

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.
2451. |
A satellite is orbiting around the earth in a circular orbit. Its orbital speed is `V_(0).` A rocket on thrust to the satellite directed radially away from the centre of the earth. The uration of the engine burn is negligible so that it can be considered instantaneous. Due to this thrust a velocity variation `DeltaV` is imparted to the satellite. Find the minimum value of the ratio `(DeltaV)/(V_(0)` for which the satellite will escape out of the gravitational field of the earth. |
Answer» Correct Answer - -1 |
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2452. |
The gravitational field in a region is given by `vec(g)=(2hat(i)+3hat(j))` N/kg. The work done in moving a particle of mass 1 kg from (1, 1) to `(2, (1)/(3))` along the line 3y + 2x = 5 isA. ZeroB. 20 JC. `-15 J`D. 18 J |
Answer» Correct Answer - A | |
2453. |
Two particles of masses `m_(1)` and `m_(2)` initially at rest a infinite distance from each other, move under the action of mutual gravitational pull. Show that at any instant therir relative velocity of approach is `sqrt(2G(m_(1)+m_(2))//R)` where `R` is their separation at that instant. |
Answer» The gravitational force of attraction on `m_(1)` due to `m_(2)` at a separation r is `F_(1) = (Gm_(1)m_(2))/(r^(2))` Therefore the acceleration of `m_(1)` is `a_(1) = (F_(1))/(m_(1)) = (Gm_(2))/(r^(2))` Similarly the accleration of `m_(2)` due to `m_(1)` is `a_(2) = - (Gm_(1))/(r^(2))` the negative sign being put as `a_(2)` is directed opposite to `a_(1)`. The relative acclleration of approach is `a = a_(1) - a_(2) = (G(m_(1) + m_(2)))/(r^(2))`...(1) If `v` is the relative velocity then `a = (dv)/(dt) = (dv)/(dr) (dr)/(dt)` `But - (dr)/(dt) =v` (negative sign shows that r decreases with increasing t) `:. a = - (dv)/(dr) v`...(2) From (1) and (2) we have `vdv = - (G(m_(1) + m_(2)))/r^(2)dr` Integrating we get `(v^(2))/(2) = (G(m_(1)+m_(2)))/(r) + C` At `r = oo`, v = 0 (given) and so C = 0 `therefore v^(2) = (2G (m_(1)+m_(2)))/(r)` Let `v = v_(R)` when `r = R` Then `v_(R)=sqrt(((2G(m_(1)+m_(2)))/(R)))` . |
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2454. |
Two particles of masses `m_(1)` and `m_(2)` initially at rest a infinite distance from each other, move under the action of mutual gravitational pull. Show that at any instant therir relative velocity of approach is `sqrt(2G(m_(1)+m_(2))//R)` where `R` is their separation at that instant.A. `sqrt((G(m_(1)+m_(2)))/(r))`B. `sqrt((2G(m_(1)+m_(2)))/(r))`C. `sqrt((Gm_(1)m_(2))/((m_(1)+m_(2))r))`D. `sqrt((G(m_(1)+m_(2)))/(2r))` |
Answer» Correct Answer - B |
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2455. |
Two masses `90kg` and `160 kg` are `5m` apart. The gravitational field intensity at a point `3m` from `90kg` and `4m` from `160 kg` isA. `10G`B. `5G`C. `5sqrt(2)G`D. `10sqrt(2)G` |
Answer» Correct Answer - D `E_(R)=sqrt(E_(1)^(2)+E_(2)^(2))` |
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2456. |
Consider an infinite distribution of point masses (each of mass m) placed on x-axis as shown in the diagram. What is the gravitational force acting on the point mass placed at the origin ? A. `(4Gm^(2))/(3r^(2))`B. `(Gm^(2))/(3r^(2))`C. `(4Gm^(2))/(r^(2))`D. `(Gm^(2))/(r^(2))` |
Answer» Correct Answer - A | |
2457. |
A plenet moving along an elliptical orbit is closest to the sun at a distance `r_(1)` and farthest away at a distance of `r_(2)`. If `v_(1)` and `v_(2)` are the linear velocities at these points respectively, then the ratio `(v_(1))/(v_(2))` isA. `(r_(1))/(r_(2))`B. `((r_(1))/(r_(2)))^(2)`C. `(r_(2))/(r_(1))`D. `((r_(2))/(r_(1)))^(2)` |
Answer» Correct Answer - C `mv_(1)r_(1)=mv_(2)r_(2) implies (v_(1))/(v_(2))=(r_(2))/(r_(1))` |
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2458. |
