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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.
| 2251. |
The effect of rotation of the eath on the value of acceleration deu to gravity g isA. maximum at the equator and minimum at the polesB. minimum at the equator and maximum at the polesC. maximum at both polesD. minimum at both poles |
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Answer» Correct Answer - A As `g=g-omega^(2)R cos^(2) lambda` The latitude is denoted by `lambda` for poles `lambda=90^(@)` and for equator `lambda=0^(@)` (i) substituting `lambda=90^(@)` in the above expression we get `g_("poles")=g-omega^(2)R cos^(2) 90^(@) rarr g_("pole")=g` i.e there is no effect of rotational motion of the earth on the value of g at the poles (ii) substituting `lambda=0^(@)` in the above expression we get `g_("equator") =g-omega^(2) cos^(2) 0^(@)=g-omega^(2)R` i.e the effect of rotation of earth on the value of g at the equator is maximum |
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| 2252. |
The height at which the acceleration due to gravity becomes `g//9` in terms of R the radius of the earth isA. 2RB. `(R )/sqrtr(3)`C. `(R )/(2)`D. `sqrt(2)R` |
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Answer» Correct Answer - A `g=(GM)/(r+h)^(2)` acceleration due to gravity at height h ? `(g)/(g)=(GM)/(R^(2)).(R^(2))/(R+h)^(2)=g(r )/(R+h)^(2)` `rarr 1/9=(R )/(R+n)^(2) rarr (R )/(R+h)=1/3` `rarr 3R =R +h rarr 2R =h` |
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| 2253. |
The height at which the acceleration due to gravity becomes `(g)/(9)` (where g =the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is : |
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Answer» Given `g/9=g(R/(R+h))^(2)rArr R/(R+h)=1/3` `3R=R+hrArr 2R=h` |
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| 2254. |
The height at which the acceleration due to gravity becomes `(g)/(9)` (where g =the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :A. `(R )/(sqrt(2))`B. R/2C. `sqrt(2)R`D. 2R |
| Answer» Correct Answer - D | |
| 2255. |
Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is:A. zeroB. `-(4Gm)/(r)`C. `-(6Gm)/(r)`D. `-(9Gm)/(r)` |
| Answer» Correct Answer - D | |
| 2256. |
Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is:A. zeroB. `-(4Gm)/(r )`C. `-(6Gm)/(r )`D. `-(9Gm)/(r )` |
| Answer» Correct Answer - D | |
| 2257. |
A body of mass m is placed on the surface of earth. Find work required to lift this body by a height (i). `h=(R_(e))/(1000)` (ii). `h=R_(e)` |
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Answer» (i). `h=(R_(e))/(1000)` as `hltltR_(e)` so we apply `W_(ext)=mgh` `W_(ext)=(m)((GM_(e))/(R_(e)^(2)))((R_(e))/(1000))=(GM_(e)m)/(1000R_(e))` (ii). `h=R_(e)` in this case h is not very less than `R_(e)` so we cannot apply `DeltaU=mgh` `W_(ext)=U_(f)-U_(i)=m(V_(f)-V_(i))` `W_(ext)=m[(-(GM_(e))/(R_(e)+R_(e)))-(-(GM_(e))/(R_(e)))]` `W_(ext)=+(GM_(e)m)/(2R_(e))` |
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| 2258. |
A triple star system consists of two stars each of mass m in the same circular orbit about central star with mass `M=2xx10^(33)kg`. The two outer stars always lie at opposite ends of a diameter of their common circular orbit the radius of the circular orbit is `r=10^(11)` m and the orbital period each star is `1.6xx10^(7)s` [take `pi^(2)=10` and `G=(20)/(3)xx10^(-11)Mn^(2)kg^(-2)`] Q. The total mechanical energy of the system isA. `-1375/64xx10^(35) J`B. `-1375/64xx10^(38) J`C. `-1375/64xx10^(34) J`D. `-1375/64xx10^(37) J` |
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Answer» Correct Answer - B Total mechanical energy `=K.E.+P.E.` `=2(1/2mv^(2))-(2GMm)/r-(Gm^(2))/(2r)` `=m[(G(4M+m))/(4r)-(2GM)/r-(Gm)/r]` `=-(Gm)/r[M+m/4](20/3xx10^(-11))(11/8xx10^(30))xx1/(10^(11))` `(2xx10^(30)+11/32xx10^(30))=-1375/64xx10^(38)J`. |
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| 2259. |
