359850
A mass \(m\) is placed inside a hollow sphere of mass \(M\). The gravitation force on mass \(m\) is
1 \(\frac{{GMm}}{{{R^2}}}\)
2 \(\frac{{GMm}}{{{r^2}}}\)
3 \(\frac{{GMm}}{{{{(R - r)}^2}}}\)
4 Zero
Explanation:
Inside a shell, field strength is zero. Therefore force on a particle is zero
PHXI08:GRAVITATION
359851
The magnitudes of gravitational field at distances \(r_{1}\) and \(r_{2}\) from the centre of a uniform sphere of radius \(R\) and mass \(M\) are \(I_{1}\) and \(I_{2}\), respectively. Find the ratio \(\left(I_{1} / I_{2}\right)\) if \(r_{1}>R\) and \(r_{2} < R\).
1 \(\dfrac{R^{2}}{r_{1} r_{2}}\)
2 \(\dfrac{R^{3}}{r_{1} r_{2}^{2}}\)
3 \(\dfrac{R^{3}}{r_{1}^{2} r_{2}}\)
4 \(\dfrac{R^{4}}{r_{1}^{2} r_{2}^{2}}\)
Explanation:
The magnitude of gravitational field at a distance \(r\) from the centre of a uniform sphere of radius \(R\) and mass \(M\) is as follows: (i) For \(r < R, I=\dfrac{G M}{R^{3}} r\) (ii) For \(r>R, I=\dfrac{G M}{R^{2}}\) \(\therefore I_{1}=\dfrac{G M}{r_{1}^{2}}\) and \(I_{2}=\dfrac{G M r_{2}}{R^{3}}\) Their corresponding ratio is \(\frac{{{I_1}}}{{{I_2}}} = \frac{{GM}}{{r_1^2}} \times \frac{{{R^3}}}{{GM{r_2}}} = \frac{{{R^3}}}{{r_1^2{r_2}}}\).
PHXI08:GRAVITATION
359852
The gravitional field, due to 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point, \(P\), located as shown, would be (nearly) :
1 \(\dfrac{5}{6} \dfrac{G M}{x^{2}}\)
2 \(\dfrac{8}{9} \dfrac{G M}{x^{2}}\)
3 \(\dfrac{7}{8} \dfrac{G M}{x^{2}}\)
4 \(\dfrac{6}{7} \dfrac{G M}{x^{2}}\)
Explanation:
Let mass of smaller sphere (which has to be removed ) is \(m\) Radius \(=\dfrac{R}{2}\) (from figure) As densities are same \(\begin{aligned}& \dfrac{M}{\dfrac{4}{3} \pi R^{3}}=\dfrac{m}{\dfrac{4}{3} \pi\left(\dfrac{R}{2}\right)^{3}} \\& \Rightarrow m=\dfrac{M}{8}\end{aligned}\) Mass of the left over part of the sphere \(M^{\prime}=M-\dfrac{M}{8}=\dfrac{7}{8} M\) A finite mass behaves like a point mass at large distances. The gravitational field due to the left over part of the sphere \(=\dfrac{7}{8} \dfrac{G M}{x^{2}}\)
JEE - 2013
PHXI08:GRAVITATION
359853
Assuming that the earth is a sphere of radius \(R\) with uniform density, the distance from its centre at which the acceleration due to gravity is equal to \(\dfrac{g}{3}\) ( \(\mathrm{g}\) is the acceleration due to gravity on the surface of earth) is
1 \(\dfrac{2 R}{3}\)
2 \(\dfrac{R}{4}\)
3 \(\dfrac{R}{2}\)
4 \(\dfrac{R}{3}\)
Explanation:
The acceleration due to gravity inside the earth is \(g^{\prime}=E=\dfrac{G M}{R^{3}} r\) \(\dfrac{g}{3}=\dfrac{G M}{R^{3}} r=\dfrac{g}{R} r \quad \Rightarrow r=\dfrac{R}{3}\)
359850
A mass \(m\) is placed inside a hollow sphere of mass \(M\). The gravitation force on mass \(m\) is
1 \(\frac{{GMm}}{{{R^2}}}\)
2 \(\frac{{GMm}}{{{r^2}}}\)
3 \(\frac{{GMm}}{{{{(R - r)}^2}}}\)
4 Zero
Explanation:
Inside a shell, field strength is zero. Therefore force on a particle is zero
PHXI08:GRAVITATION
359851
The magnitudes of gravitational field at distances \(r_{1}\) and \(r_{2}\) from the centre of a uniform sphere of radius \(R\) and mass \(M\) are \(I_{1}\) and \(I_{2}\), respectively. Find the ratio \(\left(I_{1} / I_{2}\right)\) if \(r_{1}>R\) and \(r_{2} < R\).
