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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.
| 351. |
Two satellites are revolving around the earth in circular orbits of same radii. Mass of one satellite is `100` times that of the other. Then their periods of revolutions are in the ratioA. `100:1`B. `1:100`C. `10:1`D. `1:1` |
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Answer» Correct Answer - d `(r_(1))/(r_(2))=((T_(1))/(T_(2)))^(2//3)=((1)/(1))^(2//3)=1:1` |
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| 352. |
A simple pendulum is transferred from the earth to the moon. Assuming its time period is inversely proportional to the square root of acceleration due to gravity, it willA. Become fasterB. Become slowerC. Remain the sameD. Be sometimes faster, sometimes slower |
| Answer» Correct Answer - B | |
| 353. |
Two satellites A and B go around a planet in circular orbits of radii 4 R and R respectively. If the speed of the satellite A is 3 V, then the speed of the satellite B will beA. 12 VB. 3 VC. 6 VD. `(3V)/(2)` |
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Answer» Correct Answer - C `(V_(2))/(V_(1))=sqrt((GM)/(r_(2))xx(r_(1))/(GM))=sqrt((4)/(1))=2` `therefore V_(2)=2V_(1)=6V` |
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| 354. |
The radii of circular orbits of two satellite `A` and `B` of the earth are `4R` and `R`, respectively. If the speed of satellite `A` is `3v`, then the speed of satellite `B` will beA. `3v//4`B. `6v`C. `12v`D. `3v//2` |
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Answer» Correct Answer - B Orbital velocity of satellite `v=sqrt((GM)/r)` `implies (v_(A))/(v_(B))=sqrt((r_(B))/(r_(A)))=sqrt(R/(4R))=1/2` `:. (v_(A))/(v_(B))=(3v)/(v_(B))=1/2` ` :. v_(B)=6v` |
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| 355. |
Two satellites P and Q go round a planet in circular orbits of radii 9R and R respectively. If the speed of the satellite P is 4V, then speed of satellite Q will beA. 6VB. 12VC. 2VD. 36V |
| Answer» Correct Answer - B | |
| 356. |
The radii of circular orbits of two satellite `A` and `B` of the earth are `4R` and `R`, respectively. If the speed of satellite `A` is `3v`, then the speed of satellite `B` will beA. `12V`B. `6V`C. `4V `D. `3V` |
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Answer» Correct Answer - B `v=sqrt((GM)/R)rArr (v_(A))/(v_(B))=sqrt((R_(B))/(R_(A)))` |
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| 357. |
A solid sphere of uniform density and radius `R` applies a gravitational force attraction equal to `F_(1)` on a particle placed at `P`, distance `2R` from the centre `O` of the sphere. A spherical cavity of radius `R//2` is now made in the sphere as shows in figure. The particle with cavity now applies a gravitational force `F_(2)` on same particle placed at `P`. The radio `F_(2)//F_(1)` will be A. `1//2`B. `7//9`C. `3`D. `7` |
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Answer» Correct Answer - B `C rarr cavity`, `T rarr` Total, `R rarr` remaining `F_(1) = (GMm)/((2R)^(2))`..(i) `F_(R ) = F_(2) = F_(T) - F_(C) = F_(1) - F_(C)` or `F_(2) = (GMm)/((2R)^(2)) - (G((M)/(8)) m)/((3R//2)^(2))` or `F_(2) = (14 GMm)/(72 R^(2))` From Eqs. (i) and (ii) we get, `(F_(2))/(F_(1)) = (7)/(9)` |
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| 358. |
Two spheres of masses `M_(1)` and `M_(2)` and separated by a distance d are situated in air and the gravitational force of attraction between them is F. If the two spheres are kept in a liquid of specific gravity 5 then the gravitational force between them will beA. 5 FB. `(F)/(5)`C. FD. `sqrt(F)` |
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Answer» Correct Answer - C The gravitational force (F) between the spheres will be = F because F is independent of the medium, in which the two bodies are placed. |
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| 359. |
Two satellites A and B go round a planet pin circular orbits having radius 4R and R respectively. If the speed of the satellite A is 3v, find the speed of the satellite B. |
