QBManager

PHXI08:GRAVITATION

359996 One can easily weigh the earth by calculating the mass of earth by using formula

1

\frac{G R_{e}^{2}}{g}

2

\frac{g R_{e}^{2}}{G}

3

\frac{g R_{e}}{G}

4

\frac{G R_{e}^{3}}{g}

Explanation:

$F = \frac{G M_{e} m}{R_{e}^{2}}$
$m g = \frac{G M_{e} m}{R_{e}^{2}}$
$g = \frac{G M_{e}}{R_{e}^{2}}$
$M_{e} = \frac{g R_{e}^{2}}{G}$

PHXI08:GRAVITATION

359997 For a body projected at an angle with the horizontal from the ground, choose the correct statement.

1 The kinetic energy (KE) is zero at the highest point of projectile motion.

2 The horizontal component of velocity is zero at the highest point.

3 The vertical component of momentum is maximum at the highest point.

4 Gravitational potential energy is maximum at the highest point.

Explanation:

(1),(2) At the highest point of the projectile, horizontal component of velocity is not zero, so $K . E$ is not zero.
(3) As vertical component of velocity is zero at highest point, vertical component of momentum is zero.
(4) Gravitational potential energy depends on height.

JEE - 2023

PHXI08:GRAVITATION

359998 If there was a reduction in gravitational effect, then which of the following forces do you think would change in some respect?

1 Magnetic force

2 Electrostatic force

3 Viscous force

4 Archimedes uplift

PHXI08:GRAVITATION

359999 A projectile of mass $m$ is fired from the surface of earth at an angle $α = 60^{\circ}$ from the vertical. The initial speed of the projectile is $u = \sqrt{\frac{G M}{R}}$ , where $M$ and $R$ are mass and radius of the earth. Find the height by the projectile in kilometer. Neglect air resistance and the rotation of earth. (Given $R = 6400 k m$ )
supporting img

1

3200 k m

2

2350 k m

3

7375 k m

4

1250 k m

Explanation:

Let $v$ be the speed of projectile at the highest point. By conservation of angular momentum and conservation of mechanical energy, we have
$m u R \sin α = m v r_{max}$ and $\frac{1}{2} m u^{2} - \frac{G M m}{R} = \frac{1}{2} m v^{2} - \frac{G M m}{r_{max}}$
Solving these two equations, we get $r_{max} = \frac{3 R}{2}$
$∴ h_{max} = r_{max} - R = \frac{R}{2} = 3200 k m$

PHXI08:GRAVITATION

359996 One can easily weigh the earth by calculating the mass of earth by using formula

1

\frac{G R_{e}^{2}}{g}

2

\frac{g R_{e}^{2}}{G}

3

\frac{g R_{e}}{G}

4

\frac{G R_{e}^{3}}{g}

Explanation:

$F = \frac{G M_{e} m}{R_{e}^{2}}$
$m g = \frac{G M_{e} m}{R_{e}^{2}}$
$g = \frac{G M_{e}}{R_{e}^{2}}$
$M_{e} = \frac{g R_{e}^{2}}{G}$

PHXI08:GRAVITATION

359997 For a body projected at an angle with the horizontal from the ground, choose the correct statement.

1 The kinetic energy (KE) is zero at the highest point of projectile motion.

2 The horizontal component of velocity is zero at the highest point.

3 The vertical component of momentum is maximum at the highest point.

4 Gravitational potential energy is maximum at the highest point.

Explanation:

(1),(2) At the highest point of the projectile, horizontal component of velocity is not zero, so $K . E$ is not zero.
(3) As vertical component of velocity is zero at highest point, vertical component of momentum is zero.
(4) Gravitational potential energy depends on height.

JEE - 2023

PHXI08:GRAVITATION

359998 If there was a reduction in gravitational effect, then which of the following forces do you think would change in some respect?

1 Magnetic force

2 Electrostatic force

3 Viscous force

4 Archimedes uplift

PHXI08:GRAVITATION

359999 A projectile of mass $m$ is fired from the surface of earth at an angle $α = 60^{\circ}$ from the vertical. The initial speed of the projectile is $u = \sqrt{\frac{G M}{R}}$ , where $M$ and $R$ are mass and radius of the earth. Find the height by the projectile in kilometer. Neglect air resistance and the rotation of earth. (Given $R = 6400 k m$ )
supporting img

1

3200 k m

2

2350 k m

3

7375 k m

4

1250 k m

Explanation:

Let $v$ be the speed of projectile at the highest point. By conservation of angular momentum and conservation of mechanical energy, we have
$m u R \sin α = m v r_{max}$ and $\frac{1}{2} m u^{2} - \frac{G M m}{R} = \frac{1}{2} m v^{2} - \frac{G M m}{r_{max}}$
Solving these two equations, we get $r_{max} = \frac{3 R}{2}$
$∴ h_{max} = r_{max} - R = \frac{R}{2} = 3200 k m$

