NEET Practice Questions, NEET Important Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers

Pratcice NEET questions from all chapters from huge question bank for free. All MCQs are based on NCERT syllabus. To practice from a specific subject and chapter, select a subject below. Login to practice in a structured way with explanations, bookmarks, lists, notes etc. Click here to Login or Sign up for free.

Physics Chemistry Botany Zoology

Question bank:

The distance of a geo-stationary satellite from the centre of the earth (Radius R = 6400 km) is nearest to -

5 R
7 R
10 R
18 R

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be (g=10 ${ms}^{- 2}$ and radius of earth is 6400 kms)

0 rad/sec

\frac{1}{800}

rad/sec

\frac{1}{80}

rad/sec

\frac{1}{8}

rad/sec

A simple pendulum has a time period $T_{1}$ when on the earth’s surface and $T_{2}$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of is $T_{2} / T_{1}$ is

\sqrt{2}

4
2

A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will be

mgR
$\frac{1}{2}$ mgR
2 mgR

\frac{1}{4}

mgR

Periodic time of a satellite revolving above Earth’s surface at a height equal to R, radius of Earth, is
[g is acceleration due to gravity at Earth’s surface]

2 π \sqrt{\frac{2 R}{g}}

4 \sqrt{2} π \sqrt{\frac{R}{g}}

2 π \sqrt{\frac{R}{g}}

8 π \sqrt{\frac{R}{g}}

An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy $E_{0}$ . Its potential energy is

- E_{0}

1.5 E_{0}

2 E_{0}

E_{0}

Given radius of Earth ‘R’ and length of a day ‘T’ , the height of a geostationary satellite is [G–Gravitational Constant, M–Mass of Earth]

{(\frac{4 π^{2} GM}{T^{2}})}^{\frac{1}{3}}

{(\frac{4 πGM}{R^{2}})}^{\frac{1}{3}} - R

{(\frac{{GMT}^{2}}{4 π^{2}})}^{\frac{1}{3}} - R

{(\frac{{GMT}^{2}}{4 π^{2}})}^{\frac{1}{3}} + R

A rocket of mass M is launched vertically from the surface of the earth with an initial speed V. Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is

R / (\frac{gR}{2 V^{2}} - 1)

R (\frac{gR}{2 V^{2}} - 1)

R / (\frac{2 gR}{V^{2}} - 1)

R (\frac{2 gR}{V^{2}} - 1)

Two bodies of masses $m_{1}$ and $m_{2}$ are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is

{[2 G \frac{(m_{1} - m_{2})}{r}]}^{1 / 2}

{[\frac{2 G}{r} (m_{1} + m_{2})]}^{1 / 2}

{[\frac{r}{2 G (m_{1} m_{2)}}]}^{1 / 2}

{[\frac{2 G}{r} m_{1} m_{2}]}^{1 / 2}

The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be

83 minutes

83 \times \sqrt{8}

minutes
664 minutes
249 minutes

A projectile is projected with velocity ${kv}_{e}$ in vertically upward direction from the ground into the space. ( $v_{e}$ is escape velocity and k<1). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be : (R = radius of earth)

\frac{R}{k^{2} + 1}

\frac{R}{k^{2} - 1}

\frac{R}{1 - k^{2}}

\frac{R}{k + 1}

A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is $R_{0}$ and mass of the earth M, the angular momentum about the centre of the earth is

m \sqrt{G M R_{0}}

M \sqrt{G m R_{0}}

m \sqrt{\frac{G M}{R_{0}}}

M \sqrt{\frac{G M}{R_{0}}}

If the distance between the earth and the sun becomes half its present value, the number of days in a year would have been

64.5
129

182.5
730

According to Kepler, the period of revolution of a planet (T) and its mean distance from the sun (r) are related by the equation

T^{3} r^{3} = c o n s \tan t

T^{2} r^{‐ 3} = c o n s \tan t

T r^{3} = c o n s \tan t

T^{2} r = c o n s \tan t

A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a satellite orbiting a few hundred kilometres above the earth’s surface ( $R_{earth}$ = 6400 km) will approximately be -

1/2 h
1 h
2 h
4 h

NEET Questions

Advanced Filters

Question bank:

The distance of a geo-stationary satellite from the centre of the earth (Radius R = 6400 km) is nearest to -

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be (g=10 ${ms}^{- 2}$ and radius of earth is 6400 kms)

A simple pendulum has a time period $T_{1}$ when on the earth’s surface and $T_{2}$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of is $T_{2} / T_{1}$ is

A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will be

Periodic time of a satellite revolving above Earth’s surface at a height equal to R, radius of Earth, is
[g is acceleration due to gravity at Earth’s surface]

An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy $E_{0}$ . Its potential energy is

Given radius of Earth ‘R’ and length of a day ‘T’ , the height of a geostationary satellite is [G–Gravitational Constant, M–Mass of Earth]

A rocket of mass M is launched vertically from the surface of the earth with an initial speed V. Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is

Two bodies of masses $m_{1}$ and $m_{2}$ are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is

The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be

A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is $R_{0}$ and mass of the earth M, the angular momentum about the centre of the earth is

If the distance between the earth and the sun becomes half its present value, the number of days in a year would have been

According to Kepler, the period of revolution of a planet (T) and its mean distance from the sun (r) are related by the equation

A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a satellite orbiting a few hundred kilometres above the earth’s surface ( $R_{earth}$ = 6400 km) will approximately be -

NEET Questions

Advanced Filters

Question bank:

The distance of a geo-stationary satellite from the centre of the earth (Radius R = 6400 km) is nearest to -

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be (g=10 ms-2 and radius of earth is 6400 kms)

A simple pendulum has a time period T1 when on the earth’s surface and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of is T2/T1 is

A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will be

Periodic time of a satellite revolving above Earth’s surface at a height equal to R, radius of Earth, is [g is acceleration due to gravity at Earth’s surface]

An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy E0 . Its potential energy is

Given radius of Earth ‘R’ and length of a day ‘T’ , the height of a geostationary satellite is [G–Gravitational Constant, M–Mass of Earth]

A rocket of mass M is launched vertically from the surface of the earth with an initial speed V. Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is

Two bodies of masses m1 and m2 are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is

The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be

A projectile is projected with velocity kve in vertically upward direction from the ground into the space. ( ve is escape velocity and k<1). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be : (R = radius of earth)

A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is R0 and mass of the earth M, the angular momentum about the centre of the earth is

If the distance between the earth and the sun becomes half its present value, the number of days in a year would have been

According to Kepler, the period of revolution of a planet (T) and its mean distance from the sun (r) are related by the equation

A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a satellite orbiting a few hundred kilometres above the earth’s surface (Rearth= 6400 km) will approximately be -

Login Required

Quota Exceeded

Unlock Unlimited Learning with MagicAI Premium

Just ₹249/month for 1000 queries!

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be (g=10 ${ms}^{- 2}$ and radius of earth is 6400 kms)

A simple pendulum has a time period $T_{1}$ when on the earth’s surface and $T_{2}$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of is $T_{2} / T_{1}$ is

Periodic time of a satellite revolving above Earth’s surface at a height equal to R, radius of Earth, is
[g is acceleration due to gravity at Earth’s surface]

An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy $E_{0}$ . Its potential energy is

Two bodies of masses $m_{1}$ and $m_{2}$ are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is

A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is $R_{0}$ and mass of the earth M, the angular momentum about the centre of the earth is

A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a satellite orbiting a few hundred kilometres above the earth’s surface ( $R_{earth}$ = 6400 km) will approximately be -