QBManager

PHXI08:GRAVITATION

359792 Correct formula for height of a satellite from earth's surface is

1

{(\frac{T^{2} R^{2} g}{4 π^{2}})}^{1 / 3} - R

2

{(\frac{T^{2} R^{2}}{4 π^{2} g})}^{1 / 3} - R

3

{(\frac{T^{2} R^{2} g}{4 π^{2}})}^{- 1 / 3} + R

4

{(\frac{T^{2} R^{2} g}{4 π})}^{1 / 2} - R

Explanation:

By balancing gravitational force and centripetal force, we get
$\frac{m v^{2}}{r} = \frac{G M m}{r^{2}}$
$v = \sqrt{\frac{G M}{R + h}} (∵ r = R + h)$
supporting img

$T = \frac{2 π (R + h)}{v} = \frac{2 π (R + h)^{3 / 2}}{\sqrt{G M}}$
$T^{2} = \frac{4 π^{2} (R + h)^{3}}{G M}$
$(R + h)^{3} = \frac{T^{2} G M}{4 π^{2}}$
$R + h = {(\frac{T^{2} G M}{4 π^{2}})}^{1 / 3}$
$∴ h = {(\frac{T^{2} g R^{2}}{4 π^{2}})}^{1 / 3} - R$
$(∵ G M = g R^{2})$

JEE - 2024

PHXI08:GRAVITATION

359793 If orbital velocity of a planet is given by $v = G^{a} M^{b} R^{c}$ , then what is the value of $\frac{2 a + b - 3 c}{3 b}$ ? [where, $G =$ gravitational constant, $M =$ mass of planet, $R =$ Radius of orbit]

1 2

2 5

3 1

4 7

Explanation:

$v = \sqrt{\frac{G M}{R}} \Rightarrow a = \frac{1}{2}; b = \frac{1}{2}; c = - \frac{1}{2}$
Now, $\frac{2 a + b - 3 c}{3 b} = \frac{2 (\frac{1}{2}) + (\frac{1}{2}) - 3 (- \frac{1}{2})}{3 (\frac{1}{2})} = 2$

PHXI08:GRAVITATION

359794 Assertion :
An astronaut experiences weightlessness in a space satellite.
Reason :
A body falls freely in space satellite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI08:GRAVITATION

359795 The rotation of the earth having radius $R$ about its axis speeds up to a value such that a man at latitude angle $60^{\circ}$ feels weightlessness. The duration of the day in such a case

1

π \sqrt{\frac{R}{g}}

2

\frac{π}{2} \sqrt{\frac{R}{g}}

3

\frac{π}{3} \sqrt{\frac{R}{g}}

4

π \sqrt{\frac{g}{R}}

Explanation:

$0 = g - R ω^{2}$ or $ω^{2} = \frac{4 g}{R}$ or $ω^{2} = 2 \sqrt{\frac{g}{R}}$
$\frac{2 π}{T} = 2 \sqrt{\frac{g}{R}}$
$∴ T = π \sqrt{\frac{g}{R}}$

PHXI08:GRAVITATION

359792 Correct formula for height of a satellite from earth's surface is

1

{(\frac{T^{2} R^{2} g}{4 π^{2}})}^{1 / 3} - R

2

{(\frac{T^{2} R^{2}}{4 π^{2} g})}^{1 / 3} - R

3

{(\frac{T^{2} R^{2} g}{4 π^{2}})}^{- 1 / 3} + R

4

{(\frac{T^{2} R^{2} g}{4 π})}^{1 / 2} - R

Explanation:

By balancing gravitational force and centripetal force, we get
$\frac{m v^{2}}{r} = \frac{G M m}{r^{2}}$
$v = \sqrt{\frac{G M}{R + h}} (∵ r = R + h)$
supporting img