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Answer» As, vo = \(\sqrt{\frac{GM}{R}}\) So, For A 3v = \(\sqrt{\frac{GM}{4R}}\) ................(1) And for B v' = \(\sqrt{\frac{GM}{R}}\) ......................(2) Dividing (2) by (1) we get, \(\therefore\) \(\frac{v'}{3v}\) = 2 or v' = 6v |
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| 360. |
Two satellite A and B go round a planet P in circular orbits having radii 4R and R respectively. If the speed of the satellite A is 3v, the speed of the satellite B will beA. 12 vB. 6 vC. `(4)/(3)` vD. `(3)/(2)` v |
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Answer» Correct Answer - B Orbital velocity, `v=sqrt((GM)/(R))rArr(v_(A))/(v_(B))=sqrt((R_(B))/(R_(A)))=sqrt((R)/(4R))=(1)/(2)` `:. (v_(A))/(v_(B))=(3v)/(v_(B))=(1)/(2)rArr v_(B)=6v`. |
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| 361. |
A solid sphere of uniform density and radius `R` applies a gravitational force of attraction equal to `F_(1)` on a particle placed at `P`, distance `2R` from the centre `O` of the sphere. A spherical cavity of radius `R//2` is now made in the sphere as shown in figure. The particle with cavity now applies a gravitational force `F_(2)` on same particle placed at `P`. The radio `F_(2)//F_(1)` will be A. `9/7`B. `7/9`C. `1/2`D. `4/3` |
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Answer» Correct Answer - A |
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| 362. |
A solid sphere of uniform density and radius `R` applies a gravitational force of attraction equal to `F_(1)` on a particle placed at `P`, distance `2R` from the centre `O` of the sphere. A spherical cavity of radius `R//2` is now made in the sphere as shown in figure. The particle with cavity now applies a gravitational force `F_(2)` on same particle placed at `P`. The radio `F_(2)//F_(1)` will be A. `(5)/(9)`B. `(7)/(8)`C. `(3)/(4)`D. `(7)/(9)` |
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Answer» Correct Answer - D `F_(1)=(GMm)/((2R)^(2))=(GMm)/(4R^(2))` `F_(2)=F_(1)-F_("cavity")=(GMm)/(4R^(2))-(G((M)/(8))(m))/(((3)/(2)R)^(2))` `rArr F_(2)=(7GMn)/(36 R^(2))rArr (F_(2))/(F_(1))=(7)/(9)`. |
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| 363. |
A person sitting in a satellite feels weightlessness because A) the earth does not attract the objects in a satellite. B) the normal force by the chair on the person balances the earth’s attraction. C) the normal force is zero. D) the person in satellite is not accelerated. |
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Answer» C) the normal force is zero |
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| 364. |
A person sitting in a chair in a satellite feels weightless becauseA. the earth does not attract the objects inside a satelliteB. the normal force by the chair on the person balances the earth attractionC. the normal force is zeroD. the person in satellite is not accelerated |
| Answer» We are conscious of our weight due to the normal reaction force acting on our body to balance the weight of our body since inside a satellite there is condition of weightlessness so normal force is zero | |
| 365. |
What are the conditions under which a rocket fired from earth, launches an artificial satellite of earth? |
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Answer» Following are the basic conditions: (i) The rocket must take the satellite to a suitable height above the surface of earth for ease of propulsion. (ii)From the desired height, the satellite must be projected with a suitable speed, called orbital speed. (iii)In the orbital path of satellite, the air resistance should be negligible so that its speed does not decrease and it does not burn due to the heat produced. |
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| 366. |
Assertion: A person sitting in an artificial satellite revolving around the earth feels weightless. Reason: There is no gravitational force on the satellite.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - C |
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| 367. |
