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Gravitation

138290 The weight of a body at the surface of earth is $18 N$ . The weight of the body at an altitude of $3200 km$ above the earth's surface is (given, radius of earth $R_{e} = 6400 km$ )

1

19.6 N

2

9.8 N

3

4.9 N

4

8 N

Explanation:

D Given that, weight of body $(W_{B}) = mg = 18 N$ Radius of earth $(R_{e}) = 6400 km$ , altitude of body $(h) =$ $3200 km$

Acceleration due to gravity at height (h) -
$g^{'} = \frac{g}{{[1 + \frac{h}{R}]}^{2}} = \frac{g}{{[1 + \frac{3200}{6400}]}^{2}} = \frac{4 g}{9}$
So, body at an altitude of $3200 km$
$W_{B}^{'} = {mg}^{'} = \frac{4 mg}{9} = \frac{4}{9} \times 18 = 8 N$

JEE Main-24.01.2023

Gravitation

138292 T is the time period of simple pendulum on the earth's surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be

1

\frac{1}{4}

2 4

3

\frac{1}{2}

4 2

Explanation:

D We know that,
Time period of simple pendulum $(T) = 2 π \sqrt{\frac{l}{g}}$
And, $g^{'} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$
Radius of earth is equal to height of planet,
$g^{'} = \frac{g}{{(1 + \frac{R}{R})}^{2}} = \frac{g}{2^{2}} = \frac{g}{4}$
Now, time period of simple pendulum $(T^{'}) = 2 π \sqrt{\frac{l}{g^{'}}}$
$∴ T^{'} = 2 π \sqrt{\frac{l}{g / 4}} = 2 π \sqrt{\frac{4 l}{g}} = 4 π \sqrt{\frac{l}{g}}$
From equation (i) and equation (ii), we get -
$T^{'} = 2 \times 2 π \sqrt{\frac{l}{g}}$
$T^{'} = 2 T$
$X = 2$

JEE Main-25.01.2023

Gravitation

138293 At a certain depth " $d$ " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $3 R$ above earth surface. Where $R$ is Radius of earth (Take $R = 6400 Km$ ). The depth $d$ is equal to

1

4800 km

2

640 km

3

2560 km

4

5260 km

Explanation:

A Given that, radius of earth $(R_{e}) = 6400 km$ Change in gravitational acceleration due to depth is equal to four times at above $3 R$ .
Since, $g_{d} = 4 g_{3 R}$
$g (1 - \frac{d}{R}) = 4 \times \frac{GMe}{(R + 3 R)^{2}}$
$g (1 - \frac{d}{R}) = \frac{GMe}{4 R^{2}} (\frac{{GM}_{e}}{R^{2}} = g)$
$g (1 - \frac{d}{R}) = \frac{g}{4}$
$1 - \frac{d}{R} = \frac{1}{4}$
$d = \frac{3 R}{4} = \frac{3 \times 6400}{4} = 4800 km$

JEE Main-31.01.2023

Gravitation

138294 Two planets $A$ and $B$ have the same average density. Their radii $R_{A}$ and $R_{B}$ are such that $R_{A} : R_{B} = 3 : 1$ . If $g_{A}$ and $g_{B}$ are the acceleration due to gravity at the surfaces of the planets, the $g_{A} : g_{B}$ equals

1

3 : 1

2

1 : 3

3

9 : 1

4

1 : 9

5

\sqrt{3} : 1

Explanation:

