2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 49 InterviewSolution

2401.	The imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero is equal to [Take g `= 10 m//s^(2)` for the acceleration due to gravity if the earth were at rest and radius of earth equal to 6400 km.]A. `1.25xx10^(-3)` rad/sB. `2.50xx10^(-3)` rad/sC. `3.75xx10^(-3)` rad/sD. `5.0xx10^(-3)` rad/s
Answer» Correct Answer - A

Discussion

2402.	As we go from pole to the equator, the effective value of acceleration due to gravity decreases due toA. rotation of the earth onlyB. shape of the eath onlyC. both rotation and shape of the earthD. neither rotation nor shape of the eearth
Answer» Correct Answer - c

Discussion

2403.	An object thrown vertically upwards reaches a height of 500m. What was its initial velcoity? How long will the object take to come back to the Earth? Assume `g=10m//s^(2)`
Answer» Given: Displacement `(s)=500m` Gravitational acceleration `(g)=-10m//s^(2)` Final velocity `(v)=0m//s` To find: Initial velocity `(u)=?` `t_("total")=?` Formulae: `v^(2)=u^(2)+2gs, t_("total")=2t` `v=u+"gt"` `:.v,g` and s is given, to find u we use 3rd equation Solution: `v^(2)=u^(2)+2gs` `0=u^(2)-10000` `10000=u^(2)` `u=100m//s` Since, we have v,u and g to find t we use 1st equation `v=u+"gt"` `0=100+(-10)xxt` `10t=100` `t=10s` Time of ascent `=` Time of descent `:.t_("total")=2xx10` `:.t_("total")=20s` Initial velocity was 100 m/s and time taken to come back to Earth is 20s.

Discussion

2404.	Shown are several curves (Fig (a), (b), (c) ,(d), (e) and (f), Explain with reason which ones amongst them can be possible trajectories traced by a projectile (neglect air friction)(a) (b) (c) (d) (e) (f)
Answer» The trajectory of particle under gravitational force of the earth will be a conic section (for motion outside the earth) with the centre of the earth as a focus. Only (c) meets this requirement because in which centre of earth is the focus of trajectory.

Discussion

2405.	Shown are several cuves (fig. (a), (b), (c), (d), (e), (f)]. Explain with reason, which ones amongst them can be possible trajectories traced by a projectile (neglect air friction).
Answer» The trajectory of a particle under gravitational force of the earth will be a conic section (for motion outside the earth) with the centre of the earth as a focus. Only (c) meets this requirement Note. The trajectory of the particle depends upon the velocity of projection. Depending upon the magnitude and direction of velocity it may be parabolic or elliptical

Discussion

2406.	If the Earth has no tilt, what happens to the seasons of the Earth?
Answer» If the earth weren’t tilted on its axis, there would be no seasons. And humanity would suffer.

Discussion

2407.	A gravity meter can detect change in acceleration due to gravity (g) of the order of `10^(-9) %.` Calculate the smallest change in altitude near the surface of the earth that results in a detectable change in g. Radius of the earth `R = 6.4 x× 10^(6)m.`
Answer» Correct Answer - 32`mum`

Discussion

2408.	Planet Mars has two moons, phobos and delmos. Phobos has a period of 7 hours 39 minutes and orbital radius of 9.4 × 103 km. Calculate the mass of Mars. (G = 6.67 × 10-11 Nm2-kg-2)
Answer» We have, T² = \(\frac{4\pi^2}{GMm}R^3\) or M_m = \(\frac{4\pi^2}{G}.\frac{R^3}{T^2}\) \(\therefore\) M_m= \(\frac{4\times(3.15)^2\times(9.4\times10^3)^3}{6.67\times10^{-11}\times(459\times60)^2}\) = 6.48 x 10²³ kg

2408.

