QBManager

Gravitation

138396 The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$ , the radius of the planet would be

1

4 R

2

2 R

3

R / 2

4

R / 4

Explanation:

C Given,
$ρ_{P} = 2 ρ_{e}$
$g_{e} = g_{P}$
Mass $=$ volume $\times$ density
$M_{e} = \frac{4}{3} π R_{e}^{3} ρ_{e}$
From equation of acceleration due to gravity
$g_{e} = \frac{{GM}_{e}}{R_{e}^{2}} = \frac{G (4 / 3) π R_{e}^{3}}{R_{e}^{2}} ρ_{e}$
Here,
$g_{e} \propto R_{e} ρ_{e}$
Now, acceleration due to gravity of given planet, $g_{P} \propto R_{P} ρ_{P}$ given,
$g_{e} = g_{P}$
$R_{e} ρ_{e} = R_{P} ρ_{P}$
$R_{e} ρ_{e} = R_{P} 2 ρ_{e} (ρ_{P} = 2 ρ_{e})$
Hence, $R_{P} = \frac{R_{e}}{2} = \frac{R}{2} {∵ R_{e} = R}$

AMU-2016

Gravitation

138397 Calculate the acceleration due to gravity on the surface of a pulsar of mass $M = 1.98 \times 10^{30} kg$ and radius $R = 12 km$ rotating with time period $T = 0.041$ seconds. $(G = 6.67 \times 10^{- 11} m^{3} {kg}^{- 1} s^{- 2})$

1

9.2 \times 10^{11} m / s^{2}

2

8.15 \times 10^{11} m / s^{2}

3

7.32 \times 10^{11} m / s^{2}

4

6.98 \times 10^{11} m / s^{2}

Explanation:

A Given,
$Mass (M) = 1.98 \times 10^{30} Kg$
$Radius (R) = 12 km = 12000 m$
We know,
$g = \frac{GM}{R^{2}}$
$g = \frac{6.67 \times 10^{- 11} \times 1.98 \times 10^{30}}{(12000)^{2}} [∵ G = 6.67 \times 10^{- 11} m^{3} {Kg}^{- 1} s^{- 2}]$
$g = 9.2 \times 10^{11} m / s^{2}$

AMU-2012

Gravitation

138399 The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be sphere of radius $R$ of uniform density is shown in figure below.
The correct figure is

1

2

3

4

Explanation:

D At depth, when $r < R$
$g^{'} = g (1 - \frac{d}{R})$
$g^{'} = g (\frac{R - d}{R})$
$g^{'} = g (\frac{r}{R})$
$Where r = R - d$
$\Rightarrow g^{'} \propto r$
Case-1 when $r = 0$
$g^{'} = 0$
Case-2 when $r = R$
$g^{'} = g$
At height $h$ , when $r > R$
$g^{'} = \frac{{gR}^{2}}{(R + h)^{2}}$
$g^{'} = \frac{{gR}^{2}}{r^{2}} Where r = R + h$
$g^{'} \propto \frac{1}{r^{2}}$
Hence, the required graph is-

JEE-MAIN 26.06.2022 Shift-I

Gravitation

138400 If the earth stops rotating, the value of $g$ at the equator

1 Increases

2 decreases

3 no effect

4 none of these

Explanation:

A When earth is rotating,
$g_{1}^{'} = g - ω^{2} R \cos^{2} α$
When earth stop rotating,
$g_{2}^{'} = g$
$So, g_{2}^{'} > g_{1}^{'}$
If the earth stops rotating, the value of $g$ at equator will increase by amount $R ω^{2} \cos^{2} α$ . with $α = 0^{\circ}$ at equator. The effect of rotation of the earth on acceleration due to gravity is to decrease its value. Therefore if the earth stops rotating, the value of $g$ will increase.

