InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2151. |
The angular velocity of rotation of star (of mass M and radius R ) at which the matter start to escape from its equator will beA. `sqrt((2GR)/(M))`B. `sqrt((2GM)/(R^(3)))`C. `sqrt((2GM)/(R))`D. `sqrt(((2GM)^(2))/(R))` |
|
Answer» Correct Answer - b `v_(e)=sqrt((2GM)/(R))` `R_(omega)=sqrt((2GM)/(R)) " "therefore omega=sqrt((2GM)/(R^(3)))` |
|
| 2152. |
The angular velocity of rotation of star (of mass M and radius R ) at which the matter start to escape from its equator will beA. `sqrt((2GM^(2))/R)`B. `sqrt((2GM)/g)`C. `sqrt((2GM)/R^(2))`D. `sqrt((2GR)/M)` |
|
Answer» Correct Answer - C |
|
| 2153. |
The angular velocity of rotation of a planet of mass M and radius R, at which the matter start to escape from its equator isA. `sqrt((2GM)/(R^(3)))`B. `sqrt((2GM)/(R ))`C. `sqrt((2G^(2)M)/(R ))`D. `sqrt((2GM^(2))/(R ))` |
|
Answer» Correct Answer - A `V_(E )=sqrt((2GM)/(R ))` `therefore omega=(V_(E ))/(R )=(1)/(R )sqrt((2GM)/(R ))=sqrt((2GM)/(R^(3)))` |
|
| 2154. |
The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just nearly provides the centripetal force needed for rotation. The corresponding shortest period of rotation is (If `rho` is the density of the earth)A. `sqrt((3pi)/(Grho))`B. `sqrt((3pirho)/(G))`C. `sqrt((3piG)/(rho))`D. `sqrt((Grho)/(3pi))` |
|
Answer» Correct Answer - a `mRomega^(2)=(GMm)/(R^(2))` `omega^(2)=(GM)/(R^(3))` `omega=sqrt((GM)/(R^(3)))=sqrt((Grho)/(K^(3)))(4pi)/(3)R^(3)=sqrt((4pi)/(3)Grho)` `T=(2pi)/(omega)=(2pi)/(sqrt((4pi G rho)/(3)))=sqrt((3xx4pi^(2))/(4piG rho))=sqrt((3pi)/(Grho)).` |
|
| 2155. |
Speed of a planet in an ellioptical orbit with semimajor axis a about sun of mass M at a distance r from sun isA. `sqrt(GM((2)/(r)-(1)/(a))`B. `sqrt(GM((1)/(r)-(1)/(a))`C. `sqrt(GM((1)/(r)-(2)/(a))`D. `sqrt((GMr)/(2a^(2))` |
| Answer» Correct Answer - A | |
| 2156. |
As the altitude of a body increases, do the weight and mass both vary ? |
| Answer» Weight of the body varies with altitude; mass of an object is constant. | |
| 2157. |
On what factor/factors does the weight of a body depend? |
|
Answer» The weight of a body depends on the mass and on the acceleration due to gravity which varies from place to place. |
|
| 2158. |
On what factors does the weight of a body depend ? |
| Answer» Weight of a body depends on mass of the body and accel eration due ro gravity. | |
| 2159. |
On what factor/factors does the weight of a body depend ? |
|
Answer» The weight of a body is directly proportional to its mass. It also depends on the acceleration due to gravity which varies from place to place. |
|
| 2160. |
An iron ball and a wooden ball of the same radius are released from a height ‘ h ’ in vacuum. The time taken by both of them to reach the ground isA. UnequalB. Exactly equalC. Roughly equalD. Zero |
|
Answer» Correct Answer - B |
|
| 2161. |
The diagram showing the variation of gravitational potential of earth with distance from the centre of earth isA. B. C. D. |
|
Answer» Correct Answer - C `V_("in")=(-Gm)/(2R)[3-((r)/(R))^(2)],V_("surface")=(-GM)/(R),V_("out")=(-GM)/(r)` |
|
| 2162. |
Three uniform spheres each having a mass M and radius a are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any of the spheres due to the other two.A. `(Gm^(2))/(r^(2))`B. `(Gm^(2))/(4r^(2))`C. `sqrt(2)(Gm^(2))/(4r^(2))`D. `sqrt(3)(Gm^(2))/(4r^(2))` |
|
Answer» Correct Answer - D The given system may be regarded as a system of three particles located at the three vertices of an equilateral triangle of side `2r`. Now, `F_(A)=F_(B)` `=(Gm^(2))/((2r)^(2))=(Gm^(2))/(4r^(2))` `F_(A)` and `F_(B)` are inclined to each at an angle of `60^(@)`. if `F` is the resultant of `F_(A)` and `F_(B)`, then `F=sqrt(3)xx(Gm^(2))/(4r^(2))` |