1 \(\dfrac{R^{2}}{r_{1} r_{2}}\)
2 \(\dfrac{R^{3}}{r_{1} r_{2}^{2}}\)
3 \(\dfrac{R^{3}}{r_{1}^{2} r_{2}}\)
4 \(\dfrac{R^{4}}{r_{1}^{2} r_{2}^{2}}\)
Explanation:
The magnitude of gravitational field at a distance \(r\) from the centre of a uniform sphere of radius \(R\) and mass \(M\) is as follows: (i) For \(r < R, I=\dfrac{G M}{R^{3}} r\) (ii) For \(r>R, I=\dfrac{G M}{R^{2}}\) \(\therefore I_{1}=\dfrac{G M}{r_{1}^{2}}\) and \(I_{2}=\dfrac{G M r_{2}}{R^{3}}\) Their corresponding ratio is \(\frac{{{I_1}}}{{{I_2}}} = \frac{{GM}}{{r_1^2}} \times \frac{{{R^3}}}{{GM{r_2}}} = \frac{{{R^3}}}{{r_1^2{r_2}}}\).
PHXI08:GRAVITATION
359852
The gravitional field, due to 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point, \(P\), located as shown, would be (nearly) :
1 \(\dfrac{5}{6} \dfrac{G M}{x^{2}}\)
2 \(\dfrac{8}{9} \dfrac{G M}{x^{2}}\)
3 \(\dfrac{7}{8} \dfrac{G M}{x^{2}}\)
4 \(\dfrac{6}{7} \dfrac{G M}{x^{2}}\)
Explanation:
Let mass of smaller sphere (which has to be removed ) is \(m\) Radius \(=\dfrac{R}{2}\) (from figure) As densities are same \(\begin{aligned}& \dfrac{M}{\dfrac{4}{3} \pi R^{3}}=\dfrac{m}{\dfrac{4}{3} \pi\left(\dfrac{R}{2}\right)^{3}} \\& \Rightarrow m=\dfrac{M}{8}\end{aligned}\) Mass of the left over part of the sphere \(M^{\prime}=M-\dfrac{M}{8}=\dfrac{7}{8} M\) A finite mass behaves like a point mass at large distances. The gravitational field due to the left over part of the sphere \(=\dfrac{7}{8} \dfrac{G M}{x^{2}}\)
JEE - 2013
PHXI08:GRAVITATION
359853
Assuming that the earth is a sphere of radius \(R\) with uniform density, the distance from its centre at which the acceleration due to gravity is equal to \(\dfrac{g}{3}\) ( \(\mathrm{g}\) is the acceleration due to gravity on the surface of earth) is
1 \(\dfrac{2 R}{3}\)
2 \(\dfrac{R}{4}\)
3 \(\dfrac{R}{2}\)
4 \(\dfrac{R}{3}\)
Explanation:
The acceleration due to gravity inside the earth is \(g^{\prime}=E=\dfrac{G M}{R^{3}} r\) \(\dfrac{g}{3}=\dfrac{G M}{R^{3}} r=\dfrac{g}{R} r \quad \Rightarrow r=\dfrac{R}{3}\)
359850
A mass \(m\) is placed inside a hollow sphere of mass \(M\). The gravitation force on mass \(m\) is
1 \(\frac{{GMm}}{{{R^2}}}\)
2 \(\frac{{GMm}}{{{r^2}}}\)
3 \(\frac{{GMm}}{{{{(R - r)}^2}}}\)
4 Zero
Explanation:
Inside a shell, field strength is zero. Therefore force on a particle is zero
PHXI08:GRAVITATION
359851
The magnitudes of gravitational field at distances \(r_{1}\) and \(r_{2}\) from the centre of a uniform sphere of radius \(R\) and mass \(M\) are \(I_{1}\) and \(I_{2}\), respectively. Find the ratio \(\left(I_{1} / I_{2}\right)\) if \(r_{1}>R\) and \(r_{2} < R\).
