138597
If the distance between the earth and the sun is half its present value, the number of days in a year would have been
1 730
2 182.5
3 129
4 64.5
Explanation:
C Given that, If the distance between the earth and the sun is half its present value. $\therefore \quad \mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{2}$ And $\mathrm{T}_{1}=365$ days By Kepler's law- $\mathrm{T}_{1}^{2} \propto \mathrm{R}_{1}^{3}$ $\text { And } \quad \mathrm{T}_{2}^{2} \propto \mathrm{R}_{2}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}} =\left(\frac{\mathrm{R}_{1}}{\frac{\mathrm{R}_{1}}{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}} =(2)^{3}$ $\mathrm{~T}_{2}^{2} =\frac{\mathrm{T}_{1}^{2}}{(2)^{3}}$ $\mathrm{~T}_{2} =\sqrt{\frac{\mathrm{T}_{1}^{2}}{(2)^{3}}}$ $\mathrm{~T}_{2} =\frac{\mathrm{T}_{1}}{2 \sqrt{2}}$ $\mathrm{~T}_{2} =\frac{365}{2 \sqrt{2}}$ $\mathrm{~T}_{2} =129 \text { days } \quad\left[\because \mathrm{T}_{1}=365 \text { days }\right]$
UPSEE - 2013
Gravitation
138598
The rotation of the earth having radius $R$ about its axis speeds upto a value such that a man at latitude angle $60^{\circ}$ feels weightless. The duration of the day in such case will be:
1 $8 \pi \sqrt{\frac{R}{g}}$
2 $8 \pi \sqrt{\frac{g}{R}}$
3 $\pi \sqrt{\frac{R}{g}}$
4 $4 \pi \sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$
Explanation:
C Given that, $\theta=60^{\circ}$ We know that, $\mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R} \omega^{2} \cos ^{2} \theta$ In weightless conditions- $g^{\prime}=0$ $0=g-R \omega^{2} \cos ^{2} \theta$ $g=R \omega^{2} \cos ^{2} 60^{\circ}$ $g=R \omega^{2}\left(\frac{1}{2}\right)^{2}$ $g=R \omega^{2} \frac{1}{4}$ $\omega=2 \sqrt{\frac{g}{R}}$ Again we know that, $\omega=2 \pi n$ $\omega=\frac{2 \pi}{\mathrm{T}} \quad\left[\because \mathrm{n}=\frac{1}{\mathrm{~T}}\right]$ $\frac{2 \pi}{\mathrm{T}}=2 \sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$ $\frac{\mathrm{T}}{\pi}=\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$ $\mathrm{T}=\pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$
BITSAT-2009
Gravitation
138599
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius ' $R$ ' around the sun will be proportional to
C Suppose that, Mass of Planet $=\mathrm{m}$ Mass of Sun $=M$ Speed of Planet $=\mathrm{v}$ Distance $=\mathrm{R}$ Given, $\quad \mathrm{F}_{\mathrm{g}} \propto \frac{1}{\mathrm{R}^{\mathrm{n}}}$ $\mathrm{F}_{\mathrm{g}}=\frac{\mathrm{GmM}}{\mathrm{R}^{\mathrm{n}}}$ The centripetal force $\left(\mathrm{F}_{\mathrm{c}}\right)=\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ Equating equation (i) and (ii) we get, $\frac{\mathrm{mv}^{2}}{\mathrm{R}} =\frac{\mathrm{GmM}}{\mathrm{R}^{\mathrm{n}}}$ $\mathrm{v} =\sqrt{\frac{\mathrm{GM}}{\mathrm{R}^{\mathrm{n}-1}}}$ Now, time period, $T=\frac{2 \pi R}{v}$ Put the value $v$ in equation (iii) we get, Then, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}^{\mathrm{n}-1}}}} \mathrm{~T}=\frac{2 \pi \mathrm{R}\left(\mathrm{R}^{\mathrm{n}-1}\right)^{1 / 2}}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}\left(\mathrm{R}^{\frac{\mathrm{n}}{2}-\frac{1}{2}}\right)}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}^{\frac{\mathrm{n}}{2}-\frac{1}{2}+1}\right)}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}^{\left(\frac{\mathrm{n}+1}{2}\right)}\right)}{\sqrt{\mathrm{GM}}}$ Therefore, $\mathrm{T} \propto \mathrm{R}^{\left(\frac{\mathrm{n}+1}{2}\right)}$