A plenet moving along an elliptical orbit is closest to the sun at a distance `r_(1)` and farthest away at a distance of `r_(2)`. If `v_(1)` and `v_(2)` are the linear velocities at these points respectively, then the ratio `(v_(1))/(v_(2))` isA. `r_(2)//r_(1)`B. `(r_(2)//r_(1))^(2)`C. `r_(1)//r_(2)`D. `(r_(1)//r_(2))^(2)` |
Answer» Correct Answer - A From the law of coservation of angular momentum `m_(1)v_(1)=mr_(2)v_(2)` `r_(1)v_(1)=r_(2)v_(2)` `(v_(1))/(v_(2))=(r_(2))/(r_(1))` |
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2459. |
Two masses `90 kg` and `160 kg` are at a distance `5 m` apart. Compute the magnitude of intensity of the gravitational field at a point distance `3 m` from the `90 kg` and `4m` from the `160 kg` mass. `G=6.67xx10^(-11)kg^(-2)` |
Answer» Correct Answer - `9.43xx10^(-10) Nkg ^(-1)` |
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2460. |
Two masses, 800 kg and 600 kg, are at a distance 0.25 m apart. Compute the magnitude of the intensity of the gravitational field at a point distant 0.20 m from the 800 kg mass and 0.15 m from the 600 kg mass. |
Answer» Correct Answer - `2.2 xx 10^(-6) N Kg^(-1)` perpendicular to the line the mass ;- `5.3 xx10 ^(-7) J kg^(-1)` |
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2461. |
Figure shows a system of point masses placed on X-axis. Find the net gravitational field intensity at the origin Take sum of an infinite GP as `S = (a)/(1-r)` where a= first term and r = Least common ratio. |
Answer» Net gravitational field intensity at the origin, `E_("net")=E_(1)+E_(2)+E_(3)+E_(4)+....infty` terms `=(GM)/((1)^(2))hat(i)+(GM)/((2)^(2))hat(i)+(GM)/((4)^(2))hat(i)+.....infty` terms `=GM hat(i)(1+(1)/(4)+(1)/(16)+....infty)" "("Here", a=1 and r=(1)/(4))` So, `E_("net")=GM((a)/(1-r))hat(i)=GM hat(i)((1)/(1-1//4))=GMhat(i)((1)/(3//4))` `rArr E_("net") =(4)/(3)GM hat(i)`. |
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2462. |
A plenet moving along an elliptical orbit is closest to the sun at a distance `r_(1)` and farthest away at a distance of `r_(2)`. If `v_(1)` and `v_(2)` are the linear velocities at these points respectively, then the ratio `(v_(1))/(v_(2))` isA. `(r_("max"))/(r_("min"))`B. `(r_("min"))/(r_("max"))`C. `(r_("min")+r_("max"))/(r_("max")-r_("min"))`D. none of these |
Answer» Correct Answer - A Conserving angular momentum `mv_(1)v_("min")=mv_(2)r_("max")` or `(v_(1))/(v_(2))=r_("max")/r_("min")` |
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2463. |
The additional kinetic energy to be provided to a satellite of mass `m` revolving around a planet of mass `M`, to transfer it forms a circular orbit of radius `R_(1)` to another of radius `R_(2)(R_(2)gtR_(1))` isA. `GmM (1/R_(1)-1/R_(2))`B. `2 Gm M (1/R_(1) - 1/R_(2))`C. `1/2 GmM (1/R_(1)-1/R_(2))`D. `GmM(1/R_(1)^(2)-1/R_(2)^(2))` |
Answer» Correct Answer - C | |
2464. |
A satellite revolving round the earth in a circular orbit with orbital velocity `v_(0)`. It has kinetic energy E. The additional kinetic energy required to be given to it so that it esscapes from the earth isA. 4 EB. 3 EC. 2 ED. E |
Answer» Correct Answer - d `(E_(2))/(E_(1))=((1)/(2)mv_(e)^())/((1)/(2)mv_(e)^(2))=((2GM)/(R))/((GM)/(R))` `(E_(1))/(E_(1))-1=2-1=1` `E_(2)-E_(1)=E` `DeltaE=E` |
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2465. |
A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of coordinates. Two sphere of equal radii 1 unit, with their centres at A(-2,0 ,0) and B(2,0,0) respectively, are taken out of the solid leaving behind spherical cavities as shown if fig Then: A. The gravitational force due to this object at the origin is zeroB. The gravitational force at the point B (2, 0, 0) is zeroC. The gravitational potential is the same at all points of the circle `y^(2)+z^(2)=36`D. The gravitational potential is the same at all points on the circle `y^(2)+z^(2)=4` |
