138654
The period of revolution of the planet $A$ round the sun is 8 times that of $B$. The distance of $A$ from the sun is how many times greater than that of $B$ from the sun?
1 5
2 4
3 3
4 2
Explanation:
B Given, The period of revolution of planet A around the sun is 8 times that of $\mathrm{B}$ From kepler's $3^{\text {rd }}$ law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\mathrm{~T}_{\mathrm{A}}=8 \mathrm{~T}_{\mathrm{B}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{8}{1}$ $\frac{\mathrm{T}_{\mathrm{A}}^{2}}{\mathrm{~T}_{\mathrm{B}}^{2}}=\frac{\mathrm{R}_{\mathrm{A}}^{3}}{\mathrm{R}_{\mathrm{B}}^{3}}$ $\because \quad \frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{8}{1}$ After putting all the values $\left(\frac{8}{1}\right)^{2}=\frac{\mathrm{R}_{\mathrm{A}}^{3}}{\mathrm{R}_{\mathrm{B}}^{3}}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=4$ $\mathrm{R}_{\mathrm{A}}=4 \mathrm{R}_{\mathrm{B}}$ Hence, radius of $A$ is 4 times the radius of $B$.
AIPMT- 1997
Gravitation
138655
A satellite $A$ of mass $m$ is at a distance $r$ from the surface of the earth. Another satellite $B$ of mass $2 \mathrm{~m}$ is at a distance of $2 \mathrm{r}$ from the earth's surface. Their time periods are in the ratio of
1 $1: 2$
2 $1: 16$
3 $1: 32$
4 $1: 2 \sqrt{2}$
Explanation:
D Mass of satellite does not affect time period. But it depends upon ' $\mathrm{r}$ ' the orbiting radius. $\therefore \quad \mathrm{T}^{2} \propto \mathrm{r}^{3}$ Then, $\quad \mathrm{T}_{\mathrm{A}}^{2} \propto \mathrm{r}_{1}^{3}$ $\mathrm{T}_{\mathrm{A}} \propto \mathrm{r}_{1}^{3 / 2}$ $\text { and } \mathrm{T}_{\mathrm{B}} \propto \mathrm{r}_{2}^{3 / 2}$ $\therefore \frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{\mathrm{r}}{2 \mathrm{r}}\right)^{3 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left[\left(\frac{1}{2}\right)^{3}\right]^{\frac{1}{2}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{1}{8}\right)^{1 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{1}{2 \sqrt{2}}$ $\therefore \quad \mathrm{T}_{\mathrm{A}}: \mathrm{T}_{\mathrm{B}}=1: 2 \sqrt{2}$
AIPMT- 1993
Gravitation
138656
A planet moving along an elliptical orbit is closest to the sun at a distance $r_{1}$ and farthest away at a distance of $r_{2}$. If $v_{1}$ and $v_{2}$ are the linear velocities at these points respectively, then the ratio
A Angular momentum of planet moving around sun remains conserved because the gravitational force and there is no external torque acting on the system. By conservation of angular momentum $\mathrm{mv}_{1} \mathrm{r}_{1}=\mathrm{mv}_{2} \mathrm{r}_{2}$ $\therefore \quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}$
AIPMT- 2011
Gravitation
138657
The period of revolution of Jupiter around the sun is $\mathbf{1 2}$ times the period of revolution of the earth around the sun. The distance between the Jupiter and sun is $n$ times the distance between the earth and sun, and then the value of $\boldsymbol{n}$ is
1 $(144)^{3 / 2}$
2 $(144)^{2 / 3}$
3 $\sqrt[3]{144}$
4 $\sqrt[4]{144}$
Explanation:
C Suppose, the time of revolution of Jupiter is $\mathrm{T}_{\mathrm{j}}$ and the radius of orbit of revolution of Jupiter $\mathrm{r}_{\mathrm{j}}$. The time of revolution of earth is $T_{e}$ and the radius of orbit of revolution is $r_{\mathrm{e}}$. Kepler's $3^{\text {rd }}$ law, So, $\quad T_{j}^{2} \propto r_{j}^{3}$ $\mathrm{T}_{\mathrm{e}}^{2} \propto \mathrm{r}_{\mathrm{e}}^{3}$ On dividing equation (i) and equation (ii), we get - $\therefore \quad \frac{\mathrm{r}_{\mathrm{j}}^{3}}{\mathrm{r}_{\mathrm{e}}^{3}}=\frac{\mathrm{T}_{\mathrm{j}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}$ $\frac{r_{j}}{r_{e}}=\left(\frac{T_{j}}{T_{e}}\right)^{2 / 3}$ $\frac{\mathrm{r}_{\mathrm{j}}}{\mathrm{r}_{\mathrm{e}}}=(12)^{2 / 3} {\left[\therefore \frac{\mathrm{T}_{\mathrm{j}}}{\mathrm{T}_{\mathrm{e}}}=12\right]}$ $\frac{\mathrm{r}_{\mathrm{j}}}{\mathrm{r}_{\mathrm{e}}} =\sqrt[3]{144}$
138654
The period of revolution of the planet $A$ round the sun is 8 times that of $B$. The distance of $A$ from the sun is how many times greater than that of $B$ from the sun?
