138634
If $A$ is the areal velocity of a planet of mass $M$, its angular momentum is
1 $\frac{\mathrm{M}}{\mathrm{A}}$
2 $2 \mathrm{MA}$
3 $\mathrm{A}^{2} \mathrm{M}$
4 $\mathrm{AM}^{2}$
Explanation:
B According to Kepler's Second law$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{L}}{2 \mathrm{M}}$ Where, $\mathrm{L}=$ angular momentum of planet $\mathrm{dx}=$ area swept by planet in time interval $\mathrm{dt}$ $\mathrm{M}=$ mass of a planet $\therefore \quad \mathrm{A}=\frac{\mathrm{L}}{2 \mathrm{M}}$ $\mathrm{L}=2 \mathrm{MA}$
EAMCET-2002
Gravitation
138635
The angular diameter of a planet measured from Earth is 90". If the diameter of the planet is $\pi \times 10^{6} \mathrm{~m}$, then its distance form the Earth is
1 $3.6 \times 10^{9} \mathrm{~m}$
2 $7.2 \times 10^{9} \mathrm{~m}$
3 $3.6 \times 10^{6} \mathrm{~m}$
4 $7.2 \times 10^{6} \mathrm{~m}$
5 $1.8 \times 10^{8} \mathrm{~m}$
Explanation:
B We know that, $1^{\prime \prime}=\frac{1}{3600} \times \frac{\pi}{180} \mathrm{rad}$ Angular diameter $(\theta)=90 \times \frac{\pi}{3600 \times 180} \mathrm{rad}$ \(\tan \theta=\frac{\mathrm{D}}{\mathrm{r}} \quad\left[\begin{array}{l}\because \theta \text { is very small } \\ \text { So, } \tan \theta=\theta\end{array}\right]\) $\theta=\frac{D}{r}$ $r=\frac{D}{\theta}=\frac{\pi \times 10^{6} \times 3600 \times 180}{90 \times \pi}$ $r=7.2 \times 10^{9} \mathrm{~m}$ Hence, its distance from the Earth is $7.2 \times 10^{9} \mathrm{~m}$.
Kerala CEE 2020
Gravitation
138636
Suppose two planets $A$ and $B$ revolve around a Sun in the galaxy. The semi-major axis of A and $B$ are 1 and 5 AU (astronomical unit) respectively. If the period of revolution of $A$ is 1 year, the period of revolution of $B$ is
1 1 year
2 5 year
3 11 year
4 25 year
5 125 year
Explanation:
C Given, semi- major axis of planets A and B $\mathrm{R}_{\mathrm{A}}=1 \mathrm{AU}, \mathrm{R}_{\mathrm{B}}=5 \mathrm{AU}$, Period of revolution of planet A and planet $B$ $\mathrm{T}_{\mathrm{A}}=1$ year, $\mathrm{T}_{\mathrm{B}}=$ ? According to Kepler's third law- $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\because \quad\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{2}=\left(\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}\right)^{3}$ $\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{2}=\left(\frac{1}{5}\right)^{3}$ $\left(\frac{1}{\mathrm{~T}_{\mathrm{B}}}\right)^{2}=\frac{1}{125}$ $\mathrm{T}_{\mathrm{B}}=\sqrt{125}$ $\mathrm{T}_{\mathrm{B}}=11.18 \approx 11 \text { year }$
Kerala CEE-2019
Gravitation
138637
If the radius of the Earth suddenly decreases by half of its present value. Then the time duration of one day will be