Planet Mars has two moons, phobos and delmos. Phobos has a period of 7 hours 39 minutes and orbital radius of 9.4 × 103 km. Calculate the mass of Mars. (G = 6.67 × 10-11 Nm2-kg-2)

Answer»

We have,

T² = \(\frac{4\pi^2}{GMm}R^3\)

or M_m = \(\frac{4\pi^2}{G}.\frac{R^3}{T^2}\)

\(\therefore\) M_m= \(\frac{4\times(3.15)^2\times(9.4\times10^3)^3}{6.67\times10^{-11}\times(459\times60)^2}\)

= 6.48 x 10²³ kg

Discussion

2409.	Calculate the minimum speed required by a rocket to pull out of the gravitational force of Mars. Given that the earth has a mass `9` times and radius twice of the planet Mars. Escape speed on the surface of earth is `11.2 km s^(-1)`.
Answer» Correct Answer - `5.28 kms^(-1)`

Discussion

2410.	The time period `T` of the moon of planet mars (mass `M_(m)`) is related to its orbital radius `R` as (`G`=gravitational constant)A. `T^(2)=(4pi^(2)R^(3))/(GM_(m))`B. `T^(2)=(4pi^(2)GR^(3))/(M_(m))`C. `T^(2)=(4piR^(3)G)/(M_(m))`D. `T^(2)=4piM_(m)GR^(3)`
Answer» Correct Answer - A (a) Time period, `T=(2pi R)/(sqrt((GM_(m))/( R)))=(2piR^(3//2))/(sqrt(GM_(m))` where the symbols have their meanings as given in the question. Squaring both sides, we get, `T^(2)=(4pi^(2)R^(3))/(GM_(m))`

Discussion

2411.	If Earth has mass nine times and radius twice that of the planet Mars, calculate the velocity required by a rocket to pull out of the gravitational force of Mars. Take escape speed on surface of Earth to be `11.2km//s`
Answer» Here,`M_(e)=9M_(m)`, and `R_(e)=2R_(m)` `v_(e)`(escape speed on surface of Earth)=`11.2 km//s` Let `V_(m)` be the speed required to pull out of the gravitational force of mars. We know that `v_(e)=sqrt((2GM_(e))/(R_(e))` and `v_(m)=sqrt((2GM_(m))/(R_(m)))` Dividing, we get `(v_(m))/(v_(e))=sqrt((2GM_(m))/(R_(m))xx(R_(e))/(2GM_(e)))` `=sqrt((M_(m))/(M_(e))xx(R_(e))/(R_(m)))=sqrt(1/9xx2)=(sqrt(2))/3` `rArr v_(m)=(sqrt(2))/3(11.2km//s)=5.3 km//s`

Discussion

2412.	The mass of planet is 1/64 of the earth but the gravitational pull is 1/9 of the earth. It is due to this reasonA. radius of that planet is 64/9 of the earthB. radius of the earth is 8/3 of that planetC. radius of the earth is 3/8 of that planetD. none
Answer» Correct Answer - b `(g_(2))/(g_(1))=(M_(2))/(M_(1))xx(R_(1)^(2))/(R_(2)^(2))` `therefore" "(R_(1))/(R_(2))=sqrt((g_(1))/(g_(1))xx(M_(1))/(M_(2)))=sqrt((1)/(9)xx64)=(8)/(3)R_(2).`

Discussion

2413.	(a) Write down an equation that defines density . (b) 5 kg of meterial A occupy `20 cm ^(2)` whereas 20 kg of material B occupy `90 cm^(2)`.Which has the greater density : A or B
Answer» Correct Answer - (b) A

Discussion

2414.	Calculate the density of an object of volume `3 m ^(3)` and mass 9 kg .State whether this object will float or sink in water or more in water .Give reason for your answer .
Answer» Correct Answer - `3 kg //m^(3) ; Float in water ;beacause the density of object is less than the density of water

Discussion

2415.	Why does an object float or sink when placed on the surface of water?
Answer» When an object is put in water, then two forces act on it: (i) weight of the object acting downwards (which tends to pull down the object), and (ii) buoyant force (or upthrust) acting upwards (which tends to push up the object) Now, whether an object will float or sink in water will depend on the relative magnitudes of these two forces (weight and buoyant force) acting on the object in opposite directions. If the buoyant force (or upthrust) exerted by water is equal to or greater than the weight of the object, will float in water. If the buoyant force (or upthrust) exerted by water is less than the weight of the object,will sink.