Assertion: A person sitting in an artificial satellite revolving around the earth feels weightless. Reason: There is no gravitational force on the satellite.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - C A person feels his weight only when the surface on which he is standing exerts a reactionary force on him. Because the acceleration of the person and that of the satellite revolving round the earth are equal `(=g)`, hence acceleration of the person with respect to the satellite is zero. Therefore person feels weightless on satellite, although the gravitational force is acting on a satellite. |
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| 368. |
A person sitting in a chair in a satellite feels weightless because (a) the earth does not attract the objects in a satellite (b) the normal force by the chair on the person balances the earth's attraction (c) the normal force is zero (d) the person in satellite is not accelerated. |
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Answer» (c) the normal force is zero EXPLANATION: Only (c) is correct. The weight and the centrifugal force balance each other, hence the normal force is zero. |
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| 369. |
The escape velocity from the surface of earth is `V_(e)`. The escape velocity from the surface of a planet whose mass and radius are 3 times those of the earth will beA. `V_(e)`B. `3 v_(e)`C. `9 V_(e)`D. `27 V_(e)` |
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Answer» Correct Answer - A |
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| 370. |
The escape velocity for the earth is `v_(e)`. The escape velocity for a planet whose radius is four times and density is nine times that of the earth, isA. `36 v_(e)`B. `12 v_(e)`C. `6 v_(e)`D. `20 v_(e)` |
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Answer» Correct Answer - B |
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| 371. |
Does a rocket really need the escape velocity of 11.2 km/s initially to escape from the earth? |
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Answer» No, a rocket can have any initial velocity at the start but its velocity should continue to increase. It will escape from the earth only if its velocity becomes 11.2 km/s. |
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| 372. |
What will be the escape speed from a planet having mass 8 times that of earth and diameter 4 times that of the earth ? `[v_(e)=11.2 km//s]` |
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Answer» `v_(p) = ?` Now, `v=sqrt((2GM)/(R ))` `therefore (v_(p))/(v_(e))=sqrt((M_(p))/(M_(e))xx(R_(e))/(R_(p)))=sqrt((8M_(e))/(M_(e))xx(R_(e))/(4R_(e)))` `therefore v_(p)=sqrt(2)v_(e)` `=sqrt(2)xx11.2` `=15.84 km//s` . |
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| 373. |
The escape velocity for a rocket from earth is 11.2 km / sec . Its value on a planet where acceleration due to gravity is double that on the earth and diameter of the planet is twice that of earth will be in km / secA. 11.2B. 5.6C. 22.4D. 53.6 |
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Answer» Correct Answer - C |
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| 374. |
The escape velocity for the earth is 11.2 km / sec . The mass of another planet is 100 times that of the earth and its radius is 4 times that of the earth. The escape velocity for this planet will beA. 56.0 km/sB. 280 km/sC. 112 km/sD. 11.2 km/s |
| Answer» Correct Answer - A | |
| 375. |
The escape velocity from the surface of the earth is `V_(e)`. The escape velcotiy from the surface of a planet whose mass and radius are three times those of the earth, will beA. `v_(e)`B. `3v_(e)`C. `9v_(e)`D. `(1)/(3v_(e))` |
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Answer» Correct Answer - A |
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| 376. |
Two masses `m_(1)` and `m_(2)` at an infinite distance from each other are initially at rest, start interacting gravitationally. Find their velocity of approach when they are at a distance `r` apart. |
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Answer» Correct Answer - `sqrt([2G(m_(1)+m_(2))//d])` |
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| 377. |
