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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.
| 1401. |
Peridic-time of satellite revolving around the earth is `(rho` density of earth) .A. Proportional to `(1)/(rho)`B. Proportiona`(1)/sqrtrho`C. Proportional `rho`D. does not depend on `rho` |
| Answer» Correct Answer - B | |
| 1402. |
A system consists of a thin ring of radius R and a very long uniform wire oriented along axis of the ring with one of its ends coinciding with the centre of the ring. If mass of ring be M and mass of wire be `lambda` per unit length, calculate interaction force between the ring and the wire. |
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Answer» Correct Answer - `[(GMlambda)/(R)]` |
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| 1403. |
A body of mass `m` is kept at a small height `h` above the ground. If the radius of the earth is R and its mass is M, the potential energy of the body and earth system (with `h=infty` being the reference position) isA. `(GMm)/(R)+mgh`B. `(-GMm)/(R)+mgh`C. `(GMm)/(R)-mgh`D. `(-GMm)/(R)-mgh` |
| Answer» Correct Answer - B | |
| 1404. |
Two identical spherical masses are kept at some distance. Potential energy when a mass `m` is taken from the surface of one sphere to the otherA. increases continuouslyB. decreases continuouslyC. first increases then decreasesD. first decreases then increases |
| Answer» Correct Answer - C | |
| 1405. |
Two identical spherical masses are kept at some distance. Potential energy when a mass `m` is taken from the surface of one sphere to the otherA. increases continuouslyB. decreases continuouslyC. first increases,then decreasesD. first decreases,then increases |
| Answer» Correct Answer - C | |
| 1406. |
Let `V_G` and `E_G` denote gravitational potential and field respectively, then choose the wrong statement.A. `V_(G)=0,E_(G)=0`B. `V_(G)ne0,E_(G)=0`C. `V_(G)=0,E_(G)ne0`D. `V_(G)ne0,E_(G)ne0` |
| Answer» Correct Answer - C | |
| 1407. |
Tidal waves in the sea are primarily due toA. The gravitational effect of the sun on the earthB. the gravitational effect of the moon on the earthC. The rotation of the earthD. The atmospheric effect of the earth it self |
| Answer» Correct Answer - B | |
| 1408. |
The intensity of the gravitational field of the earth is maximum atA. centre of earthB. equatorC. polesD. same everwhere |
| Answer» Correct Answer - C | |
| 1409. |
The work done by an external agent to shift a point mass from infinity to the centre of the earth is `W`. Then choose the correct relation.A. `=0`B. `gt 0`C. `lt 0`D. `le 0` |
| Answer» Correct Answer - C | |
| 1410. |
The work done by an external agent to shift a point mass from infinity to the centre of the earth is `W`. Then choose the correct relation.A. `W=0`B. `Wgt0`C. `Wlt0`D. `Wle0` |
| Answer» Correct Answer - C | |
| 1411. |
During the journey of space ship from earth to moon and back, the maximum fuel is consumed :-A. Against the gravitation of earth in retum journeyB. Against the gravitation of earth in onward journeyC. Against the gravitation of moon while reaching the moonD. None of the above |
| Answer» Correct Answer - B | |
| 1412. |
A person brings a mass `2kg` from `A` to `B`. The increase in kinetic energy of mass is `4J` and work done by the person on the mass is `-10J`. The potential difference between `B` and `A` is ......`J//kg`A. `4J//kg`B. `7J//kg`C. `-3J//kg`D. `-7J//kg` |
| Answer» Correct Answer - D | |
| 1413. |
A person brings a mass `2kg` from `A` to `B`. The increase in kinetic energy of mass is `4J` and work done by the person on the mass is `-10J`. The potential difference between `B` and `A` is ......`J//kg`A. `4`B. `7`C. `-3`D. `-7` |
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Answer» Correct Answer - D `W=m(DeltaV)+DeltaKE` |
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| 1414. |
Energy required to move a body of mass m from an orbit of radius 2R to 3R isA. `(GMm)/(12R)`B. `(GMm)/(3R^(2))`C. `(GMm)/(8R)`D. `(GMm)/(6R)` |
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Answer» Correct Answer - A `W=T.E_(2)-T.E_(1)=(GMm)/2(1/(r_(1))-1/(r_(2)))` |
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| 1415. |
The earth moves around the Sun in an elliptical orbit as shown in Fig. The ratio `OA//OB=x`. The ratio of the speed of the earth at `B` to that at `A` is nearly A. `sqrt(x)`B. `x`C. `xsqrt(x)`D. `x^(2)` |
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Answer» Correct Answer - B from conservation of angular momentum `mvr=const., v_(1)r_(1)=v_(2)r_(2)` |
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| 1416. |
The earth moves around the Sun in an elliptical orbit as shown in Fig. The ratio `OA//OB=x`. The ratio of the speed of the earth at `B` to that at `A` is nearly A. `sqrtx`B. `x`C. `xsqrtx`D. `x^(2)` |
