InterviewSolution
Saved Bookmarks
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.
| 1951. |
The weight of a body is 50 N. What is its mass? `(g = 9 .8 m//s^2)` |
|
Answer» Correct Answer - 5.102 kg |
|
| 1952. |
If the weight of a body on the earth is 6 N, what will it be on the moon? |
|
Answer» Correct Answer - About 1 N |
|
| 1953. |
Which is more fundamental the mass of a body or its weight ? Why ? |
|
Answer» Correct Answer - Mass |
|
| 1954. |
The field which artifical satellite are useful stelites are useful for practical purpose isA. telecommunicationB. geophysicsC. meterologyD. all of these |
|
Answer» Correct Answer - D Since 19657 advance in technology have enabled many countries includeing india to launch artifical earth satellites for practical use in fields like elecommunicatin geiphysics and meterorology |
|
| 1955. |
What would be the angular speed of earth, so that bodies lying on equator may experience weightlessness ? ( g = `10m//s^(2)` and radius of earth = 6400 km)A. `1.25xx10^(-3) rad//sec`B. `1.56xx10^(-3) rad//sec`C. `1.25xx10^(-1) rad//sec`D. `1.56 rad//sec` |
|
Answer» Correct Answer - A |
|
| 1956. |
What would be the angular speed of earth, so that bodies lying on equator may experience weightlessness ? ( g = `10m//s^(2)` and radius of earth = 6400 km)A. `1/800 rad//s`B. `1/400 rad//s`C. `1/600 rad//s`D. `1/100 rad//s` |
|
Answer» Correct Answer - A |
|
| 1957. |
What woluld be the length of a day. If angular speed of earth is increased such that bodies typing on the equator by off? |
|
Answer» Correct Answer - `1.3` hr. |
|
| 1958. |
A lauching vehicle carrying an artificial satellite of mass `m` is set for launch on the surface of the earth of mass `M` and radius `R`. If the satellite intended to move in a circular orbit of radius `7R`, the minimum energy required to be spent by the launching vehicle on the satellite is |
|
Answer» Here `r=R+h=7R` Orbital velocity `V_(0)=sqrt((GM)/r)` `TE` of satellite in its orbit `=PE+KE` `=-(GMm)/r+1/2mV_(0)^(2)` `=-(GMm)/r+1/2m((GM)/r)` `=-(GMm)/(7R)+1/2m((GM)/(7R))=-(GMm)/(14R)` `TE` of satellite on the earth `=-(GMm)/R` If the minimum energy to launch the satellite in to its orbit is `E_(min)`, then, `-(GMm)/R+E_(min)=-(GMm)/(14R)rArr E_(min)=(13GMm)/(14R)` |
|
| 1959. |
A lauching vehicle carrying an artificial satellite of mass `m` is set for launch on the surface of the earth of mass `M` and radius `R`. If the satellite intended to move in a circular orbit of radius `7R`, the minimum energy required to be spent by the launching vehicle on the satellite isA. `(GMm)/(R )`B. `-(13 GM m)/(14R )`C. `(GMm)/(7R)`D. `(GMm)/(14R)` |
|
Answer» Correct Answer - B The energy of artificial satellite at the surface of the earth `E_(1)=(GMm)/(R )` when the satellite is intended to move in a circular orbit of radius 7R then energy of artifical satellite `E_(2)=-1/2 (GMm)/(7R)` The minimum enegy required `E=E_(1)-E_(2)=-(GMm)/(R )+1/2(GMm)/(7R)` `=(-14GMm+GMm)/(14R)=(-13 G?Mm)/(14 R)` |
|
| 1960. |
A communication satellite is put in parking orbit. What is the time taken by a wave to go to satellite and come back to earth in its checking mode. |
|
Answer» Correct Answer - `0.25 s` |
|
| 1961. |
The mean radius of the orbit of a satellite is `4` times as great as that of the parking orbit of the earth. Then its period of revolution around the earth isA. `4` daysB. `8` daysC. `16` daysD. `96` days |