Computer the mass and density of the moon if acceleration due to gravity on its surfac is `1.62 m//s^(2)` and its radius is `1.74 xx 10^(6)m[G = 6.67 xx 10^(-11) MKS` units]. |
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Answer» Correct Answer - `7.35 xx 10^(22) kg, 3.3 xx 10^(3) kg//m^(3)` |
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| 378. |
To estimate the height of a bridge over a river, a stone is dropped freely on the river from the bridge. The stone takes 2 s to touch the water surface in the river. Calculate the height of the bridge from the water level. Take `g=9.8m//s^(2)` |
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Answer» The stone is being dropped freely from rest ,so the initeal velocity of the stone u=0.Again the velocity of stone is increasing as it comes down ,so the acceleration due to gravity g is to be taken as positive . Now Initial velocity of stone u=p0n Time taken t=2 s Acceleration due to gravity `g=9.8m//s^(2)` (stone come down ) Height of the brigdge h=? (to be calculated ) We know that for a freely falline body Height ` h=ut+1/2gt^(2)` Putting the above values in this formula we get `h=0xx2+1/2xx9.8xx4` ` h=1/2xx9.8xx4` h=19.6 Thus the height of bridge above the water level is 19.6 |
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| 379. |
A ball thrown up vertically returns to the thrower after 6s. Find (a) the velocity with which it was thrown up. (b) the maximum height it reaches, and (c) its position after 4s. |
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Answer» Here, time of ascent =time of descent,`t=6/2=3s` (a) `u=?,v=0, a=-g=-9.8 m//s^(2)` From `v=u+a t, 0=u-9.8xx3,u=29.4 m//s` (b) From `v^(2)-u^(2)=2 as , 0-(29.4)^(2)=2(-9.8)h, h=(29.4xx29.4)/(2xx9.8)=44.1m` (c ) At `t=3 s` , ball is at maximum height In the next 1 sec `(=4s-3s)` From `s=ut+1/2at^(2), h=0+1/2xx9.8(1)^(2)=4.9m` (below the top) |
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| 380. |
A ball trown up vertically returns to the thrower after 10 second. Calculate (i) the velocity with which it was thrown up. (ii) the maximum height it reachesand (iii) its poisition after 7 second. |
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Answer» (i) Time of ascent=time of descent`=(10)/(2)=5s`. From `v=u+at` `0=u-9.8xx5 , u=49 m//s`. This is the velocity with which it was thrown up. (i) Max.height `=u^(2)/2g=(49xx49)/(2xx9.8)=122.5m` (iii) From `s=ut+(1)/(2)at^(2)` `x=0+(1)/(2)xx9.8(2)^(2)=19.6m`,from the top. position of the ball after `7 sec=122.5-19.6=102.9`m above the ground. |
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| 381. |
A ball thrown up vertically returns to the thrower after 8 second. Calculate (i) velocity with which it was thrown (ii) maximum height it acquired. (iii) velocity with which it hit the ground.Given `g=9.8 m//s^(2)`. |
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Answer» As time of ascent =time of descent`=(1)/(2)` total time. `:. t=(8)/(2)=4s` (i)`u=?,v=0,a=-g9.8m//s^(2)`. From `v=u+at` `0=u-9.8xx4` or `u=39.2m//s` (ii)From `v^(2)-u^(2)=2a s` `0-(39.2)^(2)=2(-9.8)h` `h=(-(39.2)^(2))/(-2xx9.98)=78.4m` (iii) velocity with which body hits the ground = velocity of projection of body `=39.2m//s`. |
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| 382. |
What is uniform circular motion? |
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Answer» Uniform circular motion is a motion of the body with a constant speed in a circular path |
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| 383. |
What is centripetal acceleration? |
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Answer» The acceleration which can change only the direction of the velocity of a body is called centripetal acceleration. |
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| 384. |
What is centripetal force? |
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Answer» The net force which can change only the direction of the velocity of a body is called centripetal force directed towards the centre of the circle. Centripetal force F = mv2/R |
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| 385. |