|
| 2163. |
Weight of a body is maximum atA. polesB. equatorC. centre of earthD. at latitude `45^(@)` |
|
Answer» Correct Answer - A At poles, value of g is maximum. So, there is no effect of rotation of earth. |
|
| 2164. |
A particle is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done per unit mass against the gravitational force between them, to take the particle far away from the sphere (you may take `h=6.67 xx 10^(-11) "Nm"^(2) "kg"^(-2)`)A. `13.34 xx 10^(-10) J`B. `3.33 xx 10^(-10) J`C. `6.67 xx 10^(-9)J`D. `6.67 xx 10^(-8) J` |
|
Answer» Correct Answer - D Gravitational potential `V_(i)=-(GM)/(r)` `V_(i)=-(6.67xx10^(-11)xx100)/(0.1)` `V_(i)=-(6.67xx10^(-9))/(0.1)=-6.67xx10^(-8)J` `because V_(f)=0` `:.` Work done per unit mass, `W=Delta V=(V_(f)-V_(i))=6.67xx10^(-8)J`. |
|
| 2165. |
Two concentric shells of masses `M_(1)` and `M_(2)` are concentric as shown. Calculate the gravitational force on m due to `M_(1)` and `M_(2)` at points P,Q and `R`. A. `F_(P) =0`B. `F_(Q)=(GM_(1)m)/(b^(2))`C. `F_(R)=(G(M_(1)+M_(2)))/(c^(2))`D. All are correct |
|
Answer» Correct Answer - D At P, F = 0 At Q, `F=(GM_(1)m)/(b^(2))` At R, `F=(G(M_(1)+M_(2))m)/(c^(2))`. |
|
| 2166. |
Two concentric shells of masses `M_(1)` and `M_(2)` are concentric as shown. Calculate the gravitational force on m due to `M_(1)` at points P,Q and `R`. |
|
Answer» At P,`F = 0` At Q, `F = (GM_(1)m)/(b^(2))` At `R`, `F = (G(M_(1))+(M_(2))m)/(c^(2)`. |
|
| 2167. |
Consider three concentric shells of masses `M_(1),M_(2) and M_(3)` having radii a,b and c respectively are situated as shown in Gravitational field at a point located at Q and P isA. `(G(M_(1)+M_(2)))/(y^(2)),(G(M_(1)+M_(2)))/(y^(2))`B. `(G(M_(1)+M_(2)+M_(3)))/(Y^(2)),(G(M_(1)+M_(2)))/(x^(2))`C. `(G(M_(1)+M_(2)+M_(3)))/(a^(2)),(GM_(1))/(a^(2))`D. `(G(M_(1)+M_(2)+M_(3)))/(C^(2)),(GM_(2))/(b^(2))` |
|
Answer» Correct Answer - B Gravitational field at an external point due to spherical shell of mass M is `((GM)/(r^(2)))` while at an internal point is zero (i) Point Q is external to shell, `M_(1),M_(2) and M_(3)` So, field at Q will be `E_(Q)=(GM_(1))/(y^(2))+(GM_(2))/(y^(2))+(GM_(3))/(y^(2))=(G)/(y^(2))(M_(1)+M_(2)+M_(3))` (ii) Field at P will be `F_(p)=(GM_(1))/(x^(2))+(GM_(2))/(x^(2))+0=(G)/(x^(2))(M_(1)+M_(2))`. |
|
| 2168. |
A satellite is orbiting the earth in a circular orbit of radius r. Its period of revolution varies asA. `sqrtr`B. rC. `r^(3//2)`D. `r^(2)` |
|
Answer» Correct Answer - c `T^(2) prop r^(3)" "therefore T prop r^(3//2)` |
|
| 2169. |
What should be the orbit radius of a communication satellite that it can cover 75% of the surface area of earth during its revolution. |
|
Answer» Correct Answer - `[1.515 R_(e )]` |
|
| 2170. |
The earth satellite has an orbit radius which is 4 times that of a communication satellite. Then the period of revolution of will beA. 4 daysB. 8 daysC. 16 daysD. 32 days |
|
Answer» Correct Answer - b `T^(2)prop R^(3) therefore [(T_(s))/(T_(c))]^(2)=(4)^(3)` `therefore" "T_(s)="8 days"` |
|
| 2171. |
Orbital radius of a satellite S of earth is four times that s communication satellite C. Period of revolution of S is :-A. 4 daysB. 8 daysC. 12 daysD. 2 days |
|
Answer» Correct Answer - B `(T_(s))/(T_(c ))=((r_(s))/(r_(c )))^(3//2)=8` `therefore T_(s)=8xx1=8` days |
|
| 2172. |
A mass `m` is placed inside a hollow sphere of mass M as shown in figure. The gravitaional force on mass `m` is A. `(GMm)/(R^(2))`B. `(GMm)/(r^(2))`C. `(GMm)/((R-r)^(2))`D. zero |
|
Answer» Correct Answer - D Inside a shell, field strength is zero. Therefore, force on a particle is zero. |
|
| 2173. |
If the distance between two objects increases 5 times, the gravitational force becomes _____________ times.A. 5B. 15C. `1/25`D. 25 |
|
Answer» Correct Answer - C `1/25` |