A triple star system consists of two stars each of mass m in the same circular orbit about central star with mass `M=2xx10^(33)kg`. The two outer stars always lie at opposite ends of a diameter of their common circular orbit the radius of the circular orbit is `r=10^(11)` m and the orbital period each star is `1.6xx10^(7)s` [take `pi^(2)=10` and `G=(20)/(3)xx10^(-11)Mn^(2)kg^(-2)`] Q. The mass m of the outer stars is:A. `16/15xx10^(30) kg`B. `11/8xx10^(30) kg`C. `15/16xx10^(30) kg`D. `8/11xx10^(30) kg` |
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Answer» Correct Answer - B `F_(m m)`=Gravitational force between two outer stars `=(Gm^(2))/(4r)`. `F_(mM)`=Gravitational force between central star and outer star=`(GmM)/(r^(2))` for circular motion of outer star, `(mv^(2))/r=F_(m m)+F_(m M)` `:. V^(2)=(G(m+4M))/(4r)` `T`=period of orbtal motion `=(2pir)/v` `m=(16pi^(2)r^(2))/(GT^(2))-4M=(150/16-8)10^(30)=11/8xx10^(30)kg` |
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| 2260. |
A triple stars system consists of two stars each of mass m in the same circular orbit about central star with mass `M=2xx10^(33)kg`. The two outer stars always lie at opposite ends of a diameter of their common circular orbit the radius of the circular orbit is `r=10^(11)` m and the orbital period each star is `1.6xx10^(7)s` [take `pi^(2)=10` and `G=(20)/(3)xx10^(-11)Mn^(2)kg^(-2)`] Q. The orbital velocity of each star isA. `5/4sqrt(10) xx10^(3) m//s`B. `5/4sqrt(10) xx10^(5) m//s`C. `5/4sqrt(10)xx10^(2) m//s`D. `5/4 sqrt(10)xx10^(4) m//s` |
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Answer» Correct Answer - D `T=(2pir)/vrArr v=(2pir)/T=((2)(sqrt(10))(10^(11)))/(1.6xx10^(7))` `=5/4sqrt(10)xx10^(4)ms^(-1)`. |
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| 2261. |
A solid sphere of mass M and radius R is surrounded by a spherical shell of same mass and radius 2R as shown. A small particle of mass m is relased from rest from a height h`(lt lt R)` above the shell. There is a hole in the shell. Q. With what approximate speed will it collide at B?A. `sqrt((2GM)/(R))`B. `sqrt((GM)/(2R))`C. `sqrt((3GM)/(2R))`D. `sqrt((GM)/(R))` |
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Answer» Correct Answer - D COME `implies0-(G(2M)m)/((2R+h))=(1)/(2)mv^(2)-(GMm)/(R)-(GMm)/(2R)` `impliesvcongsqrt((GM)/(R))` |
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| 2262. |
A solid sphere of mass M and radius R is surrounded by a spherical shell of same mass and radius 2R as shown. A small particle of mass m is relased from rest from a height h`(lt lt R)` above the shell. There is a hole in the shell. Q. What time will it take to move from A to B?A. `=(4R^(2))/(sqrt(GMR))`B. `gt(4R^(2))/(sqrt(GMR))`C. `lt(4R^(2))/(sqrt(GMR))`D. None of these |
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Answer» Correct Answer - C Given that `(h lt lt R)` so the velocity at A is also zero we can see here that the acceleration always increases from 2 R to R its value must be greater than `a=(GM)/(4R^(2))` (at A) `thereforetlt(v)/(a)impliestltsqrt((GM)/(R))xx(4R^(2))/(GM)impliestlt(4R^(2))/(sqrt(GMR))` |
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| 2263. |
A triple star ststem consists of two stars each of mass m in the same circular orbit about central star with mass `M=2xx10^(33)kg`. The two outer stars always lie at opposite ends of a diameter of their common circular orbit the radius of the circular orbit is `r=10^(11)` m and the orbital period each star is `1.6xx10^(7)s` [take `pi^(2)=10` and `G=(20)/(3)xx10^(-11)Mn^(2)kg^(-2)`] Q. The total machanical energy of the system isA. `-(1375)/(64)xx10^(35)J`B. `-(1375)/(64)xx10^(38)J`C. `-(1375)/(64)xx10^(34)J`D. `-(1375)/(64)xx10^(37)J` |
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Answer» Correct Answer - B Total machanical energy `=K.E+P.E` `=2((1)/(2)mv^(2))-(2GMm)/(r)-(Gm^(2))/(2r)=m[(G(4M+m))/(4r)-(2GM)/(r)-(Gm)/(2)]=-(Gm)/(r)[M+(m)/(4)]` `=-((20)/(3)xx10^(-11))((11)/(8)xx10^(30))xx(1)/(10^(11))(2xx10^(30)+(11)/(32)xx10^(30))=-(1375)/(64)xx10^(38)J` |
