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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.
| 401. |
The planets with radii `R_(1)` and `R_(2)` have densities `p_(1),p_(2)` respectively. Their atmospheric pressues are `p_(1)` and `p_(2)` respectively.Therefore, the ratio of masses of their atmospheres, neglecting variation of g within the limits of atmoshpere isA. `p_(1)R_(2)rho_(1)//p_(2)R_(1)rho_(2)`B. `p_(1)R_(2)rho_(2)//p_(2)R_(1)rho_(1)`C. `p_(1)R_(1)rho_(1)//p_(2)R_(2)rho_(2)`D. `p_(1)R_(1)rho_(2)//p_(2)R_(2)rho_(1)` |
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Answer» Correct Answer - D |
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| 402. |
If mass of the earth is `M`, radius is `R`, and gravitational constant is `G`, then workdone to take `1kg` mass from earth surface to infinity will beA. `sqrt((GM)/(2R))`B. `(GM)/R`C. `sqrt((2GM)/R)`D. `(GM)/(2R)` |
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Answer» Correct Answer - B |
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| 403. |
The mass of earth is 80 times that of moon. Their diameters are 12800 km and 3200 km respectively. The value of g on moon will be, if its value on earth is `980cm//s^(2)`A. `98cm//s^(2)`B. `196cm//s^(2)`C. `100cm//s^(2)`D. `294cm//s^(2)` |
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Answer» Correct Answer - b `(g_(2))/(g_(1))=(m_(2))/(m_(1))xx((R_(1))/(R_(2)))^(2)` `=(1)/(80)xx((12800)/(3200))^(2)=(1)/(80)xx16=(1)/(5)` `(g_(2))/(g_(1))=(980)/(5)=196cm//s^(2)`. |
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| 404. |
The two planets with radii `R_(1), R_(2)` have densities `rho_(1), rho_(2)`, and atmospheric pressures `p_(1)` and `p_(2)`, respectively. Therefore, the ratio of masses of their atmospheres, neglecting variation of `g` and `rho` within the limits of atmosphere, isA. `(p_(1)R_(2)rho_(1))/(p_(2)R_(1)rho_(2))`B. `(p_(1)R_(2)rho_(2))/(p_(2)R_(1)rho_(1))`C. `(p_(1)R_(1)rho_(1))/(p_(2)R_(2)rho_(2))`D. `(p_(1)R_(1)rho_(2))/(p_(2)R_(2)rho_(1))` |
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Answer» Correct Answer - D `p=rho_(atm)gh=m/(4/3pi[(R+h)^(3)-R] (GM)/(R^(2))h` `implies p=m/(4/3piR^(3)[(1+h/R)^(3)-1)]xx(G4/3piR^(3)l_(0))/(R^(2))xxh` `m/(4/3piR^(3)[(1+(3h)/h-1]xxG4/3piRl_(0)h` (Here `M=4/3piR^(3)l_(0)`) Here `p=(mGrho)/(3R)` `=:. Mprop(pR)/rho` |
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| 405. |
Two planets have radii `r_(1) and 2r_(1)` and densities are `rho_(1) and 4rho_(1)` respectively. The ratio of their acceleration due to gravities isA. `1:8`B. `8:1`C. `4:1`D. `1:4` |
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Answer» Correct Answer - a `(g_(1))/(g_(2))=(rho_(1))/(rho_(2))xx(R_(1))/(R_(2))` `=(1)/(4)xx(1)/(2)=(1)/(8)` |
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| 406. |
A satellite is orbiting around the earth in an orbitin equatorial plane of radius `2R_(e)` where `R_(e)` is the radius of earth. Find the area on earth, this satellite covers for communication purpose in its complete revolution.A. `sqrt(3) piR_(e)^(2)`B. `2R_(e)^(2)`C. `2sqrt(3)piR_(e)^(2)`D. `sqrt(2)piR_(e)^(2)` |
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Answer» Correct Answer - C Gravitational potential , `V=(-GM)/r(rgtR)` `V=(-GM(3R^(2)-r^(2)))/(2R^(3))(rltR)` Potential energy, `U=(-GMm)/r(rgtR)` `U=(-GMm(3R^(2)-r^(2)))/(2R^(3))(rltR)` `KE="Loss of "PE=O-U` `KE=U` `:. KE` will very with `r` exactly as that of `V` with sign changed. |
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| 407. |
The escape velocity for a body projected vertically upwards from the surface of earth is 11km/s. If the body is projected at an angle of `45^@` with the vertical, the escape velocity will beA. `11km//s`B. `11sqrt(3)km//s`C. `11/(sqrt(3))km//s`D. `33 km//s` |
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Answer» Correct Answer - A Escape velocity does not depends upon the angle of projection. |
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| 408. |
The escape velocity for a body projected vertically upwards from the surface of earth is 11km/s. If the body is projected at an angle of `45^@` with the vertical, the escape velocity will beA. `11sqrt(2)km//s`B. `22 km//s`C. `11 km//s`D. `11sqrt(2)m//s` |
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Answer» Correct Answer - 3 Escape velocity is independent of direction of projection. |
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| 409. |