$T = \frac{2 π (R + h)}{v} = \frac{2 π (R + h)^{3 / 2}}{\sqrt{G M}}$
$T^{2} = \frac{4 π^{2} (R + h)^{3}}{G M}$
$(R + h)^{3} = \frac{T^{2} G M}{4 π^{2}}$
$R + h = {(\frac{T^{2} G M}{4 π^{2}})}^{1 / 3}$
$∴ h = {(\frac{T^{2} g R^{2}}{4 π^{2}})}^{1 / 3} - R$
$(∵ G M = g R^{2})$

JEE - 2024

PHXI08:GRAVITATION

359793 If orbital velocity of a planet is given by $v = G^{a} M^{b} R^{c}$ , then what is the value of $\frac{2 a + b - 3 c}{3 b}$ ? [where, $G =$ gravitational constant, $M =$ mass of planet, $R =$ Radius of orbit]

1 2

2 5

3 1

4 7

Explanation:

$v = \sqrt{\frac{G M}{R}} \Rightarrow a = \frac{1}{2}; b = \frac{1}{2}; c = - \frac{1}{2}$
Now, $\frac{2 a + b - 3 c}{3 b} = \frac{2 (\frac{1}{2}) + (\frac{1}{2}) - 3 (- \frac{1}{2})}{3 (\frac{1}{2})} = 2$

PHXI08:GRAVITATION

359794 Assertion :
An astronaut experiences weightlessness in a space satellite.
Reason :
A body falls freely in space satellite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI08:GRAVITATION

359795 The rotation of the earth having radius $R$ about its axis speeds up to a value such that a man at latitude angle $60^{\circ}$ feels weightlessness. The duration of the day in such a case

1

π \sqrt{\frac{R}{g}}

2

\frac{π}{2} \sqrt{\frac{R}{g}}

3

\frac{π}{3} \sqrt{\frac{R}{g}}

4

π \sqrt{\frac{g}{R}}

Explanation:

$0 = g - R ω^{2}$ or $ω^{2} = \frac{4 g}{R}$ or $ω^{2} = 2 \sqrt{\frac{g}{R}}$
$\frac{2 π}{T} = 2 \sqrt{\frac{g}{R}}$
$∴ T = π \sqrt{\frac{g}{R}}$

PHXI08:GRAVITATION

359792 Correct formula for height of a satellite from earth's surface is

1

{(\frac{T^{2} R^{2} g}{4 π^{2}})}^{1 / 3} - R

2

{(\frac{T^{2} R^{2}}{4 π^{2} g})}^{1 / 3} - R

3

{(\frac{T^{2} R^{2} g}{4 π^{2}})}^{- 1 / 3} + R

4

{(\frac{T^{2} R^{2} g}{4 π})}^{1 / 2} - R

Explanation:

By balancing gravitational force and centripetal force, we get
$\frac{m v^{2}}{r} = \frac{G M m}{r^{2}}$
$v = \sqrt{\frac{G M}{R + h}} (∵ r = R + h)$
supporting img

$T = \frac{2 π (R + h)}{v} = \frac{2 π (R + h)^{3 / 2}}{\sqrt{G M}}$
$T^{2} = \frac{4 π^{2} (R + h)^{3}}{G M}$
$(R + h)^{3} = \frac{T^{2} G M}{4 π^{2}}$
$R + h = {(\frac{T^{2} G M}{4 π^{2}})}^{1 / 3}$
$∴ h = {(\frac{T^{2} g R^{2}}{4 π^{2}})}^{1 / 3} - R$
$(∵ G M = g R^{2})$

JEE - 2024

PHXI08:GRAVITATION

359793 If orbital velocity of a planet is given by $v = G^{a} M^{b} R^{c}$ , then what is the value of $\frac{2 a + b - 3 c}{3 b}$ ? [where, $G =$ gravitational constant, $M =$ mass of planet, $R =$ Radius of orbit]

1 2

2 5

3 1

4 7

Explanation:

$v = \sqrt{\frac{G M}{R}} \Rightarrow a = \frac{1}{2}; b = \frac{1}{2}; c = - \frac{1}{2}$
Now, $\frac{2 a + b - 3 c}{3 b} = \frac{2 (\frac{1}{2}) + (\frac{1}{2}) - 3 (- \frac{1}{2})}{3 (\frac{1}{2})} = 2$