|
| 2174. |
What will happen to the gravitational force between two bodies if they are brought closer by half of their initial separation ?A. increasesB. decreasesC. remains the sameD. becomes zero |
| Answer» Correct Answer - a | |
| 2175. |
If the distance between two bodies becomes half, the gravitational force between them becomes_____________.A. halfB. one forthC. 4 timesD. 2 times |
|
Answer» Correct Answer - C 4 times |
|
| 2176. |
The gravitational force of attraction between two objects is given by_______________.A. `F alpha(m_(1)m_(2))/(d^(2))`B. `F alpha(d^(2))/(m_(1)m_(2))`C. `F alpha(m_(1)m_(2))/(sqrt(d^(2)))`D. `F alpha(m_(1)m_(2))/(d^(3))` |
|
Answer» Correct Answer - A `F alpha(m_(1)m_(2))/(d^(2))` |
|
| 2177. |
Mass `M` is split into two parts `m` and `(M-m)`, which are then separated by a certain distance. What is the ratio of `(m//M)` which maximises the gravitational force between the parts ?A. `1/3`B. `1/2`C. `1/4`D. `1/5` |
|
Answer» Correct Answer - B |
|
| 2178. |
Two astronauts have deserted their spaceship in a region of space far from the gravitational attraction of any other body. Each has a mass of `100 kg` and they are `100 m` apart. They are initially at rest relative to one another. How long will it be before the gravitational attraction brings them `1 cm` closer together?A. 2.52 daysB. 1.41 daysC. 0.70 dayD. 0.41 day |
|
Answer» Correct Answer - B Here `m_(1)=m_(2)=100 kg , r=100 m` Accelerations of first and second astronauts are `a_(1)=(Gm_(1)m_(2))/(r^(2))xx(1)/(m_(1))=(Gm_(2))/(r^(2))` and `a_(2)=(Gm_(1)m(2))/(r^(2))xx(1)/(m^(2))=(Gm_(1))/(r^(2))` Net acceleration of approach `a=a_(1)+a_(2)=(GM_(2))/(r^(2))+(Gm_(1))/(r^(2))=(2Gm_(1))/(r^(2))` `=(2xx(6.67xx10^(-11))xx100)/(100)^(2)=2xx6.67xx10^(-13) ms^(-2)` on solving we get t = 1.41 days |
|
| 2179. |
Two objects of massses m and 4 m are at rest at infinite separtion they move twoards each other under mutula gravitonal attraction then at a separation r which of the following is true ?A. the total energy of the system is not zeroB. the force between them is not zeroC. the centre of mass of the system is at restD. all the above are true |
|
Answer» Correct Answer - D When they are at infinite distance their PE =0 and KE -0 and gravitational force `F=(Gm_(1)m_(2))/(d^(2))` is also zero when the distance of separation between the two masses becomes r their total energy is finite (ii) force of attraction between them is finite and (iii) the control of mass of the system is at rest because initially the masses are at rest and there is no external force acting on the system |
|
| 2180. |
The gravitational acceleration on the surface of earth of radius R and mean density `rho` isA. `(4pi//3)GR^(2)rho`B. `(4pi^(2)//3)GR^(2)rho`C. `(2pi^(2)//3)GR^(2)rho`D. `(4pi//3)GR rho` |
|
Answer» Correct Answer - d `g=(GM)/(R^(2))=(Grho(4pi)/(3)R^(3))/(R^(2))` `g=Grho(4pi)/(3)R=(4pi)/(3)GRrho.` |
|
| 2181. |
The acceleration due to gravity g and density of the earth `rho` are related by which of the following relations? (where G is the gravitational constant and `R_(E)` is the radius of the earth)A. `rho=(4piGR_(E))/(3g)`B. `rho=(3g)/(4piGR_(E))`C. `rho=(3g)/(4pigR_(E))`D. `rho=(4pigR_(E))/(3G)` |
|
Answer» Correct Answer - B Accleration due to gravity of earth is `g=(GM_(E))/(R_(E)^(2))......(i)` As `rho=(M_(E))/(4/3piR_(E)^(3)) implies M_(E)=rho4/3piR_(E)^(2)` Subtituting this value in Eq(i), we get `g=(G(rho4/3piR_(E)^(3)))/(R_(E)^(2))=4/3pirhoGR_(E) or rho=(3g)/(4piGR_(E))` |
|
| 2182. |
The gravitational potential of a body on the surface of the earth is proportional to (R = radius of earth, `rho`=density)A. radius of the earthB. square of density of earthC. the product of radius and densityD. the product `R^(2)rho` |