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| 2264. |
Calculate the force of gravitation between the bodies , each of mass 100 kg and 1 m apart on the surface of the earth . Will the force of attraction be different if the same bodies are taken on the moon , their separation remaining constant ? |
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Answer» Correct Answer - `6.67 xx 10^(-7)` N , No |
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| 2265. |
What will happen if the distance of a body from the centre of the earth increases ? A) Force of gravity increasesB) Force of gravity decreases C) Force of gravity does not change D) None |
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Answer» B) Force of gravity decreases |
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| 2266. |
1) Force of gravitation acts between any two bodies in the universe. 2) Force of gravitation acts between objects nearer to earth and earth only.3) Fgravity = G \(\cfrac{M_1M_2}{d^2}\)4) G = 6.67 x 10-11 Nm2 Kg-2 The wrong statement in the above is A) 1 B) 2 C) 3 D) 2 and 4 |
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Answer» Correct option is B) 2 |
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| 2267. |
A body has constant speed while rotating in a circular path. Its direction of velocity A) constant B) changesC) A or B D) 0 |
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Answer» Correct option is B) changes |
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| 2268. |
Once while Sir Isaac Newton was sitting under a tree, an apple fell to the ground. Then some questions arose in his mind. Which questions would be those ? a) Why did the apple fall to the ground ? b) Does the apple eatable ? c) What is the colour of the apple ? d) What makes the apple to move ? A) a B) b and c C) a and d D) All |
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Answer» Correct option is C) a and d |
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| 2269. |
If ΔV is decreases, then the path of the body is A) in circular motion B) in polygon motion C) in spherical motion D) none |
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Answer» A) in circular motion |
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| 2270. |
An unmanned satellite `A` and a spacecraft `B` are orbiting around the earth in the same circular orbit as shown. The spacecraft is ahead of the satellite by some time. Let us consider that some technical problem has arisen in the satellite and the astronaut from `B` has made it correct. For this to be done docking of two (`A` and `B`) is required (in layman terms connecting `A` and `B`). To achieve this, the rockets of `A` have been fired in forward direction and docking takes place as shown in the figure below: Take mass of the earth `=5.98xx10^(24)kg` Radius of the earth `=6400 km` Orbital radius `=9600km` Mass of satellite `A=320 kg` Mass of spacecraft `=3200 kg` Assume that initially spacecraft `B` leads satellite `A` by `100s`, i.e., `A` arrives at any particle position after `100 s` of `B’s` arrival. Based on the above information answer the following questions. After once returning to the original point, i.e., the place from where the rockets have been fired, in which direction and with what extent the rockets have to be fired from the satellite to again come back in the original orbit?A. Forward direction with the same extent.B. Backward direction with the same extent.C. Forward direction with the higher extent.D. Backward direction with the higher extent. |
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Answer» Correct Answer - B To bring back the satellite in the same orbit, we have to make the speed of the satellite as initial one for which we have to increase `v_(f)` by such amont so that it becomes equal to `v_(i)` for which we have to fire the rocket in backward direction with the same extent. |
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| 2271. |