1 6 hours
2 8 hours
3 12 hours
4 24 hours
5 48 hours
Explanation:
A Given, $\mathrm{T}_{1}=24 \mathrm{~h}, \mathrm{r}_{1}=\mathrm{R}$ and $\mathrm{r}_{2}=\frac{\mathrm{R}}{2}$ As we know in the absence of external torque angular momentum is constant. So, $\mathrm{L}_{\text {initial }}=\mathrm{L}_{\text {final }}$ $\mathrm{m}_{1}\left(\mathrm{r}_{1} \times \mathrm{v}_{1}\right)=\mathrm{m}_{2}\left(\mathrm{r}_{2} \times \mathrm{v}_{2}\right)$ $\mathrm{m}_{1} \omega_{1} \mathrm{r}_{1}^{2}=\mathrm{m}_{2} \omega_{2} \mathrm{r}_{2}^{2} \quad(\because \mathrm{v}=\mathrm{r} \omega)$ For Earth, mass remain constant (i.e $\mathrm{m}_{1}=\mathrm{m}_{2}$ ). $\omega_{1} \mathrm{r}_{1}^{2}=\omega_{2} \mathrm{r}_{2}^{2} \quad\left(\because \omega=\frac{2 \pi}{\mathrm{T}}\right)$ $\frac{\mathrm{r}_{1}^{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{r}_{2}^{2}}{\mathrm{~T}_{2}}$ $\mathrm{T}_{2}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2} \mathrm{~T}_{1}$ Putting the value, $r_{1}, r_{2}$ and $T_{1}$ in above equation- $\mathrm{T}_{2}=\left(\frac{\mathrm{R} / 2}{\mathrm{R}}\right)^{2} \times 24=\frac{1}{4} \times 24$ $\mathrm{~T}_{2}=6 \mathrm{~h}$
138634
If $A$ is the areal velocity of a planet of mass $M$, its angular momentum is
1 $\frac{\mathrm{M}}{\mathrm{A}}$
2 $2 \mathrm{MA}$
3 $\mathrm{A}^{2} \mathrm{M}$
4 $\mathrm{AM}^{2}$
Explanation:
B According to Kepler's Second law$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{L}}{2 \mathrm{M}}$ Where, $\mathrm{L}=$ angular momentum of planet $\mathrm{dx}=$ area swept by planet in time interval $\mathrm{dt}$ $\mathrm{M}=$ mass of a planet $\therefore \quad \mathrm{A}=\frac{\mathrm{L}}{2 \mathrm{M}}$ $\mathrm{L}=2 \mathrm{MA}$
EAMCET-2002
Gravitation
138635
The angular diameter of a planet measured from Earth is 90". If the diameter of the planet is $\pi \times 10^{6} \mathrm{~m}$, then its distance form the Earth is
1 $3.6 \times 10^{9} \mathrm{~m}$
2 $7.2 \times 10^{9} \mathrm{~m}$
3 $3.6 \times 10^{6} \mathrm{~m}$
4 $7.2 \times 10^{6} \mathrm{~m}$
5 $1.8 \times 10^{8} \mathrm{~m}$
Explanation:
B We know that, $1^{\prime \prime}=\frac{1}{3600} \times \frac{\pi}{180} \mathrm{rad}$ Angular diameter $(\theta)=90 \times \frac{\pi}{3600 \times 180} \mathrm{rad}$ \(\tan \theta=\frac{\mathrm{D}}{\mathrm{r}} \quad\left[\begin{array}{l}\because \theta \text { is very small } \\ \text { So, } \tan \theta=\theta\end{array}\right]\) $\theta=\frac{D}{r}$ $r=\frac{D}{\theta}=\frac{\pi \times 10^{6} \times 3600 \times 180}{90 \times \pi}$ $r=7.2 \times 10^{9} \mathrm{~m}$ Hence, its distance from the Earth is $7.2 \times 10^{9} \mathrm{~m}$.