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Answer» Correct Answer - B Applying conservation of angular momentum at position `A` and `B` `mv_(A)xxOA=mv_(B)xxOB` Hence, `(v_(B))/(v_(A))(OA)/(OB)=x` |
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| 1417. |
Figure shows two shells of masses `m_(1)` and `m_(2)`. The shells are concentric. At which point, a particle of mass `m` shall experience zero force? A. `A`B. `B`C. `C`D. `D` |
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Answer» Correct Answer - D the gravitational field intensity at a point inside the spherical shell is zero. |
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| 1418. |
Figure shows two shells of masses `m_(1)` and `m_(2)`. The shells are concentric. At which point, a particle of mass `m` shall experience zero force? A. `A`B. `B`C. `C`D. `D` |
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Answer» Correct Answer - D The gravitational field intensity at a point inside the spherical shell is zero. |
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| 1419. |
Radius of earth is ………………….. A) 6.4 × 106 km B) 6.4 × 106 m C) 6.4 × 106 cm D) 6.4 × 106 mm |
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Answer» B) 6.4 × 106 m |
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| 1420. |
Acceleration of bodies on earth surface is A) 981 m/s2B) 981 cm/s2 C) 9.81 cm/s2 D) 0.98 cm/s2 |
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Answer» B) 981 cm/s2 |
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| 1421. |
Acceleration of moon towards earth is A) 0.27 cm/s2 B)1.63 m/s2 C) 9.8 m/s2 D) 27.4 m/s2 |
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Answer» A) 0.27 cm/s2 |
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| 1422. |
The value of universal gravitational constant is found by …………. A) Henry Cavendish B) Berzelius C) Avagadro D) Boyle |
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Answer» A) Henry Cavendish |
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| 1423. |
Binding energy of satellite is `4xx10^(8)J`. Its potential energy isA. `-4xxx10^(8) j`B. `8xx10^(8) j`C. `8xx10^(8) j`D. `4xx10^(8) J` |
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Answer» Binding energy `E_(B)=4xx10^(8) j` `therefore` PE of satellite `=-2E_(B)=-2xx4xx10^(8)=-8xx10^(8) J` |
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| 1424. |
Assuming the radius of the earth to be `6.4xx10^(6)` m calculate the time period T of a satellite for equatorial orbit at `1.4xx10^(3)` km above the surface of the earth and the speed of the satellite in this orbit ?A. 6831 and 7200 `ms^(-1)`B. 6850 s and 7151 `ms^(-1)`C. 68321 s and 7190 `ms^(-1)`D. 6850 and 7400 `ms^(-1)` |
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Answer» Given radius of earth `R_(e )=6.4xx10^(3) m` height `h=1.4xx10^(3)km` Mass of the earth `M_(e )=5.98xx10^(24) kg` Gravitational constant `G=6.67xx10^(11) Nm^(2)(kg)^(2)` `r=R_(e )+h` `=6.4xx10^(6)+1.4xx10^(6)m=7.8xx10^(6)` m Time period `T=[(4 pi^(2)r^(12))/(Gm_(e ))]^(1//2)` `=[(4xx(3.14)^(2))xx(7.8xx10^(6))/(6.67xx10^(-11)xx5.98xx10^(24))]=6853.4` s speed of satellite `v=sqrt(GM_(e ))/(r )` `=[(6.67xx10^(11)xx5.98xx10^(24))/(7.8xx10^(6))]^(1//2) =7151 ms^(-1)` |
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| 1425. |
Assuming the radius of the earth to be `6.5 xx 10^(6)` m. What is the time period T and speed of satellite for equatorial orbit at `1.4 xx 10^(3)` km above the surface of the earth.A. `6831 s and 7174 " ms"^(-1)`B. `34155 s and 3204 " ms"^(-1)`C. `6831 s and 2144 " ms"^(-1)`D. `2431 s and 3514 "ms"^(-1)` |
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Answer» Correct Answer - A `r=R_("earth")+h=(6.4xx10^(6)+14xx10^(6))"m"` `rArr r=7.8 xx 10^(6)"m"` `T=[(4pi^(2)r^(3))/(GM_("earth"))]=6831s` Speed of satellite, `v=sqrt((GM_("earth"))/(r))=7174 " ms"^(-1)`. |
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| 1426. |
A small asteroid is approaching a planet of mass M and radius R from a large distance. Initially its velocity (u) is along a tangent to the surface of the planet. It fall on the surface making an angle of `30^(2)` with the vertical. Calculate u. |
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Answer» Correct Answer - `u=sqrt((2GM)/(3R))` |
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| 1427. |
An asteroid of mass `m` is approaching earth, initially at a distance `10R_(E)` with speed `v_(i)`. It hits earth with a speed `v_(f)` (`R_(E)` and `M_(E)` are radius and mass of earth),. ThenA. `v_(f)^(2)=v_(i)^(2)+(2Gm)/(R_(E))(1+(1)/(10))`B. `v_(f)^(2)=v_(i)^(2)+(2GM_(E))/(R_(E))(1+(1)/(10))`C. `v_(f)^(2)=v_(i)^(2)+(2GM_(E))/(R_(E))(1-(1)/(10))`D. `v_(f)^(2)=v_(i)^(2)+(2Gm)/(R_(E))(1-(1)/(10))` |