|
Answer» Correct Answer - B `v_(0)=sqrt((GM)/(R+h))` and `T=(2pir)/(v_(0))` |
|
| 1962. |
A body is orbiting around earth at a mean radius which is two times as greater as the parking orbit of a satellite, the period of body isA. 4daysB. 16 daysC. `2sqrt(2)`daysD. 64 days |
|
Answer» Correct Answer - C From kepler third law of planetary motion `T^(2) prop r^(3)` Given `T_(1)=1` day (geosatationary) `therefore (T_(1))/(T_(2))^(2)=(r_(1))/(r_(2))^(3)` `rarr T_(2)^(2)=(r_(2))/(r_(1))^(3)xxT_(1)^(2)=(2r)/(R )^(3)xx1=8` `rarr T_(2)=2sqrt(2)`days |
|
| 1963. |
In previous problem what is the angular speed of `S_(2)` is the observed by an astronaut in `S_(1)` when they are closet :A. `(pi)/(2)`B. `(pi)/(3)`C. `(pi)/(4)`D. `(pi)/(6)` |
|
Answer» Correct Answer - B |
|
| 1964. |
In previous problem, find the work done by external agent.A. `40J`B. `-40J`C. zeroD. `+10J` |
|
Answer» Correct Answer - B Since, particle is shifted slowly. So, change in kinetic energy is zero. According to work-energy theorem `W +W_(ext) = 0` `W_(ext) =- W =- 40 J` |
|
| 1965. |
In the previous problem, the minimum speed with which the body must be thrown from the surface of the earth so as to reach a height of `R//4` isA. `sqrt((gR)/2)`B. `sqrt(gR)`C. `sqrt((gR)/5)`D. `sqrt((2gR)/5)` |
|
Answer» Correct Answer - D By energy conservation `K_(1)+U_(1)=K_(2)+U_(2)` `1/2mv^(2)-mgR=0-4/5 mgR` `v=sqrt((2gR)/5)` |
|
| 1966. |
The figure shows four arrangement of three particles of equal masses. Rank the arrangement according to the magnitude of the net gravitational force on the particle m, greatest first. A. `1`, tie of `2` and `4`, then `3`B. `1,4,3,2`C. `2,3,4,1`D. `4,3,1,2` |
|
Answer» Correct Answer - A In arrangement 1, both forces act in the same direction. In arrangement 3, both forces act in opposite direction. This alone decides in favour of option (a), |
|
| 1967. |
Out of mass and weight of a body, which is scaler and which is vector ? |
| Answer» Mass is scaler and weight is vector. | |
| 1968. |
The orbit of a geostationary satellite isA. very close to the surface of the earthB. in any plane around the earthC. in the equatorial plane of the earthD. any of the above |
| Answer» Correct Answer - c | |
| 1969. |
If `S_1` is surface satellite and `S_2` is geostationary satellite, with time periods `T_1` and `T_2`, orbital velocities `V_1` and `V_2`,A. `T_(1) gt T_(2), v_(1) gt v_(2)`B. `T_(1) gt T_(2), v_(1) lt v_(2)`C. `T_(1) lt T_(2), v_(1) lt v_(2)`D. `T_(1) lt T_(2), v_(1) gt v_(2)` |
| Answer» Correct Answer - d | |
| 1970. |
A planet moves around the sun. at a given point `P`, it is closest from the sun at a distance `d_(1)`, and has a speed `V_(1)`. At another point `Q`, when it is farthest from the sun at a distance `d_(2)`, its speed will beA. `(d_(1)^(2)v_(1))/(d_(2)^(2))`B. `(d_(2)v_(1))/(d_(1))`C. `(d_(1)v_(1))/(d_(2))`D. `(d_(2)^(2)v_(1))/(d_(1)^(2))` |
|
Answer» Correct Answer - c `mv_(1)d_(1)=mv_(2)d_(2) rArr v_(2)=(d_(1)v_(1))/(d_(2))` |
|
| 1971. |
A planet moves around the sun. at a given point `P`, it is closest from the sun at a distance `d_(1)`, and has a speed `V_(1)`. At another point `Q`, when it is farthest from the sun at a distance `d_(2)`, its speed will beA. `(d_(1)^(2) v_(1))/d_(2)^(2)`B. `(d_(2)v_(1))/d_(1)`C. `(d_(1)v_(1))/d_(2)`D. `(d_(2)^(2) v_(1))/d_(1)^(2)` |
|
Answer» Correct Answer - C |
|
| 1972. |