DCE-2007

Gravitation

138401 The acceleration due to gravity at a height (take $g = 10 {ms}^{- 2}$ on earth surface) above earth's surface is $9 {ms}^{- 2}$ . It value at a point, at an equal distance below the surface of the earth is

1

9.7 {ms}^{- 2}

2

8.5 {ms}^{- 2}

3

7 . {ms}^{- 2}

4

5.5 {ms}^{- 2}

Explanation:

A Given,
$g_{h} = 9 m / s^{2}$
$g = 10 m / s^{2}$
We know,
$g_{h} = g [1 - \frac{2 h}{R}]$
$9 = 10 [1 - \frac{2 h}{R}]$
$[1 - \frac{2 h}{R}] = \frac{9}{10}$
$\frac{1}{10} = \frac{2 h}{R}$
$h = \frac{R}{20}$
Now, The value of $g$ below the earth's surface
$g_{d} = g [1 - \frac{d}{R}] = 10 [1 - \frac{R}{R}]$
$= 10 [1 - \frac{1}{20}]$
$g_{d} = \frac{10 \times 19}{20} = 9.5 {ms}^{- 2} \approx 9.7 {ms}^{- 2}$

EAMCET-1991

Gravitation

138396 The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$ , the radius of the planet would be

1

4 R

2

2 R

3

R / 2

4

R / 4

Explanation:

C Given,
$ρ_{P} = 2 ρ_{e}$
$g_{e} = g_{P}$
Mass $=$ volume $\times$ density
$M_{e} = \frac{4}{3} π R_{e}^{3} ρ_{e}$
From equation of acceleration due to gravity
$g_{e} = \frac{{GM}_{e}}{R_{e}^{2}} = \frac{G (4 / 3) π R_{e}^{3}}{R_{e}^{2}} ρ_{e}$
Here,
$g_{e} \propto R_{e} ρ_{e}$
Now, acceleration due to gravity of given planet, $g_{P} \propto R_{P} ρ_{P}$ given,
$g_{e} = g_{P}$
$R_{e} ρ_{e} = R_{P} ρ_{P}$
$R_{e} ρ_{e} = R_{P} 2 ρ_{e} (ρ_{P} = 2 ρ_{e})$
Hence, $R_{P} = \frac{R_{e}}{2} = \frac{R}{2} {∵ R_{e} = R}$

AMU-2016

Gravitation

138397 Calculate the acceleration due to gravity on the surface of a pulsar of mass $M = 1.98 \times 10^{30} kg$ and radius $R = 12 km$ rotating with time period $T = 0.041$ seconds. $(G = 6.67 \times 10^{- 11} m^{3} {kg}^{- 1} s^{- 2})$

1

9.2 \times 10^{11} m / s^{2}

2

8.15 \times 10^{11} m / s^{2}

3

7.32 \times 10^{11} m / s^{2}

4

6.98 \times 10^{11} m / s^{2}

Explanation:

A Given,
$Mass (M) = 1.98 \times 10^{30} Kg$
$Radius (R) = 12 km = 12000 m$
We know,
$g = \frac{GM}{R^{2}}$
$g = \frac{6.67 \times 10^{- 11} \times 1.98 \times 10^{30}}{(12000)^{2}} [∵ G = 6.67 \times 10^{- 11} m^{3} {Kg}^{- 1} s^{- 2}]$
$g = 9.2 \times 10^{11} m / s^{2}$

AMU-2012

Gravitation

138399 The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be sphere of radius $R$ of uniform density is shown in figure below.
The correct figure is

1

2

3

4

Explanation:

D At depth, when $r < R$
$g^{'} = g (1 - \frac{d}{R})$
$g^{'} = g (\frac{R - d}{R})$
$g^{'} = g (\frac{r}{R})$
$Where r = R - d$
$\Rightarrow g^{'} \propto r$
Case-1 when $r = 0$
$g^{'} = 0$
Case-2 when $r = R$
$g^{'} = g$
At height $h$ , when $r > R$
$g^{'} = \frac{{gR}^{2}}{(R + h)^{2}}$
$g^{'} = \frac{{gR}^{2}}{r^{2}} Where r = R + h$
$g^{'} \propto \frac{1}{r^{2}}$
Hence, the required graph is-

JEE-MAIN 26.06.2022 Shift-I

Gravitation

138400 If the earth stops rotating, the value of $g$ at the equator

1 Increases

2 decreases

3 no effect

4 none of these

Explanation:

A When earth is rotating,
$g_{1}^{'} = g - ω^{2} R \cos^{2} α$
When earth stop rotating,
$g_{2}^{'} = g$
$So, g_{2}^{'} > g_{1}^{'}$
If the earth stops rotating, the value of $g$ at equator will increase by amount $R ω^{2} \cos^{2} α$ . with $α = 0^{\circ}$ at equator. The effect of rotation of the earth on acceleration due to gravity is to decrease its value. Therefore if the earth stops rotating, the value of $g$ will increase.