An unmanned satellite `A` and a spacecraft `B` are orbiting around the earth in the same circular orbit as shown. The spacecraft is ahead of the satellite by some time. Let us consider that some technical problem has arisen in the satellite and the astronaut from `B` has made it correct. For this to be done docking of two (`A` and `B`) is required (in layman terms connecting `A` and `B`). To achieve this, the rockets of `A` have been fired in forward direction and docking takes place as shown in the figure below: Take mass of the earth `=5.98xx10^(24)kg` Radius of the earth `=6400 km` Orbital radius `=9600km` Mass of satellite `A=320 kg` Mass of spacecraft `=3200 kg` Assume that initially spacecraft `B` leads satellite `A` by `100s`, i.e., `A` arrives at any particle position after `100 s` of `B’s` arrival. Based on the above information answer the following questions. By doing the above operation, now by how much time is the satellite leading the spacecraft?A. `38s`B. `138s`C. Lags by `38 s`D. Lags by `138 s` |
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Answer» Correct Answer - A Initially the time periods of the satellite and the spacedraft were same equal to `9358 s` and after firing rockets of `A`, it acquires, slower and faster orbit and `A` will come to initial position in time period of `9220 s` while `B` will take `9358 s` to come to its initial position or `(9358-100)s` to come to the original position of `A` as `100 s` is the initial time gap between the two. It means now `A` leads `B` by `(9358-100-9220)=38s`. |
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| 2272. |
An unmanned satellite `A` and a spacecraft `B` are orbiting around the earth in the same circular orbit as shown. The spacecraft is ahead of the satellite by some time. Let us consider that some technical problem has arisen in the satellite and the astronaut from `B` has made it correct. For this to be done docking of two (`A` and `B`) is required (in layman terms connecting `A` and `B`). To achieve this, the rockets of `A` have been fired in forward direction and docking takes place as shown in the figure below: Take mass of the earth `=5.98xx10^(24)kg` Radius of the earth `=6400 km` Orbital radius `=9600km` Mass of satellite `A=320 kg` Mass of spacecraft `=3200 kg` Assume that initially spacecraft `B` leads satellite `A` by `100s`, i.e., `A` arrives at any particle position after `100 s` of `B’s` arrival. Based on the above information answer the following questions The initial total energy and time period of satellite are, respectively,A. `-6.65xx10^(10)J, 9358s`B. `-6.65xx10^(9)J, 9358s`C. `-6.65xx10^(10)J, 9140s`D. `-6.65xx10^(9)J, 9140s` |
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Answer» Correct Answer - B Time period of `A` and `B` in a given circular orbit is `T=(2pir^(3/2))//(sqrt(GM))` where `r=9600km, G=6.67xx10^(-11)Nm^(2)kg^(-2), M=5.98xx10^(24)kg,` which gives `T=9357.79s=9358s` Initially, total energy of` A` is `E=-(Gmm)/(2r)=-6.65xx10^(9)` |
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| 2273. |
For a low altitude orbit if `r ~= r_(p)`, where `r_(p)` is planet radius, show that for a given average planetary density, the orbital period of satellite is independent of the size of the planet. Calculate its value average density is `rho`. |
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Answer» Correct Answer - `[sqrt((3pi)/(G rho))]` |
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| 2274. |
Three masses ,100kg 200kg and 500 kg are placed at the vetices of an equilateral triangle with sides 10 m they are rearranged by an agent on the verticws of a bigger triangle of siders 15 m and with the same in -centre Calcualte the work dine by the agent. |
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Answer» Correct Answer - `3.77xx10^(7)J` |
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| 2275. |
A body weighs 200 N on the surface of the Earth. How much will it weigh half way down to the centre of the Earth?(A) 250 N (B) 100 N (C) 150 N (D) 200 N |