The escape velocity for a body projected vertically upwards from the surface of earth is 11km/s. If the body is projected at an angle of `45^@` with the vertical, the escape velocity will beA. `11/sqrt(2)` km/sB. `11sqrt(2)` km/sC. 22 km/sD. 11 km/s |
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Answer» Correct Answer - D |
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| 410. |
The escape velocity for a body projected vertically upwards from the surface of earth is 11km/s. If the body is projected at an angle of `45^@` with the vertical, the escape velocity will beA. `11sqrt2km//s`B. `22km//s`C. `11m//s`D. `11sqrt2m//s` |
| Answer» Correct Answer - C | |
| 411. |
The escape velocity for a body projected vertically upwards from the surface of earth is 11km/s. If the body is projected at an angle of `45^@` with the vertical, the escape velocity will beA. `11sqrt(2) km//s`B. `22 km//s`C. `11 km//s`D. `22sqrt(2)km//s` |
| Answer» Correct Answer - C | |
| 412. |
Determine the speed with which the Earth has to rotate so that the weight of a person on the equator is 2/3rd of the current weight. Assume radius of Earth is 6.4 × 106 m. |
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Answer» Let m be the mass of the person, then, Original weight = mg and new weight = mg' Given that \(\frac {{mg}'}{mg} \) = \(\frac {2}{3}\) ⇒ g' = \(\frac{2}{3}\)g We know that, g' = g - Rω2 (for equatorial plane) \(\frac{2}{3}\) g = g - Rω2 ⇒ ω = \(\sqrt\frac{g}{3R}\) ω = \(\sqrt \frac {9.8}{3 \times 6.4 \times 10^6}\) ω = 5.104 × 10-4 rad s-1. |
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| 413. |
The angular speed of rotation of earth about its axis at which the weight of man standing on the equator becomes half of his weight at the poles is given byA. `0.034 rad s^(-1)`B. `8.75 xx 10^(-4) rad s^(-1)`C. `1.23 xx 10^(-2) rad s^(-1)`D. `7.65 xx 10^(-7) rad s^(-1)` |
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Answer» Correct Answer - B `g_("equator") = g - R omega^(2)` and `g_("pole") - R omega^(2)` `:. (g_("pole"))/(2)` or `(g)/(2) = g - R omega^(2)` or `omega sqrt((g)/(2R))` `= sqrt((9.8)/(2 xx 6.4 xx 10^(6))` `= 8.75 xx 10^(-4) rad//s` |
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| 414. |
When a body is taken from the equator to the poles, its weightA. remains constantB. increasesC. decreasesD. increase at n-pole and decrease at s-pole |
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Answer» Correct Answer - B When a body is taken from equator to poles, its weight increases, because acceleration due to gravity increases. |
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| 415. |
If earth were to rotate on its own axis such that the weight of a person at the equator becomes half the weight at the poles, then its time period of rotation is (g = acceleration due to gravity near the poles and R is the radius of earth) (Ignore equatorial bulge)A. `2 pi sqrt((2R)/(g))`B. `2pi sqrt((R)/(2g))`C. `2 pi sqrt((R)/(3g))`D. `2 pi sqrt((R)/(g))` |
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Answer» Correct Answer - A Here, `g_(0)=g_(90)-Romega^(2)` Given, `g_(0)=(g)/(2)and g_(90)=g` So, `(g)/(2)=g-R omega^(2),omega=sqrt((g)/(2R))` We have, `T=2pi(r)/(v)rArr T=(2pi)/(omega)" "(because omega=(v)/(r))` Hence, `T=2pi sqrt((2R)/(g))` |
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| 416. |
If suddenly the gravitational force of attraction between earth and satellite become zero, what would happen to the satellite ? |
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Answer» The satellite will move tangentially to the original orbit with a velocity with which it was revolving. |
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| 417. |
When a body is taken from the equator to the poles, its weightA. remains constantB. increasesC. decreasesD. increases at N-pole and decreases at S-pole |
| Answer» Correct Answer - b | |
| 418. |
When a body is taken from the equator to the poles, its weightA. remains the sameB. increasesC. decreasesD. may increase or decrease |
| Answer» Correct Answer - b | |
| 419. |
A nut becomes loose and gets detached from a satellite revolving around the earth. Will it land on the earth? If yes, where will it land? If no, how can an astronaut make it land on the earth? |
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Answer» No it will not land on the earth. It will orbit the earth like the satellite. The astronaut will have to make its speed roughly towards the earth instead of the tangential speed. |