A Given, $ρ_{A} = ρ_{B} = ρ, R_{A} : R_{B} = 3 : 1$
$\frac{R_{A}}{R_{B}} = \frac{3}{1} \Rightarrow R_{A} = 3 R_{B}$
$∵$ Two planets A and B have the same average density-
$∴$ Mass $= ρ \times$ volume
$M_{A} = ρ \times \frac{4}{3} π R_{A}^{3}$
$M_{B} = ρ \times \frac{4}{3} π R_{B}^{3}$
Acceleration due to gravity for A -
$g_{A} = \frac{G M_{A}}{{(R_{A})}^{2}}$
$g_{A} = \frac{G \times \frac{4}{3} π R_{A}^{3} ρ}{{(R_{A})}^{2}}$
$g_{A} = \frac{G \times 4 π R_{A} ρ}{3}$
Acceleration due to gravity for B -
$g_{B} = \frac{G_{B}}{{(R_{B})}^{2}}$
$= \frac{G \times \frac{4}{3} π R_{B}^{3} ρ}{{(R_{B})}^{2}}$
$= \frac{G \times 4 π R_{B} ρ}{3}$
On dividing equation (i) by (ii), we get -
$\frac{g_{A}}{g_{B}} = \frac{R_{A}}{R_{B}} = \frac{3 R_{B}}{R_{B}}$
$\frac{g_{A}}{g_{B}} = \frac{3}{1}$
$g_{A} : g_{B} = 3 : 1$

Kerala CEE -2018

Gravitation

138295 A uniform rod of length of $1 m$ and mass of 2 $kg$ is attached to a side support at $O$ as shown in the figure. The rod is at equilibrium due to upward force $T$ acting at $P$ . Assume the acceleration due to gravity as $10 m / s^{2}$ . The value of $T$ is

1 0

2

2 N

3

5 N

4

10 N

5

20 N

Explanation:

D Given, mass of uniform rod $(m) = 2 kg$ , length $= 1 m$ , acceleration due to gravity $= 10 m / s^{2}$
The one end of the uniform rod is fixed and force $T$ is acting in upward direction at point $P$ .
$∵ T = \frac{mg}{2} = \frac{2 \times 10}{2} = 10 N$

Kerala CEE - 2017

Gravitation

138290 The weight of a body at the surface of earth is $18 N$ . The weight of the body at an altitude of $3200 km$ above the earth's surface is (given, radius of earth $R_{e} = 6400 km$ )

1

19.6 N

2

9.8 N

3

4.9 N

4

8 N

Explanation:

D Given that, weight of body $(W_{B}) = mg = 18 N$ Radius of earth $(R_{e}) = 6400 km$ , altitude of body $(h) =$ $3200 km$

Acceleration due to gravity at height (h) -
$g^{'} = \frac{g}{{[1 + \frac{h}{R}]}^{2}} = \frac{g}{{[1 + \frac{3200}{6400}]}^{2}} = \frac{4 g}{9}$
So, body at an altitude of $3200 km$
$W_{B}^{'} = {mg}^{'} = \frac{4 mg}{9} = \frac{4}{9} \times 18 = 8 N$

JEE Main-24.01.2023

Gravitation

138292 T is the time period of simple pendulum on the earth's surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be

1

\frac{1}{4}

2 4

3

\frac{1}{2}

4 2

Explanation:

D We know that,
Time period of simple pendulum $(T) = 2 π \sqrt{\frac{l}{g}}$
And, $g^{'} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$
Radius of earth is equal to height of planet,
$g^{'} = \frac{g}{{(1 + \frac{R}{R})}^{2}} = \frac{g}{2^{2}} = \frac{g}{4}$
Now, time period of simple pendulum $(T^{'}) = 2 π \sqrt{\frac{l}{g^{'}}}$
$∴ T^{'} = 2 π \sqrt{\frac{l}{g / 4}} = 2 π \sqrt{\frac{4 l}{g}} = 4 π \sqrt{\frac{l}{g}}$
From equation (i) and equation (ii), we get -
$T^{'} = 2 \times 2 π \sqrt{\frac{l}{g}}$
$T^{'} = 2 T$
$X = 2$

JEE Main-25.01.2023

Gravitation

138293 At a certain depth " $d$ " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $3 R$ above earth surface. Where $R$ is Radius of earth (Take $R = 6400 Km$ ). The depth $d$ is equal to

1

4800 km

2

640 km

3

2560 km

4

5260 km

Explanation:

A Given that, radius of earth $(R_{e}) = 6400 km$ Change in gravitational acceleration due to depth is equal to four times at above $3 R$ .
Since, $g_{d} = 4 g_{3 R}$
$g (1 - \frac{d}{R}) = 4 \times \frac{GMe}{(R + 3 R)^{2}}$
$g (1 - \frac{d}{R}) = \frac{GMe}{4 R^{2}} (\frac{{GM}_{e}}{R^{2}} = g)$
$g (1 - \frac{d}{R}) = \frac{g}{4}$
$1 - \frac{d}{R} = \frac{1}{4}$
$d = \frac{3 R}{4} = \frac{3 \times 6400}{4} = 4800 km$

JEE Main-31.01.2023

Gravitation

138294 Two planets $A$ and $B$ have the same average density. Their radii $R_{A}$ and $R_{B}$ are such that $R_{A} : R_{B} = 3 : 1$ . If $g_{A}$ and $g_{B}$ are the acceleration due to gravity at the surfaces of the planets, the $g_{A} : g_{B}$ equals

1

3 : 1

2

1 : 3

3

9 : 1

4

1 : 9

5

\sqrt{3} : 1

Explanation:

A Given, $ρ_{A} = ρ_{B} = ρ, R_{A} : R_{B} = 3 : 1$
$\frac{R_{A}}{R_{B}} = \frac{3}{1} \Rightarrow R_{A} = 3 R_{B}$
$∵$ Two planets A and B have the same average density-
$∴$ Mass $= ρ \times$ volume
$M_{A} = ρ \times \frac{4}{3} π R_{A}^{3}$
$M_{B} = ρ \times \frac{4}{3} π R_{B}^{3}$
Acceleration due to gravity for A -
$g_{A} = \frac{G M_{A}}{{(R_{A})}^{2}}$
$g_{A} = \frac{G \times \frac{4}{3} π R_{A}^{3} ρ}{{(R_{A})}^{2}}$
$g_{A} = \frac{G \times 4 π R_{A} ρ}{3}$
Acceleration due to gravity for B -
$g_{B} = \frac{G_{B}}{{(R_{B})}^{2}}$
$= \frac{G \times \frac{4}{3} π R_{B}^{3} ρ}{{(R_{B})}^{2}}$
$= \frac{G \times 4 π R_{B} ρ}{3}$
On dividing equation (i) by (ii), we get -
$\frac{g_{A}}{g_{B}} = \frac{R_{A}}{R_{B}} = \frac{3 R_{B}}{R_{B}}$
$\frac{g_{A}}{g_{B}} = \frac{3}{1}$
$g_{A} : g_{B} = 3 : 1$

Kerala CEE -2018

Gravitation

138295 A uniform rod of length of $1 m$ and mass of 2 $kg$ is attached to a side support at $O$ as shown in the figure. The rod is at equilibrium due to upward force $T$ acting at $P$ . Assume the acceleration due to gravity as $10 m / s^{2}$ . The value of $T$ is

1 0

2

2 N

3

5 N

4

10 N

5

20 N

Explanation:

D Given, mass of uniform rod $(m) = 2 kg$ , length $= 1 m$ , acceleration due to gravity $= 10 m / s^{2}$
The one end of the uniform rod is fixed and force $T$ is acting in upward direction at point $P$ .
$∵ T = \frac{mg}{2} = \frac{2 \times 10}{2} = 10 N$

Kerala CEE - 2017

Gravitation

138290 The weight of a body at the surface of earth is $18 N$ . The weight of the body at an altitude of $3200 km$ above the earth's surface is (given, radius of earth $R_{e} = 6400 km$ )

1

19.6 N

2

9.8 N

3

4.9 N

4

8 N

Explanation:

D Given that, weight of body $(W_{B}) = mg = 18 N$ Radius of earth $(R_{e}) = 6400 km$ , altitude of body $(h) =$ $3200 km$

Acceleration due to gravity at height (h) -
$g^{'} = \frac{g}{{[1 + \frac{h}{R}]}^{2}} = \frac{g}{{[1 + \frac{3200}{6400}]}^{2}} = \frac{4 g}{9}$
So, body at an altitude of $3200 km$
$W_{B}^{'} = {mg}^{'} = \frac{4 mg}{9} = \frac{4}{9} \times 18 = 8 N$