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Answer» (B) 100 N Acceleration due to gravity at depth d, gd = g (1 – \(\frac{d}{R}\)) = g(1 – \(\frac{1}{2}\)) …(∵ d = \(\frac{1}{2}\)) ∴ gd = \(\frac{g}{2}\) Weight of the body at depth d = R/2, Wd = mgd = m × g/2 = \(\frac{1}{2}\) × 200 ∴ Wd = 100 N |
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| 2276. |
Assertion : In planetary motion angular momentum of planet about centre of sun remains constant. But linear momentum of system does not remain constant. Reason : Net torque on planet any point is zero.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explantion of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D Force on planet is always centre of sun. Hence, torque is zero only about centre of sun. |
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| 2277. |
In planetary motion the areal velocity of possition vector of a planet depends of angular velocity `(omega)` and the distance of the planet from sun (r). If so the correct relation for areal velocity isA. `(dA)/(dt)prop omegar`B. `(dA)/(dt) prop omega^(2) r`C. `(dA)/(dt) prop omega r^(2)`D. `(dA)/(dt) prop sqrt(omega r)` |
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Answer» Correct Answer - C `(dA)/(dt)=L/(2m)rArr (dA)/(dt)prop vr prop omegar^(2)` |
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| 2278. |
A planet of mass `m` is the elliptical orbit about the sun `(mlt ltM_("sun"))` with an orbital period `T`. If `A` be the area of orbit, then its angular momentum would be:A. `(2mA)/T`B. `mAT`C. `(mA)/(2T)`D. `2mAT` |
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Answer» Correct Answer - A `(dA)/(dt)=L/(2m)` |
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| 2279. |
The masses of two planets are in the ratio `1 : 2`. Their radii are in the ratio `1 : 2`. The acceleration due to gravity on the planets are in the ratioA. `1 : 2`B. `2 : 1`C. `3 : 5`D. `5 : 3` |
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Answer» Correct Answer - B |
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| 2280. |
The radius of a planet is twice that of the earth but its average density is the same. If the escape speed at the planet and at the earth are `v_(p)` and `v_(e)`, respectively, then prove that `v_(p)=2v_(e)` |
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Answer» `v_(e)sqrt((2GM)/R)=sqrt((2G)/Mxx4/3piR^(3)rho)=Rsqrt((8pi)/3GKrho)` `(v_(p))/(v_(e))=(R_(p))/(R_(e))=(2R_(e))/(R_(e))=2` or `v_(p)=2v_(e)` |
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| 2281. |
The ratio of escape velocity at earth `(v_(e))` to the escape velocity at a planet `(v_(y))` whose radius and density are twiceA. `1:sqrt(2)`B. `1:2`C. `1:2sqrt(2)`D. `1:4` |
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Answer» Correct Answer - C `ve=sqrt((2GM)/R) =sqrt((2G)/R. (4/3 piR^(3) rho)) prop Rsqrt(rho)` `:. Ratio =1:2sqrt(2)` |
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| 2282. |
The escape velocity on the surface of the earth is 11.2 `kms^(-1)`. If mass and radius of a planet is 4 and 2 tims respectively than that of the earth, what is the escape velocity from the planet?A. 11.2 km/sB. 1.112 km/sC. 15.8 km/sD. 22.4 km/s |
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Answer» Correct Answer - c `v_(e)=sqrt(2gR)=sqrt((2GM)/(R))=sqrt((2xxGxx4M)/(2R))` `=sqrt2sqrt((2GM)/(R))=sqrt2=11.2=15.8km//s` |
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| 2283. |
A cord of length 4 m is used to connect a 100 kg astronaut to spaceship whose mass is much larger than that of the astronaut. Estimat the value of the tension in the cord. Assume that the spaceship is orbiting near earth surface. A ssume that the spaceship and th eastronaut fall on a straight line from the earth centre. The radius of the earth is 6400 km. |