PHXI08:GRAVITATION

359996 One can easily weigh the earth by calculating the mass of earth by using formula

1

\frac{G R_{e}^{2}}{g}

2

\frac{g R_{e}^{2}}{G}

3

\frac{g R_{e}}{G}

4

\frac{G R_{e}^{3}}{g}

Explanation:

$F = \frac{G M_{e} m}{R_{e}^{2}}$
$m g = \frac{G M_{e} m}{R_{e}^{2}}$
$g = \frac{G M_{e}}{R_{e}^{2}}$
$M_{e} = \frac{g R_{e}^{2}}{G}$

PHXI08:GRAVITATION

359997 For a body projected at an angle with the horizontal from the ground, choose the correct statement.

1 The kinetic energy (KE) is zero at the highest point of projectile motion.

2 The horizontal component of velocity is zero at the highest point.

3 The vertical component of momentum is maximum at the highest point.

4 Gravitational potential energy is maximum at the highest point.

Explanation:

(1),(2) At the highest point of the projectile, horizontal component of velocity is not zero, so $K . E$ is not zero.
(3) As vertical component of velocity is zero at highest point, vertical component of momentum is zero.
(4) Gravitational potential energy depends on height.

JEE - 2023

PHXI08:GRAVITATION

359998 If there was a reduction in gravitational effect, then which of the following forces do you think would change in some respect?

1 Magnetic force

2 Electrostatic force

3 Viscous force

4 Archimedes uplift

PHXI08:GRAVITATION

359999 A projectile of mass $m$ is fired from the surface of earth at an angle $α = 60^{\circ}$ from the vertical. The initial speed of the projectile is $u = \sqrt{\frac{G M}{R}}$ , where $M$ and $R$ are mass and radius of the earth. Find the height by the projectile in kilometer. Neglect air resistance and the rotation of earth. (Given $R = 6400 k m$ )
supporting img

1

3200 k m

2

2350 k m

3

7375 k m

4

1250 k m

Explanation:

Let $v$ be the speed of projectile at the highest point. By conservation of angular momentum and conservation of mechanical energy, we have
$m u R \sin α = m v r_{max}$ and $\frac{1}{2} m u^{2} - \frac{G M m}{R} = \frac{1}{2} m v^{2} - \frac{G M m}{r_{max}}$
Solving these two equations, we get $r_{max} = \frac{3 R}{2}$
$∴ h_{max} = r_{max} - R = \frac{R}{2} = 3200 k m$

PHXI08:GRAVITATION

359996 One can easily weigh the earth by calculating the mass of earth by using formula

1

\frac{G R_{e}^{2}}{g}

2

\frac{g R_{e}^{2}}{G}

3

\frac{g R_{e}}{G}

4

\frac{G R_{e}^{3}}{g}

Explanation:

$F = \frac{G M_{e} m}{R_{e}^{2}}$
$m g = \frac{G M_{e} m}{R_{e}^{2}}$
$g = \frac{G M_{e}}{R_{e}^{2}}$
$M_{e} = \frac{g R_{e}^{2}}{G}$

PHXI08:GRAVITATION

359997 For a body projected at an angle with the horizontal from the ground, choose the correct statement.

1 The kinetic energy (KE) is zero at the highest point of projectile motion.

2 The horizontal component of velocity is zero at the highest point.

3 The vertical component of momentum is maximum at the highest point.

4 Gravitational potential energy is maximum at the highest point.

Explanation:

(1),(2) At the highest point of the projectile, horizontal component of velocity is not zero, so $K . E$ is not zero.
(3) As vertical component of velocity is zero at highest point, vertical component of momentum is zero.
(4) Gravitational potential energy depends on height.

JEE - 2023

PHXI08:GRAVITATION

359998 If there was a reduction in gravitational effect, then which of the following forces do you think would change in some respect?

1 Magnetic force

2 Electrostatic force

3 Viscous force

4 Archimedes uplift

PHXI08:GRAVITATION

359999 A projectile of mass $m$ is fired from the surface of earth at an angle $α = 60^{\circ}$ from the vertical. The initial speed of the projectile is $u = \sqrt{\frac{G M}{R}}$ , where $M$ and $R$ are mass and radius of the earth. Find the height by the projectile in kilometer. Neglect air resistance and the rotation of earth. (Given $R = 6400 k m$ )
supporting img

1

3200 k m

2

2350 k m

3

7375 k m

4

1250 k m

Explanation:

Let $v$ be the speed of projectile at the highest point. By conservation of angular momentum and conservation of mechanical energy, we have
$m u R \sin α = m v r_{max}$ and $\frac{1}{2} m u^{2} - \frac{G M m}{R} = \frac{1}{2} m v^{2} - \frac{G M m}{r_{max}}$
Solving these two equations, we get $r_{max} = \frac{3 R}{2}$
$∴ h_{max} = r_{max} - R = \frac{R}{2} = 3200 k m$