PHXI08:GRAVITATION

359794 Assertion :
An astronaut experiences weightlessness in a space satellite.
Reason :
A body falls freely in space satellite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI08:GRAVITATION

359795 The rotation of the earth having radius $R$ about its axis speeds up to a value such that a man at latitude angle $60^{\circ}$ feels weightlessness. The duration of the day in such a case

1

π \sqrt{\frac{R}{g}}

2

\frac{π}{2} \sqrt{\frac{R}{g}}

3

\frac{π}{3} \sqrt{\frac{R}{g}}

4

π \sqrt{\frac{g}{R}}

Explanation:

$0 = g - R ω^{2}$ or $ω^{2} = \frac{4 g}{R}$ or $ω^{2} = 2 \sqrt{\frac{g}{R}}$
$\frac{2 π}{T} = 2 \sqrt{\frac{g}{R}}$
$∴ T = π \sqrt{\frac{g}{R}}$

PHXI08:GRAVITATION

359792 Correct formula for height of a satellite from earth's surface is

1

{(\frac{T^{2} R^{2} g}{4 π^{2}})}^{1 / 3} - R

2

{(\frac{T^{2} R^{2}}{4 π^{2} g})}^{1 / 3} - R

3

{(\frac{T^{2} R^{2} g}{4 π^{2}})}^{- 1 / 3} + R

4

{(\frac{T^{2} R^{2} g}{4 π})}^{1 / 2} - R

Explanation:

By balancing gravitational force and centripetal force, we get
$\frac{m v^{2}}{r} = \frac{G M m}{r^{2}}$
$v = \sqrt{\frac{G M}{R + h}} (∵ r = R + h)$
supporting img

$T = \frac{2 π (R + h)}{v} = \frac{2 π (R + h)^{3 / 2}}{\sqrt{G M}}$
$T^{2} = \frac{4 π^{2} (R + h)^{3}}{G M}$
$(R + h)^{3} = \frac{T^{2} G M}{4 π^{2}}$
$R + h = {(\frac{T^{2} G M}{4 π^{2}})}^{1 / 3}$
$∴ h = {(\frac{T^{2} g R^{2}}{4 π^{2}})}^{1 / 3} - R$
$(∵ G M = g R^{2})$

JEE - 2024

PHXI08:GRAVITATION

359793 If orbital velocity of a planet is given by $v = G^{a} M^{b} R^{c}$ , then what is the value of $\frac{2 a + b - 3 c}{3 b}$ ? [where, $G =$ gravitational constant, $M =$ mass of planet, $R =$ Radius of orbit]

1 2

2 5

3 1

4 7

Explanation:

$v = \sqrt{\frac{G M}{R}} \Rightarrow a = \frac{1}{2}; b = \frac{1}{2}; c = - \frac{1}{2}$
Now, $\frac{2 a + b - 3 c}{3 b} = \frac{2 (\frac{1}{2}) + (\frac{1}{2}) - 3 (- \frac{1}{2})}{3 (\frac{1}{2})} = 2$

PHXI08:GRAVITATION

359794 Assertion :
An astronaut experiences weightlessness in a space satellite.
Reason :
A body falls freely in space satellite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI08:GRAVITATION

359795 The rotation of the earth having radius $R$ about its axis speeds up to a value such that a man at latitude angle $60^{\circ}$ feels weightlessness. The duration of the day in such a case

1

π \sqrt{\frac{R}{g}}

2

\frac{π}{2} \sqrt{\frac{R}{g}}

3

\frac{π}{3} \sqrt{\frac{R}{g}}

4

π \sqrt{\frac{g}{R}}

Explanation:

$0 = g - R ω^{2}$ or $ω^{2} = \frac{4 g}{R}$ or $ω^{2} = 2 \sqrt{\frac{g}{R}}$
$\frac{2 π}{T} = 2 \sqrt{\frac{g}{R}}$
$∴ T = π \sqrt{\frac{g}{R}}$

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