A planet moves around the sun. at a given point `P`, it is closest from the sun at a distance `d_(1)`, and has a speed `V_(1)`. At another point `Q`, when it is farthest from the sun at a distance `d_(2)`, its speed will beA. `(v_(1))/(d_(1)).d_(2)`B. `v_(1). (d_(1))/(d_(2))`C. `v_(1).sqrt((d_(2))/(d_(1))`D. `v_(1).sqrt((d_(1))/(d_(2)))` |
|
Answer» Correct Answer - B |
|
| 1973. |
A tunnel is made inside earth passing throgh centre of earth. A particle is dropped from the surface of earth. Select the correct statement : A. Kinetic energy of particle is maximum at centre and its potential energ is zero at centerB. Velocity of particle is proportional to x [where x is distance of particle from center of earthC. Kinetic energy of particle is maximum when it reaches on the other side of tunnelD. Kinetic energy of particle is maximum at center |
|
Answer» Correct Answer - D |
|
| 1974. |
Two particle A and B (of masses m and `4m`) are released from rest in the two tunnels as shown in the figure-6.93. Which particle will cross the equatorial plane first? A. AB. BC. Both simultaneouslyD. Data insufficient |
|
Answer» Correct Answer - C |
|
| 1975. |
The gravitational force of attraction between two objects is x. Keeping the masses of the objects unchanged, if the distance between the objects is halved, then the magnitude of gravitational force between them will become:A. x/4B. x/2C. 2xD. 4x |
|
Answer» Correct Answer - d |
|
| 1976. |
In the relation `F=G M m//d^(2)`, the quantity GA. depends on the value of g at the place of observationB. is used only when the earth is one of the two massesC. is the greatest on the surface of the earthD. is of the same value irrespective of the place of observation |
|
Answer» Correct Answer - d |
|
| 1977. |
An object is put in three liquids having different densities, one.The object floats with `1/9,2/11, and 3/7` parts of its volume outside the surface of liquid of densities,`d_(1),d_(2) and d_(3)` respectively.Which of the following is the correct order of the densities of the three liquids ?A. `d_(1)gtd_(2)gtd_(3)`B. `d_(2)gtd_(3)gtd_(1)`C. `d_(1)ltd_(2)ltd_(3)`D. `d_(3)gtd_(2)gtd_(1)` |
|
Answer» Correct Answer - C |
|
| 1978. |
Consider the following equation of motion and rewrite them for a free fall body v = u + at s = ut +1/2 at2 |
|
Answer» For a free fall body u = 0; a = g and s i) v = u + at ⇒ v = 0 + gt ⇒ v = gt ii) s = ut + 1/2 at2 ⇒ h = 0 + 1/2 gt2 ⇒ h = 1/2 gt2 |
|
| 1979. |
The maximum vertical distance through which a person can jump on the earth (1/2)m. Then the maximum vertical distance through which he can jumb on the moon is [Initial velocity is same is both cases and acceleration due to gravity on the moon is `(1//6)^("th")` that on the earth]A. 2 mB. 12 mC. 6 mD. 3 m |
|
Answer» Correct Answer - d `PE_(2)=PE_(1)` `mg_(1)h_(2)=mg_(1)h_(1)` `(1)/(6)gh_(2)=gxx(1)/(2)` `h_(2)=(6)/(2)=3m` |
|
| 1980. |
Find the value of acceleration due to gravity in a mine at a depth of 80 km from the surface of the earth . Radius of the earth = 6400 km . |
|
Answer» Correct Answer - `9.68 ms^(-2)` |
|
| 1981. |
If the earth suddenly shrinks (without changing mass) to half of its present radius, the acceleration due to gravity will beA. g/2B. 4gC. g/4D. 2g |
|
Answer» Correct Answer - b `g=(GM)/(R^(2))` |
|
| 1982. |
If the earth suddenly shrinks (without changing mass) to half of its present radius, the acceleration due to gravity will beA. `g//2`B. `4 g`C. `g//4`D. `2g` |