JEE Main-24.01.2023

Gravitation

138292 T is the time period of simple pendulum on the earth's surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be

1

\frac{1}{4}

2 4

3

\frac{1}{2}

4 2

Explanation:

D We know that,
Time period of simple pendulum $(T) = 2 π \sqrt{\frac{l}{g}}$
And, $g^{'} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$
Radius of earth is equal to height of planet,
$g^{'} = \frac{g}{{(1 + \frac{R}{R})}^{2}} = \frac{g}{2^{2}} = \frac{g}{4}$
Now, time period of simple pendulum $(T^{'}) = 2 π \sqrt{\frac{l}{g^{'}}}$
$∴ T^{'} = 2 π \sqrt{\frac{l}{g / 4}} = 2 π \sqrt{\frac{4 l}{g}} = 4 π \sqrt{\frac{l}{g}}$
From equation (i) and equation (ii), we get -
$T^{'} = 2 \times 2 π \sqrt{\frac{l}{g}}$
$T^{'} = 2 T$
$X = 2$

JEE Main-25.01.2023

Gravitation

138293 At a certain depth " $d$ " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $3 R$ above earth surface. Where $R$ is Radius of earth (Take $R = 6400 Km$ ). The depth $d$ is equal to

1

4800 km

2

640 km

3

2560 km

4

5260 km

Explanation:

A Given that, radius of earth $(R_{e}) = 6400 km$ Change in gravitational acceleration due to depth is equal to four times at above $3 R$ .
Since, $g_{d} = 4 g_{3 R}$
$g (1 - \frac{d}{R}) = 4 \times \frac{GMe}{(R + 3 R)^{2}}$
$g (1 - \frac{d}{R}) = \frac{GMe}{4 R^{2}} (\frac{{GM}_{e}}{R^{2}} = g)$
$g (1 - \frac{d}{R}) = \frac{g}{4}$
$1 - \frac{d}{R} = \frac{1}{4}$
$d = \frac{3 R}{4} = \frac{3 \times 6400}{4} = 4800 km$

JEE Main-31.01.2023

Gravitation

138294 Two planets $A$ and $B$ have the same average density. Their radii $R_{A}$ and $R_{B}$ are such that $R_{A} : R_{B} = 3 : 1$ . If $g_{A}$ and $g_{B}$ are the acceleration due to gravity at the surfaces of the planets, the $g_{A} : g_{B}$ equals

1

3 : 1

2

1 : 3

3

9 : 1

4

1 : 9

5

\sqrt{3} : 1

Explanation:

A Given, $ρ_{A} = ρ_{B} = ρ, R_{A} : R_{B} = 3 : 1$
$\frac{R_{A}}{R_{B}} = \frac{3}{1} \Rightarrow R_{A} = 3 R_{B}$
$∵$ Two planets A and B have the same average density-
$∴$ Mass $= ρ \times$ volume
$M_{A} = ρ \times \frac{4}{3} π R_{A}^{3}$
$M_{B} = ρ \times \frac{4}{3} π R_{B}^{3}$
Acceleration due to gravity for A -
$g_{A} = \frac{G M_{A}}{{(R_{A})}^{2}}$
$g_{A} = \frac{G \times \frac{4}{3} π R_{A}^{3} ρ}{{(R_{A})}^{2}}$
$g_{A} = \frac{G \times 4 π R_{A} ρ}{3}$
Acceleration due to gravity for B -
$g_{B} = \frac{G_{B}}{{(R_{B})}^{2}}$
$= \frac{G \times \frac{4}{3} π R_{B}^{3} ρ}{{(R_{B})}^{2}}$
$= \frac{G \times 4 π R_{B} ρ}{3}$
On dividing equation (i) by (ii), we get -
$\frac{g_{A}}{g_{B}} = \frac{R_{A}}{R_{B}} = \frac{3 R_{B}}{R_{B}}$
$\frac{g_{A}}{g_{B}} = \frac{3}{1}$
$g_{A} : g_{B} = 3 : 1$