Discussion

2416.	Define the Buoyancy? State two factors on which buoyant force depends.
Answer» Buoyancy The upward force experienced by an object when it is immersed into a fluid is called force of buoyancy. It acts in upward direction and it depends on the density of the fluid. Factors affecting buoyant force: (i) Volume of object immersed in the liquid, (ii) Density of the liquid.

2416.

Define the Buoyancy? State two factors on which buoyant force depends.

Answer»

Buoyancy

The upward force experienced by an object when it is immersed into a fluid is called force of buoyancy. It acts in upward direction and it depends on the density of the fluid.

Factors affecting buoyant force:

(i) Volume of object immersed in the liquid,

(ii) Density of the liquid.

Discussion

2417.	(a) Define buoyant force. Name two factors on which buoyant force depends.(b) What is the cause of buoyant force ?(c) When a boat is partially immersed in water, it displaces 600 kg of water. How much is the buoyant force acting on the boat in newtons ? (g = 10 m s-2 )
Answer» (a) The upward force acting on an object immersed in a liquid is called buoyant force. Factors affecting buoyant force: (i) Volume of object immersed in the liquid, (ii) Density of the liquid. (b) The cause of buoyant force is the greater upward pressure exerted by water underneath the object.. (c) Mass of water displaced = 600kg Weight of water displaced, W =m x g =600 x 10 =6000N Since, the weight of water displaced by the boat is 6000N, therefore the buoyant force acting on the boat will also be 6000N.

2417.

(a) Define buoyant force. Name two factors on which buoyant force depends.(b) What is the cause of buoyant force ?(c) When a boat is partially immersed in water, it displaces 600 kg of water. How much is the buoyant force acting on the boat in newtons ? (g = 10 m s-2 )

Answer»

(a) The upward force acting on an object immersed in a liquid is called buoyant force. Factors affecting buoyant force:

(i) Volume of object immersed in the liquid,

(ii) Density of the liquid.

(b) The cause of buoyant force is the greater upward pressure exerted by water underneath the object..

(c) Mass of water displaced = 600kg

Weight of water displaced, W =m x g

=600 x 10 =6000N

Since, the weight of water displaced by the boat is 6000N, therefore the buoyant force acting on the boat will also be 6000N.

Discussion

2418.	An object weighs 500 grams in air. This object is then fully immersed in water. State whether it will weigh less in water or more in water. Give reason for your answer.
Answer» The object will weigh less in water because an upward force (buoyant force) equal to the weight of water displaced acts on the object when immersed in water which reduces its weight apparently.

Discussion

2419.	An object weighs 10N in air. When immersed fully in liquid, it weighs only 8N. The weight of the liquid displaced by the object will be:A. 2 NB. 8 NC. 10 ND. 12 N
Answer» Correct Answer - A

Discussion

2420.	In the given figure `L=1` metre if total gravitational force on 4 kg mass is `verF_(1)` and on 2 kg mass is `vecF_(2)` then find out the ratio of `\|(vecF_(2))/(vecF_(2))\|`
Answer» Correct Answer - `sqrt(2)` `F_(1)=\|F_(42)+vecF_(41)+vecF_(41)\|` `=(G8)/(4L^(2))+(G4)/(2L^(2))(2cos45)=((2+2sqrt(2))G)/(L^(2))` `F_(2)=\|vecF_(24)+vecF_(21)+vecF_(21)\|=(G8)/(4L^(2))+(G2)/(2L^(2))(2cos45)` `=(G)/(L^(2))(2+sqrt(2))implies(F_(1))/(F_(2))=sqrt(2)`

Discussion

2421.	A particle moves in a circular path with decreasing speed . Choose the correct statement.A. its angular momentum remains constantB. its resultant acceleration is towards the centreC. particle moves in a spiral path with decreasing radiusD. the direction of angular momentum remains constant
Answer» Correct Answer - D The direction of angular momentum is along the axis. It remains constant.