Answer» Correct Answer - A::C::D |
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2466. |
A satellite is revolving round the earth with orbital speed `v_(0)` if it is imagined to stop suddenly the speed with which it will strike the surface of the earth would be `(v_(e)` - escape speed of a body from earth s surface)A. `(v_(e)^(2))/(v_(0))`B. `v_(0)`C. `sqrt(v_(e)^(2)-v_(0)^(2))`D. `sqrt(v_(e)^(2)-2v_(0)^(2))` |
Answer» Correct Answer - D | |
2467. |
Two bodies of mass `m_(1) and m_(2)` are initially at rest placed infinite distance apart. They are then allowed to move towards each other under mutual gravitational attaction. Show that their relative velocity of approach at separation r betweeen them is `v=sqrt(2G(m_(1)+m_(2)))/(r)`A. `sqrt((2G(m_(1)+m_(2)))/(r)`B. `sqrt((2Gm_(1)m_(2))/((m_(1)+m_(2))r)`C. `sqrt((G(m_(1)+m_(2)))/(r))`D. `sqrt((Gm_(1)m_(2))/((m_(1)+m_(2))r)` |
Answer» Correct Answer - A | |
2468. |
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. `[2G((m_(1)-m_(2)))/r]^(1//2)`B. `[(2G)/r (m_(1)+m_(2))]^(1//2)`C. `[r/(2G(m_(1)m_(2)))]^(1//2)`D. `[(2G)/r m_(1)m_(2)]^(1//2)` |
Answer» Correct Answer - B |
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2469. |
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. `[2G((m_(1)-m_(2)))/r]^(1//2)`B. `[(2G)/r (m_(1)-m_(2))]^(1//2)`C. `[r/(2G(m_(1)m_(2)))]^(1//2)`D. `[(2G)/rm_(1)m_(2)]^(1//2)` |
Answer» Correct Answer - B `1/2muv^(2)=(Gm_(1)m_(2))/r`, `mu=((m_(1)+m_(2))/(m_(1)+m))`: reduced mass `1/2(m_(1)m_(2))/(m_(1)+m_(2))v^(2)=(Gm_(1)m_(2))/r` `v=sqrt((2G(m_(1)+m_(2)))/r)` |
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2470. |
Energy required to move a body of mass m from an orbit of radius 2R to 3R isA. `GMm//3R^(2)`B. `GMm//12R^(2)`C. `GMm//8R`D. `GMm//6R` |
Answer» Correct Answer - D |
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2471. |
Calculate the energy required to move an earth satellite of mass `10^(3)` kg from a circular orbit of radius 2R to that of radius 3R. Given mass of the earth,`M=5.98xx10^(24)` kg and radius of the earth, `R=6.37xx10^(6)`m. |
Answer» Given M = 5.98 x 1024 kg m = 103 kg R = 6.37 x 106 m Centripetal acceleration of the satellite is provided by the gravitational force exerted by Earth. \(\frac{mv^2}{r}=\frac{GMm}{r^2}\) v2 = GM/r Total energy of the system = K.E. + Gravitational Potential energy = 1/2 mv2 - GMm/r = 1/2 m GM/r - GMm/r = - 1/2 GMm/r Therefore, energy required to move on Earth satellite radius 2R to orbital radius 3R. E = GMm/2 (1/2R - 1/3R) E = GMm/12R \(E=\frac{6.67\times10^{-11}\times5.98\times10^{24}\times10^3}{12\times6.37\times10^6}\) \(E=\frac{39.88\times10^{10}}{76.44}\) ⇒ 0.5217 x 1010 J Correct Answer - `5.02xx10^(9)`J |
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2472. |
Two bodies of masses m and M are placed at distance d apart. The gravitational potential (V) at the position where the gravitational field due to them is zero V isA. `V=-G/d (m+M)`B. `V=- G/d`C. `V=- (GM)/d`D. `V= -G/d (sqrt(m)+sqrt(M))^(2)` |
Answer» Correct Answer - D | |
2473. |
Two bodies of masses m and M are placed at distance d apart. The gravitational potential (V) at the position where the gravitational field due to them is zero V isA. `V=-(G)/(d)(m+M)`B. `V=-(G)/(d)`C. `V=-(GM)/(d)`D. `V=-(G)/(d)sqrt(sqrt(m)+sqrt(M))^(2)` |
Answer» Correct Answer - D Equilibrium position of the neutral point from mass m is `=((sqrt(m))/(sqrt(m)+sqrt(M)))d` `V_(1)=(-Gm_(1))/(r_(1)),V_(2)=(-Gm_(2))/(r_(2))` `V_(1)=(-Gm)/(sqrt(m)d)(sqrt(M)+sqrt(m)),V_(2)=(-GM)/(sqrt(M)d)(sqrt(M)+sqrt(m))` `V_(1)=(-G)/(d)sqrt(m)(sqrt(M)+sqrt(m)),V_(2)=(-G)/(d)sqrt(M)(sqrt(M)+sqrt(m))` `V=(-G)/(d)(sqrt(M)+sqrt(m))^(2)` |
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2474. |