1 \(\dfrac{R^{2}}{r_{1} r_{2}}\)
2 \(\dfrac{R^{3}}{r_{1} r_{2}^{2}}\)
3 \(\dfrac{R^{3}}{r_{1}^{2} r_{2}}\)
4 \(\dfrac{R^{4}}{r_{1}^{2} r_{2}^{2}}\)
Explanation:
The magnitude of gravitational field at a distance \(r\) from the centre of a uniform sphere of radius \(R\) and mass \(M\) is as follows: (i) For \(r < R, I=\dfrac{G M}{R^{3}} r\) (ii) For \(r>R, I=\dfrac{G M}{R^{2}}\) \(\therefore I_{1}=\dfrac{G M}{r_{1}^{2}}\) and \(I_{2}=\dfrac{G M r_{2}}{R^{3}}\) Their corresponding ratio is \(\frac{{{I_1}}}{{{I_2}}} = \frac{{GM}}{{r_1^2}} \times \frac{{{R^3}}}{{GM{r_2}}} = \frac{{{R^3}}}{{r_1^2{r_2}}}\).
PHXI08:GRAVITATION
359852
The gravitional field, due to 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point, \(P\), located as shown, would be (nearly) :
1 \(\dfrac{5}{6} \dfrac{G M}{x^{2}}\)
2 \(\dfrac{8}{9} \dfrac{G M}{x^{2}}\)
3 \(\dfrac{7}{8} \dfrac{G M}{x^{2}}\)
4 \(\dfrac{6}{7} \dfrac{G M}{x^{2}}\)
Explanation:
Let mass of smaller sphere (which has to be removed ) is \(m\) Radius \(=\dfrac{R}{2}\) (from figure) As densities are same \(\begin{aligned}& \dfrac{M}{\dfrac{4}{3} \pi R^{3}}=\dfrac{m}{\dfrac{4}{3} \pi\left(\dfrac{R}{2}\right)^{3}} \\& \Rightarrow m=\dfrac{M}{8}\end{aligned}\) Mass of the left over part of the sphere \(M^{\prime}=M-\dfrac{M}{8}=\dfrac{7}{8} M\) A finite mass behaves like a point mass at large distances. The gravitational field due to the left over part of the sphere \(=\dfrac{7}{8} \dfrac{G M}{x^{2}}\)
JEE - 2013
PHXI08:GRAVITATION
359853
Assuming that the earth is a sphere of radius \(R\) with uniform density, the distance from its centre at which the acceleration due to gravity is equal to \(\dfrac{g}{3}\) ( \(\mathrm{g}\) is the acceleration due to gravity on the surface of earth) is
1 \(\dfrac{2 R}{3}\)
2 \(\dfrac{R}{4}\)
3 \(\dfrac{R}{2}\)
4 \(\dfrac{R}{3}\)
Explanation:
The acceleration due to gravity inside the earth is \(g^{\prime}=E=\dfrac{G M}{R^{3}} r\) \(\dfrac{g}{3}=\dfrac{G M}{R^{3}} r=\dfrac{g}{R} r \quad \Rightarrow r=\dfrac{R}{3}\)
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PHXI08:GRAVITATION
359850
A mass \(m\) is placed inside a hollow sphere of mass \(M\). The gravitation force on mass \(m\) is
1 \(\frac{{GMm}}{{{R^2}}}\)
2 \(\frac{{GMm}}{{{r^2}}}\)
3 \(\frac{{GMm}}{{{{(R - r)}^2}}}\)
4 Zero
Explanation:
Inside a shell, field strength is zero. Therefore force on a particle is zero
PHXI08:GRAVITATION
359851
The magnitudes of gravitational field at distances \(r_{1}\) and \(r_{2}\) from the centre of a uniform sphere of radius \(R\) and mass \(M\) are \(I_{1}\) and \(I_{2}\), respectively. Find the ratio \(\left(I_{1} / I_{2}\right)\) if \(r_{1}>R\) and \(r_{2} < R\).