Kerala CEE -2018

Gravitation

138295 A uniform rod of length of $1 m$ and mass of 2 $kg$ is attached to a side support at $O$ as shown in the figure. The rod is at equilibrium due to upward force $T$ acting at $P$ . Assume the acceleration due to gravity as $10 m / s^{2}$ . The value of $T$ is

1 0

2

2 N

3

5 N

4

10 N

5

20 N

Explanation:

D Given, mass of uniform rod $(m) = 2 kg$ , length $= 1 m$ , acceleration due to gravity $= 10 m / s^{2}$
The one end of the uniform rod is fixed and force $T$ is acting in upward direction at point $P$ .
$∵ T = \frac{mg}{2} = \frac{2 \times 10}{2} = 10 N$

Kerala CEE - 2017

Gravitation

138290 The weight of a body at the surface of earth is $18 N$ . The weight of the body at an altitude of $3200 km$ above the earth's surface is (given, radius of earth $R_{e} = 6400 km$ )

1

19.6 N

2

9.8 N

3

4.9 N

4

8 N

Explanation:

D Given that, weight of body $(W_{B}) = mg = 18 N$ Radius of earth $(R_{e}) = 6400 km$ , altitude of body $(h) =$ $3200 km$

Acceleration due to gravity at height (h) -
$g^{'} = \frac{g}{{[1 + \frac{h}{R}]}^{2}} = \frac{g}{{[1 + \frac{3200}{6400}]}^{2}} = \frac{4 g}{9}$
So, body at an altitude of $3200 km$
$W_{B}^{'} = {mg}^{'} = \frac{4 mg}{9} = \frac{4}{9} \times 18 = 8 N$

JEE Main-24.01.2023

Gravitation

138292 T is the time period of simple pendulum on the earth's surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be

1

\frac{1}{4}

2 4

3

\frac{1}{2}

4 2

Explanation:

D We know that,
Time period of simple pendulum $(T) = 2 π \sqrt{\frac{l}{g}}$
And, $g^{'} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$
Radius of earth is equal to height of planet,
$g^{'} = \frac{g}{{(1 + \frac{R}{R})}^{2}} = \frac{g}{2^{2}} = \frac{g}{4}$
Now, time period of simple pendulum $(T^{'}) = 2 π \sqrt{\frac{l}{g^{'}}}$
$∴ T^{'} = 2 π \sqrt{\frac{l}{g / 4}} = 2 π \sqrt{\frac{4 l}{g}} = 4 π \sqrt{\frac{l}{g}}$
From equation (i) and equation (ii), we get -
$T^{'} = 2 \times 2 π \sqrt{\frac{l}{g}}$
$T^{'} = 2 T$
$X = 2$

JEE Main-25.01.2023

Gravitation

138293 At a certain depth " $d$ " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $3 R$ above earth surface. Where $R$ is Radius of earth (Take $R = 6400 Km$ ). The depth $d$ is equal to

1

4800 km

2

640 km

3

2560 km

4

5260 km

Explanation:

A Given that, radius of earth $(R_{e}) = 6400 km$ Change in gravitational acceleration due to depth is equal to four times at above $3 R$ .
Since, $g_{d} = 4 g_{3 R}$
$g (1 - \frac{d}{R}) = 4 \times \frac{GMe}{(R + 3 R)^{2}}$
$g (1 - \frac{d}{R}) = \frac{GMe}{4 R^{2}} (\frac{{GM}_{e}}{R^{2}} = g)$
$g (1 - \frac{d}{R}) = \frac{g}{4}$
$1 - \frac{d}{R} = \frac{1}{4}$
$d = \frac{3 R}{4} = \frac{3 \times 6400}{4} = 4800 km$

JEE Main-31.01.2023

Gravitation

138294 Two planets $A$ and $B$ have the same average density. Their radii $R_{A}$ and $R_{B}$ are such that $R_{A} : R_{B} = 3 : 1$ . If $g_{A}$ and $g_{B}$ are the acceleration due to gravity at the surfaces of the planets, the $g_{A} : g_{B}$ equals