| Answer» Correct Answer - d | |
| 2183. |
The volume of 500g sealed packet is 350 cm-3, will the packet float or sink in water if the density of water is 1gm cm-3? What will be the mass of the water displaced by this packet? |
|
Answer» Mass of packet = 500g Volume of packet = 350 cm-3 ∴ Density of packet = \(\frac{Mass}{Volume}\) = \(\frac{500}{350}\) = 1.43 gm ∴ ie the density of packet is greater than density of water, so it will sink in water. Mass of water displaced by the packet = Volume of packet × Density of water = 350 × 1 = 350 gm. |
|
| 2184. |
An earth satellite is movewd from one stable circular orbit to another larger and stable circular orbit. The following quantities increase for the satellite a result of this changeA. gravitational potential energyB. angular velocityC. linera orbital velocityD. centripetal acceleration |
|
Answer» Correct Answer - A `U=-(GMm)/(r )` ltbRgt `omega= sqrt((GM)/(r_(3)))` `v= sqrt((GM)/(r )), a_(1) = r omega^(2)` so only option (A) is correct. |
|
| 2185. |
A uniform solid of valume mass density `rho` and radius R is shown in figure. (a) Find the gravitational field at a point P inside the sphere at a distance r from the centre of the sphere. Represent the gravitational field vector `vec(l)` in terms of radius vector `vec(r )` of point P. (b) Now a spherical cavity is made inside the solid sphere in such a way that the point P comes inside the cavity. The centre is at a distance a from the centre of solid sphere and point P is a distance of b from the centre of the cavity. Find the gravitational field `vec(E )` at point P in vector formulationand interpret the result. |
|
Answer» (a) `(4pi G rho r)/(3), -(4)/(3)pi G rho vec(r )` (b) `-(4)/(3)pi G rho vec(a)`, where `vec(a)` is position vector of centre of cavity w.r.t. centre of solid sphere. |
|
| 2186. |
Three particles, each of the mass `m` are situated at the vertices of an equilateral triangle of side `a`. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation `a`. Find the initial velocity that should be given to each particle and also the time period of the circular motion. `(F=(Gm_(1)m_(2))/(r^(2)))` |
|
Answer» Correct Answer - A `F_(A) = F _(AB) +F_(AC)` `= 2[(Gm^(2))/(a^(2))]cos30^(@) = sqrt 3[(Gm^(2))/(a^(2))]` `r = (a)/(sqrt3)`, `(m upsilon^(2))/(r) = F` or `(sqrt3m upsilon ^(2))/(a) = (sqrt3Gm^(2))/(a)` `upsilon = sqrt ((Gm)/(a))`. |
|
| 2187. |
When a planet is orbiting around the sun in an elliptical orbit, `r_(1) and r_(2)` denote its closest and farthest distances from the sun. Then the ratio of the magnitudes of maximum and minimum gravitational forces between them isA. `r_(1):r_(2)`B. `r_(1)^(2):r_(2)^(2)`C. `r_(2):r_(1)`D. `r_(2)^(2):r_(1)^(2)` |
|
Answer» Correct Answer - d `(F_("max"))/(F_("min"))=((r_(2))/(r_(1)))^(2)=(r_(2)^(2))/(r_(1)^(2)).` |
|
| 2188. |
The weight of a person on the Earth is `80 kgwt`. What will be his weight on the Moon ? Mass of the Moon `= 7.34 xx 10^(22) kg`, radius `= 1.75 xx 10^(6) m` and gravitational constant `= 6.67 xx 10^(-11) N m^(2) kg^(-2)`. What will be the mass of the person at the Moon and acceleration due to gravity there ? If this person can jump `2 m` high on the Earth, how much high can he jump at the Moon ? |
|
Answer» Correct Answer - `128 N , 80 kg , 1.6 ms^(-2)` , about 12 m |
|
| 2189. |
STATEMENT -1 : If a body is projected from a satellite so as to escape then its velocity of projection with respect to the satellite depends upon the direction of projection. STATEMENT -2 : Speed of a planet revolving around the sun in an elliptical orbit is maximum at perigee and minimum at apogee. STATEMENT - 3 : Gravitational field is conservative in nature.A. F F FB. T T FC. T F FD. F T T |
| Answer» Correct Answer - A | |