1 \(\dfrac{R^{2}}{r_{1} r_{2}}\)
2 \(\dfrac{R^{3}}{r_{1} r_{2}^{2}}\)
3 \(\dfrac{R^{3}}{r_{1}^{2} r_{2}}\)
4 \(\dfrac{R^{4}}{r_{1}^{2} r_{2}^{2}}\)
Explanation:
The magnitude of gravitational field at a distance \(r\) from the centre of a uniform sphere of radius \(R\) and mass \(M\) is as follows: (i) For \(r < R, I=\dfrac{G M}{R^{3}} r\) (ii) For \(r>R, I=\dfrac{G M}{R^{2}}\) \(\therefore I_{1}=\dfrac{G M}{r_{1}^{2}}\) and \(I_{2}=\dfrac{G M r_{2}}{R^{3}}\) Their corresponding ratio is \(\frac{{{I_1}}}{{{I_2}}} = \frac{{GM}}{{r_1^2}} \times \frac{{{R^3}}}{{GM{r_2}}} = \frac{{{R^3}}}{{r_1^2{r_2}}}\).
PHXI08:GRAVITATION
359852
The gravitional field, due to 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point, \(P\), located as shown, would be (nearly) :
1 \(\dfrac{5}{6} \dfrac{G M}{x^{2}}\)
2 \(\dfrac{8}{9} \dfrac{G M}{x^{2}}\)
3 \(\dfrac{7}{8} \dfrac{G M}{x^{2}}\)
4 \(\dfrac{6}{7} \dfrac{G M}{x^{2}}\)
Explanation:
Let mass of smaller sphere (which has to be removed ) is \(m\) Radius \(=\dfrac{R}{2}\) (from figure) As densities are same \(\begin{aligned}& \dfrac{M}{\dfrac{4}{3} \pi R^{3}}=\dfrac{m}{\dfrac{4}{3} \pi\left(\dfrac{R}{2}\right)^{3}} \\& \Rightarrow m=\dfrac{M}{8}\end{aligned}\) Mass of the left over part of the sphere \(M^{\prime}=M-\dfrac{M}{8}=\dfrac{7}{8} M\) A finite mass behaves like a point mass at large distances. The gravitational field due to the left over part of the sphere \(=\dfrac{7}{8} \dfrac{G M}{x^{2}}\)
JEE - 2013
PHXI08:GRAVITATION
359853
Assuming that the earth is a sphere of radius \(R\) with uniform density, the distance from its centre at which the acceleration due to gravity is equal to \(\dfrac{g}{3}\) ( \(\mathrm{g}\) is the acceleration due to gravity on the surface of earth) is
1 \(\dfrac{2 R}{3}\)
2 \(\dfrac{R}{4}\)
3 \(\dfrac{R}{2}\)
4 \(\dfrac{R}{3}\)
Explanation:
The acceleration due to gravity inside the earth is \(g^{\prime}=E=\dfrac{G M}{R^{3}} r\) \(\dfrac{g}{3}=\dfrac{G M}{R^{3}} r=\dfrac{g}{R} r \quad \Rightarrow r=\dfrac{R}{3}\)