|
Answer» Correct Answer - B |
|
| 1983. |
What would be the accelearation due to gravity at a depth of 200 km from ythe earth surface assuming that earth has uniform denisty [Take ,R =6400 km]A. 4.37 `ms^(-2)`B. 6.737 `ms^(-2)`C. 5.709 `ms^(-2)`D. 4.751 `ms^(-2)` |
|
Answer» Given depth d= 2000km =`=2x10^(6)m` Radius of earth R=6400 km `=6.4xx10^(6)`m The acceleration due to gravity at a point at a depth d from the earth surface is given by `g_(d)=g[1-(d)/(R )]` `therefore g_(d)=9.8[(1-(2xx10^(6))/(6.4xx10^(6)))]` `=(9.8xx4.4)/(6.4)=6.737 ms^(-2)` |
|
| 1984. |
Three particle each of mass `m`, are located at the vertices of an equilateral triangle of side a. At what speed must they move if they all revolve under the influence of their gravitational force of attraction in a circular orbit circumscribing the triangle while still preserving the equilateral triangle ? A. `sqrt((Gm)/(a))`B. `sqrt((2Gm)/(a))`C. `sqrt((3Gm)/(a))`D. `sqrt((4Gm)/(a))` |
|
Answer» Correct Answer - A From the figure, `F_(A)=F_(AB)+F_(AC)=2[(GM^(2))/(a^(2))]cos 30^(@) =[(Gm^(2))/(a^(2)),sqrt(3)]` Also, `r=(a)/(sqrt(3))`, Now, `(mv^(2))/(r)=F|F_(A)|or (mv^(2)sqrt(3))/(a)=(Gm)/(a)=(Gm)/(a^(2))sqrt(3)` `v=sqrt((Gm)/(a))`. |
|
| 1985. |
Assuming the earth to be a sphere of uniform density, the acceleration due to gravityA. at a point outside the earth is inversely proportional to the square of its distance from the centreB. at a point outside the earth is inversely proportional to its disatance from the centreC. at any point inside is zeroD. at a point inside is proportional to its distance from the centre. |
|
Answer» Correct Answer - A::D `g=(Gm)/(R^(3))xxr` for `r lt R` `g = (Gm)/(r^(2))` for `r gt R` so choice (A) and (D) are correct |
|
| 1986. |
A monkey is descending from the branch of a tree with constant acceleration. If the breaking strength is 75% of the weight of the monkey, then find the maximum acceleration with which the monkey can slide down without breaking the branch. |
|
Answer» The tension in the branch of the tree is T = mg - ma But, T = 75% of the weight = 75/100 x mg = 3/4 mg Hence, 3/4 mg = mg - ma or, a = g - 3/4 g = g/4 = 9.8/4 = 2.45 ms-2 |
|
| 1987. |
Suppose the gravitational force varies inversely as the `n^(th) `power of distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to-A. `R^(((n+1)/(2)))`B. `R^(((n-1)/(2)))`C. `R^(n)`D. `R^(((n-2)/(2)))` |
|
Answer» Correct Answer - A `F=(GMm)/(r^(n))=momega^(2)r` `impliesomega=sqrt((GM)/(r^(n+1)))impliesT=(2pi)/(omega)=(r^((n+1)/(2)))` |
|
| 1988. |
A uniform thin rod of mass `m` and length `R` is placed normally on surface of earth as shown. The mass of earth is `M` and its radius `R`. Then the magnitude of gravitational force exerted by earth on the rod is A. `(GMm)/(4R^(2))`B. `(GMm)/(2R^(2))`C. `(4GMm)/(9R^(2))`D. `(GMm)/(8R^(2))` |
|
Answer» Correct Answer - B |
|
| 1989. |
The gravitational force between two objects is proportional to `1//R` (and not as `1//R^(2)`) where `R` is separation between them, then a particle in circular orbit under such a force would have its orbital speed `v` proportional toA. `1/(R^(2))`B. `R^(0)`C. `R^(1)`D. `1/R` |
|
Answer» Correct Answer - B `F_(G)=(GMm)/R` (as `F_(G)prop1/R` given) So `(mv^(2))/R=(GMm)/Rimpliesv^(2)propR^(0)` |
|
| 1990. |
The orbital speed of Jupiter isA. Greater than the orbital speed of earthB. Less than the orbital speed of earthC. Equal to the orbital speed of earthD. Zero |