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Answer» Correct Answer - C (c ) Initial energy of the asteroid is `E_(i)=K_(i)+U_(i)=(1)/(2)mv_(i)^(2)-(GM_(E)m)/(10R_(E))` Final energy of the asteroid is `E_(f)=(1)/(2)mv_(f)^(2)-(GM_(E)m)/(R_(E))` According to law of conservation of energy, `E_(i)=E_(f)` `(1)/(2)mv_(i)^(2)-(GM_(E)m)/(10R_(E))=(1)/(2)mv_(f)^(2)-(GM_(E)m)/(R_(E))` `v_(f)^(2)-(2GM_(E))/(R_(E))=v_(i)^(2)-(2GM_(E))/(10R_(E))` `v_(f)^(2)=v_(i)^(2)+(2GM_(E))/(R_(E))(1-(1)/(10))` |
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| 1428. |
An asteroid of mass `m` is approaching earth, initially at a distance `10R_(E)` with speed `v_(i)`. It hits earth with a speed `v_(f)` (`R_(E)` and `M_(E)` are radius and mass of earth),. ThenA. `v_(f)^(2)=v_(i)^(2)+(2GM)/(R_(E)) (1+1/10)`B. `v_(f)^(2)=v_(i)^(2)+(2Gm_(E))/(R_(E)) (1+1/10)`C. `v_(f)^(2)=v_(i)^(2)+(2Gm_(E))/(R_(E)) (1-1/10)`D. `v_(f)^(2)=v_(i)^(2)+(2GM)/(R_(E)) (1-1/10)` |
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Answer» Correct Answer - C Initial energy of the asteroid is `E_(i)=K_(i)+U_(i)=1/2mv_(i)^(2)-(GM_(E)m)/(10R_(E))` Final energy of the asteroid is `E_(f)=1/2mv_(f)^(2)-(GM_(E)m)/(R_(E))` According to law of conservation of energy, `E_(i)=E_(f)` `1/2mv_(i)^(2)-(GM_(E)m)/(10R_(E))=1/2mv_(f)^(2)-(GM_(E)m)/(R_(E))` `v_(f)^(2)-(2GM_(E))/(R_(E)) =v_(i)^(2)-(2GM_(E))/(10R_(E))` `v_(f)^(2)=v_(i)^(2)+(GM_(E))/(R_(E))(1-1/10)` |
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| 1429. |
An asteroid of mass `m` is approaching earth, initially at a distance `10R_(E)` with speed `v_(i)`. It hits earth with a speed `v_(f)` (`R_(E)` and `M_(E)` are radius and mass of earth),. ThenA. `v_(f)^(2)=v_(i)^(2)+(2Gm)/(M_(e)R)(1-1/10)`B. `v_(f)^(2)=v_(i)^(2)+(2Gm_(e))/(R_(e))(1+1/10)`C. `v_(f)^(2)=v_(i)^(2)+(2Gm_(e))/(R_(e))(1-1/10)`D. `v_(f)^(2)=v_(i)^(2)+(2Gm)/(R_(e))(1-1/10)` |
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Answer» Correct Answer - C `1/2mv_(i)^(2)-(GM_(e)m)/((10 R_(e)))=1/2mv_(f)^(2)-(GM_(e)m)/(( R_(e)))` `v_(f)^(2)=v_(i)^(2)+(2GM_(e))/(R_(e))(1-1/10)` |
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| 1430. |
The point where total weight appears to act is called A) centre of gravity B) centre of equilibrium C) mid point D) equilibrium point |
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Answer» A) centre of gravity |
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| 1431. |
An asteroid was fast approaching the earth. Scientists fired a rocket which hit the asteroid at a distance of 5 R from the centre of the earth (R = radius of the earth). mmediately after the hit the asteroid’s velocity `(V_(0))` was making an angle of `theta=30^(@)` with the line joining the centre of the earth to the asteroid. The asteroid just grazed past the surface of the earth. Find `V_(0)` [Mass of the earth = M] |
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Answer» Correct Answer - `V_(0) =sqrt(32/105 (GM)/R)` |
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| 1432. |
If the line through the centre of gravity falls inside the base of the object, then the object will be A) weightless B) unstable C) stable D) freely falling |
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Answer» Correct option is C) stable |
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| 1433. |
If the line through the centre of gravity falls out side the base of the object, then the object will be A) weightless B) unstable C) stable D) freely falling |
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Answer» Correct option is B) unstable |
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| 1434. |
A rocket is fired from the earth to the moon. The distance between the earth and the moon is r and the mass of the earth is 81 times the mass of the moon. The gravitational force on the rocket will be zero when its distance from the moon isA. `(r)/(20)`B. `(r)/(15)`C. `(r)/(10)`D. `(r)/(5)` |
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Answer» Correct Answer - c Let the rocket be at a distance x from the moon the gravitational force on it is zero. Its distance from earth `=r-x`. Gravitational force on rocket due to earth is `F_(e)=(GmM_(e))/((r-x)^(2))` Gravitational force on rocket due to moon is `F_(m)=(GmM_(m))/(x^(2))` `F_(e)=F_(m)` `(r-x)/(x)=sqrt((M_(e))/(M_(m)))=sqrt(81)=9` `x=(r)/(10)` |
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| 1435. |
The value of the acceleration due to gravity g on earth depends uponA. the mass of the earthB. the average radius of the earthC. the average density of the earthD. mass, density and radius of earth |
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Answer» Correct Answer - d `g=(GM)/(R^(2))` |
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| 1436. |