Two lead spheres of `20 cm` and `2 cm` diametre respectively are planet with centres `100 cm` apart. Calculate the attraction between them, given the radius of the Earth as `6.37 xx 10^(8) cm` and its mean density as `5.53 xx 10^(3) kg m^(-3)`. Speciffic gravity of lead `= 11.5`. If the lead spheres are replaced by bress sphere of the same radii, would the force of attraction be the same? |
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Answer» Correct Answer - `1.5 xx 10^(-10)N` |
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| 386. |
Two identical spheres each of radius `R` are placed with their centres at a distance `nR`, where `n` is integer greater than `2`. The gravitational force between them will be proportional toA. `1//4R^(4)`B. `1//R^(2)`C. `R^(2)`D. `R^(4)` |
| Answer» Correct Answer - d | |
| 387. |
Two identical spheres each of radius `R` are placed with their centres at a distance `nR`, where `n` is integer greater than `2`. The gravitational force between them will be proportional toA. `1//R^(4)`B. `1//R^(2)`C. `R^(2)`D. `R^(4)` |
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Answer» Correct Answer - D `F=(Gm_(1)m_(2))/(R^(2)),` here `m=4/3 piR^(3)rho` |
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| 388. |
Six particles each of mass`m` are placed at the corners of a regular hexagon of edge length `a`. If a point mass `m_0` is placed at the centre of the hexagon, then the net gravitational force on the point mass isA. `(6Gm^(2))/(a^(2))`B. `(6Gmm_(0))/(a^(2))`C. zeroD. None |
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Answer» Correct Answer - C If equal masses are placed at respective corners of a regular polygon. Then, the gravitational force on a point mass placed at centre of the regular hexagon is zero. |
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| 389. |
A proton moving with a velocity of `10^(6) m//s` collides with a neutron at rest. If collision is prefectliy elastic, then after collsion, what are the velocities of porton and neutron? |
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Answer» As mass of porton and neutron are (almost) equel, therefore on perfectly elastic collsion between then. Their velocities are exchange, i.e., velocity of neutron `=10^(6) m//s`, and velocity of porton=zero. |
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| 390. |
i) What are Kepler’s laws? ii) What is the shape of the orbits of planets? |
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Answer» 1. The Kepler’s laws are:
2. The orbits of the planet are elliptical in shape. |
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| 391. |
Mention some main characteristics of gravitational force. |
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Answer» Characteristics of gravitational force: 1. Every massive object in the universe experiences gravitational force. 2. It is the force of mutual attraction between any two objects by virtue of their masses. 3. It is always an attractive force with infinite range. 4. It does not depend upon the intervening medium. 5. It is much weaker than other fundamental forces. Gravitational force is 10 times weaker than strong nuclear force. |
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| 392. |
What is a central force? |
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Answer» A central force on an object is a force which is always directed along the line joining the position of object and a fired point usually taken to the origin of the coordinate system. |
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| 393. |
If two planets of radii `R_(1) and R_(2)` have densities `d_(1) and d_(2)`, then the ratio of their respective acceleration due to gravity isA. `R_(1)d_(1):R_(2)d_(2)`B. `R_(1)^(2)d_(1):R_(2)^(2)d_(2)`C. `R_(1)^(3)d_(1):R_(2)^(3)d_(2)`D. `R_(1)d_(1)^(2):R_(2)^(2)d_(2)^(2)` |
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Answer» Correct Answer - A We know the gravity, `g = (4)/(3)pi GRd` Here given densities `d_(1) and d_(2)` and radii `R_(1) and R_(2)` So, `(g_(1))/(g_(2))=((4)/(3)piGR_(1)d_(1))/((4)/(3)piGP_(2)d_(2))rArr(g_(1))/(g_(2))=(R_(1)d_(1))/(R_(2)d_(2))`. |