|
Answer» Correct Answer - B |
|
| 1991. |
The gravitational force between two objects is proportional to `1//R` (and not as `1//R^(2)`) where `R` is separation between them, then a particle in circular orbit under such a force would have its orbital speed `v` proportional toA. `1//R^(2)`B. `R^(0)`C. `R^(1)`D. `1//R` |
|
Answer» Correct Answer - b `F=(K)/(R)=(Mv^(2))/(R)" hence "v prop R^(0)` |
|
| 1992. |
If `a` and `b` are the nearest and farthest distances of a planet from the sun and the planet is revolving in an elliptical orbit, then square of the time period of revolution of that planets is directly proportional toA. `a^(3)`B. `b^(3)`C. `(a+b)^(3)`D. `(a-b)^(3)` |
|
Answer» Correct Answer - b `T^(2) prop ((a+b)/(2))^(3) prop (a+b)^(3)`. |
|
| 1993. |
Time period of pendulum, on a satellite orbiting the earth, isA. TB. 2TC. zeroD. infinite |
|
Answer» Correct Answer - D `g=0 " " therefore T = 2pi sqrt((l)/(g))to oo` |
|
| 1994. |
Two satellites of mass m1 and m2 (m1 > m2 ) are revolving round the earth in circular orbits of r1 and r2 (r1 > r2 ) respectively. Which of the following statement is true regarding their speeds v1 and v2 ……..(a) v1 = v2(b) v1 < v2(c) v1 > v2(d) \(\frac{v_1}{r_1}\) = \(\frac{v_2}{r_2}\) |
|
Answer» Correct answer is (b) v1 < v2 |
|
| 1995. |
An artificial satellite is revolving round the Earth in a circular orbit, its velocity is half the escape velocity. Its height from the Earth surface is …. km.(a) 6400 (b) 12800 (c) 3200 (d) 1600 |
|
Answer» Correct answer is (a) 6400 |
|
| 1996. |
A planet is revolving in an elliptical orbit about the sun. Its closest distance is R. If `K_(1) and K_(2)` are the maximum and minimum kinetic energies of the planet, `K_(1)//K_(2)`=A. `R//r`B. `sqrt(R//r)`C. `R^(2)//r^(2)`D. 1 |
|
Answer» Correct Answer - c `(k_(1))/(k_(2))=((1)/(2)mv_(1)^(2))/((1)/(2)mv_(2)^(2))=((v_(1))/(v_(2)))^(2)` `(k_(1))/(k_(2))=((R)/(r))^(2)` |
|
| 1997. |
As astronaut orbiting the earth in a circular orbit 120 km above the surface of Earth, gently drops a spoon out of space-ship. The spoon will ……….(a) fall vertically down to the Earth (b) move towards the moon (c) will move along with space-ship (d) will move in an irregular way then fall down to Earth. |
|
Answer» (c) will move along with space-ship The velocity of the spoon will be equal to the orbital velocity when dropped out of the spaceship |
|
| 1998. |
A satellite revolves around the earth in an elliptical orbit. Its speed isA. is the same at all points in the orbitB. is greatest when it is closest to the earthC. is greatst when it is farthest from the earthD. goes on increasing or decreasing continuously depending upon the mass of the satellite |
|
Answer» Correct Answer - b `v=R_(e)sqrt((g)/(R_(e)+h))` |
|
| 1999. |
orbital velocity of an artificial satellite does not depend upon ....... (a) mass of Earth (b) mass of satellite (c) radius of Earth(d) acceleration due to gravity |
|
Answer» (b) mass of satellite |
|
| 2000. |
A planet is revolving in an elliptical orbit around the sun. Its closest distance from the sun is r and the farthest distance is R. If the velocity of the planet nearest to the sun be v and that farthest away from the sun be V. then v/V isA. `x//y`B. `y//x`C. `x^(2)//y^(2)`D. `y^(3)//x^(2)` |
|
Answer» Correct Answer - a `v_("max")r_("min")=v_("min")r_("max")` `ux=vy` `therefore" "(v)/(u)=(x)/(y)` |
|