Kerala CEE 2020
Gravitation
138636
Suppose two planets $A$ and $B$ revolve around a Sun in the galaxy. The semi-major axis of A and $B$ are 1 and 5 AU (astronomical unit) respectively. If the period of revolution of $A$ is 1 year, the period of revolution of $B$ is
1 1 year
2 5 year
3 11 year
4 25 year
5 125 year
Explanation:
C Given, semi- major axis of planets A and B $\mathrm{R}_{\mathrm{A}}=1 \mathrm{AU}, \mathrm{R}_{\mathrm{B}}=5 \mathrm{AU}$, Period of revolution of planet A and planet $B$ $\mathrm{T}_{\mathrm{A}}=1$ year, $\mathrm{T}_{\mathrm{B}}=$ ? According to Kepler's third law- $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\because \quad\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{2}=\left(\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}\right)^{3}$ $\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{2}=\left(\frac{1}{5}\right)^{3}$ $\left(\frac{1}{\mathrm{~T}_{\mathrm{B}}}\right)^{2}=\frac{1}{125}$ $\mathrm{T}_{\mathrm{B}}=\sqrt{125}$ $\mathrm{T}_{\mathrm{B}}=11.18 \approx 11 \text { year }$
Kerala CEE-2019
Gravitation
138637
If the radius of the Earth suddenly decreases by half of its present value. Then the time duration of one day will be
1 6 hours
2 8 hours
3 12 hours
4 24 hours
5 48 hours
Explanation:
A Given, $\mathrm{T}_{1}=24 \mathrm{~h}, \mathrm{r}_{1}=\mathrm{R}$ and $\mathrm{r}_{2}=\frac{\mathrm{R}}{2}$ As we know in the absence of external torque angular momentum is constant. So, $\mathrm{L}_{\text {initial }}=\mathrm{L}_{\text {final }}$ $\mathrm{m}_{1}\left(\mathrm{r}_{1} \times \mathrm{v}_{1}\right)=\mathrm{m}_{2}\left(\mathrm{r}_{2} \times \mathrm{v}_{2}\right)$ $\mathrm{m}_{1} \omega_{1} \mathrm{r}_{1}^{2}=\mathrm{m}_{2} \omega_{2} \mathrm{r}_{2}^{2} \quad(\because \mathrm{v}=\mathrm{r} \omega)$ For Earth, mass remain constant (i.e $\mathrm{m}_{1}=\mathrm{m}_{2}$ ). $\omega_{1} \mathrm{r}_{1}^{2}=\omega_{2} \mathrm{r}_{2}^{2} \quad\left(\because \omega=\frac{2 \pi}{\mathrm{T}}\right)$ $\frac{\mathrm{r}_{1}^{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{r}_{2}^{2}}{\mathrm{~T}_{2}}$ $\mathrm{T}_{2}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2} \mathrm{~T}_{1}$ Putting the value, $r_{1}, r_{2}$ and $T_{1}$ in above equation- $\mathrm{T}_{2}=\left(\frac{\mathrm{R} / 2}{\mathrm{R}}\right)^{2} \times 24=\frac{1}{4} \times 24$ $\mathrm{~T}_{2}=6 \mathrm{~h}$
138634
If $A$ is the areal velocity of a planet of mass $M$, its angular momentum is
1 $\frac{\mathrm{M}}{\mathrm{A}}$
2 $2 \mathrm{MA}$
3 $\mathrm{A}^{2} \mathrm{M}$
4 $\mathrm{AM}^{2}$
Explanation:
B According to Kepler's Second law$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{L}}{2 \mathrm{M}}$ Where, $\mathrm{L}=$ angular momentum of planet $\mathrm{dx}=$ area swept by planet in time interval $\mathrm{dt}$ $\mathrm{M}=$ mass of a planet $\therefore \quad \mathrm{A}=\frac{\mathrm{L}}{2 \mathrm{M}}$ $\mathrm{L}=2 \mathrm{MA}$
EAMCET-2002
Gravitation
138635
The angular diameter of a planet measured from Earth is 90". If the diameter of the planet is $\pi \times 10^{6} \mathrm{~m}$, then its distance form the Earth is
1 $3.6 \times 10^{9} \mathrm{~m}$
2 $7.2 \times 10^{9} \mathrm{~m}$
3 $3.6 \times 10^{6} \mathrm{~m}$
4 $7.2 \times 10^{6} \mathrm{~m}$
5 $1.8 \times 10^{8} \mathrm{~m}$
Explanation:
B We know that, $1^{\prime \prime}=\frac{1}{3600} \times \frac{\pi}{180} \mathrm{rad}$ Angular diameter $(\theta)=90 \times \frac{\pi}{3600 \times 180} \mathrm{rad}$ \(\tan \theta=\frac{\mathrm{D}}{\mathrm{r}} \quad\left[\begin{array}{l}\because \theta \text { is very small } \\ \text { So, } \tan \theta=\theta\end{array}\right]\) $\theta=\frac{D}{r}$ $r=\frac{D}{\theta}=\frac{\pi \times 10^{6} \times 3600 \times 180}{90 \times \pi}$ $r=7.2 \times 10^{9} \mathrm{~m}$ Hence, its distance from the Earth is $7.2 \times 10^{9} \mathrm{~m}$.