Two bodies of masses m and M are placed at distance d apart. The gravitational potential (V) at the position where the gravitational field due to them is zero V isA. `V=-(G)/(d)(m+M)`B. `V=-(Gm)/(d)`C. `V=-(GM)/(d)`D. `V=-(G)/(d)(sqrt(m)+sqrt(M))^(2)` |
Answer» Correct Answer - D `E_("net")=0` `:. (Gm)/(x^(2))=(GM)/((d-x)^(2)),x` is distance from `m` `:. (x)/(d-x)=(sqrt(m))/(sqrt(M))rArrx=(sqrt(m))/(sqrt(m)+sqrt(M)),d` and `d-x=(sqrt(M))/(sqrt(m)+sqrt(M)).d` Since, gravitational potential between two bodies of masses M and m respectively, is given by `V=-(Gm)/(x)-(GM)/(d-x)` `=-(GM(sqrt(m)+sqrt(M)))/(sqrt(m).d)-(GMsqrt(m)+sqrt(M))/(sqrt(M).d)=-(G)/(d)(sqrt(m)+sqrt(M))^(2)`. |
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2475. |
Two bodies of masses m and M are placed at distance d apart. The gravitational potential (V) at the position where the gravitational field due to them is zero V isA. `V = - (G)/(d) (m + M)`B. `V = - (Gm)/(d)`C. `V = - (GM)/(d)`D. `V = - (G)/(d) (sqrt(m) + sqrt(M))^(2)` |
Answer» Correct Answer - D |
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2476. |
A sky lab of mass `2 xx 10^(3)kg` is first lauched from the surface of earth in a circular orbit of radius `2R` and them it is shifted from this circular orbit to another circular orbit of radius `3R`. Calculate the energy required (a) to place the lab in the first orbit, (b) to shift the lab from first orbit to the second orbit. `(R = 6400 km`, `g = 10 m//s^(2))` |
Answer» Correct Answer - A::B (a) `W =` energy of satellite in first orbit - energy of satellite on the surface of earth `= - (GMm)/(2(2R)) - ((-Gmm)/(R))` `= (3)/(4) (GMm)/(R )` `= (3)/(4) mgR = (3)/(4) xx 2 xx 10^(3) xx 10 xx 6.4 xx 10^(6)` `= 9.6 xx 10^(10) J` (b) `W =` enargy of satellite in second orbit - energy in first orbit ` = - (GMm)/(2(3R)) - [(-Gmm)/(2(2R))]` `= (1)/(2) (GMm)/(R) = (1)/(2) mgR` `= (1)/(2) xx 2 xx 10^(3) xx 10 xx 6.4 xx 10^(6)` `= 1.07 xx 10^(10)J` |
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2477. |
A skylab of mass `m` kg is first launched from the surface of the earth in a circular orbit of radius `2R` (from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius `3R`. The minimum energy required to place the lab in the first orbit and to shift the lab from first orbit to the second orbit areA. `3/4mgR,(mgR)/6`B. `3/4mgR,(mgR)/12`C. `mgR,mgR`D. `2mgR,mgR` |
Answer» Correct Answer - B Energy of the skylab in the first orbit is `-(GMm)/(2(2R))=-(GMm)/(4R)` Total energy required to place the skylab into the orbt of radius `2R` frm the surface of earth is `=(GMm)/(4R)-(-(GMm)/R)=(3G/Mm)/(4R)` `=(3gR^(2)m)/(4R)=3/4mgR` Energy of the skylab in the second orbit `=-(GMm)//6R` Energy needed to shift the skylab from the first orbit to the second orbit is `-(GMm)/(4R)-(GMm)/(6R)=(GMm)/Rxx2/24=(mgR)/12` |
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2478. |
Two spherical balls of mass 10 kg each are placed 10 cm apart. Find the gravitational force of attraction between them. |
Answer» Correct Answer - `[6.67xx10^(-7)N]` |
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2479. |
Two spherical balls of mass 10 kg each are placed 10 cm apart. Find the gravitational force f attraction between them. |
Answer» Correct Answer - A Gravitational force of attraction `F=(GMm)/r^2` `=(6.67xx10^-11xx10xx10)/((0.1)^2)` `=6.67xx10^-7N` |
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2480. |
Universal gravitational constant G = A) 6.67 × 10-11 Nm2 kg-2B) 6.67 × 10-11 N2 m2 kg-2C) 6.67 × 10-11 N2 mkg-2D) 6.67 × 10-11 Nm2 kg2 |
Answer» A) 6.67 × 10-11 Nm2 kg-2 |
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2481. |
On Earth value of `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`. What is its value on Moon, where `g` is nearly one-sixth than that of Earth? |