1 5
2 4
3 3
4 2
Explanation:
B Given, The period of revolution of planet A around the sun is 8 times that of $\mathrm{B}$ From kepler's $3^{\text {rd }}$ law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\mathrm{~T}_{\mathrm{A}}=8 \mathrm{~T}_{\mathrm{B}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{8}{1}$ $\frac{\mathrm{T}_{\mathrm{A}}^{2}}{\mathrm{~T}_{\mathrm{B}}^{2}}=\frac{\mathrm{R}_{\mathrm{A}}^{3}}{\mathrm{R}_{\mathrm{B}}^{3}}$ $\because \quad \frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{8}{1}$ After putting all the values $\left(\frac{8}{1}\right)^{2}=\frac{\mathrm{R}_{\mathrm{A}}^{3}}{\mathrm{R}_{\mathrm{B}}^{3}}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=4$ $\mathrm{R}_{\mathrm{A}}=4 \mathrm{R}_{\mathrm{B}}$ Hence, radius of $A$ is 4 times the radius of $B$.
AIPMT- 1997
Gravitation
138655
A satellite $A$ of mass $m$ is at a distance $r$ from the surface of the earth. Another satellite $B$ of mass $2 \mathrm{~m}$ is at a distance of $2 \mathrm{r}$ from the earth's surface. Their time periods are in the ratio of
1 $1: 2$
2 $1: 16$
3 $1: 32$
4 $1: 2 \sqrt{2}$
Explanation:
D Mass of satellite does not affect time period. But it depends upon ' $\mathrm{r}$ ' the orbiting radius. $\therefore \quad \mathrm{T}^{2} \propto \mathrm{r}^{3}$ Then, $\quad \mathrm{T}_{\mathrm{A}}^{2} \propto \mathrm{r}_{1}^{3}$ $\mathrm{T}_{\mathrm{A}} \propto \mathrm{r}_{1}^{3 / 2}$ $\text { and } \mathrm{T}_{\mathrm{B}} \propto \mathrm{r}_{2}^{3 / 2}$ $\therefore \frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{\mathrm{r}}{2 \mathrm{r}}\right)^{3 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left[\left(\frac{1}{2}\right)^{3}\right]^{\frac{1}{2}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{1}{8}\right)^{1 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{1}{2 \sqrt{2}}$ $\therefore \quad \mathrm{T}_{\mathrm{A}}: \mathrm{T}_{\mathrm{B}}=1: 2 \sqrt{2}$
AIPMT- 1993
Gravitation
138656
A planet moving along an elliptical orbit is closest to the sun at a distance $r_{1}$ and farthest away at a distance of $r_{2}$. If $v_{1}$ and $v_{2}$ are the linear velocities at these points respectively, then the ratio
A Angular momentum of planet moving around sun remains conserved because the gravitational force and there is no external torque acting on the system. By conservation of angular momentum $\mathrm{mv}_{1} \mathrm{r}_{1}=\mathrm{mv}_{2} \mathrm{r}_{2}$ $\therefore \quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}$