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| 420. |
The weight of the body at poles is greater than the weight at the equator. Is it the actual weight or the apparent weight we are talking about? Does your answer depend on whether only the earth's rotation is taken into account or the flattening of the earth at the poles is also taken into account? |
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Answer» The weight at the equator is the apparent weight which is affected by the earth's rotation and flattening of the earth at the poles. At poles, it is maximum and unaffected by these two reasons. |
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| 421. |
If the radius of the earth decreases by 1% without changing its mass, will the acceleration due to gravity at the surface of the earth increase or decrease? If so by what percent? |
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Answer» The mass of the earth can be considered concentrated at its center for calculation of the force between the earth and the object on or beyond its surface. With the mass of the earth being the same if the radius decreases by 1% it means the distance between the earth and the object reduces by 1%, other things being the same. Since the acceleration due to gravity is inversely proportional to the square of the radius of the earth R, the decrease in R will result in an increase in g. The 1% decrease in R means it decreases by 0.99 times. So, the increase in g will be 1/(0.99)² = 1.02 times. That means the value of g increases by 2%. |
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| 422. |
When the radius of earth is reduced by 1% without changing the mass, then the acceleration due to gravity willA. increase by 2%B. decrease by 1.5%C. increase by 1%D. decrease by 1% |
| Answer» Correct Answer - A | |
| 423. |
The angular speed of rotation of the earth is `omega=7.27xx10^(-5)rads^(-1)`and its radius is `R=6.37xx10^(6)m.` Calculate the acceleration of a man standing at a place at `40^(@)` latitude. `[cos40^(@)=10.77]` If the earth suddenly stops rotating, the acceleration due to gravity on its surface will become`g_(0)=9.82ms^(-2).` Find the effective value of acceleration due to gravity (g) at `40^(@)` latitude taking into account the rotation of the earth. |
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Answer» Correct Answer - 0.026`ms^(-2)` 9.80 `ms^(-2)` |
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| 424. |
A body is projected vertically upward from the surface of the earth with escape velocity. Calculate the time in which it will be at a height (measured from the surface of the arth) 8 time the radius of the earth (R). Acceleration due to gravity on the surface of the earth is g. |
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Answer» Correct Answer - `t=52/3 sqrt(R/(2g))` |
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| 425. |
A satellite is in a circular polar orbit of altitude `300 km`. Determine the sparation d at the equator between the ground tracks associated with two successive overhead passes of the satellite. |
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Answer» Correct Answer - `d = 2520 km` |
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| 426. |
An astronaut on the surface of the moon throws a piece of lunar rock (mass m) directly towards the earth at a great speed such that the rock reaches the earth. Mass of the earth = M, Mass of the moon `=(M)/(81)` Radius of the earth = R, Distance between the centre of the earth and the moon = 60R In the course of its journey calculate themaximum gravitational potential energy of the rock Find the minimum possible speed of the rock when it enters the atmosphere of the earth. |
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Answer» Correct Answer - (a) `-(GMm)/(243R)` (b) `14045/14337 (GMm)/R` |
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| 427. |
A lunar probe is rocketed from earth directly toward the moon in such a way that it always between the earth and the moon. The probe narrowly misses the moon and continues to travel beyond it on an extension of the line segement described above. At what distance from the center of the earth will the force due to the earth be equal to the force due to the moon? |
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Answer» Correct Answer - `432 Mm` |
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| 428. |
