QBManager

5 RBTS PAPER

163994 The value of ' $g$ ' at a place is $9.8 {ms}^{- 2}$ , if the earth shrinks uniformly (without changing its mass or centre) to half its size the value of ' $g$ ' at the same point will now be :

1

9.8 m s^{- 2}

2

19.6 m s^{- 2}

3

4.9 m s^{- 2}

4 zero

Explanation:

Since the point remains the same
$∴ M$ and $R$ are same and $g = \frac{GM}{R^{2}}$ remains the same

NCERT-XI- I -133

5 RBTS PAPER

163995 The weight of the body at the centre of earth is :

1 zero

2 infinity

3 9.8 times the weight on the surface

4 None of these

Explanation:

The weight of the body is $W = m g$ .
At centre of earth the value of $g$ becomes zero
So weight becomes zero at centre of earth

NCERT-XI- I -134

5 RBTS PAPER

163996 To what height the body is to be raised so that force of gravity on it is $1 %$ of the value on the surfaace of earth? (Radius of earth is $6400 km$ ) :

1

6400 km

2

12800 km

3

57600 km

4

3200 km

Explanation:

$\frac{{mg}^{'}}{mg} = \frac{1}{100}$ or $\frac{g^{'}}{g} = \frac{1}{100} = \frac{(R)^{2}}{(R + h)^{2}}$ or
$\frac{R}{R + h} = \frac{1}{10}$ or $9 R = h$
or $h = 6400 \times 9 = 57600 km$

NCERT-XI- I -133

5 RBTS PAPER

163997 The period of satellite in a circular orbit of radius $R_{0}$ is $T$ , the period of another satellite in circular orbit of radius $4 R_{0}$ is :

1

T / 4

2

T / 8

3

4 T

4

8 T

Explanation:

$T \propto R_{0}^{3 / 2}$ and ${(4 R_{0})}^{3 / 2}$
or $\frac{T^{'}}{T} = (4)^{3 / 2} = (64)^{1 / 2} = 8$

NCERT-XI- I -137

5 RBTS PAPER

163998 The time period of simple pendulum in a satellite is :

1 infinite

2 zero

3

2 \sec

4 Cannot be calculated

Explanation:

$T \propto \frac{1}{\sqrt{g}}$ inside a satellite $g = 0 ∴ T = \infty$

Modified NLI Expert

5 RBTS PAPER

163994 The value of ' $g$ ' at a place is $9.8 {ms}^{- 2}$ , if the earth shrinks uniformly (without changing its mass or centre) to half its size the value of ' $g$ ' at the same point will now be :

1

9.8 m s^{- 2}

2

19.6 m s^{- 2}

3

4.9 m s^{- 2}

4 zero

Explanation:

Since the point remains the same
$∴ M$ and $R$ are same and $g = \frac{GM}{R^{2}}$ remains the same

NCERT-XI- I -133

5 RBTS PAPER

163995 The weight of the body at the centre of earth is :

1 zero

2 infinity

3 9.8 times the weight on the surface

4 None of these

Explanation:

The weight of the body is $W = m g$ .
At centre of earth the value of $g$ becomes zero
So weight becomes zero at centre of earth

NCERT-XI- I -134

5 RBTS PAPER

163996 To what height the body is to be raised so that force of gravity on it is $1 %$ of the value on the surfaace of earth? (Radius of earth is $6400 km$ ) :

1

6400 km

2

12800 km

3

57600 km

4

3200 km

Explanation:

$\frac{{mg}^{'}}{mg} = \frac{1}{100}$ or $\frac{g^{'}}{g} = \frac{1}{100} = \frac{(R)^{2}}{(R + h)^{2}}$ or
$\frac{R}{R + h} = \frac{1}{10}$ or $9 R = h$
or $h = 6400 \times 9 = 57600 km$