AIPMT- 2011
Gravitation
138657
The period of revolution of Jupiter around the sun is $\mathbf{1 2}$ times the period of revolution of the earth around the sun. The distance between the Jupiter and sun is $n$ times the distance between the earth and sun, and then the value of $\boldsymbol{n}$ is
1 $(144)^{3 / 2}$
2 $(144)^{2 / 3}$
3 $\sqrt[3]{144}$
4 $\sqrt[4]{144}$
Explanation:
C Suppose, the time of revolution of Jupiter is $\mathrm{T}_{\mathrm{j}}$ and the radius of orbit of revolution of Jupiter $\mathrm{r}_{\mathrm{j}}$. The time of revolution of earth is $T_{e}$ and the radius of orbit of revolution is $r_{\mathrm{e}}$. Kepler's $3^{\text {rd }}$ law, So, $\quad T_{j}^{2} \propto r_{j}^{3}$ $\mathrm{T}_{\mathrm{e}}^{2} \propto \mathrm{r}_{\mathrm{e}}^{3}$ On dividing equation (i) and equation (ii), we get - $\therefore \quad \frac{\mathrm{r}_{\mathrm{j}}^{3}}{\mathrm{r}_{\mathrm{e}}^{3}}=\frac{\mathrm{T}_{\mathrm{j}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}$ $\frac{r_{j}}{r_{e}}=\left(\frac{T_{j}}{T_{e}}\right)^{2 / 3}$ $\frac{\mathrm{r}_{\mathrm{j}}}{\mathrm{r}_{\mathrm{e}}}=(12)^{2 / 3} {\left[\therefore \frac{\mathrm{T}_{\mathrm{j}}}{\mathrm{T}_{\mathrm{e}}}=12\right]}$ $\frac{\mathrm{r}_{\mathrm{j}}}{\mathrm{r}_{\mathrm{e}}} =\sqrt[3]{144}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138654
The period of revolution of the planet $A$ round the sun is 8 times that of $B$. The distance of $A$ from the sun is how many times greater than that of $B$ from the sun?
1 5
2 4
3 3
4 2
Explanation:
B Given, The period of revolution of planet A around the sun is 8 times that of $\mathrm{B}$ From kepler's $3^{\text {rd }}$ law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\mathrm{~T}_{\mathrm{A}}=8 \mathrm{~T}_{\mathrm{B}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{8}{1}$ $\frac{\mathrm{T}_{\mathrm{A}}^{2}}{\mathrm{~T}_{\mathrm{B}}^{2}}=\frac{\mathrm{R}_{\mathrm{A}}^{3}}{\mathrm{R}_{\mathrm{B}}^{3}}$ $\because \quad \frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{8}{1}$ After putting all the values $\left(\frac{8}{1}\right)^{2}=\frac{\mathrm{R}_{\mathrm{A}}^{3}}{\mathrm{R}_{\mathrm{B}}^{3}}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=4$ $\mathrm{R}_{\mathrm{A}}=4 \mathrm{R}_{\mathrm{B}}$ Hence, radius of $A$ is 4 times the radius of $B$.