DCE-2007

Gravitation

138401 The acceleration due to gravity at a height (take $g = 10 {ms}^{- 2}$ on earth surface) above earth's surface is $9 {ms}^{- 2}$ . It value at a point, at an equal distance below the surface of the earth is

1

9.7 {ms}^{- 2}

2

8.5 {ms}^{- 2}

3

7 . {ms}^{- 2}

4

5.5 {ms}^{- 2}

Explanation:

A Given,
$g_{h} = 9 m / s^{2}$
$g = 10 m / s^{2}$
We know,
$g_{h} = g [1 - \frac{2 h}{R}]$
$9 = 10 [1 - \frac{2 h}{R}]$
$[1 - \frac{2 h}{R}] = \frac{9}{10}$
$\frac{1}{10} = \frac{2 h}{R}$
$h = \frac{R}{20}$
Now, The value of $g$ below the earth's surface
$g_{d} = g [1 - \frac{d}{R}] = 10 [1 - \frac{R}{R}]$
$= 10 [1 - \frac{1}{20}]$
$g_{d} = \frac{10 \times 19}{20} = 9.5 {ms}^{- 2} \approx 9.7 {ms}^{- 2}$

EAMCET-1991

Gravitation

138396 The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$ , the radius of the planet would be

1

4 R

2

2 R

3

R / 2

4

R / 4

Explanation:

C Given,
$ρ_{P} = 2 ρ_{e}$
$g_{e} = g_{P}$
Mass $=$ volume $\times$ density
$M_{e} = \frac{4}{3} π R_{e}^{3} ρ_{e}$
From equation of acceleration due to gravity
$g_{e} = \frac{{GM}_{e}}{R_{e}^{2}} = \frac{G (4 / 3) π R_{e}^{3}}{R_{e}^{2}} ρ_{e}$
Here,
$g_{e} \propto R_{e} ρ_{e}$
Now, acceleration due to gravity of given planet, $g_{P} \propto R_{P} ρ_{P}$ given,
$g_{e} = g_{P}$
$R_{e} ρ_{e} = R_{P} ρ_{P}$
$R_{e} ρ_{e} = R_{P} 2 ρ_{e} (ρ_{P} = 2 ρ_{e})$
Hence, $R_{P} = \frac{R_{e}}{2} = \frac{R}{2} {∵ R_{e} = R}$

AMU-2016

Gravitation

138397 Calculate the acceleration due to gravity on the surface of a pulsar of mass $M = 1.98 \times 10^{30} kg$ and radius $R = 12 km$ rotating with time period $T = 0.041$ seconds. $(G = 6.67 \times 10^{- 11} m^{3} {kg}^{- 1} s^{- 2})$

1

9.2 \times 10^{11} m / s^{2}

2

8.15 \times 10^{11} m / s^{2}

3

7.32 \times 10^{11} m / s^{2}

4

6.98 \times 10^{11} m / s^{2}

Explanation:

A Given,
$Mass (M) = 1.98 \times 10^{30} Kg$
$Radius (R) = 12 km = 12000 m$
We know,
$g = \frac{GM}{R^{2}}$
$g = \frac{6.67 \times 10^{- 11} \times 1.98 \times 10^{30}}{(12000)^{2}} [∵ G = 6.67 \times 10^{- 11} m^{3} {Kg}^{- 1} s^{- 2}]$
$g = 9.2 \times 10^{11} m / s^{2}$

AMU-2012

Gravitation

138399 The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be sphere of radius $R$ of uniform density is shown in figure below.
The correct figure is

1

2

3

4

Explanation:

D At depth, when $r < R$
$g^{'} = g (1 - \frac{d}{R})$
$g^{'} = g (\frac{R - d}{R})$
$g^{'} = g (\frac{r}{R})$
$Where r = R - d$
$\Rightarrow g^{'} \propto r$
Case-1 when $r = 0$
$g^{'} = 0$
Case-2 when $r = R$
$g^{'} = g$
At height $h$ , when $r > R$
$g^{'} = \frac{{gR}^{2}}{(R + h)^{2}}$
$g^{'} = \frac{{gR}^{2}}{r^{2}} Where r = R + h$
$g^{'} \propto \frac{1}{r^{2}}$
Hence, the required graph is-

JEE-MAIN 26.06.2022 Shift-I

Gravitation

138400 If the earth stops rotating, the value of $g$ at the equator

1 Increases

2 decreases

3 no effect

4 none of these

Explanation:

A When earth is rotating,
$g_{1}^{'} = g - ω^{2} R \cos^{2} α$
When earth stop rotating,
$g_{2}^{'} = g$
$So, g_{2}^{'} > g_{1}^{'}$
If the earth stops rotating, the value of $g$ at equator will increase by amount $R ω^{2} \cos^{2} α$ . with $α = 0^{\circ}$ at equator. The effect of rotation of the earth on acceleration due to gravity is to decrease its value. Therefore if the earth stops rotating, the value of $g$ will increase.

DCE-2007

Gravitation

138401 The acceleration due to gravity at a height (take $g = 10 {ms}^{- 2}$ on earth surface) above earth's surface is $9 {ms}^{- 2}$ . It value at a point, at an equal distance below the surface of the earth is

1

9.7 {ms}^{- 2}

2

8.5 {ms}^{- 2}

3

7 . {ms}^{- 2}

4

5.5 {ms}^{- 2}

Explanation:

A Given,
$g_{h} = 9 m / s^{2}$
$g = 10 m / s^{2}$
We know,
$g_{h} = g [1 - \frac{2 h}{R}]$
$9 = 10 [1 - \frac{2 h}{R}]$
$[1 - \frac{2 h}{R}] = \frac{9}{10}$
$\frac{1}{10} = \frac{2 h}{R}$
$h = \frac{R}{20}$
Now, The value of $g$ below the earth's surface
$g_{d} = g [1 - \frac{d}{R}] = 10 [1 - \frac{R}{R}]$
$= 10 [1 - \frac{1}{20}]$
$g_{d} = \frac{10 \times 19}{20} = 9.5 {ms}^{- 2} \approx 9.7 {ms}^{- 2}$

EAMCET-1991

Gravitation

138396 The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$ , the radius of the planet would be

1

4 R

2

2 R

3

R / 2

4

R / 4

Explanation:

C Given,
$ρ_{P} = 2 ρ_{e}$
$g_{e} = g_{P}$
Mass $=$ volume $\times$ density
$M_{e} = \frac{4}{3} π R_{e}^{3} ρ_{e}$
From equation of acceleration due to gravity
$g_{e} = \frac{{GM}_{e}}{R_{e}^{2}} = \frac{G (4 / 3) π R_{e}^{3}}{R_{e}^{2}} ρ_{e}$
Here,
$g_{e} \propto R_{e} ρ_{e}$
Now, acceleration due to gravity of given planet, $g_{P} \propto R_{P} ρ_{P}$ given,
$g_{e} = g_{P}$
$R_{e} ρ_{e} = R_{P} ρ_{P}$
$R_{e} ρ_{e} = R_{P} 2 ρ_{e} (ρ_{P} = 2 ρ_{e})$
Hence, $R_{P} = \frac{R_{e}}{2} = \frac{R}{2} {∵ R_{e} = R}$

AMU-2016

Gravitation

138397 Calculate the acceleration due to gravity on the surface of a pulsar of mass $M = 1.98 \times 10^{30} kg$ and radius $R = 12 km$ rotating with time period $T = 0.041$ seconds. $(G = 6.67 \times 10^{- 11} m^{3} {kg}^{- 1} s^{- 2})$

1

9.2 \times 10^{11} m / s^{2}

2

8.15 \times 10^{11} m / s^{2}

3

7.32 \times 10^{11} m / s^{2}

4

6.98 \times 10^{11} m / s^{2}

Explanation:

A Given,
$Mass (M) = 1.98 \times 10^{30} Kg$
$Radius (R) = 12 km = 12000 m$
We know,
$g = \frac{GM}{R^{2}}$
$g = \frac{6.67 \times 10^{- 11} \times 1.98 \times 10^{30}}{(12000)^{2}} [∵ G = 6.67 \times 10^{- 11} m^{3} {Kg}^{- 1} s^{- 2}]$
$g = 9.2 \times 10^{11} m / s^{2}$

AMU-2012

Gravitation

138399 The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be sphere of radius $R$ of uniform density is shown in figure below.
The correct figure is

1

2

3

4

Explanation:

D At depth, when $r < R$
$g^{'} = g (1 - \frac{d}{R})$
$g^{'} = g (\frac{R - d}{R})$
$g^{'} = g (\frac{r}{R})$
$Where r = R - d$
$\Rightarrow g^{'} \propto r$
Case-1 when $r = 0$
$g^{'} = 0$
Case-2 when $r = R$
$g^{'} = g$
At height $h$ , when $r > R$
$g^{'} = \frac{{gR}^{2}}{(R + h)^{2}}$
$g^{'} = \frac{{gR}^{2}}{r^{2}} Where r = R + h$
$g^{'} \propto \frac{1}{r^{2}}$
Hence, the required graph is-

JEE-MAIN 26.06.2022 Shift-I

Gravitation

138400 If the earth stops rotating, the value of $g$ at the equator

1 Increases

2 decreases

3 no effect

4 none of these

Explanation:

A When earth is rotating,
$g_{1}^{'} = g - ω^{2} R \cos^{2} α$
When earth stop rotating,
$g_{2}^{'} = g$
$So, g_{2}^{'} > g_{1}^{'}$
If the earth stops rotating, the value of $g$ at equator will increase by amount $R ω^{2} \cos^{2} α$ . with $α = 0^{\circ}$ at equator. The effect of rotation of the earth on acceleration due to gravity is to decrease its value. Therefore if the earth stops rotating, the value of $g$ will increase.

DCE-2007

Gravitation

138401 The acceleration due to gravity at a height (take $g = 10 {ms}^{- 2}$ on earth surface) above earth's surface is $9 {ms}^{- 2}$ . It value at a point, at an equal distance below the surface of the earth is

1

9.7 {ms}^{- 2}

2

8.5 {ms}^{- 2}

3

7 . {ms}^{- 2}

4

5.5 {ms}^{- 2}

Explanation:

A Given,
$g_{h} = 9 m / s^{2}$
$g = 10 m / s^{2}$
We know,
$g_{h} = g [1 - \frac{2 h}{R}]$
$9 = 10 [1 - \frac{2 h}{R}]$
$[1 - \frac{2 h}{R}] = \frac{9}{10}$
$\frac{1}{10} = \frac{2 h}{R}$
$h = \frac{R}{20}$
Now, The value of $g$ below the earth's surface
$g_{d} = g [1 - \frac{d}{R}] = 10 [1 - \frac{R}{R}]$
$= 10 [1 - \frac{1}{20}]$
$g_{d} = \frac{10 \times 19}{20} = 9.5 {ms}^{- 2} \approx 9.7 {ms}^{- 2}$

EAMCET-1991

Gravitation

138396 The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$ , the radius of the planet would be

1

4 R

2

2 R

3

R / 2

4

R / 4

Explanation:

C Given,
$ρ_{P} = 2 ρ_{e}$
$g_{e} = g_{P}$
Mass $=$ volume $\times$ density
$M_{e} = \frac{4}{3} π R_{e}^{3} ρ_{e}$
From equation of acceleration due to gravity
$g_{e} = \frac{{GM}_{e}}{R_{e}^{2}} = \frac{G (4 / 3) π R_{e}^{3}}{R_{e}^{2}} ρ_{e}$
Here,
$g_{e} \propto R_{e} ρ_{e}$
Now, acceleration due to gravity of given planet, $g_{P} \propto R_{P} ρ_{P}$ given,
$g_{e} = g_{P}$
$R_{e} ρ_{e} = R_{P} ρ_{P}$
$R_{e} ρ_{e} = R_{P} 2 ρ_{e} (ρ_{P} = 2 ρ_{e})$
Hence, $R_{P} = \frac{R_{e}}{2} = \frac{R}{2} {∵ R_{e} = R}$