Gravitational field intensity at the centre of the semi circle formed by a thin wire `AB` of mass `m` and length `L` is A. `(Gm)/(l^(2))"along+xaxis"`B. `(Gm)/(pil^(2))"along+xaxis"`C. `(2piGm)/(l^(2))"along+xaxis"`D. `(2piGm)/(l^(2))"along+xaxis"` |
| Answer» Correct Answer - D | |
| 429. |
Let gravitation field in a space be given as `E = - (k//r)` If the reference point is a distance d where potential is `V_(1)` then relation for potential is .A. `V = kl n(1)/(V_(i)) + 0`B. `V =k /n (r)/(d_(i)) + V`C. `V = "ln" (r)/(d_(i)) + KV`D. `V = "ln" (r)/(d_(i)) + (V_(i))/(k)` . |
| Answer» Correct Answer - B | |
| 430. |
Relation between force of attraction and distance between two objects is ………A) F ∝ \(\cfrac{1}R\)B) F ∝ \(\cfrac{1}{R^2}\)C) F ∝ \(\cfrac{1}{R^3}\)D) F ∝R |
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Answer» B) F ∝ \(\cfrac{1}{R^2}\) |
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| 431. |
What does not change in the field of central forceA) Potential energyB) Kinetic energyC) Linear momentumD) Angular momentum |
| Answer» D) Angular momentum | |
| 432. |
If the radius of the earth was half of its present value and its mass `(1//8)^("th")` of the present mass, the g value would have been reduced toA. `(1)/(8)g`B. `(1)/(2)g`C. `(1)/(3)g`D. g |
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Answer» Correct Answer - b `(g_(2))/(g_(1))=(m_(2))/(m_(1))xx((R_(1))/(R_(2)))^(2)=(1)/(8)xx4=(1)/(2)` `g_(2)=(1)/(2)g_(1).` |
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| 433. |
If the distance between the earth and moon were doubled, the gravitational force between them will beA. halvedB. doubledC. quadrupledD. reduced to `(1//4)^("th")` |
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Answer» Correct Answer - d `(F_(2))/(F_(1))=((r_(1))/(r_(2)))^(2)=((r)/(2r))^(2)=(1)/(4)` `F_(2)=(F)/(4).` |
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| 434. |
A satellite of mass m revolves around the earth of radius R at a hight x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite isA. `gx`B. `(gR)/(R-x)`C. `(gR^(2))/(R+x)`D. `((gR^(2))/(R+x))^(1/2)` |
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Answer» Correct Answer - D For the satellite the gravitational force provides the necessary centripetal force i.e., `(GM_(e)m)/((R+x)^(2))=(mv_(2)^(2))/((R+x))` or `(GM_(e))/(R^(2))=g` `v_(0)=sqrt((gR^(2))/(R+x))` |
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| 435. |
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` daysD. `1.41 s` |
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Answer» Correct Answer - B `a_(1)` acceleration of first `=(Gm_(1)m_(2))/(r_(2))xx1/m-1=(Gm_2)/(r^(2))=(6.67xx10^-11xx100)/((100)^(2))` `=6.67xx10^(-13)ms^(-1)` `a_(2)=` acceleration of second `=(Gm_(1))/(r^(2))=(6.67xx10^(-11)xx100)/((100)^(2))=6.67xx10^(-13)ms^(-1)` Net acceleration of approach `a=a_(1)+a_(2)=2xx6.67xx10^(-13)ms^(-1)` Now `s=1/2at^(2)` `1xx10^(-2)=1/2xx6.67xx10^(-13)xx2xxt^(2)` Solving we get `t=1.41`days. |
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| 436. |
On a planet whose size is the same and mass four times as that of our earth, find the amount of work done to lift `3 kg` mass vertically upwards through `3 m` distance on the planet. The value of `g` on the surface of earth is `10 ms^(-2)` |
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Answer» Here `R_(P)=R_(e),M_(P)=4,M_(e)` `M=3kg, h=em,g_(e)=10m//s^(2)` On the surface of the earth `g_(e)=(GM_(e))/(R_(e)^(2)` On the surface of the planet `g_(P)=(GM_()P)/(R_(P)^(2))=(G4M_(e))/(R_(e)^(2))=3(GM_(e))/(R^(2))=4g_(e)` `g_(P)=4xx10=40m//s^(2)` work done `=mg_(p)h=3xx40xx3=360J` |
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| 437. |
In the problem discussed in the text, find the values of `E` and `V` at `p` dua to the remaining mass. |
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Answer» Correct Answer - A::B::C::D Total mass is `M` of volume `(4)/(3) pi R^(3)`. Therefore, mass of cavity of volume `(4)/(3) pi (R/(4))^(3)` will be `(M)/(64)`. Further `C_(1)P = 2 R +( R)/(4) = (9R)/(4)` Using Eq. (i), `E_(R) = (GM)/((3R)^(2)) (-hati) - (G((M)/(64)))/((9R//4)^(2) )(-hati)` `:. E_(R) = (35)/(324) (GM)/(R^(2)) (-hat i)` `:.` Field strength is `(35 GM)/(324 R^(2))` towards `C`. USing Eq. (ii), we have `V_(R) = - (GM)/(3R) - [(-G((M)/(64)))/((9R//4))]` `= - (47)/(144) (GM)/(R)` |
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| 438. |