AIPMT- 1997
Gravitation
138655
A satellite $A$ of mass $m$ is at a distance $r$ from the surface of the earth. Another satellite $B$ of mass $2 \mathrm{~m}$ is at a distance of $2 \mathrm{r}$ from the earth's surface. Their time periods are in the ratio of
1 $1: 2$
2 $1: 16$
3 $1: 32$
4 $1: 2 \sqrt{2}$
Explanation:
D Mass of satellite does not affect time period. But it depends upon ' $\mathrm{r}$ ' the orbiting radius. $\therefore \quad \mathrm{T}^{2} \propto \mathrm{r}^{3}$ Then, $\quad \mathrm{T}_{\mathrm{A}}^{2} \propto \mathrm{r}_{1}^{3}$ $\mathrm{T}_{\mathrm{A}} \propto \mathrm{r}_{1}^{3 / 2}$ $\text { and } \mathrm{T}_{\mathrm{B}} \propto \mathrm{r}_{2}^{3 / 2}$ $\therefore \frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{\mathrm{r}}{2 \mathrm{r}}\right)^{3 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left[\left(\frac{1}{2}\right)^{3}\right]^{\frac{1}{2}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{1}{8}\right)^{1 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{1}{2 \sqrt{2}}$ $\therefore \quad \mathrm{T}_{\mathrm{A}}: \mathrm{T}_{\mathrm{B}}=1: 2 \sqrt{2}$
AIPMT- 1993
Gravitation
138656
A planet moving along an elliptical orbit is closest to the sun at a distance $r_{1}$ and farthest away at a distance of $r_{2}$. If $v_{1}$ and $v_{2}$ are the linear velocities at these points respectively, then the ratio
A Angular momentum of planet moving around sun remains conserved because the gravitational force and there is no external torque acting on the system. By conservation of angular momentum $\mathrm{mv}_{1} \mathrm{r}_{1}=\mathrm{mv}_{2} \mathrm{r}_{2}$ $\therefore \quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}$
AIPMT- 2011
Gravitation
138657
The period of revolution of Jupiter around the sun is $\mathbf{1 2}$ times the period of revolution of the earth around the sun. The distance between the Jupiter and sun is $n$ times the distance between the earth and sun, and then the value of $\boldsymbol{n}$ is
1 $(144)^{3 / 2}$
2 $(144)^{2 / 3}$
3 $\sqrt[3]{144}$
4 $\sqrt[4]{144}$
Explanation:
C Suppose, the time of revolution of Jupiter is $\mathrm{T}_{\mathrm{j}}$ and the radius of orbit of revolution of Jupiter $\mathrm{r}_{\mathrm{j}}$. The time of revolution of earth is $T_{e}$ and the radius of orbit of revolution is $r_{\mathrm{e}}$. Kepler's $3^{\text {rd }}$ law, So, $\quad T_{j}^{2} \propto r_{j}^{3}$ $\mathrm{T}_{\mathrm{e}}^{2} \propto \mathrm{r}_{\mathrm{e}}^{3}$ On dividing equation (i) and equation (ii), we get - $\therefore \quad \frac{\mathrm{r}_{\mathrm{j}}^{3}}{\mathrm{r}_{\mathrm{e}}^{3}}=\frac{\mathrm{T}_{\mathrm{j}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}$ $\frac{r_{j}}{r_{e}}=\left(\frac{T_{j}}{T_{e}}\right)^{2 / 3}$ $\frac{\mathrm{r}_{\mathrm{j}}}{\mathrm{r}_{\mathrm{e}}}=(12)^{2 / 3} {\left[\therefore \frac{\mathrm{T}_{\mathrm{j}}}{\mathrm{T}_{\mathrm{e}}}=12\right]}$ $\frac{\mathrm{r}_{\mathrm{j}}}{\mathrm{r}_{\mathrm{e}}} =\sqrt[3]{144}$
138654
The period of revolution of the planet $A$ round the sun is 8 times that of $B$. The distance of $A$ from the sun is how many times greater than that of $B$ from the sun?
1 5
2 4
3 3
4 2
Explanation:
B Given, The period of revolution of planet A around the sun is 8 times that of $\mathrm{B}$ From kepler's $3^{\text {rd }}$ law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\mathrm{~T}_{\mathrm{A}}=8 \mathrm{~T}_{\mathrm{B}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{8}{1}$ $\frac{\mathrm{T}_{\mathrm{A}}^{2}}{\mathrm{~T}_{\mathrm{B}}^{2}}=\frac{\mathrm{R}_{\mathrm{A}}^{3}}{\mathrm{R}_{\mathrm{B}}^{3}}$ $\because \quad \frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{8}{1}$ After putting all the values $\left(\frac{8}{1}\right)^{2}=\frac{\mathrm{R}_{\mathrm{A}}^{3}}{\mathrm{R}_{\mathrm{B}}^{3}}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=4$ $\mathrm{R}_{\mathrm{A}}=4 \mathrm{R}_{\mathrm{B}}$ Hence, radius of $A$ is 4 times the radius of $B$.