1

3 : 1

2

1 : 3

3

9 : 1

4

1 : 9

5

\sqrt{3} : 1

Explanation:

A Given, $ρ_{A} = ρ_{B} = ρ, R_{A} : R_{B} = 3 : 1$
$\frac{R_{A}}{R_{B}} = \frac{3}{1} \Rightarrow R_{A} = 3 R_{B}$
$∵$ Two planets A and B have the same average density-
$∴$ Mass $= ρ \times$ volume
$M_{A} = ρ \times \frac{4}{3} π R_{A}^{3}$
$M_{B} = ρ \times \frac{4}{3} π R_{B}^{3}$
Acceleration due to gravity for A -
$g_{A} = \frac{G M_{A}}{{(R_{A})}^{2}}$
$g_{A} = \frac{G \times \frac{4}{3} π R_{A}^{3} ρ}{{(R_{A})}^{2}}$
$g_{A} = \frac{G \times 4 π R_{A} ρ}{3}$
Acceleration due to gravity for B -
$g_{B} = \frac{G_{B}}{{(R_{B})}^{2}}$
$= \frac{G \times \frac{4}{3} π R_{B}^{3} ρ}{{(R_{B})}^{2}}$
$= \frac{G \times 4 π R_{B} ρ}{3}$
On dividing equation (i) by (ii), we get -
$\frac{g_{A}}{g_{B}} = \frac{R_{A}}{R_{B}} = \frac{3 R_{B}}{R_{B}}$
$\frac{g_{A}}{g_{B}} = \frac{3}{1}$
$g_{A} : g_{B} = 3 : 1$

Kerala CEE -2018

Gravitation

138295 A uniform rod of length of $1 m$ and mass of 2 $kg$ is attached to a side support at $O$ as shown in the figure. The rod is at equilibrium due to upward force $T$ acting at $P$ . Assume the acceleration due to gravity as $10 m / s^{2}$ . The value of $T$ is

1 0

2

2 N

3

5 N

4

10 N

5

20 N

Explanation:

D Given, mass of uniform rod $(m) = 2 kg$ , length $= 1 m$ , acceleration due to gravity $= 10 m / s^{2}$
The one end of the uniform rod is fixed and force $T$ is acting in upward direction at point $P$ .
$∵ T = \frac{mg}{2} = \frac{2 \times 10}{2} = 10 N$

Kerala CEE - 2017

Gravitation

138290 The weight of a body at the surface of earth is $18 N$ . The weight of the body at an altitude of $3200 km$ above the earth's surface is (given, radius of earth $R_{e} = 6400 km$ )

1

19.6 N

2

9.8 N

3

4.9 N

4

8 N

Explanation:

D Given that, weight of body $(W_{B}) = mg = 18 N$ Radius of earth $(R_{e}) = 6400 km$ , altitude of body $(h) =$ $3200 km$

Acceleration due to gravity at height (h) -
$g^{'} = \frac{g}{{[1 + \frac{h}{R}]}^{2}} = \frac{g}{{[1 + \frac{3200}{6400}]}^{2}} = \frac{4 g}{9}$
So, body at an altitude of $3200 km$
$W_{B}^{'} = {mg}^{'} = \frac{4 mg}{9} = \frac{4}{9} \times 18 = 8 N$

JEE Main-24.01.2023

Gravitation

138292 T is the time period of simple pendulum on the earth's surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be

1

\frac{1}{4}

2 4

3

\frac{1}{2}

4 2

Explanation:

D We know that,
Time period of simple pendulum $(T) = 2 π \sqrt{\frac{l}{g}}$
And, $g^{'} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$
Radius of earth is equal to height of planet,
$g^{'} = \frac{g}{{(1 + \frac{R}{R})}^{2}} = \frac{g}{2^{2}} = \frac{g}{4}$
Now, time period of simple pendulum $(T^{'}) = 2 π \sqrt{\frac{l}{g^{'}}}$
$∴ T^{'} = 2 π \sqrt{\frac{l}{g / 4}} = 2 π \sqrt{\frac{4 l}{g}} = 4 π \sqrt{\frac{l}{g}}$
From equation (i) and equation (ii), we get -
$T^{'} = 2 \times 2 π \sqrt{\frac{l}{g}}$
$T^{'} = 2 T$
$X = 2$