A planet whose size is the same and mass `4` times as that of Earth, find the amount of energy needed to lift a `2 kg` mass vertically upwards through `2m` distance on the planet. The value of `g` on the surface of Earth is `10 ms^(-2)`. |
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Answer» Correct Answer - 160 J |
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| 439. |
Gravitational potential at a height R from the surface of the earth will be [Take M = mass of the earth, R = radius of the earth]A. `(-GM)/(2R)`B. `(-GM)/(R )`C. `(-GM)/(4R)`D. `-GM` |
| Answer» Correct Answer - A | |
| 440. |
The gravitational potential at the centre of a square of side a when four point masses m each are kept at its vertices will beA. `4sqrt(2)(Gm)/(a)`B. `-4sqrt(2)(Gm)/(a)`C. `2sqrt(2)(Gm)/(a)`D. `-2sqrt(2)(Gm)/(a)` |
| Answer» Correct Answer - B | |
| 441. |
The gravitational potential energy of a system of three particles of mass m each kept at the vertices of equilateral triangle of side x will beA. `-(Gm^(2))/(x)`B. `-(Gm^(2))/(3x)`C. `-(3Gm^(2))/(x)`D. `(3Gm^(2))/(x^(2))` |
| Answer» Correct Answer - C | |
| 442. |
Three particles each of mass `m` are kept at vertices of an equilateral triangle of side L. The gravitational field at centre due to these particle isA. zeroB. `(Gm^(2))/(a^(2))`C. `(2Gm^(2))/(a^(2))`D. `(3Gm^(2))/(a^(2))` |
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Answer» Correct Answer - a `V=-(GM)/(r//2)-(GM)/(r//2)=-(4GM)/(r)` |
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| 443. |
Three particles each of mass `m` are kept at the vertices of an euilateral triangle of side `L`. The gravitational field at the centre due to these particle isA. ZeroB. `(3GM)/L^(2)`C. `(9GM)/L^(2)`D. `12/sqrt(2) (GM)/L^(2)` |
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Answer» Correct Answer - A |
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| 444. |
Three particles each of mass `m` are kept at the vertices of an euilateral triangle of side `L`. The gravitational field at the centre due to these particle isA. zeroB. `(3GM)/(L^(2))`C. `(9GM)/(L^(2))`D. `12/(sqrt(3))(GM)/(L^(2))` |
| Answer» Correct Answer - A | |
| 445. |
Why is Newton's law of gravitation called a universal law? |
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Answer» Newton's law of gravitation holds good irrespective of the nature of two bodies, i.e., big or small, at all times at all locations and for all distances in the universe. |
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| 446. |
Imagine an astronaut inside a satellite going around the earth in a circular orbit at a speed of `sqrt((gR)/(2))` where R is radius of the earth and g is acceleration due to gravity on the surface of the earth. What is weight experienced by the astronaut inside the satellite? Assume that an alien demon stops the satellite and holds it at rest. What is eightexperienced by the astronaut now? The demon now releases the satellite (from rest). What is weight experienced by the astronaut now? |
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Answer» Correct Answer - Zero 1(mg)/(4)` zero |
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| 447. |
What is the time period of a goestationary satellite? |
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Answer» The time period of a goestationary satellite is 24 hours. |
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| 448. |
What is the maximum value of gravitational potential energy and at where? |
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Answer» The value of gravitational potential energy increases as we move away from the earth and becomes maximum (in fact zero) at infinity. |
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| 449. |
An astronaut, while revolving in a circular orbit happens to throw a ball outside. Will the ball reach the surface of the earth? |
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Answer» The ball will never reach the surface of the earth. Actually it will continue to move in the same circular orbit and chase the astronaut. |
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| 450. |
A mass `m` is placed in the cavity inside hollow sphere of mass `M` as shown in the figure. The gravitational force on `m` is A. `(GMm)/(R^(2))`B. `(GMm)/(r^(2))`C. `(GMm)/((R-r)^(2))`D. zero |
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Answer» Correct Answer - D Gravitational force is zero to symmetry |
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