BITSAT-2010
Gravitation
138601
A system of binary stars of masses $m_{A}$ and $m_{B}$ are moving in circular orbits of radii $r_{A}$ and $r_{B}$ respectively. If $T_{A}$ and $T_{B}$ are the time periods of masses $m_{A}$ and $m_{B}$ respectively, then
D Since, $\mathrm{C}$ is the centre of mass. $\therefore \quad \mathrm{m}_{\mathrm{A}} \mathrm{r}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{r}_{\mathrm{B}}$ For the rotation of stars, $\mathrm{F}_{\text {gravitational }}=\mathrm{F}_{\text {centripetal }}$ $\frac{\mathrm{Gm}_{\mathrm{A}} \mathrm{m}_{\mathrm{B}}}{\left(\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}\right)^{2}}=\frac{\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}^{2}}{\mathrm{r}_{\mathrm{B}}}=\frac{\mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{A}}}$ Put the value $m_{B}$ from equation (i) we get Then, $\frac{m_{A} r_{A} v_{B}^{2}}{r_{B}^{2}}=\frac{m_{A} v_{A}^{2}}{r_{A}}$ $\frac{v_{A}^{2}}{v_{B}^{2}}=\frac{r_{A}^{2}}{r_{B}^{2}}$ $\frac{r_{A} v_{B}}{r_{B} v_{A}}=1$ Time periods of stars, $\mathrm{T}_{\mathrm{A}}=\frac{2 \pi \mathrm{r}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{A}}}$ $\mathrm{T}_{\mathrm{B}}=\frac{2 \pi \mathrm{r}_{\mathrm{B}}}{\mathrm{v}_{\mathrm{B}}}$ Dividing equation (iii) by (iv) we get - $\frac{T_{A}}{T_{B}}=\frac{r_{A} \cdot v_{B}}{r_{B} \cdot v_{A}}$ $\frac{T_{A}}{T_{B}}=1$ $\therefore \quad \mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}}$
138597
If the distance between the earth and the sun is half its present value, the number of days in a year would have been
1 730
2 182.5
3 129
4 64.5
Explanation:
C Given that, If the distance between the earth and the sun is half its present value. $\therefore \quad \mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{2}$ And $\mathrm{T}_{1}=365$ days By Kepler's law- $\mathrm{T}_{1}^{2} \propto \mathrm{R}_{1}^{3}$ $\text { And } \quad \mathrm{T}_{2}^{2} \propto \mathrm{R}_{2}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}} =\left(\frac{\mathrm{R}_{1}}{\frac{\mathrm{R}_{1}}{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}} =(2)^{3}$ $\mathrm{~T}_{2}^{2} =\frac{\mathrm{T}_{1}^{2}}{(2)^{3}}$ $\mathrm{~T}_{2} =\sqrt{\frac{\mathrm{T}_{1}^{2}}{(2)^{3}}}$ $\mathrm{~T}_{2} =\frac{\mathrm{T}_{1}}{2 \sqrt{2}}$ $\mathrm{~T}_{2} =\frac{365}{2 \sqrt{2}}$ $\mathrm{~T}_{2} =129 \text { days } \quad\left[\because \mathrm{T}_{1}=365 \text { days }\right]$
UPSEE - 2013
Gravitation
138598
The rotation of the earth having radius $R$ about its axis speeds upto a value such that a man at latitude angle $60^{\circ}$ feels weightless. The duration of the day in such case will be:
1 $8 \pi \sqrt{\frac{R}{g}}$
2 $8 \pi \sqrt{\frac{g}{R}}$
3 $\pi \sqrt{\frac{R}{g}}$
4 $4 \pi \sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$
Explanation:
C Given that, $\theta=60^{\circ}$ We know that, $\mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R} \omega^{2} \cos ^{2} \theta$ In weightless conditions- $g^{\prime}=0$ $0=g-R \omega^{2} \cos ^{2} \theta$ $g=R \omega^{2} \cos ^{2} 60^{\circ}$ $g=R \omega^{2}\left(\frac{1}{2}\right)^{2}$ $g=R \omega^{2} \frac{1}{4}$ $\omega=2 \sqrt{\frac{g}{R}}$ Again we know that, $\omega=2 \pi n$ $\omega=\frac{2 \pi}{\mathrm{T}} \quad\left[\because \mathrm{n}=\frac{1}{\mathrm{~T}}\right]$ $\frac{2 \pi}{\mathrm{T}}=2 \sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$ $\frac{\mathrm{T}}{\pi}=\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$ $\mathrm{T}=\pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$
BITSAT-2009
Gravitation
138599
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius ' $R$ ' around the sun will be proportional to
C Suppose that, Mass of Planet $=\mathrm{m}$ Mass of Sun $=M$ Speed of Planet $=\mathrm{v}$ Distance $=\mathrm{R}$ Given, $\quad \mathrm{F}_{\mathrm{g}} \propto \frac{1}{\mathrm{R}^{\mathrm{n}}}$ $\mathrm{F}_{\mathrm{g}}=\frac{\mathrm{GmM}}{\mathrm{R}^{\mathrm{n}}}$ The centripetal force $\left(\mathrm{F}_{\mathrm{c}}\right)=\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ Equating equation (i) and (ii) we get, $\frac{\mathrm{mv}^{2}}{\mathrm{R}} =\frac{\mathrm{GmM}}{\mathrm{R}^{\mathrm{n}}}$ $\mathrm{v} =\sqrt{\frac{\mathrm{GM}}{\mathrm{R}^{\mathrm{n}-1}}}$ Now, time period, $T=\frac{2 \pi R}{v}$ Put the value $v$ in equation (iii) we get, Then, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}^{\mathrm{n}-1}}}} \mathrm{~T}=\frac{2 \pi \mathrm{R}\left(\mathrm{R}^{\mathrm{n}-1}\right)^{1 / 2}}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}\left(\mathrm{R}^{\frac{\mathrm{n}}{2}-\frac{1}{2}}\right)}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}^{\frac{\mathrm{n}}{2}-\frac{1}{2}+1}\right)}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}^{\left(\frac{\mathrm{n}+1}{2}\right)}\right)}{\sqrt{\mathrm{GM}}}$ Therefore, $\mathrm{T} \propto \mathrm{R}^{\left(\frac{\mathrm{n}+1}{2}\right)}$
BITSAT-2010
Gravitation
138601
A system of binary stars of masses $m_{A}$ and $m_{B}$ are moving in circular orbits of radii $r_{A}$ and $r_{B}$ respectively. If $T_{A}$ and $T_{B}$ are the time periods of masses $m_{A}$ and $m_{B}$ respectively, then
D Since, $\mathrm{C}$ is the centre of mass. $\therefore \quad \mathrm{m}_{\mathrm{A}} \mathrm{r}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{r}_{\mathrm{B}}$ For the rotation of stars, $\mathrm{F}_{\text {gravitational }}=\mathrm{F}_{\text {centripetal }}$ $\frac{\mathrm{Gm}_{\mathrm{A}} \mathrm{m}_{\mathrm{B}}}{\left(\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}\right)^{2}}=\frac{\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}^{2}}{\mathrm{r}_{\mathrm{B}}}=\frac{\mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{A}}}$ Put the value $m_{B}$ from equation (i) we get Then, $\frac{m_{A} r_{A} v_{B}^{2}}{r_{B}^{2}}=\frac{m_{A} v_{A}^{2}}{r_{A}}$ $\frac{v_{A}^{2}}{v_{B}^{2}}=\frac{r_{A}^{2}}{r_{B}^{2}}$ $\frac{r_{A} v_{B}}{r_{B} v_{A}}=1$ Time periods of stars, $\mathrm{T}_{\mathrm{A}}=\frac{2 \pi \mathrm{r}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{A}}}$ $\mathrm{T}_{\mathrm{B}}=\frac{2 \pi \mathrm{r}_{\mathrm{B}}}{\mathrm{v}_{\mathrm{B}}}$ Dividing equation (iii) by (iv) we get - $\frac{T_{A}}{T_{B}}=\frac{r_{A} \cdot v_{B}}{r_{B} \cdot v_{A}}$ $\frac{T_{A}}{T_{B}}=1$ $\therefore \quad \mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}}$
138597