| 2190. |
STATEMENT -1 : If a body is projected from a satellite so as to escape then its velocity of projection with respect to the satellite depends upon the direction of projection. STATEMENT -2 : Speed of a planet revolving around the sun in an elliptical orbit is maximum at perigee and minimum at apogee. STATEMENT - 3 : Gravitational field is conservative in nature.A. T T TB. T F TC. F T FD. F F F |
| Answer» Correct Answer - A | |
| 2191. |
The escape speed of a body from the earth depends uponA. the mass of the bodyB. the direction of projectionC. the latitude and altitude of the location, from where the body is launchedD. none of the above |
|
Answer» Correct Answer - C `v_(e)=sqrt((2GM)/(R ))` It is independent of the mass of the body and the direction of projection. It depends upon the gravitational potential (V) at the point of projection. `V=-(GM)/(r )` and it depends upon the latitude and altitude (height of the point). Thus `v_(e)` depends upon latitude and altitude. |
|
| 2192. |
Choose the correct alternative:The energy required to rocket an orbiting satellite out of Earth's gravitational influence is more/less than the energy require to project a stationary object at the same height (as the satellite) out of Earth's influence. |
|
Answer» The answer is less |
|
| 2193. |
Which of the following statements are correct about a planet rotating around the sun in an elleptical orbitA. its mechanical energy is constant .B. its angular momentum about the sun is constantC. its areal velocity about the sun is constant .D. its time period is proportional to `r^3` |
| Answer» Correct Answer - A::B::C::D | |
| 2194. |
Choose the correct alternative:If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy. |
|
Answer» The answer is Kinetic energy |
|
| 2195. |
If a satellite is moved from one stable circular orbit to a farther stable circular orbit, then the following quantity increasesA. Linear orbital speedB. Gravitational forceC. Centripetal accelerationsD. Gravitational P.E. |
|
Answer» Correct Answer - D Orbital speed `v_(c )=sqrt((GM)/(r ))`, Gravitational force `prop (1)/(r^(2))`, c.p. acceleration `=(v^(2))/(r )` All the three decrease as the orbital radius is increased. But the gravitational P.E. `= -(GMm)/(r )`. It is given that `r_(2)gt r_(1)`. Hence the P.E. will increase, because of the -ve sign. |
|
| 2196. |
In side a hollow isolated spherical shell-A. everywhere gravitational potential is zeroB. everywhere gravitational potential field is zeroC. everywhere gravitational potential is sameD. everywhere gravitational field is same |
|
Answer» Correct Answer - b,c,d Inside shell `V=-(GM)/R,E_(g)=0` So choice `(B)(C)` and `(D)` are correct. |
|
| 2197. |
The energy requierd to remove an earth satellite of mass `m` from its orbit of radius `r` to infinity isA. `(GMm)/r`B. `(-GMm)/(2r)`C. `(GMm)/(2r)`D. `(Mm)/(2r)` |
| Answer» Correct Answer - C | |
| 2198. |
If a satellite is moved from one stable circular orbit to a farther stable circular orbit, then the following quantity increasesA. Gravitational forceB. Gravitational `P.E.`C. linear orbital speedD. Centripetal acceleration |
| Answer» Correct Answer - B | |
| 2199. |
In side a hollow isolated spherical shell-A. The gravitation potential is not zeroB. The gravitational field is not zeroC. The gravitational potential is same everywhereD. The gravitational field is same everywhere |
| Answer» Correct Answer - A::B::D | |
| 2200. |
The distance of Neptune and Saturn from the Sun are respectively `10^(13)` and `10^(12)` meters and their periodic times are respectively `T_n` and `T_s`. If their orbits are circular, then the value of `T_n//T_s` isA. `sqrt(100)`B. `100`C. `10sqrt(10)`D. `1//sqrt(10)` |
|
Answer» Correct Answer - C |
|