Discussion

2422.	Which of the diagrams shown in figure. Most closely shows the variation inkinetic energy of the earth as it moves once around the sun in its elliptical orbit ?A. B. C. D.
Answer» Correct Answer - D When the earth moves one around the sun in its elliptical orbit, its K.E. does not remain constant. It is maximum when it is cloest to the sun and minimum when it is farthest from the sun. But it is never zero during its motion. Hence (d) is the correct graph.

Discussion

2423.	A shell of mass M and radius R has point mass m placed at a distance r from its centre. The gravitational potential energy U(r) vs r will beA. B. C. D.
Answer» Correct Answer - D The gravitational potential energy U (r ) has a constant value inside the shell for `r le R`. and it is given by `U (r ) = -(GMm)/(r )` For `r gr R, U (r )=-(GMm)/(r )` It decreases as r increases.

Discussion

2424.	Infinite number of bodies, each of mass `2kg`, are situated on `x`-axis at distance `1m,2m,4m,8m.........` respectively, from the origin. The resulting gravitational potential the to this system at the origing will beA. `-G`B. `-8/3G`C. `-4/3G`D. `-4G`
Answer» Correct Answer - D `V=-2G[1/1+1/2+1/4+1/8........]` `=-2G[1+1/2+1/(2^(2))+1/(2^(3)).......]` `=-2G1/((1-1/2))=-4G`

Discussion

2425.	Infinite number of bodies, each of mass 2kg are situated on X-axis at distance `1m,2m,4m,8m,` respectively from the origin, What is the resulting gravitational potential due to this system at the origin ?
Answer» The resulting gravitational potential `V=-2G((1)/(1)+(1)/(2)+(1)/(4)+(1)/(8)....)=-2G(1+(1)/(2)+(1)/(2^(2))+(1)/(2^(3)).....)` `rArr V=(-2G)/((1-(1)/(2)))=(-2G)/((1)/(2))=-4G`.

Discussion

2426.	A body weighs 72 N on the surface of the earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surfaceA. 16 NB. 28 NC. 32 ND. 72 N
Answer» Correct Answer - C ( c) Weight of the body on the surface of the earth = mg= 72 N Acceleration due to gravity at height h is `g_(h)=(gR_(E)^(2))/((R_(E)+h)^(2))=(gR_(E)^(2))/((R_(E)+(R_(E))/(2))^(2))=(4)/(9)g( :.h=(R_(E))/(2))` Gravitational force on body at height h is `F=mg_(h)=mxx(4)/(9) g= (4)/(9)xx mg=(4)/(9)xx72=32 N`

Discussion

2427.	An object of mass 2 kg is moved from infinity to a point P. Initially that object was at rest but on reaching P its speed is 2 m/s. The work done in moving that object is `-16J`. Then potential at P isA. 8 KJ/kgB. `-8J//kg`C. `4J//kg`D. `-4J//kg`
Answer» Correct Answer - b `"PE = work done"` `"E = -16J` `PE=G.P.xx"mass"` `16=G.P.xx2` `G.P.=(16)/(2)="8J/kg."`

Discussion

2428.	Two point masses 100 kg and 25 kg are situated at two points 2 m apart, then the gravitational potential midway between them will beA. `-228xx10^(-11)J//kg`B. `-25xx10^(-11)J//kg`C. `-8xx10^(-10J//kg`D. `-825xx10^(-11)J//kg`
Answer» Correct Answer - d `V=V_(1)+V_(2)` `=-(Gm_(1))/(R)-(Gm_(2))/(R)=-(G)/(R)(m_(1)+m_(2))`

Discussion

2429.	There are two bodies of masses 100 kg and 10000 kg separated by a distance 1 m . At what distance from the smaller body, the intensity of gravitational field will be zeroA. `(1)/(9)m`B. `(1)/(10)m`C. `(1)/(11)m`D. `(10)/(11)m`
Answer» Correct Answer - c `(Gxx100)/(x^(2))=(Gxx10000)/((1-x)^(2))` `therefore" "(10)/(x)=(100)/(1-x)" "therefore x=(1)/(11)m`

