mathrm{~km} / \mathrm{sec}$
4 $5.8 \mathrm{~km} / \mathrm{sec}$
Explanation:
B We know that, $\text { Escape velocity }\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Where, $\quad \mathrm{G}=$ gravitational constant $\mathrm{M}_{\mathrm{e}}=\text { mass of earth }$ $\mathrm{R}_{\mathrm{e}}=\text { radius of earth }$ For planet $\quad \mathrm{R}_{\mathrm{P}}=2 \mathrm{R}_{\mathrm{e}}$ Same density of planet and earth, $\rho_{\mathrm{P}}=\rho_{\mathrm{e}}=\rho$ $\therefore$ Escape veloctiy of earth - \(\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}_{\mathrm{e}}} \times \frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^3 \rho} \quad\left(\begin{array}{rl}\mathrm{M}_{\mathrm{e}} & =\mathrm{V} \times \rho \\ & =\frac{4}{3} \pi \mathrm{R}^3 \cdot \rho\end{array}\right) \ldots\) Escape velocity for planet - $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{G}}{2 \mathrm{R}_{\mathrm{e}}} \times \frac{4}{3} \pi\left(2 \mathrm{R}_{\mathrm{e}}\right)^{3} \rho}$ Dividing eq. (i) by eq. (ii) we get $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{P}}} =\frac{1}{2}$ $\mathrm{v}_{\mathrm{P}} =2 \mathrm{v}_{\mathrm{e}}$ $=2 \times 11.2=22.4 \mathrm{~km} / \mathrm{s}$
MP PMT 1994
Gravitation
138726
If $v_{0}$ be the orbital velocity of a satellite in a circular orbit close to the earth's surface and $v_{e}$ is the escape velocity from the earth, then relation between the two is
C Given that, Orbital velocity of satellite $=\mathrm{v}_{\mathrm{o}}$ Escape velocity form the earth $=v_{e}$ As we know that, $\mathrm{v}_{\mathrm{o}}=\sqrt{\mathrm{gR}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}$ Now, ratio between orbital velocity and escape velocity $\frac{\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{\mathrm{gR}}{2 \mathrm{gR}}}=\frac{1}{\sqrt{2}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{2} \mathrm{v}_{\mathrm{o}}$
AIIMS-2002
Gravitation
138728
Knowing that the mass of the moon is $1 / 81$ times that of earth and its radius is $1 / 4$ the radius of earth. If the escape velocity at the surface of the earth is $11.2 \mathrm{~km} / \mathrm{sec}$, then the value of escape velocity at the surface of the moon is
1 $2.5 \mathrm{~km} / \mathrm{sec}$
2 $0.14 \mathrm{~km} / \mathrm{sec}$
3 $5 \mathrm{~km} / \mathrm{sec}$
4 $8 \mathrm{~km} / \mathrm{sec}$
Explanation:
A Given that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=11.2 \mathrm{~km} / \mathrm{s}$ Mass of moon $(\mathrm{M})=\frac{\mathrm{M}}{81}$ Radius of moon $(\mathrm{R})=\frac{\mathrm{R}}{4}$ So, escape velocity at the surface of the moon is - $\mathrm{v}_{\mathrm{e}}^{\prime}=\sqrt{\frac{2 \mathrm{GM}^{\prime}}{\mathrm{R}^{\prime}}}=\sqrt{\frac{2 \mathrm{GM}}{81(\mathrm{R} / 4)}}=\frac{2}{9} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}^{\prime}=\frac{2}{9} \times 11.2 \mathrm{~km} / \mathrm{s}=2.488 \approx 2.5 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{e}}^{\prime}=2.5 \mathrm{~km} / \mathrm{s}$
138724
The velocity with which a projectile must be fired so that it escapes earth's gravitation does not depend on:
1 mass of the earth
2 mass of the projectile
3 radius of the projectile's orbit
4 gravitational constant
Explanation:
B The kinetic energy given to the body should be equal to potential energy for body to escape i.e. Potential energy $=$ kinetic energy $\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}}=\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Where $\mathrm{m}=$ mass of projectile, $\mathrm{M}_{\mathrm{e}}=$ mass of earth, $\mathrm{G}=$ gravitational constant $\mathrm{R}_{\mathrm{e}}=$ Radius of earth $\therefore \quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From the above formula it is clear that escape velocity is independent of the mass of the projectile.
