Gravitation CBSE Questions & Answers
Gravitation
This is Physics Class 11 Gravitation CBSE Questions & Answers. There are 15 questions in this test with each question having around four answer choices.
Questions & Answers
1
According to Kepler’s Law of orbits,
- AAll planets move in parabolic orbits with the Earth situated at one of the foci of the parabola
- BAll planets move in elliptical orbits with the earth situated at one of the foci of the ellipse
- CAll planets move in parabolic orbits with the Sun situated at one of the foci of the parabola
- DAll planets move in elliptical orbits with the Sun situated at one of the foci of the ellipseCorrect
2
According to Kepler’s Law of areas,
- AOnly the line that joins mars to the earth sweeps equal areas in equal intervals of time
- BThe line that joins any planet to the earth sweeps equal areas in equal intervals of time
- COnly the line that joins earth to the sun sweeps equal areas in equal intervals of time
- DThe line that joins any planet to the sun sweeps equal areas in equal intervals of timeCorrect
3
According to Kepler’s Law of periods,
- AThe cube of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet
- BThe square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.Correct
- CThe square of the time period of revolution of a planet is proportional to the cube of the semi-minor axis of the ellipse traced out by the planet
- DThe cube of the time period of revolution of a planet is proportional to the square of the semi-major axis of the ellipse traced out by the planet
4
The area \(\Delta \)A swept out by a planet of mass m in time interval \(\Delta \)t is related to the angular momentum by (g is acceleration due to gravity and G is the universal gravitational constant)
- A\(\Delta {\bf{A}}/\Delta {\rm{t}}\) = 2L / m
- B\(\Delta {\bf{A}}/\Delta {\rm{t}}\) = gL / (2 m)
- C\(\Delta {\bf{A}}/\Delta {\rm{t}}\) = GL / (2 m)
- D\(\Delta {\bf{A}}/\Delta {\rm{t}}\) = L / (2 m)Correct
5
Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses m1 and m2 separated by a distance r has the magnitude
- A\({\rm{F}} = {\rm{G}}{{{{\rm{m}}_2}} \over {{{\rm{r}}^2}}}\)
- B\({\rm{F}} = {\rm{G}}{{{{\rm{m}}_1}{{\rm{m}}_2}} \over {{{\rm{r}}^2}}}\)Correct
- C\({\rm{F}} = {{{{\rm{m}}_1}{{\rm{m}}_2}} \over {{{\rm{r}}^2}}}\)
- D\({\rm{F}} = {\rm{G}}{{{{\rm{m}}_1}} \over {{{\rm{r}}^2}}}\)
6
The direction of the universal gravitational force between particles of masses \({{\rm{m}}_{\rm{1}}}\) and \({{\rm{m}}_{\rm{2}}}\)
- Adirected towards the center of the earth
- Bdirected towards \({{\rm{m}}_{\rm{1}}}\)
- Cdirected towards \({{\rm{m}}_{\rm{2}}}\)
- Ddirected towards \({{\text{m}}_{\text{2}}}\) on \({{\text{m}}_{\text{1}}}\) and directed towards \({{\rm{m}}_{\rm{1}}}\) on \({{\rm{m}}_{\rm{2}}}\)Correct
7
A ‘central’ force is always directed
- Aalong the position vector of the point of application of the force with respect to the fixed pointCorrect
- Bperpendicular to the position vector of the point of application of the force with respect to the fixed point
- Cat a varying angle to the position vector of the point of application of the force with respect to the fixed point
- Dat a fixed angle to the position vector of the point of application of the force with respect to the fixed point
8
Gravitational force on point mass m1 is
- Athe vector sum of the gravitational forces exerted by \({{\rm{m}}_{\rm{2}}}\), \({{\rm{m}}_{\rm{3}}}\) and \({{\rm{m}}_{\rm{4}}}\)Correct
- Bthe difference of gravitational forces exerted by \({{\rm{m}}_{\rm{4}}}\), \({{\rm{m}}_{\rm{3}}}\) and \({{\rm{m}}_{\rm{2}}}\)
- Cthe scalar sum of the gravitational forces exerted by \({{\rm{m}}_{\rm{2}}}\), \({{\rm{m}}_{\rm{3}}}\) and \({{\rm{m}}_{\rm{4}}}\)
- Dthe difference of gravitational forces exerted by \({{\rm{m}}_{\rm{2}}}\), \({{\rm{m}}_{\rm{3}}}\) and \({{\rm{m}}_{\rm{4}}}\)
9
To find the resultant gravitational force acting on the particle m due to a number of masses we need to use
- Athe principle of no action
- Bthe principle of least action
- Cthe principle of superpositionCorrect
- Dthe principle of maximal action
10
The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is
- Aconcentrated at the centre of the shell.Correct
- Bequally concentrated at three points on a triangle of the shell.
- Cequally concentrated at opposite ends of the diameter of the sphere
- Dequally concentrated at four points on a square of the shell.
11
The force of attraction due to a hollow spherical shell of uniform density, on a point mass situated inside it is
- Anone of these
- BzeroCorrect
- Cnegative
- Dpositive
12
In Cavendish’s experiment,
- Atorque on bar AB having two small lead spheres due to gravitational forces is balanced by the restoring torque of the wireCorrect
- Bnet force on bar AB having two small lead spheres due to gravitational forces is negative
- Cnet torque on bar AB having two small lead spheres due to gravitational forces is positive
- Dnet force on bar AB having two small lead spheres due to gravitational forces is positive
13
The value of the gravitational constant G is
- AG = 6.67\( \times \)\({\rm{1}}{0^{ - {\rm{11}}}}\) N m\(^{\rm{2}}\)/kg
- BG = 6.67\( \times \)\({\rm{1}}{0^{ - {\rm{11}}}}\)N m\(^{\rm{2}}\)/kg\(^{\rm{2}}\)Correct
- CG = 6.67\( \times \)\({\rm{1}}{0^{ - {\rm{11}}}}\) m\(^{\rm{2}}\)/kg\(^{\rm{2}}\)
- DG = 6.67\( \times \)\({\rm{1}}{0^{ - {\rm{11}}}}\) N m/kg\(^{\rm{2}}\)
14
The acceleration due to gravity at a height h in terms of mass of earth \({{\text{M}}_{\text{E}}}\) and radius of the earth \({{\rm{R}}_{\rm{E}}}\) and gravitational constant G is
- A\(g\left( h \right) = {{G{M_E}} \over {{{\left( {{R_E} + h} \right)}^2}}}\)Correct
- B\(g\left( h \right) = {{{M_E}} \over {{{\left( {{R_E} + h} \right)}^2}}}\)
- C\(g\left( h \right) = {{G{M_E}} \over {{{\left( {{R_E} + 2h} \right)}^2}}}\)
- D\(g\left( h \right) = {{Gm{M_E}} \over {{{\left( {{R_E} + h} \right)}^2}}}\)
15
The acceleration due to gravity at a depth d in terms of g the acceleration due to gravity at radius of the earth \({{\rm{R}}_{\rm{E}}}\) and \({{\rm{R}}_{\rm{E}}}\) is
- A\(g\left( d \right) = g\left( {1 + d/{R_E}} \right)\)
- B\(g\left( d \right) = g\left( {1 - d/{R_E}} \right)\)Correct
- C\(g\left( d \right) = g\left( {1 + d/{R_E}} \right)\)
- D\(g\left( d \right) = g\left( {1 - 2d/{R_E}} \right)\)