Discussion

2430.	Assertion : The centre of semicircular ring of mass `m` and radius R is the origin O. The potential at origin is `-(Gm)/(R)` Reason : The gravitational field strength is towards Y-axis.A. If both Assertin and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are correct but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is falseD. If Assertion is false but Reason is true
Answer» Correct Answer - B

Discussion

2431.	The escape velocity for a planet is `v_e`. A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will beA. `sqrt(1.5) v_(e)`B. `v_(e)/sqrt(2)`C. `v_(e)`D. zero
Answer» Correct Answer - A From mechanical energy conservation, `0+0=1/2 mv^(2) - (3GMm)/(2R) implies v= sqrt((3GM)/R)=sqrt(1.5) v_(e)`.

Discussion

2432.	The escape velocity for a planet is `v_e`. A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will beA. `sqrt(1.5)v_(e)`B. `(v_(e))/(sqrt(2))`C. `v_(e)`D. zero
Answer» Correct Answer - A From mechanical energy conservation `0+0=(1)/(2)mv^(2)-(3GMm)/(2R)impliesv=sqrt((3GM)/(R))=sqrt(1.5)v_(e)`

Discussion

2433.	There are two bodies of masses 100 kg and 10000 kg separated by a distance 1 m . At what distance from the smaller body, the intensity of gravitational field will be zeroA. `1/9 m`B. `1/10 m`C. `1/11 m`D. `10/11 m`
Answer» Correct Answer - C

Discussion

2434.	Assertion : The principle of superposition is not valid for gravitational force. Reason : Gravitational force is a conservative force.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are truebut reason is not the correct explanation of assertionC. If assertion is true but reason is falseD. If both assertion and reason are false.
Answer» (d) Gravitational forces are conservative forces. The total gravitational force on a particle due to other surrounding particles is the vector sum of the individual forces.

Discussion

2435.	The change in the value of `g` at a height `h` above the surface of the earth is the same as at a depth d below the surface of earth. When both `d` and `h` are much smaller than the radius of earth, then which one of the following is correct?A. `x=h`B. `x=2h`C. `x=h/2`D. `x=h^(2)`
Answer» Correct Answer - B

Discussion

2436.	The time period of a simple pendulum oscillating in a freely falling lift isA. ZeroB. 2 secC. 3 secD. Infinite
Answer» Correct Answer - D

Discussion

2437.	A projectile is projected tangentially from the surface of a planet of radius R. If it is at a height of 3R at the farthest point of its trajectory, then the velocity of projection `V_(0)` is given by ( acceleration due to gravity on surface=g)A. `v_(0)=sqrt(1.5gR)`B. `v_(0)=sqrt(0.5gR)`C. `v_(0)=sqrt(1.6gR)`D. `v_(0)=sqrt(2gR//3)`
Answer» Correct Answer - C

Discussion

2438.	If the mass of a planet is `10%` less than that of the earth and the radius is `20%` greater than that of the earth, the acceleration due to gravity on the planet will beA. `5/8` times that on the surface of the earthB. `3/4` times that on the surface of the earthC. `1/2` times that on the surface of the earthD. `9/10` times that on the surface of the earth
Answer» Correct Answer - A `g_(p)=(G[M-10/100M])/([R+20/100R]^(2))=(Gxx9M)/10xx25/(36R^(2))=5/8g`

Discussion

2439.	Statement-1:Weight of an object on the Earth is more in mid-night than what it is at the noon. Statement-2: At noon gravitational pull on the object by the sun and that by the earth are oppositely directed, and in the mid night they are in the same direction.A. Statement-1 is true, statement-2 is true, Statement-2 is a correct explanation for statement-1.B. Statement-1 is true, Statement-2 is true, statement-2 is Not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false Statement-2 is True.
Answer» Correct Answer - d

Discussion

2440.	Statement-1: The plane of the orbit of an artificial satellite must contain the centre of the Earth. Statement-2: For an artificial satellite, the necessary centripetal force is provided by gravity.A. Statement-1 is true, statement-2 is true, Statement-2 is a correct explanation for statement-1.B. Statement-1 is true, Statement-2 is true, statement-2 is Not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false Statement-2 is True.
Answer» Correct Answer - b

Discussion

2441.	Statement-1:A spherical symmetric shell produces no gravitational field anywhere. Statement-2: The field due to various mass elements cancels out, everywhere inside the shell.A. Statement-1 is true, statement-2 is true, Statement-2 is a correct explanation for statement-1.B. Statement-1 is true, Statement-2 is true, statement-2 is Not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false Statement-2 is True.
Answer» Correct Answer - d Although no gravitational field is produced inside a symmetric shell, it produces a field at points outside of shell.