A package is released from an orbiting earth satellite by simply detaching it from the outer wall of the satellite. The package willA. go away from the earth and get lost in outer spaceB. fall to the surface of the earthC. continue moving along with the satellite in the same orbit and with the same velocityD. fall through a certain distance and then move in an orbit around the earth. |
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Answer» Correct Answer - c While inside the satellite, the instrument packet is orbiting the earth along with the satellite. When detached, it, therefore, has the necessary orbital velocity. |
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| 1437. |
The motion of a body with constant speed in circular path is known as …………… A) circular motion B) rotatory motion C) uniform circular motion D) none of these |
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Answer» C) uniform circular motion |
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| 1438. |
In uniform circular motion ……………….. is constant. A) speed B) velocity C) acceleration D) momentum |
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Answer» Correct option is A) speed |
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| 1439. |
Acceleration due to gravity is independent of A) speed of the object B) gravitational force on the object C) mass of the object D) mass of the earth |
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Answer» C) mass of the object |
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| 1440. |
The average position of weight distribution is simply the …………………… A) centre of velocity B) centre of momentum C) centre of gravity D) centre of acceleration |
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Answer» C) centre of gravity |
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| 1441. |
The acceleration which can change only the direction of velocity of a body is called ……………….. A) retardation B) deceleration C) negative acceleration D) centripetal acceleration |
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Answer» D) centripetal acceleration |
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| 1442. |
Acceleration due to gravity is independent of ………….. of the object.A) mass B) volume C) weight D) density |
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Answer» Correct option is A) mass |
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| 1443. |
The net force which can change only the direction of the velocity of a body is called …………………… A) gravitational force B) frictional forceC) centripetal force D) electrical force |
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Answer» C) centripetal force |
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| 1444. |
Why did the apple fall to the ground? |
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Answer» The apple fell to the ground due to the gravitational attraction of the earth. |
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| 1445. |
The moon takes ………….. for a complete revolution around the earth. A) 25 days B) 27.3 days C) 30 days D) 35 days |
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Answer» B) 27.3 days |
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| 1446. |
The direction of centripetal force is ……………….. the centre of the circle. A) towards B) away C) left side of D) right side of |
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Answer» Correct option is A) towards |
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| 1447. |
Does the body in uniform circular motion have an acceleration? What is the direction of acceleration? |
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Answer» The body in uniform circular motion have an acceleration, which is directed towards the centre of the circle, known as centripetal acceleration. |
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| 1448. |
Motion of moon around the earth is …………………. motion.A) vibrational B) circular C) rotational D) uniform circular |
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Answer» D) uniform circular |
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| 1449. |
Does the velocity of the body change in uniform circular motion? Why? |
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Answer» The velocity of the body in uniform circular motion is constant. If the velocity is not constant, the time period changes from time to time and it cannot be treated as uniform circular motion. |
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| 1450. |
Is the motion of the moon around the earth uniform motion? |
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Answer» Moon takes 27.3 days to complete one rotation around the earth, which does not change. Hence it is uniform motion. |
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