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Gravitation

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:

$\frac{g_{h}}{g} = \frac{R^{2}}{(R + h)^{2}}$ Given $h = R / 2$

Gravitation

270548 The depth at which the value of $g$ becomes $25 %$ of that at the surface of earth is (in $Km$

1 4800

2 2400

3 3600

4 1200

Explanation:

$g^{'} = g - 1 - \frac{d}{R}$ G Given $g^{'} = \frac{25}{100} g$

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 $10 {ms}^{- 2}$ and equatorial radius is $6400 km$ 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} {rads}^{- 1}

4

\frac{1}{200}

rads

^{- 1}

Explanation:

$g_{φ} = g - R ω^{2}$ Given $g_{φ} = \frac{75}{100} g$

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

1

5 : 4

2

4 : 5

3

25 : 16

4

16 : 25

Explanation:

$R_{max} = \frac{u^{2}}{g} \Rightarrow R_{max} \propto \frac{1}{g}$

Gravitation

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:

$\frac{g_{h}}{g} = \frac{R^{2}}{(R + h)^{2}}$ Given $h = R / 2$

Gravitation

270548 The depth at which the value of $g$ becomes $25 %$ of that at the surface of earth is (in $Km$

1 4800

2 2400

3 3600

4 1200

Explanation:

$g^{'} = g - 1 - \frac{d}{R}$ G Given $g^{'} = \frac{25}{100} g$

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 $10 {ms}^{- 2}$ and equatorial radius is $6400 km$ 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} {rads}^{- 1}

4

\frac{1}{200}

rads

^{- 1}

Explanation:

$g_{φ} = g - R ω^{2}$ Given $g_{φ} = \frac{75}{100} g$

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

1

5 : 4

2

4 : 5

3

25 : 16

4

16 : 25

Explanation:

$R_{max} = \frac{u^{2}}{g} \Rightarrow R_{max} \propto \frac{1}{g}$

Gravitation

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:

$\frac{g_{h}}{g} = \frac{R^{2}}{(R + h)^{2}}$ Given $h = R / 2$

Gravitation

270548 The depth at which the value of $g$ becomes $25 %$ of that at the surface of earth is (in $Km$

1 4800

2 2400

3 3600

4 1200

Explanation:

$g^{'} = g - 1 - \frac{d}{R}$ G Given $g^{'} = \frac{25}{100} g$

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 $10 {ms}^{- 2}$ and equatorial radius is $6400 km$ 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} {rads}^{- 1}

4

\frac{1}{200}

rads

^{- 1}

Explanation:

$g_{φ} = g - R ω^{2}$ Given $g_{φ} = \frac{75}{100} g$

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

1

5 : 4

2

4 : 5

3

25 : 16

4

16 : 25

Explanation:

$R_{max} = \frac{u^{2}}{g} \Rightarrow R_{max} \propto \frac{1}{g}$

Gravitation

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:

$\frac{g_{h}}{g} = \frac{R^{2}}{(R + h)^{2}}$ Given $h = R / 2$

Gravitation

270548 The depth at which the value of $g$ becomes $25 %$ of that at the surface of earth is (in $Km$

1 4800

2 2400

3 3600

4 1200

Explanation:

$g^{'} = g - 1 - \frac{d}{R}$ G Given $g^{'} = \frac{25}{100} g$

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 $10 {ms}^{- 2}$ and equatorial radius is $6400 km$ 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} {rads}^{- 1}

4

\frac{1}{200}

rads

^{- 1}

Explanation:

$g_{φ} = g - R ω^{2}$ Given $g_{φ} = \frac{75}{100} g$

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

1

5 : 4

2

4 : 5

3

25 : 16

4

16 : 25

Explanation:

$R_{max} = \frac{u^{2}}{g} \Rightarrow R_{max} \propto \frac{1}{g}$

Gravitation

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:

$\frac{g_{h}}{g} = \frac{R^{2}}{(R + h)^{2}}$ Given $h = R / 2$

Gravitation

270548 The depth at which the value of $g$ becomes $25 %$ of that at the surface of earth is (in $Km$

1 4800

2 2400

3 3600

4 1200

Explanation:

$g^{'} = g - 1 - \frac{d}{R}$ G Given $g^{'} = \frac{25}{100} g$

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 $10 {ms}^{- 2}$ and equatorial radius is $6400 km$ 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} {rads}^{- 1}

4

\frac{1}{200}

rads

^{- 1}

Explanation:

$g_{φ} = g - R ω^{2}$ Given $g_{φ} = \frac{75}{100} g$

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

1

5 : 4

2

4 : 5

3

25 : 16

4

16 : 25

Explanation:

$R_{max} = \frac{u^{2}}{g} \Rightarrow R_{max} \propto \frac{1}{g}$

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