Discussion

2442.	Two satellites `A` and `B` of the same mass are orbiting the earth at altitudes `R` and `3R` respectively, where `R` is the radius of the earth. Taking their orbit to be circular abtain the ratios of their kinetic and potential energies.
Answer» Correct Answer - A::B `r_(1) = R + h_(1) = R + r = 2R` `r_(2) = R + h_(2) = R + 3R = 4R` `KE` and `PE prop (1)/(R)` `:. (K_(1))/(K_(2)) = (U_(1))/(U_(2)) = (r_(2))/(r_(1)) = (2)/(1)`

Discussion

2443.	The kinetic energy needed to project a body of mass `m` from the earth surface (radius R) to infinity isA. `(mgR)/(2)`B. `2mgR`C. `mgR`D. `(mgR)/(2)`
Answer» Correct Answer - C The kinetic energy needed to project a body of mass m from the earth surface to infinity is given by `KE=(1)/(2)mv_(e)^(2)=(1)/(2)m 2Rg` `(because "Escape velociy", v_(e)=sqrt(2Rg)rArrv_(e)^(2)=2Rg)` `rArr KE = mRg`

Discussion

2444.	The kinetic energy needed to project a body of mass m from the eath surface (radus R) to infinity isA. mgR/2B. 2mgRC. mgRD. mgR/4
Answer» Correct Answer - C (c ) `K.E = (1)/(2) mv_e^2 where v_c = escape velocity = sqrt(2gR)` `:. K.E. = (1)/(2) mxx 2gR =mgR`

Discussion

2445.	The kinetic energy needed to project a body of mass m from the surface of the earth to infinity isA. `12xx10^(7)J//kg`B. `12.5xx10^(7)J//kg`C. `6.25xx10^(7)J//kg`D. `25xx10^(7)J//kg`
Answer» Correct Answer - c `"K E = BE"=("GMm")/(R)` `(KE)/(m)=(gR^(2))/(R)=gR` `=10xx6.4xx10^(6)=6.4xx10^(7)`

Discussion

2446.	The kinetic energy needed to project a body of mass m from the eath surface (radus R) to infinity is
Answer» Correct Answer - `[(GMm)/(R )]`

Discussion

2447.	The kinetic energy needed to project a body of mass m from the eath surface (radus R) to infinity isA. mgR /2B. 2 mgRC. mgRD. mgR /4
Answer» Correct Answer - C

Discussion

2448.	Calculate te radus of an isolated sphere of density `3.0 g cm^(-3)` from the surface of which the escape velocity be `40 ms^(-1)`
Answer» `v_(e)=sqrt((2GM)/R)=sqrt((2G(4/3pR^(3)rho))/R)` `R=v_(e)sqrt(3/(8piGrho))=30894m=31km`

Discussion

2449.	A satellite of mass `m` is orbiting the earth in a circular orbit of radius `r`. It starts losing energy due to small air resistance at the rate of `C J//s`. Then the time teken for the satellite to reach the earth is...........
Answer» Correct Answer - `[1/R-1/r]` `E=-(GMm)/(2r)` `(dE)/(dt)=(GMm)/2 1/(r^(2)) (dr)/(dt)` `int_(0)^(t)dt=(GMm)/(2C)int_(r)^(R)(dr)/(r^(2))` `t=(GMm)/(2C)[1/R-1/r]=(GMm)/(2C)[1/R=1/r]`

Discussion

2450.	A satellite of mass m is orbiting around the earth (mass M, radius R) in a circular orbital of radius 4R. It starts losing energy slowly at a constant `-(dE)/(dt)=eta`due to friction. Find the time (t) in which the satellite will spiral down to the surface of the earth.
Answer» Correct Answer - `t=(3GMm)/(8etaR)`

Discussion

Explore topic-wise InterviewSolutions in .

