270547
The height at which the value of \(g\) is half that on the surface of earth of radius \(R\) is
1 \(R\)
2 \(2 R\)
3 \(0.414 R\)
4 \(0.75 R\)
Explanation:
\(\quad \frac{g_{h}}{g}=\frac{R^{2}}{(R+h)^{2}} \quad\) Given \(\mathrm{h}=\mathrm{R} / 2\)
Gravitation
270548
The depth at which the value of \(g\) becomes \(\mathbf{2 5 \%}\) of that at the surface of earth is (in \(\mathrm{Km}\)
1 4800
2 2400
3 3600
4 1200
Explanation:
\(g^{\prime}=g-1-\frac{d}{R}\) G Given \(g^{\prime}=\frac{25}{100} g\)
Gravitation
270549
If the radius of earth decreases by \(10 \%\), the mass remaining unchanged, then the acceleration due to gravity
1 decreases by\(19 \%\)
2 increases by\(19 \%\)
3 decreases by more than\(19 \%\)
4 increases by more than\(19 \%\)
Explanation:
\(g=\frac{G M}{R^{2}} \Rightarrow g \propto \frac{1}{R^{2}}\)
Gravitation
270550
The acceleration due to gravity at the poles is \(10 \mathrm{~ms}^{-2}\) and equatorial radius is \(6400 \mathrm{~km}\) for the earth. Then the angular velocity of rotation of the earth about its axis so that the weight of a body at the equator reduces to \(75 \%\) is
1 \(\frac{1}{1600}\) rads
2 \(\frac{1}{800}\) rads \(^{-1}\)
3 \(\frac{1}{400} \mathrm{rads}^{-1}\)
4 \(\frac{1}{200}\) rads \(^{-1}\)
Explanation:
\(g_{\varphi}=g-R \omega^{2}\) Given \(g_{\varphi}=\frac{75}{100} g\)
Gravitation
270551
The maximum horizontal range of a projectile on the earth is \(R\). Then for the same velocity of projection, its maximum range on another planet is \(\frac{5 R}{4}\). Then, ratio of acceleration due to gravity on that planet and on the earth is
270547
The height at which the value of \(g\) is half that on the surface of earth of radius \(R\) is
1 \(R\)
2 \(2 R\)
3 \(0.414 R\)
4 \(0.75 R\)
Explanation:
\(\quad \frac{g_{h}}{g}=\frac{R^{2}}{(R+h)^{2}} \quad\) Given \(\mathrm{h}=\mathrm{R} / 2\)
Gravitation
270548
The depth at which the value of \(g\) becomes \(\mathbf{2 5 \%}\) of that at the surface of earth is (in \(\mathrm{Km}\)
1 4800
2 2400
3 3600
4 1200
Explanation:
\(g^{\prime}=g-1-\frac{d}{R}\) G Given \(g^{\prime}=\frac{25}{100} g\)
Gravitation
270549
If the radius of earth decreases by \(10 \%\), the mass remaining unchanged, then the acceleration due to gravity
1 decreases by\(19 \%\)
2 increases by\(19 \%\)
3 decreases by more than\(19 \%\)
4 increases by more than\(19 \%\)
Explanation:
\(g=\frac{G M}{R^{2}} \Rightarrow g \propto \frac{1}{R^{2}}\)
Gravitation
270550
The acceleration due to gravity at the poles is \(10 \mathrm{~ms}^{-2}\) and equatorial radius is \(6400 \mathrm{~km}\) for the earth. Then the angular velocity of rotation of the earth about its axis so that the weight of a body at the equator reduces to \(75 \%\) is
1 \(\frac{1}{1600}\) rads
2 \(\frac{1}{800}\) rads \(^{-1}\)
3 \(\frac{1}{400} \mathrm{rads}^{-1}\)
4 \(\frac{1}{200}\) rads \(^{-1}\)
Explanation:
\(g_{\varphi}=g-R \omega^{2}\) Given \(g_{\varphi}=\frac{75}{100} g\)
Gravitation
270551
The maximum horizontal range of a projectile on the earth is \(R\). Then for the same velocity of projection, its maximum range on another planet is \(\frac{5 R}{4}\). Then, ratio of acceleration due to gravity on that planet and on the earth is
270547
The height at which the value of \(g\) is half that on the surface of earth of radius \(R\) is
1 \(R\)
2 \(2 R\)
3 \(0.414 R\)
4 \(0.75 R\)
Explanation:
\(\quad \frac{g_{h}}{g}=\frac{R^{2}}{(R+h)^{2}} \quad\) Given \(\mathrm{h}=\mathrm{R} / 2\)
Gravitation
270548
The depth at which the value of \(g\) becomes \(\mathbf{2 5 \%}\) of that at the surface of earth is (in \(\mathrm{Km}\)
1 4800
2 2400
3 3600
4 1200
Explanation:
\(g^{\prime}=g-1-\frac{d}{R}\) G Given \(g^{\prime}=\frac{25}{100} g\)
Gravitation
270549
If the radius of earth decreases by \(10 \%\), the mass remaining unchanged, then the acceleration due to gravity
1 decreases by\(19 \%\)
2 increases by\(19 \%\)
3 decreases by more than\(19 \%\)
4 increases by more than\(19 \%\)
Explanation:
\(g=\frac{G M}{R^{2}} \Rightarrow g \propto \frac{1}{R^{2}}\)