If the distance between the earth and the sun is half its present value, the number of days in a year would have been
1 730
2 182.5
3 129
4 64.5
Explanation:
C Given that, If the distance between the earth and the sun is half its present value. $\therefore \quad \mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{2}$ And $\mathrm{T}_{1}=365$ days By Kepler's law- $\mathrm{T}_{1}^{2} \propto \mathrm{R}_{1}^{3}$ $\text { And } \quad \mathrm{T}_{2}^{2} \propto \mathrm{R}_{2}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}} =\left(\frac{\mathrm{R}_{1}}{\frac{\mathrm{R}_{1}}{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}} =(2)^{3}$ $\mathrm{~T}_{2}^{2} =\frac{\mathrm{T}_{1}^{2}}{(2)^{3}}$ $\mathrm{~T}_{2} =\sqrt{\frac{\mathrm{T}_{1}^{2}}{(2)^{3}}}$ $\mathrm{~T}_{2} =\frac{\mathrm{T}_{1}}{2 \sqrt{2}}$ $\mathrm{~T}_{2} =\frac{365}{2 \sqrt{2}}$ $\mathrm{~T}_{2} =129 \text { days } \quad\left[\because \mathrm{T}_{1}=365 \text { days }\right]$
UPSEE - 2013
Gravitation
138598
The rotation of the earth having radius $R$ about its axis speeds upto a value such that a man at latitude angle $60^{\circ}$ feels weightless. The duration of the day in such case will be:
1 $8 \pi \sqrt{\frac{R}{g}}$
2 $8 \pi \sqrt{\frac{g}{R}}$
3 $\pi \sqrt{\frac{R}{g}}$
4 $4 \pi \sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$
Explanation:
C Given that, $\theta=60^{\circ}$ We know that, $\mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R} \omega^{2} \cos ^{2} \theta$ In weightless conditions- $g^{\prime}=0$ $0=g-R \omega^{2} \cos ^{2} \theta$ $g=R \omega^{2} \cos ^{2} 60^{\circ}$ $g=R \omega^{2}\left(\frac{1}{2}\right)^{2}$ $g=R \omega^{2} \frac{1}{4}$ $\omega=2 \sqrt{\frac{g}{R}}$ Again we know that, $\omega=2 \pi n$ $\omega=\frac{2 \pi}{\mathrm{T}} \quad\left[\because \mathrm{n}=\frac{1}{\mathrm{~T}}\right]$ $\frac{2 \pi}{\mathrm{T}}=2 \sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$ $\frac{\mathrm{T}}{\pi}=\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$ $\mathrm{T}=\pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$
BITSAT-2009
Gravitation
138599
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius ' $R$ ' around the sun will be proportional to
C Suppose that, Mass of Planet $=\mathrm{m}$ Mass of Sun $=M$ Speed of Planet $=\mathrm{v}$ Distance $=\mathrm{R}$ Given, $\quad \mathrm{F}_{\mathrm{g}} \propto \frac{1}{\mathrm{R}^{\mathrm{n}}}$ $\mathrm{F}_{\mathrm{g}}=\frac{\mathrm{GmM}}{\mathrm{R}^{\mathrm{n}}}$ The centripetal force $\left(\mathrm{F}_{\mathrm{c}}\right)=\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ Equating equation (i) and (ii) we get, $\frac{\mathrm{mv}^{2}}{\mathrm{R}} =\frac{\mathrm{GmM}}{\mathrm{R}^{\mathrm{n}}}$ $\mathrm{v} =\sqrt{\frac{\mathrm{GM}}{\mathrm{R}^{\mathrm{n}-1}}}$ Now, time period, $T=\frac{2 \pi R}{v}$ Put the value $v$ in equation (iii) we get, Then, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}^{\mathrm{n}-1}}}} \mathrm{~T}=\frac{2 \pi \mathrm{R}\left(\mathrm{R}^{\mathrm{n}-1}\right)^{1 / 2}}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}\left(\mathrm{R}^{\frac{\mathrm{n}}{2}-\frac{1}{2}}\right)}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}^{\frac{\mathrm{n}}{2}-\frac{1}{2}+1}\right)}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}^{\left(\frac{\mathrm{n}+1}{2}\right)}\right)}{\sqrt{\mathrm{GM}}}$ Therefore, $\mathrm{T} \propto \mathrm{R}^{\left(\frac{\mathrm{n}+1}{2}\right)}$