NCERT-XI- I -133

5 RBTS PAPER

163997 The period of satellite in a circular orbit of radius $R_{0}$ is $T$ , the period of another satellite in circular orbit of radius $4 R_{0}$ is :

1

T / 4

2

T / 8

3

4 T

4

8 T

Explanation:

$T \propto R_{0}^{3 / 2}$ and ${(4 R_{0})}^{3 / 2}$
or $\frac{T^{'}}{T} = (4)^{3 / 2} = (64)^{1 / 2} = 8$

NCERT-XI- I -137

5 RBTS PAPER

163998 The time period of simple pendulum in a satellite is :

1 infinite

2 zero

3

2 \sec

4 Cannot be calculated

Explanation:

$T \propto \frac{1}{\sqrt{g}}$ inside a satellite $g = 0 ∴ T = \infty$

Modified NLI Expert

5 RBTS PAPER

163994 The value of ' $g$ ' at a place is $9.8 {ms}^{- 2}$ , if the earth shrinks uniformly (without changing its mass or centre) to half its size the value of ' $g$ ' at the same point will now be :

1

9.8 m s^{- 2}

2

19.6 m s^{- 2}

3

4.9 m s^{- 2}

4 zero

Explanation:

Since the point remains the same
$∴ M$ and $R$ are same and $g = \frac{GM}{R^{2}}$ remains the same

NCERT-XI- I -133

5 RBTS PAPER

163995 The weight of the body at the centre of earth is :

1 zero

2 infinity

3 9.8 times the weight on the surface

4 None of these

Explanation:

The weight of the body is $W = m g$ .
At centre of earth the value of $g$ becomes zero
So weight becomes zero at centre of earth

NCERT-XI- I -134

5 RBTS PAPER

163996 To what height the body is to be raised so that force of gravity on it is $1 %$ of the value on the surfaace of earth? (Radius of earth is $6400 km$ ) :

1

6400 km

2

12800 km

3

57600 km

4

3200 km

Explanation:

$\frac{{mg}^{'}}{mg} = \frac{1}{100}$ or $\frac{g^{'}}{g} = \frac{1}{100} = \frac{(R)^{2}}{(R + h)^{2}}$ or
$\frac{R}{R + h} = \frac{1}{10}$ or $9 R = h$
or $h = 6400 \times 9 = 57600 km$

NCERT-XI- I -133

5 RBTS PAPER

163997 The period of satellite in a circular orbit of radius $R_{0}$ is $T$ , the period of another satellite in circular orbit of radius $4 R_{0}$ is :

1

T / 4

2

T / 8

3

4 T

4

8 T

Explanation:

$T \propto R_{0}^{3 / 2}$ and ${(4 R_{0})}^{3 / 2}$
or $\frac{T^{'}}{T} = (4)^{3 / 2} = (64)^{1 / 2} = 8$

NCERT-XI- I -137

5 RBTS PAPER

163998 The time period of simple pendulum in a satellite is :

1 infinite

2 zero

3

2 \sec

4 Cannot be calculated

Explanation:

$T \propto \frac{1}{\sqrt{g}}$ inside a satellite $g = 0 ∴ T = \infty$

Modified NLI Expert

5 RBTS PAPER

163994 The value of ' $g$ ' at a place is $9.8 {ms}^{- 2}$ , if the earth shrinks uniformly (without changing its mass or centre) to half its size the value of ' $g$ ' at the same point will now be :

1

9.8 m s^{- 2}

2

19.6 m s^{- 2}

3

4.9 m s^{- 2}

4 zero

Explanation:

Since the point remains the same
$∴ M$ and $R$ are same and $g = \frac{GM}{R^{2}}$ remains the same

NCERT-XI- I -133

5 RBTS PAPER

163995 The weight of the body at the centre of earth is :

1 zero

2 infinity

3 9.8 times the weight on the surface

4 None of these

Explanation:

The weight of the body is $W = m g$ .
At centre of earth the value of $g$ becomes zero
So weight becomes zero at centre of earth