As we go from pole to the equator, the effective value of acceleration due to gravity decreases due toA. rotation of the earth onlyB. shape of the eath onlyC. both rotation and shape of the earthD. neither rotation nor shape of the eearth

An object thrown vertically upwards reaches a height of 500m. What was its initial velcoity? How long will the object take to come back to the Earth? Assume `g=10m//s^(2)`

Shown are several curves (Fig (a), (b), (c) ,(d), (e) and (f), Explain with reason which ones amongst them can be possible trajectories traced by a projectile (neglect air friction)(a) (b) (c) (d) (e) (f)

Shown are several cuves (fig. (a), (b), (c), (d), (e), (f)]. Explain with reason, which ones amongst them can be possible trajectories traced by a projectile (neglect air friction).

If the Earth has no tilt, what happens to the seasons of the Earth?

A gravity meter can detect change in acceleration due to gravity (g) of the order of `10^(-9) %.` Calculate the smallest change in altitude near the surface of the earth that results in a detectable change in g. Radius of the earth `R = 6.4 x× 10^(6)m.`

Planet Mars has two moons, phobos and delmos. Phobos has a period of 7 hours 39 minutes and orbital radius of 9.4 × 103 km. Calculate the mass of Mars. (G = 6.67 × 10-11 Nm2-kg-2)

Calculate the minimum speed required by a rocket to pull out of the gravitational force of Mars. Given that the earth has a mass `9` times and radius twice of the planet Mars. Escape speed on the surface of earth is `11.2 km s^(-1)`.

The time period `T` of the moon of planet mars (mass `M_(m)`) is related to its orbital radius `R` as (`G`=gravitational constant)A. `T^(2)=(4pi^(2)R^(3))/(GM_(m))`B. `T^(2)=(4pi^(2)GR^(3))/(M_(m))`C. `T^(2)=(4piR^(3)G)/(M_(m))`D. `T^(2)=4piM_(m)GR^(3)`

If Earth has mass nine times and radius twice that of the planet Mars, calculate the velocity required by a rocket to pull out of the gravitational force of Mars. Take escape speed on surface of Earth to be `11.2km//s`

The mass of planet is 1/64 of the earth but the gravitational pull is 1/9 of the earth. It is due to this reasonA. radius of that planet is 64/9 of the earthB. radius of the earth is 8/3 of that planetC. radius of the earth is 3/8 of that planetD. none

(a) Write down an equation that defines density . (b) 5 kg of meterial A occupy `20 cm ^(2)` whereas 20 kg of material B occupy `90 cm^(2)`.Which has the greater density : A or B

Calculate the density of an object of volume `3 m ^(3)` and mass 9 kg .State whether this object will float or sink in water or more in water .Give reason for your answer .

Why does an object float or sink when placed on the surface of water?

Define the Buoyancy? State two factors on which buoyant force depends.

(a) Define buoyant force. Name two factors on which buoyant force depends.(b) What is the cause of buoyant force ?(c) When a boat is partially immersed in water, it displaces 600 kg of water. How much is the buoyant force acting on the boat in newtons ? (g = 10 m s-2 )

An object weighs 500 grams in air. This object is then fully immersed in water. State whether it will weigh less in water or more in water. Give reason for your answer.

An object weighs 10N in air. When immersed fully in liquid, it weighs only 8N. The weight of the liquid displaced by the object will be:A. 2 NB. 8 NC. 10 ND. 12 N

In the given figure `L=1` metre if total gravitational force on 4 kg mass is `verF_(1)` and on 2 kg mass is `vecF_(2)` then find out the ratio of `|(vecF_(2))/(vecF_(2))|`

A particle moves in a circular path with decreasing speed . Choose the correct statement.A. its angular momentum remains constantB. its resultant acceleration is towards the centreC. particle moves in a spiral path with decreasing radiusD. the direction of angular momentum remains constant

Which of the diagrams shown in figure. Most closely shows the variation inkinetic energy of the earth as it moves once around the sun in its elliptical orbit ?A. B. C. D.