Answer» `G=6.67xx10^(-11)Nm^(2)//kg^(2)` The value is the same on earth and on moon. |
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2482. |
Calculate the gravitational force of attraction between two spherical bodies, each of mass 1kg placed at 10m apart `(G = 6.67 xx 10^(-11)Nm^(2)//kg^(2))`.A. `6.67 xx 10^(-13)N`B. `6.67xx 10^(-11)N`C. `6.67 xx 10^(-7)N`D. None of these |
Answer» Correct Answer - A | |
2483. |
How many times faster than its present speed the earth should rotate so that the apparent weight of an object at equator becomes zero ? Given radius of the earth = `6.37 xx 10^(6)` m . What would be the duration of the day in that case ? |
Answer» Correct Answer - 17 times faster , 1.412 h |
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2484. |
A satellite of the earth is revolving in circular orbit with a uniform velocity V. If the gravitational force suddenly disappears, the satelite willA. continue to move with the same velocity in the same orbit.B. move tangentially to the original orbit with velocity V.C. fall down with increasing velocity.D. come to a stop somewhere in its original orbit. |
Answer» Correct Answer - B | |
2485. |
A satellite of the earth is revolving in a circular orbit with a uniform speed `v`. If the gravitational force suddenly disappears, the satellite willA. Continue to move with velocity v along the original orbitB. Move with a velocity v , tangentially to the original orbitC. Fall down with increasing velocityD. Ultimately come to rest somewhere on the original orbit |
Answer» Correct Answer - B |
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2486. |
A satellite of the earth is revolving in a circular orbit with a uniform speed `v`. If the gravitational force suddenly disappears, the satellite willA. fall down with increasing velocityB. ultimately come to rest somewhere on the original orbitC. move with a velocity v, tangentially to the original orbitD. continue to move with velocity v along the original orbit |
Answer» Correct Answer - D It will move with the same velocity along the original orbit due to inertia of direction. |
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2487. |
A satellite of the earth is revolving in a circular orbit with a uniform speed `v`. If the gravitational force suddenly disappears, the satellite willA. vanish into outer spaceB. continue to move with velocity v in original orbitC. fall down with increasing velocityD. fly off tangentially from the orbit with velocity v |
Answer» Correct Answer - D | |
2488. |
An object is projected vertically upward from the surface of the earth at which above and below the surface acceleration due to gravity is the same. |
Answer» Correct Answer - `h=(sqrt(5-1))/2 R` Let at height or depth `h` from the surface of earth acceleration due to gravity is same. `g_(d)=g_(h)` `g(1-h/R)=(gR^(2))/((R+h)^(2))implies((R-h))/R=(R^(2))/((R+h)^(2))` `(R-h)(R+h)^(2)=R^(3)` `(R-h)(R^(2)+h^(2)+2hR)=R^(3)` `R^(3)+Rh^(2)+2hR^(2)-hR^(2)-h^(3)-2h^(2)R=R^(3)` `h^(3)+h^(2)R-hR^(2)=0` `h^(2)+hR-R^(2)=0` `h=(-R+-sqrt(R^(2)+4R^(2)))/2impliesh=((sqrt(5)-1)R)/2` |
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2489. |
A satellite revolves in the geostationary orbit but in a direction east to west. The time interval between its successive passing about a point on the equator is:A. `48 hrs`B. `24 hrs`C. `12 hrs`D. never |
Answer» Correct Answer - c For geostationary satallite when orbiliting west to east `omega_(s)=omega_(e)` but when orbiting east to west `T=(2pir)/(omega_(s)+omega_(e))impliesT=(2pir)/(2omega_(e))impliesT=24/2` `T=12h` |
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2490. |