Kerala CEE 2020
Gravitation
138636
Suppose two planets $A$ and $B$ revolve around a Sun in the galaxy. The semi-major axis of A and $B$ are 1 and 5 AU (astronomical unit) respectively. If the period of revolution of $A$ is 1 year, the period of revolution of $B$ is
1 1 year
2 5 year
3 11 year
4 25 year
5 125 year
Explanation:
C Given, semi- major axis of planets A and B $\mathrm{R}_{\mathrm{A}}=1 \mathrm{AU}, \mathrm{R}_{\mathrm{B}}=5 \mathrm{AU}$, Period of revolution of planet A and planet $B$ $\mathrm{T}_{\mathrm{A}}=1$ year, $\mathrm{T}_{\mathrm{B}}=$ ? According to Kepler's third law- $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\because \quad\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{2}=\left(\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}\right)^{3}$ $\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{2}=\left(\frac{1}{5}\right)^{3}$ $\left(\frac{1}{\mathrm{~T}_{\mathrm{B}}}\right)^{2}=\frac{1}{125}$ $\mathrm{T}_{\mathrm{B}}=\sqrt{125}$ $\mathrm{T}_{\mathrm{B}}=11.18 \approx 11 \text { year }$
Kerala CEE-2019
Gravitation
138637
If the radius of the Earth suddenly decreases by half of its present value. Then the time duration of one day will be
1 6 hours
2 8 hours
3 12 hours
4 24 hours
5 48 hours
Explanation:
A Given, $\mathrm{T}_{1}=24 \mathrm{~h}, \mathrm{r}_{1}=\mathrm{R}$ and $\mathrm{r}_{2}=\frac{\mathrm{R}}{2}$ As we know in the absence of external torque angular momentum is constant. So, $\mathrm{L}_{\text {initial }}=\mathrm{L}_{\text {final }}$ $\mathrm{m}_{1}\left(\mathrm{r}_{1} \times \mathrm{v}_{1}\right)=\mathrm{m}_{2}\left(\mathrm{r}_{2} \times \mathrm{v}_{2}\right)$ $\mathrm{m}_{1} \omega_{1} \mathrm{r}_{1}^{2}=\mathrm{m}_{2} \omega_{2} \mathrm{r}_{2}^{2} \quad(\because \mathrm{v}=\mathrm{r} \omega)$ For Earth, mass remain constant (i.e $\mathrm{m}_{1}=\mathrm{m}_{2}$ ). $\omega_{1} \mathrm{r}_{1}^{2}=\omega_{2} \mathrm{r}_{2}^{2} \quad\left(\because \omega=\frac{2 \pi}{\mathrm{T}}\right)$ $\frac{\mathrm{r}_{1}^{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{r}_{2}^{2}}{\mathrm{~T}_{2}}$ $\mathrm{T}_{2}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2} \mathrm{~T}_{1}$ Putting the value, $r_{1}, r_{2}$ and $T_{1}$ in above equation- $\mathrm{T}_{2}=\left(\frac{\mathrm{R} / 2}{\mathrm{R}}\right)^{2} \times 24=\frac{1}{4} \times 24$ $\mathrm{~T}_{2}=6 \mathrm{~h}$
138634
If $A$ is the areal velocity of a planet of mass $M$, its angular momentum is
1 $\frac{\mathrm{M}}{\mathrm{A}}$
2 $2 \mathrm{MA}$
3 $\mathrm{A}^{2} \mathrm{M}$
4 $\mathrm{AM}^{2}$
Explanation:
B According to Kepler's Second law$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{L}}{2 \mathrm{M}}$ Where, $\mathrm{L}=$ angular momentum of planet $\mathrm{dx}=$ area swept by planet in time interval $\mathrm{dt}$ $\mathrm{M}=$ mass of a planet $\therefore \quad \mathrm{A}=\frac{\mathrm{L}}{2 \mathrm{M}}$ $\mathrm{L}=2 \mathrm{MA}$
EAMCET-2002
Gravitation
138635
The angular diameter of a planet measured from Earth is 90". If the diameter of the planet is $\pi \times 10^{6} \mathrm{~m}$, then its distance form the Earth is
1 $3.6 \times 10^{9} \mathrm{~m}$
2 $7.2 \times 10^{9} \mathrm{~m}$
3 $3.6 \times 10^{6} \mathrm{~m}$
4 $7.2 \times 10^{6} \mathrm{~m}$
5 $1.8 \times 10^{8} \mathrm{~m}$
Explanation:
B We know that, $1^{\prime \prime}=\frac{1}{3600} \times \frac{\pi}{180} \mathrm{rad}$ Angular diameter $(\theta)=90 \times \frac{\pi}{3600 \times 180} \mathrm{rad}$ \(\tan \theta=\frac{\mathrm{D}}{\mathrm{r}} \quad\left[\begin{array}{l}\because \theta \text { is very small } \\ \text { So, } \tan \theta=\theta\end{array}\right]\) $\theta=\frac{D}{r}$ $r=\frac{D}{\theta}=\frac{\pi \times 10^{6} \times 3600 \times 180}{90 \times \pi}$ $r=7.2 \times 10^{9} \mathrm{~m}$ Hence, its distance from the Earth is $7.2 \times 10^{9} \mathrm{~m}$.