Gravitation
270550
The acceleration due to gravity at the poles is \(10 \mathrm{~ms}^{-2}\) and equatorial radius is \(6400 \mathrm{~km}\) for the earth. Then the angular velocity of rotation of the earth about its axis so that the weight of a body at the equator reduces to \(75 \%\) is
1 \(\frac{1}{1600}\) rads
2 \(\frac{1}{800}\) rads \(^{-1}\)
3 \(\frac{1}{400} \mathrm{rads}^{-1}\)
4 \(\frac{1}{200}\) rads \(^{-1}\)
Explanation:
\(g_{\varphi}=g-R \omega^{2}\) Given \(g_{\varphi}=\frac{75}{100} g\)
Gravitation
270551
The maximum horizontal range of a projectile on the earth is \(R\). Then for the same velocity of projection, its maximum range on another planet is \(\frac{5 R}{4}\). Then, ratio of acceleration due to gravity on that planet and on the earth is
270547
The height at which the value of \(g\) is half that on the surface of earth of radius \(R\) is
1 \(R\)
2 \(2 R\)
3 \(0.414 R\)
4 \(0.75 R\)
Explanation:
\(\quad \frac{g_{h}}{g}=\frac{R^{2}}{(R+h)^{2}} \quad\) Given \(\mathrm{h}=\mathrm{R} / 2\)
Gravitation
270548
The depth at which the value of \(g\) becomes \(\mathbf{2 5 \%}\) of that at the surface of earth is (in \(\mathrm{Km}\)
1 4800
2 2400
3 3600
4 1200
Explanation:
\(g^{\prime}=g-1-\frac{d}{R}\) G Given \(g^{\prime}=\frac{25}{100} g\)
Gravitation
270549
If the radius of earth decreases by \(10 \%\), the mass remaining unchanged, then the acceleration due to gravity
1 decreases by\(19 \%\)
2 increases by\(19 \%\)
3 decreases by more than\(19 \%\)
4 increases by more than\(19 \%\)
Explanation:
\(g=\frac{G M}{R^{2}} \Rightarrow g \propto \frac{1}{R^{2}}\)
Gravitation
270550
The acceleration due to gravity at the poles is \(10 \mathrm{~ms}^{-2}\) and equatorial radius is \(6400 \mathrm{~km}\) for the earth. Then the angular velocity of rotation of the earth about its axis so that the weight of a body at the equator reduces to \(75 \%\) is
1 \(\frac{1}{1600}\) rads
2 \(\frac{1}{800}\) rads \(^{-1}\)
3 \(\frac{1}{400} \mathrm{rads}^{-1}\)
4 \(\frac{1}{200}\) rads \(^{-1}\)
Explanation:
\(g_{\varphi}=g-R \omega^{2}\) Given \(g_{\varphi}=\frac{75}{100} g\)
Gravitation
270551
The maximum horizontal range of a projectile on the earth is \(R\). Then for the same velocity of projection, its maximum range on another planet is \(\frac{5 R}{4}\). Then, ratio of acceleration due to gravity on that planet and on the earth is
270547
The height at which the value of \(g\) is half that on the surface of earth of radius \(R\) is
1 \(R\)
2 \(2 R\)
3 \(0.414 R\)
4 \(0.75 R\)
Explanation:
\(\quad \frac{g_{h}}{g}=\frac{R^{2}}{(R+h)^{2}} \quad\) Given \(\mathrm{h}=\mathrm{R} / 2\)
Gravitation
270548
The depth at which the value of \(g\) becomes \(\mathbf{2 5 \%}\) of that at the surface of earth is (in \(\mathrm{Km}\)
1 4800
2 2400
3 3600
4 1200
Explanation:
\(g^{\prime}=g-1-\frac{d}{R}\) G Given \(g^{\prime}=\frac{25}{100} g\)
Gravitation
270549
If the radius of earth decreases by \(10 \%\), the mass remaining unchanged, then the acceleration due to gravity
1 decreases by\(19 \%\)
2 increases by\(19 \%\)
3 decreases by more than\(19 \%\)
4 increases by more than\(19 \%\)
Explanation:
\(g=\frac{G M}{R^{2}} \Rightarrow g \propto \frac{1}{R^{2}}\)
Gravitation
270550
The acceleration due to gravity at the poles is \(10 \mathrm{~ms}^{-2}\) and equatorial radius is \(6400 \mathrm{~km}\) for the earth. Then the angular velocity of rotation of the earth about its axis so that the weight of a body at the equator reduces to \(75 \%\) is
1 \(\frac{1}{1600}\) rads
2 \(\frac{1}{800}\) rads \(^{-1}\)
3 \(\frac{1}{400} \mathrm{rads}^{-1}\)
4 \(\frac{1}{200}\) rads \(^{-1}\)
Explanation:
\(g_{\varphi}=g-R \omega^{2}\) Given \(g_{\varphi}=\frac{75}{100} g\)
Gravitation
270551
The maximum horizontal range of a projectile on the earth is \(R\). Then for the same velocity of projection, its maximum range on another planet is \(\frac{5 R}{4}\). Then, ratio of acceleration due to gravity on that planet and on the earth is