A shell of mass M and radius R has point mass m placed at a distance r from its centre. The gravitational potential energy U(r) vs r will beA. B. C. D.

Infinite number of bodies, each of mass `2kg`, are situated on `x`-axis at distance `1m,2m,4m,8m.........` respectively, from the origin. The resulting gravitational potential the to this system at the origing will beA. `-G`B. `-8/3G`C. `-4/3G`D. `-4G`

Infinite number of bodies, each of mass 2kg are situated on X-axis at distance `1m,2m,4m,8m,` respectively from the origin, What is the resulting gravitational potential due to this system at the origin ?

A body weighs 72 N on the surface of the earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surfaceA. 16 NB. 28 NC. 32 ND. 72 N

An object of mass 2 kg is moved from infinity to a point P. Initially that object was at rest but on reaching P its speed is 2 m/s. The work done in moving that object is `-16J`. Then potential at P isA. 8 KJ/kgB. `-8J//kg`C. `4J//kg`D. `-4J//kg`

Two point masses 100 kg and 25 kg are situated at two points 2 m apart, then the gravitational potential midway between them will beA. `-228xx10^(-11)J//kg`B. `-25xx10^(-11)J//kg`C. `-8xx10^(-10J//kg`D. `-825xx10^(-11)J//kg`

There are two bodies of masses 100 kg and 10000 kg separated by a distance 1 m . At what distance from the smaller body, the intensity of gravitational field will be zeroA. `(1)/(9)m`B. `(1)/(10)m`C. `(1)/(11)m`D. `(10)/(11)m`

There are two bodies of masses 100 kg and 10000 kg separated by a distance 1 m . At what distance from the smaller body, the intensity of gravitational field will be zeroA. `1/9 m`B. `1/10 m`C. `1/11 m`D. `10/11 m`

The change in the value of `g` at a height `h` above the surface of the earth is the same as at a depth d below the surface of earth. When both `d` and `h` are much smaller than the radius of earth, then which one of the following is correct?A. `x=h`B. `x=2h`C. `x=h/2`D. `x=h^(2)`

The time period of a simple pendulum oscillating in a freely falling lift isA. ZeroB. 2 secC. 3 secD. Infinite

Two satellites `A` and `B` of the same mass are orbiting the earth at altitudes `R` and `3R` respectively, where `R` is the radius of the earth. Taking their orbit to be circular abtain the ratios of their kinetic and potential energies.

The kinetic energy needed to project a body of mass `m` from the earth surface (radius R) to infinity isA. `(mgR)/(2)`B. `2mgR`C. `mgR`D. `(mgR)/(2)`

The kinetic energy needed to project a body of mass m from the eath surface (radus R) to infinity isA. mgR/2B. 2mgRC. mgRD. mgR/4

The kinetic energy needed to project a body of mass m from the surface of the earth to infinity isA. `12xx10^(7)J//kg`B. `12.5xx10^(7)J//kg`C. `6.25xx10^(7)J//kg`D. `25xx10^(7)J//kg`

The kinetic energy needed to project a body of mass m from the eath surface (radus R) to infinity is

The kinetic energy needed to project a body of mass m from the eath surface (radus R) to infinity isA. mgR /2B. 2 mgRC. mgRD. mgR /4

Calculate te radus of an isolated sphere of density `3.0 g cm^(-3)` from the surface of which the escape velocity be `40 ms^(-1)`

A satellite of mass `m` is orbiting the earth in a circular orbit of radius `r`. It starts losing energy due to small air resistance at the rate of `C J//s`. Then the time teken for the satellite to reach the earth is...........

A satellite of mass m is orbiting around the earth (mass M, radius R) in a circular orbital of radius 4R. It starts losing energy slowly at a constant `-(dE)/(dt)=eta`due to friction. Find the time (t) in which the satellite will spiral down to the surface of the earth.