Kerala CEE 2020
Gravitation
138636
Suppose two planets $A$ and $B$ revolve around a Sun in the galaxy. The semi-major axis of A and $B$ are 1 and 5 AU (astronomical unit) respectively. If the period of revolution of $A$ is 1 year, the period of revolution of $B$ is
1 1 year
2 5 year
3 11 year
4 25 year
5 125 year
Explanation:
C Given, semi- major axis of planets A and B $\mathrm{R}_{\mathrm{A}}=1 \mathrm{AU}, \mathrm{R}_{\mathrm{B}}=5 \mathrm{AU}$, Period of revolution of planet A and planet $B$ $\mathrm{T}_{\mathrm{A}}=1$ year, $\mathrm{T}_{\mathrm{B}}=$ ? According to Kepler's third law- $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\because \quad\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{2}=\left(\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}\right)^{3}$ $\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{2}=\left(\frac{1}{5}\right)^{3}$ $\left(\frac{1}{\mathrm{~T}_{\mathrm{B}}}\right)^{2}=\frac{1}{125}$ $\mathrm{T}_{\mathrm{B}}=\sqrt{125}$ $\mathrm{T}_{\mathrm{B}}=11.18 \approx 11 \text { year }$
Kerala CEE-2019
Gravitation
138637
If the radius of the Earth suddenly decreases by half of its present value. Then the time duration of one day will be
1 6 hours
2 8 hours
3 12 hours
4 24 hours
5 48 hours
Explanation:
A Given, $\mathrm{T}_{1}=24 \mathrm{~h}, \mathrm{r}_{1}=\mathrm{R}$ and $\mathrm{r}_{2}=\frac{\mathrm{R}}{2}$ As we know in the absence of external torque angular momentum is constant. So, $\mathrm{L}_{\text {initial }}=\mathrm{L}_{\text {final }}$ $\mathrm{m}_{1}\left(\mathrm{r}_{1} \times \mathrm{v}_{1}\right)=\mathrm{m}_{2}\left(\mathrm{r}_{2} \times \mathrm{v}_{2}\right)$ $\mathrm{m}_{1} \omega_{1} \mathrm{r}_{1}^{2}=\mathrm{m}_{2} \omega_{2} \mathrm{r}_{2}^{2} \quad(\because \mathrm{v}=\mathrm{r} \omega)$ For Earth, mass remain constant (i.e $\mathrm{m}_{1}=\mathrm{m}_{2}$ ). $\omega_{1} \mathrm{r}_{1}^{2}=\omega_{2} \mathrm{r}_{2}^{2} \quad\left(\because \omega=\frac{2 \pi}{\mathrm{T}}\right)$ $\frac{\mathrm{r}_{1}^{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{r}_{2}^{2}}{\mathrm{~T}_{2}}$ $\mathrm{T}_{2}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2} \mathrm{~T}_{1}$ Putting the value, $r_{1}, r_{2}$ and $T_{1}$ in above equation- $\mathrm{T}_{2}=\left(\frac{\mathrm{R} / 2}{\mathrm{R}}\right)^{2} \times 24=\frac{1}{4} \times 24$ $\mathrm{~T}_{2}=6 \mathrm{~h}$