AMU-2016

Gravitation

138397 Calculate the acceleration due to gravity on the surface of a pulsar of mass $M = 1.98 \times 10^{30} kg$ and radius $R = 12 km$ rotating with time period $T = 0.041$ seconds. $(G = 6.67 \times 10^{- 11} m^{3} {kg}^{- 1} s^{- 2})$

1

9.2 \times 10^{11} m / s^{2}

2

8.15 \times 10^{11} m / s^{2}

3

7.32 \times 10^{11} m / s^{2}

4

6.98 \times 10^{11} m / s^{2}

Explanation:

A Given,
$Mass (M) = 1.98 \times 10^{30} Kg$
$Radius (R) = 12 km = 12000 m$
We know,
$g = \frac{GM}{R^{2}}$
$g = \frac{6.67 \times 10^{- 11} \times 1.98 \times 10^{30}}{(12000)^{2}} [∵ G = 6.67 \times 10^{- 11} m^{3} {Kg}^{- 1} s^{- 2}]$
$g = 9.2 \times 10^{11} m / s^{2}$

AMU-2012

Gravitation

138399 The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be sphere of radius $R$ of uniform density is shown in figure below.
The correct figure is

1

2

3

4

Explanation:

D At depth, when $r < R$
$g^{'} = g (1 - \frac{d}{R})$
$g^{'} = g (\frac{R - d}{R})$
$g^{'} = g (\frac{r}{R})$
$Where r = R - d$
$\Rightarrow g^{'} \propto r$
Case-1 when $r = 0$
$g^{'} = 0$
Case-2 when $r = R$
$g^{'} = g$
At height $h$ , when $r > R$
$g^{'} = \frac{{gR}^{2}}{(R + h)^{2}}$
$g^{'} = \frac{{gR}^{2}}{r^{2}} Where r = R + h$
$g^{'} \propto \frac{1}{r^{2}}$
Hence, the required graph is-

JEE-MAIN 26.06.2022 Shift-I

Gravitation

138400 If the earth stops rotating, the value of $g$ at the equator

1 Increases

2 decreases

3 no effect

4 none of these

Explanation:

A When earth is rotating,
$g_{1}^{'} = g - ω^{2} R \cos^{2} α$
When earth stop rotating,
$g_{2}^{'} = g$
$So, g_{2}^{'} > g_{1}^{'}$
If the earth stops rotating, the value of $g$ at equator will increase by amount $R ω^{2} \cos^{2} α$ . with $α = 0^{\circ}$ at equator. The effect of rotation of the earth on acceleration due to gravity is to decrease its value. Therefore if the earth stops rotating, the value of $g$ will increase.

DCE-2007

Gravitation

138401 The acceleration due to gravity at a height (take $g = 10 {ms}^{- 2}$ on earth surface) above earth's surface is $9 {ms}^{- 2}$ . It value at a point, at an equal distance below the surface of the earth is

1

9.7 {ms}^{- 2}

2

8.5 {ms}^{- 2}

3

7 . {ms}^{- 2}

4

5.5 {ms}^{- 2}

Explanation:

A Given,
$g_{h} = 9 m / s^{2}$
$g = 10 m / s^{2}$
We know,
$g_{h} = g [1 - \frac{2 h}{R}]$
$9 = 10 [1 - \frac{2 h}{R}]$
$[1 - \frac{2 h}{R}] = \frac{9}{10}$
$\frac{1}{10} = \frac{2 h}{R}$
$h = \frac{R}{20}$
Now, The value of $g$ below the earth's surface
$g_{d} = g [1 - \frac{d}{R}] = 10 [1 - \frac{R}{R}]$
$= 10 [1 - \frac{1}{20}]$
$g_{d} = \frac{10 \times 19}{20} = 9.5 {ms}^{- 2} \approx 9.7 {ms}^{- 2}$

EAMCET-1991

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