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| 394. |
Weight of a body on the surfaces of two planets is the same. If their densities are `d_(1) and d_(2)`, then ratio of their radii isA. `d_(1)//d_(2)`B. `d_(2)//d_(1)`C. `d_(1)^(2)//d_(2)^(2)`D. `d_(2)^(2)//d_(2)^(2)` |
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Answer» Correct Answer - b `W_(1)=W_(2)` `mg_(1)=mg_(2)` `(GM_(1))/(R_(1)^(2))=(GM_(2))/(R_(2)^(2))` `(Gd_(1)(4pi)/(3)R_(1)^(3))/(R_(1)^(2))=(Gd_(2)(4pi)/(3)R_(2)^(3))/(R_(2)^(2))` `d_(1)R_(1)^(1)=d_(2)R_(2)^(2)` `(d_(1))/(d_(2))=(R_(2))/(R_(1))` |
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| 395. |
Kepler’s law of equal areas is an outcome of(A) conservation of energy(B) conservation of linear momentum (C) conservation of angular momentum (D) conservation of mass |
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Answer» (C) conservation of angular momentum |
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| 396. |
A system of binary stars of mass `m_(A)` and `m_(B)` are moving in circular orbits of radii `r_(A)` and `r_(B)` respectively. If `T_(A)` and `T_(B)` are at the time periods of masses `m_(A)` and `m_(B)` respectively thenA. `T_(A)gtT_(B)(if r_(A)gtr_(B))`B. `T_(A)gtT_(B)(if m_(A)gtm_(B))`C. `((T_(A))/(T_(B)))^(2)=((r_(A))/r_(B))^(3)`D. `T_(A)=T_(B)` |
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Answer» Correct Answer - `D` In binary star system angular velocity are equal hence the time period of both the stars are equal. `omega=(2pi)/(T_(A))=(2pi)/(T_(B))` `:. T_(A)=T_(B)` |
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| 397. |
A system of binary stars of mass `m_(A)` and `m_(B)` are moving in circular orbits of radii `r_(A)` and `r_(B)` respectively. If `T_(A)` and `T_(B)` are at the time periods of masses `m_(A)` and `m_(B)` respectively thenA. `T_(A)gt T_(B)(if r_(A) gt r_(B))`B. `T_(A)gt T_(B) (if m_(A)gt m_(B))`C. `((T_(A))/(T_(B)))^(2) = ((r_(A))/(r_(B)))^(3)`D. `T_(A)=T_(B)` |
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Answer» Correct Answer - D In binary star system angular velocity are qual hence the time period of both the stars are equal. `omega = (2pi)/(T_(A)) = (2pi)/(T_(B))` `therefore T_(A) = T_(B)` |
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| 398. |
If two planets of radii `R_(1) and R_(2)` have densities `d_(1) and d_(2)`, then the ratio of their respective acceleration due to gravity isA. `r_(1)d_(1):r_(2)d_(2)`B. `r_(1)d_(2):r_(2)d_(1)`C. `r_(1)^(2)d_(1):r_(2)^(2)d_(2)`D. `r_(1):r_(2)` |
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Answer» Correct Answer - A We know that `g=(GM_(e ))/(R_(e)^(2)) and d=(M_(e ))/(4/3pi R_(e)^(3))` `rarr g=(4pi GdR_(e))/(3) rarr g prop R_(e )` `therefore (g_(1))/(g_(2))=(d_(1)r_(1))/(d_(2)r_(2))` |
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| 399. |
A body of mass `2 kg` is moving under the influence of a central force whose potential energy is given by `U = 2r^(3) J`. If the body is moving in a circular orbit of `5 m`, its energy will beA. `625 J`B. `250 J`C. `500 J`D. `125 J` |
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Answer» Correct Answer - A `F = - (dU)/(dr) = - 6r^(2)` Now `(m nu^(2))/(r) =6r^(2)` `:. (1)/(2) m nu^(2) = 3r^(3)` `E = K + U = 3r^(3) + 2r^(3) = 5r^(3)` At `r = 5m, E = 625 J` |
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| 400. |
A binary star system consists of two stars of masses `M_(1)` and `M_(2)` revolving in circular orbits of radii `R_(1)` and `R_(2)` respectively. If their respective time periods are `T_(1)` and `T_(2)`, thenA. `T_(1)gt T_(2)` if `R_(1)gt R_(2)`B. `T_(1)gt T_(2)` if `M_(1)gt M_(2)`C. `T_(1)=T_(2)`D. `(T_(1))/(T_(2))=((R_(1))/(R_(2)))^(3//2)` |
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Answer» Correct Answer - C In a binary star system, there are two stars moving under their mutual gravitational force of attraction. Their angular velocities are equal and hence their periodic times are also equal. `[because omega = (2pi)/(T)]` `therefore T_(1)=T_(2)` |
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