UPSEE 2001
Gravitation
138725
The escape velocity from the earth is $\mathbf{1 1 . 2}$ $\mathrm{km} / \mathrm{sec}$. The escape velocity from a planet having twice the radius and the same mean density as the earth, is:
1 $11.2 \mathrm{~km} / \mathrm{sec}$
2 $22.4 \mathrm{~km} / \mathrm{sec}$
3 $15.00 \mathrm{~km} / \mathrm{sec}$
4 $5.8 \mathrm{~km} / \mathrm{sec}$
Explanation:
B We know that, $\text { Escape velocity }\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Where, $\quad \mathrm{G}=$ gravitational constant $\mathrm{M}_{\mathrm{e}}=\text { mass of earth }$ $\mathrm{R}_{\mathrm{e}}=\text { radius of earth }$ For planet $\quad \mathrm{R}_{\mathrm{P}}=2 \mathrm{R}_{\mathrm{e}}$ Same density of planet and earth, $\rho_{\mathrm{P}}=\rho_{\mathrm{e}}=\rho$ $\therefore$ Escape veloctiy of earth - \(\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}_{\mathrm{e}}} \times \frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^3 \rho} \quad\left(\begin{array}{rl}\mathrm{M}_{\mathrm{e}} & =\mathrm{V} \times \rho \\ & =\frac{4}{3} \pi \mathrm{R}^3 \cdot \rho\end{array}\right) \ldots\) Escape velocity for planet - $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{G}}{2 \mathrm{R}_{\mathrm{e}}} \times \frac{4}{3} \pi\left(2 \mathrm{R}_{\mathrm{e}}\right)^{3} \rho}$ Dividing eq. (i) by eq. (ii) we get $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{P}}} =\frac{1}{2}$ $\mathrm{v}_{\mathrm{P}} =2 \mathrm{v}_{\mathrm{e}}$ $=2 \times 11.2=22.4 \mathrm{~km} / \mathrm{s}$
MP PMT 1994
Gravitation
138726
If $v_{0}$ be the orbital velocity of a satellite in a circular orbit close to the earth's surface and $v_{e}$ is the escape velocity from the earth, then relation between the two is
C Given that, Orbital velocity of satellite $=\mathrm{v}_{\mathrm{o}}$ Escape velocity form the earth $=v_{e}$ As we know that, $\mathrm{v}_{\mathrm{o}}=\sqrt{\mathrm{gR}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}$ Now, ratio between orbital velocity and escape velocity $\frac{\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{\mathrm{gR}}{2 \mathrm{gR}}}=\frac{1}{\sqrt{2}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{2} \mathrm{v}_{\mathrm{o}}$
AIIMS-2002
Gravitation
138728
Knowing that the mass of the moon is $1 / 81$ times that of earth and its radius is $1 / 4$ the radius of earth. If the escape velocity at the surface of the earth is $11.2 \mathrm{~km} / \mathrm{sec}$, then the value of escape velocity at the surface of the moon is
1 $2.5 \mathrm{~km} / \mathrm{sec}$
2 $0.14 \mathrm{~km} / \mathrm{sec}$
3 $5 \mathrm{~km} / \mathrm{sec}$
4 $8 \mathrm{~km} / \mathrm{sec}$
Explanation:
A Given that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=11.2 \mathrm{~km} / \mathrm{s}$ Mass of moon $(\mathrm{M})=\frac{\mathrm{M}}{81}$ Radius of moon $(\mathrm{R})=\frac{\mathrm{R}}{4}$ So, escape velocity at the surface of the moon is - $\mathrm{v}_{\mathrm{e}}^{\prime}=\sqrt{\frac{2 \mathrm{GM}^{\prime}}{\mathrm{R}^{\prime}}}=\sqrt{\frac{2 \mathrm{GM}}{81(\mathrm{R} / 4)}}=\frac{2}{9} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}^{\prime}=\frac{2}{9} \times 11.2 \mathrm{~km} / \mathrm{s}=2.488 \approx 2.5 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{e}}^{\prime}=2.5 \mathrm{~km} / \mathrm{s}$