A satellite of the earth is revolving in circular orbit with a uniform velocity `V`. If the gravitational force suddenly disappers, the satellite will-A. continue to move with the same velocity in the same orbitB. move tangentially to the original orbit with velocity `V`.C. fall down with increasing velocityD. come to a stop somewhere in its original orbit |
Answer» Correct Answer - b | |
2491. |
When a golf ball is lowered into a measuring cylinder containing water, the water level rises by `30 cm^3` when the ball is completely submerged. If the mass of ball in air is 33 g find its density |
Answer» Correct Answer - `1.1 g//cm^(2)` |
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2492. |
During the rotation of a giant wheel, a person experiences loss of weight on the descent. Explain why. |
Answer» Because the giant wheel is in free fall |
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2493. |
A beaker full of water is suspended from a spring balance. Will the reading of the balance change: If a cork is placed in water ? If a piece of heavy metal is placed in it ? Give the reasone for your answer |
Answer» Correct Answer - The reading of spring balance will not change if a cork is placed reading of spring balance will change if a piece of heavy metal is placed in water because heavy metal, being denser than water, sinks in water |
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2494. |
A beaker full of water is suspended from a spring balance. Will the reading of the balance change : (a) if a cork is placed in water ? (b) if a piece of heavy metal is placed in it ? Give reasons for your answer. |
Answer» The reading of spring balance will not change if a cork is placed in water because cork, being lighter than water, floats in water. (b)The reading of spring balance will change if a piece of heavy metal is placed in water because heavy metal being denser than water, sinks in water. |
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2495. |
Suppose a spring balance with a body suspended from it is allowed to fall. What will be the reading shown by the balance? |
Answer» Reading will be zero. |
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2496. |
Show the Relation between 1 kg wt and express it into Newton. |
Answer» We know that W = m × g If mass (m) = 1 kg, g = 9.8 m/s2, then W = 1 kg × 9.8 m/s2 Or 1 kg wt = 9.8 N So, the gravitational force of earth that acts on an object of mass 1 kg is called as 1 kg wt. |
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2497. |
Find out the weight of a body of mass 20 kg. Express the value in newton. |
Answer» F = mg = 20 × 9.8 = 196 N |
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2498. |
Is the acceleration due to gravity of earth g a constant ? Discuss. Calculate the acceleration due to gravity on the surface of a satellite having a mass of `7.4 xx 10 ^(22)` and a radius of `7.4 xx 10 ^(22)` kg and a radius of `1.74 xx 10^(6)m G(=6.7 xx 10 ^(11) Nm^(2)//kg ^(2)`).Which satellite do you think it could be ? |
Answer» Correct Answer - (a) No (b) `1.63 m//s^(2)` |
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2499. |
A satellite is revolving around earth in a circular orbit at a height `(R )/(2)` from the surface of earth. Which of the following are correct statements about it (M = mass of earth, R = radius of earth, m = mass of satellite) ?A. Its orbital velocity is `sqrt((2GM)/(3R))`B. Its total energy is `-(GMm)/(3R)`C. Its kinetic energy is `-(GMm)/(3R)`D. Its potential energy is `(GMm)/(3R)` |
Answer» Correct Answer - A::B::C::D | |
2500. |
Two astronauts are floating in gravitational free space after having lost contact with their spaceship.The two will:(A) keep floating at the same distance between them.(B) move towards each other.(C) move away from each other.(D) will become stationary. |
Answer» Correct option is: (B) move towards each other. |
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