BITSAT-2010
Gravitation
138601
A system of binary stars of masses $m_{A}$ and $m_{B}$ are moving in circular orbits of radii $r_{A}$ and $r_{B}$ respectively. If $T_{A}$ and $T_{B}$ are the time periods of masses $m_{A}$ and $m_{B}$ respectively, then
D Since, $\mathrm{C}$ is the centre of mass. $\therefore \quad \mathrm{m}_{\mathrm{A}} \mathrm{r}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{r}_{\mathrm{B}}$ For the rotation of stars, $\mathrm{F}_{\text {gravitational }}=\mathrm{F}_{\text {centripetal }}$ $\frac{\mathrm{Gm}_{\mathrm{A}} \mathrm{m}_{\mathrm{B}}}{\left(\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}\right)^{2}}=\frac{\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}^{2}}{\mathrm{r}_{\mathrm{B}}}=\frac{\mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{A}}}$ Put the value $m_{B}$ from equation (i) we get Then, $\frac{m_{A} r_{A} v_{B}^{2}}{r_{B}^{2}}=\frac{m_{A} v_{A}^{2}}{r_{A}}$ $\frac{v_{A}^{2}}{v_{B}^{2}}=\frac{r_{A}^{2}}{r_{B}^{2}}$ $\frac{r_{A} v_{B}}{r_{B} v_{A}}=1$ Time periods of stars, $\mathrm{T}_{\mathrm{A}}=\frac{2 \pi \mathrm{r}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{A}}}$ $\mathrm{T}_{\mathrm{B}}=\frac{2 \pi \mathrm{r}_{\mathrm{B}}}{\mathrm{v}_{\mathrm{B}}}$ Dividing equation (iii) by (iv) we get - $\frac{T_{A}}{T_{B}}=\frac{r_{A} \cdot v_{B}}{r_{B} \cdot v_{A}}$ $\frac{T_{A}}{T_{B}}=1$ $\therefore \quad \mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}}$
138597
If the distance between the earth and the sun is half its present value, the number of days in a year would have been
1 730
2 182.5
3 129
4 64.5
Explanation:
C Given that, If the distance between the earth and the sun is half its present value. $\therefore \quad \mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{2}$ And $\mathrm{T}_{1}=365$ days By Kepler's law- $\mathrm{T}_{1}^{2} \propto \mathrm{R}_{1}^{3}$ $\text { And } \quad \mathrm{T}_{2}^{2} \propto \mathrm{R}_{2}^{3}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}} =\left(\frac{\mathrm{R}_{1}}{\frac{\mathrm{R}_{1}}{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}} =(2)^{3}$ $\mathrm{~T}_{2}^{2} =\frac{\mathrm{T}_{1}^{2}}{(2)^{3}}$ $\mathrm{~T}_{2} =\sqrt{\frac{\mathrm{T}_{1}^{2}}{(2)^{3}}}$ $\mathrm{~T}_{2} =\frac{\mathrm{T}_{1}}{2 \sqrt{2}}$ $\mathrm{~T}_{2} =\frac{365}{2 \sqrt{2}}$ $\mathrm{~T}_{2} =129 \text { days } \quad\left[\because \mathrm{T}_{1}=365 \text { days }\right]$
UPSEE - 2013
Gravitation
138598
The rotation of the earth having radius $R$ about its axis speeds upto a value such that a man at latitude angle $60^{\circ}$ feels weightless. The duration of the day in such case will be:
1 $8 \pi \sqrt{\frac{R}{g}}$
2 $8 \pi \sqrt{\frac{g}{R}}$
3 $\pi \sqrt{\frac{R}{g}}$
4 $4 \pi \sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$
Explanation:
C Given that, $\theta=60^{\circ}$ We know that, $\mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R} \omega^{2} \cos ^{2} \theta$ In weightless conditions- $g^{\prime}=0$ $0=g-R \omega^{2} \cos ^{2} \theta$ $g=R \omega^{2} \cos ^{2} 60^{\circ}$ $g=R \omega^{2}\left(\frac{1}{2}\right)^{2}$ $g=R \omega^{2} \frac{1}{4}$ $\omega=2 \sqrt{\frac{g}{R}}$ Again we know that, $\omega=2 \pi n$ $\omega=\frac{2 \pi}{\mathrm{T}} \quad\left[\because \mathrm{n}=\frac{1}{\mathrm{~T}}\right]$ $\frac{2 \pi}{\mathrm{T}}=2 \sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$ $\frac{\mathrm{T}}{\pi}=\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$ $\mathrm{T}=\pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$
BITSAT-2009
Gravitation
138599
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius ' $R$ ' around the sun will be proportional to
C Suppose that, Mass of Planet $=\mathrm{m}$ Mass of Sun $=M$ Speed of Planet $=\mathrm{v}$ Distance $=\mathrm{R}$ Given, $\quad \mathrm{F}_{\mathrm{g}} \propto \frac{1}{\mathrm{R}^{\mathrm{n}}}$ $\mathrm{F}_{\mathrm{g}}=\frac{\mathrm{GmM}}{\mathrm{R}^{\mathrm{n}}}$ The centripetal force $\left(\mathrm{F}_{\mathrm{c}}\right)=\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ Equating equation (i) and (ii) we get, $\frac{\mathrm{mv}^{2}}{\mathrm{R}} =\frac{\mathrm{GmM}}{\mathrm{R}^{\mathrm{n}}}$ $\mathrm{v} =\sqrt{\frac{\mathrm{GM}}{\mathrm{R}^{\mathrm{n}-1}}}$ Now, time period, $T=\frac{2 \pi R}{v}$ Put the value $v$ in equation (iii) we get, Then, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}^{\mathrm{n}-1}}}} \mathrm{~T}=\frac{2 \pi \mathrm{R}\left(\mathrm{R}^{\mathrm{n}-1}\right)^{1 / 2}}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}\left(\mathrm{R}^{\frac{\mathrm{n}}{2}-\frac{1}{2}}\right)}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}^{\frac{\mathrm{n}}{2}-\frac{1}{2}+1}\right)}{\sqrt{\mathrm{GM}}}$ $\mathrm{T}=\frac{2 \pi\left(\mathrm{R}^{\left(\frac{\mathrm{n}+1}{2}\right)}\right)}{\sqrt{\mathrm{GM}}}$ Therefore, $\mathrm{T} \propto \mathrm{R}^{\left(\frac{\mathrm{n}+1}{2}\right)}$
BITSAT-2010
Gravitation
138601
A system of binary stars of masses $m_{A}$ and $m_{B}$ are moving in circular orbits of radii $r_{A}$ and $r_{B}$ respectively. If $T_{A}$ and $T_{B}$ are the time periods of masses $m_{A}$ and $m_{B}$ respectively, then
D Since, $\mathrm{C}$ is the centre of mass. $\therefore \quad \mathrm{m}_{\mathrm{A}} \mathrm{r}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{r}_{\mathrm{B}}$ For the rotation of stars, $\mathrm{F}_{\text {gravitational }}=\mathrm{F}_{\text {centripetal }}$ $\frac{\mathrm{Gm}_{\mathrm{A}} \mathrm{m}_{\mathrm{B}}}{\left(\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}\right)^{2}}=\frac{\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}^{2}}{\mathrm{r}_{\mathrm{B}}}=\frac{\mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^{2}}{\mathrm{r}_{\mathrm{A}}}$ Put the value $m_{B}$ from equation (i) we get Then, $\frac{m_{A} r_{A} v_{B}^{2}}{r_{B}^{2}}=\frac{m_{A} v_{A}^{2}}{r_{A}}$ $\frac{v_{A}^{2}}{v_{B}^{2}}=\frac{r_{A}^{2}}{r_{B}^{2}}$ $\frac{r_{A} v_{B}}{r_{B} v_{A}}=1$ Time periods of stars, $\mathrm{T}_{\mathrm{A}}=\frac{2 \pi \mathrm{r}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{A}}}$ $\mathrm{T}_{\mathrm{B}}=\frac{2 \pi \mathrm{r}_{\mathrm{B}}}{\mathrm{v}_{\mathrm{B}}}$ Dividing equation (iii) by (iv) we get - $\frac{T_{A}}{T_{B}}=\frac{r_{A} \cdot v_{B}}{r_{B} \cdot v_{A}}$ $\frac{T_{A}}{T_{B}}=1$ $\therefore \quad \mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}}$