NCERT-XI- I -134

5 RBTS PAPER

163996 To what height the body is to be raised so that force of gravity on it is $1 %$ of the value on the surfaace of earth? (Radius of earth is $6400 km$ ) :

1

6400 km

2

12800 km

3

57600 km

4

3200 km

Explanation:

$\frac{{mg}^{'}}{mg} = \frac{1}{100}$ or $\frac{g^{'}}{g} = \frac{1}{100} = \frac{(R)^{2}}{(R + h)^{2}}$ or
$\frac{R}{R + h} = \frac{1}{10}$ or $9 R = h$
or $h = 6400 \times 9 = 57600 km$

NCERT-XI- I -133

5 RBTS PAPER

163997 The period of satellite in a circular orbit of radius $R_{0}$ is $T$ , the period of another satellite in circular orbit of radius $4 R_{0}$ is :

1

T / 4

2

T / 8

3

4 T

4

8 T

Explanation:

$T \propto R_{0}^{3 / 2}$ and ${(4 R_{0})}^{3 / 2}$
or $\frac{T^{'}}{T} = (4)^{3 / 2} = (64)^{1 / 2} = 8$

NCERT-XI- I -137

5 RBTS PAPER

163998 The time period of simple pendulum in a satellite is :

1 infinite

2 zero

3

2 \sec

4 Cannot be calculated

Explanation:

$T \propto \frac{1}{\sqrt{g}}$ inside a satellite $g = 0 ∴ T = \infty$

Modified NLI Expert

5 RBTS PAPER

163994 The value of ' $g$ ' at a place is $9.8 {ms}^{- 2}$ , if the earth shrinks uniformly (without changing its mass or centre) to half its size the value of ' $g$ ' at the same point will now be :

1

9.8 m s^{- 2}

2

19.6 m s^{- 2}

3

4.9 m s^{- 2}

4 zero

Explanation:

Since the point remains the same
$∴ M$ and $R$ are same and $g = \frac{GM}{R^{2}}$ remains the same

NCERT-XI- I -133

5 RBTS PAPER

163995 The weight of the body at the centre of earth is :

1 zero

2 infinity

3 9.8 times the weight on the surface

4 None of these

Explanation:

The weight of the body is $W = m g$ .
At centre of earth the value of $g$ becomes zero
So weight becomes zero at centre of earth

NCERT-XI- I -134

5 RBTS PAPER

163996 To what height the body is to be raised so that force of gravity on it is $1 %$ of the value on the surfaace of earth? (Radius of earth is $6400 km$ ) :

1

6400 km

2

12800 km

3

57600 km

4

3200 km

Explanation:

$\frac{{mg}^{'}}{mg} = \frac{1}{100}$ or $\frac{g^{'}}{g} = \frac{1}{100} = \frac{(R)^{2}}{(R + h)^{2}}$ or
$\frac{R}{R + h} = \frac{1}{10}$ or $9 R = h$
or $h = 6400 \times 9 = 57600 km$

NCERT-XI- I -133

5 RBTS PAPER

163997 The period of satellite in a circular orbit of radius $R_{0}$ is $T$ , the period of another satellite in circular orbit of radius $4 R_{0}$ is :

1

T / 4

2

T / 8

3

4 T

4

8 T

Explanation:

$T \propto R_{0}^{3 / 2}$ and ${(4 R_{0})}^{3 / 2}$
or $\frac{T^{'}}{T} = (4)^{3 / 2} = (64)^{1 / 2} = 8$

NCERT-XI- I -137

5 RBTS PAPER

163998 The time period of simple pendulum in a satellite is :

1 infinite

2 zero

3

2 \sec

4 Cannot be calculated

Explanation:

$T \propto \frac{1}{\sqrt{g}}$ inside a satellite $g = 0 ∴ T = \infty$

Modified NLI Expert

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