JEE Main-25.01.2023

Gravitation

138293 At a certain depth " $d$ " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $3 R$ above earth surface. Where $R$ is Radius of earth (Take $R = 6400 Km$ ). The depth $d$ is equal to

1

4800 km

2

640 km

3

2560 km

4

5260 km

Explanation:

A Given that, radius of earth $(R_{e}) = 6400 km$ Change in gravitational acceleration due to depth is equal to four times at above $3 R$ .
Since, $g_{d} = 4 g_{3 R}$
$g (1 - \frac{d}{R}) = 4 \times \frac{GMe}{(R + 3 R)^{2}}$
$g (1 - \frac{d}{R}) = \frac{GMe}{4 R^{2}} (\frac{{GM}_{e}}{R^{2}} = g)$
$g (1 - \frac{d}{R}) = \frac{g}{4}$
$1 - \frac{d}{R} = \frac{1}{4}$
$d = \frac{3 R}{4} = \frac{3 \times 6400}{4} = 4800 km$

JEE Main-31.01.2023

Gravitation

138294 Two planets $A$ and $B$ have the same average density. Their radii $R_{A}$ and $R_{B}$ are such that $R_{A} : R_{B} = 3 : 1$ . If $g_{A}$ and $g_{B}$ are the acceleration due to gravity at the surfaces of the planets, the $g_{A} : g_{B}$ equals

1

3 : 1

2

1 : 3

3

9 : 1

4

1 : 9

5

\sqrt{3} : 1

Explanation:

A Given, $ρ_{A} = ρ_{B} = ρ, R_{A} : R_{B} = 3 : 1$
$\frac{R_{A}}{R_{B}} = \frac{3}{1} \Rightarrow R_{A} = 3 R_{B}$
$∵$ Two planets A and B have the same average density-
$∴$ Mass $= ρ \times$ volume
$M_{A} = ρ \times \frac{4}{3} π R_{A}^{3}$
$M_{B} = ρ \times \frac{4}{3} π R_{B}^{3}$
Acceleration due to gravity for A -
$g_{A} = \frac{G M_{A}}{{(R_{A})}^{2}}$
$g_{A} = \frac{G \times \frac{4}{3} π R_{A}^{3} ρ}{{(R_{A})}^{2}}$
$g_{A} = \frac{G \times 4 π R_{A} ρ}{3}$
Acceleration due to gravity for B -
$g_{B} = \frac{G_{B}}{{(R_{B})}^{2}}$
$= \frac{G \times \frac{4}{3} π R_{B}^{3} ρ}{{(R_{B})}^{2}}$
$= \frac{G \times 4 π R_{B} ρ}{3}$
On dividing equation (i) by (ii), we get -
$\frac{g_{A}}{g_{B}} = \frac{R_{A}}{R_{B}} = \frac{3 R_{B}}{R_{B}}$
$\frac{g_{A}}{g_{B}} = \frac{3}{1}$
$g_{A} : g_{B} = 3 : 1$

Kerala CEE -2018

Gravitation

138295 A uniform rod of length of $1 m$ and mass of 2 $kg$ is attached to a side support at $O$ as shown in the figure. The rod is at equilibrium due to upward force $T$ acting at $P$ . Assume the acceleration due to gravity as $10 m / s^{2}$ . The value of $T$ is

1 0

2

2 N

3

5 N

4

10 N

5

20 N

Explanation:

D Given, mass of uniform rod $(m) = 2 kg$ , length $= 1 m$ , acceleration due to gravity $= 10 m / s^{2}$
The one end of the uniform rod is fixed and force $T$ is acting in upward direction at point $P$ .
$∵ T = \frac{mg}{2} = \frac{2 \times 10}{2} = 10 N$

Kerala CEE - 2017

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