AIPMT- 1997
Gravitation
138655
A satellite $A$ of mass $m$ is at a distance $r$ from the surface of the earth. Another satellite $B$ of mass $2 \mathrm{~m}$ is at a distance of $2 \mathrm{r}$ from the earth's surface. Their time periods are in the ratio of
1 $1: 2$
2 $1: 16$
3 $1: 32$
4 $1: 2 \sqrt{2}$
Explanation:
D Mass of satellite does not affect time period. But it depends upon ' $\mathrm{r}$ ' the orbiting radius. $\therefore \quad \mathrm{T}^{2} \propto \mathrm{r}^{3}$ Then, $\quad \mathrm{T}_{\mathrm{A}}^{2} \propto \mathrm{r}_{1}^{3}$ $\mathrm{T}_{\mathrm{A}} \propto \mathrm{r}_{1}^{3 / 2}$ $\text { and } \mathrm{T}_{\mathrm{B}} \propto \mathrm{r}_{2}^{3 / 2}$ $\therefore \frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{\mathrm{r}}{2 \mathrm{r}}\right)^{3 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left[\left(\frac{1}{2}\right)^{3}\right]^{\frac{1}{2}}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\left(\frac{1}{8}\right)^{1 / 2}$ $\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=\frac{1}{2 \sqrt{2}}$ $\therefore \quad \mathrm{T}_{\mathrm{A}}: \mathrm{T}_{\mathrm{B}}=1: 2 \sqrt{2}$
AIPMT- 1993
Gravitation
138656
A planet moving along an elliptical orbit is closest to the sun at a distance $r_{1}$ and farthest away at a distance of $r_{2}$. If $v_{1}$ and $v_{2}$ are the linear velocities at these points respectively, then the ratio
A Angular momentum of planet moving around sun remains conserved because the gravitational force and there is no external torque acting on the system. By conservation of angular momentum $\mathrm{mv}_{1} \mathrm{r}_{1}=\mathrm{mv}_{2} \mathrm{r}_{2}$ $\therefore \quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}$
AIPMT- 2011
Gravitation
138657
The period of revolution of Jupiter around the sun is $\mathbf{1 2}$ times the period of revolution of the earth around the sun. The distance between the Jupiter and sun is $n$ times the distance between the earth and sun, and then the value of $\boldsymbol{n}$ is
1 $(144)^{3 / 2}$
2 $(144)^{2 / 3}$
3 $\sqrt[3]{144}$
4 $\sqrt[4]{144}$
Explanation:
C Suppose, the time of revolution of Jupiter is $\mathrm{T}_{\mathrm{j}}$ and the radius of orbit of revolution of Jupiter $\mathrm{r}_{\mathrm{j}}$. The time of revolution of earth is $T_{e}$ and the radius of orbit of revolution is $r_{\mathrm{e}}$. Kepler's $3^{\text {rd }}$ law, So, $\quad T_{j}^{2} \propto r_{j}^{3}$ $\mathrm{T}_{\mathrm{e}}^{2} \propto \mathrm{r}_{\mathrm{e}}^{3}$ On dividing equation (i) and equation (ii), we get - $\therefore \quad \frac{\mathrm{r}_{\mathrm{j}}^{3}}{\mathrm{r}_{\mathrm{e}}^{3}}=\frac{\mathrm{T}_{\mathrm{j}}^{2}}{\mathrm{~T}_{\mathrm{e}}^{2}}$ $\frac{r_{j}}{r_{e}}=\left(\frac{T_{j}}{T_{e}}\right)^{2 / 3}$ $\frac{\mathrm{r}_{\mathrm{j}}}{\mathrm{r}_{\mathrm{e}}}=(12)^{2 / 3} {\left[\therefore \frac{\mathrm{T}_{\mathrm{j}}}{\mathrm{T}_{\mathrm{e}}}=12\right]}$ $\frac{\mathrm{r}_{\mathrm{j}}}{\mathrm{r}_{\mathrm{e}}} =\sqrt[3]{144}$