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Answer» Correct Answer - `T=3xx10^(-2)N` Cemtripetal force for orbiting of sface ship is provided by gravitational force. `(GM M_(e ))/(R^(2))=Mromega^(2) rArr omega = sqrt((GM_(e ))/(R^(3)))` …(i) For orbitting of astronaut `F_(g)+T=m(R+L)omega^(2) rArr (GM_(e )m)/((R+L)^(2))+T=m(R+L)((GM_(e ))/(R^(3)))` `T=m(R+L)(GM_(e ))/(R^(3))-(GM_(e )m)/((R+L)^(2))` `T = m(R+L)(gR^(2))/(R^(3))-(mgR^(2))/((R+L)^(2)) rArr T = mg+(mgL)/(R )-mg+(2mgL)/(R )` `T = (3mgL)/(R )rArr T = (3xx100xx10xx64)/(64xx10^(5))rArr T=3xx10^(-2)N` |
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| 2284. |
The escape velocity of an object on a planet whose radius is `4` times that of the earth and `g` value `9` tims that on the earth, in `Kms^(-1)`, isA. 67.2B. 33.6C. 16.8D. 25.2 |
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Answer» Correct Answer - A |
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| 2285. |
`v_(e)` and `v_(p)` denotes the escape velocity from the earth and another planet having twice the radius and the same mean density as the earth. ThenA. `v_(e)=v_(p)`B. `v_(e)=v_(p)//2`C. `v_(e)=2v_(p)`D. `v_(e)=v_(p)//4` |
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Answer» Correct Answer - B |
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| 2286. |
The escape velocity of a particle of a particle from the surface of the earth is given byA. `(gR)^(1//2)`B. `(2gR)^(1//2)`C. `(3gR)^(1//2)`D. `(gR//2)^(1//2)` |
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Answer» Correct Answer - B Escape velocity `v_(e)=sqrt(2gR)` |
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| 2287. |
The escape velocity of an object on a planet whose radius is `4` times that of the earth and `g` value `9` tims that on the earth, in `Kms^(-1)`, isA. `33.6`B. `67.2`C. `16.8`D. `25.2` |
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Answer» Correct Answer - B `v_(e)=sqrt(2gR)rArr V_(e)propsqrt(gR)` |
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| 2288. |
The escape velocity of a sphere of mass `m` is given byA. `sqrt((2GMm)/(R_(e)))`B. `sqrt((2GM)/(R_(e)^(2)))`C. `sqrt((2GMm)/(R_(e)^(2)))`D. `sqrt((2GM)/(R_(e)))` |
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Answer» Correct Answer - D `v=sqrt((2GM)/R)` Escape velocity does not depends upon the mass of the projected body. |
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| 2289. |
The escape velocity of a body from the surface of the earth is `V_(1)` and from an altitude equal to twice the radius of the earth, is, `V_(2)`. ThenA. `V_(1)=V_(2)`B. `V_(1)=7V_(2)`C. `V_(1)=sqrt(3)V_(2)`D. `V_(1)=sqrt(2)V_(2)` |
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Answer» Correct Answer - C `1/2mv^(2)=(GmM)/((R+h))`, |
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| 2290. |
Two identical solid copper spheres of radius `R` placed in contact with each other. The gravitational attracton between them is proportional to |
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Answer» Correct Answer - `n = 4` |
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| 2291. |
Two metal spheres each of radius `r` are kept in contact with each other. If `d` is the density of the material of the sphere,then the gravitational force between those spheres is proportional toA. `d^(2)//r^(6)`B. `d^(2)r^(4)`C. `d^(2)//r^(4)`D. `r^(4)//d^(2)` |
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Answer» Correct Answer - b `F prop (m^(2))/(r^(2))=((4pi)/(3)r^(6))/(r^(2))d^(2)` `F prop r^(4)d^(2)`. |
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| 2292. |
The escape velocity on earth is 11.2 km s-1 . What is its value for a planet having double the radius and 8 times the mass of the earth? |
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Answer» 22.4 kms-1 . The escape velocity on earth is 11.2 km s-1, then the value for a planet having double the radius and 8 times the mass of the earth is 22.4 kms-1. |
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| 2293. |
If the escape velocity on the earth is `11.2 km//s`, its value for a planet having double the radius and `8` times the mass of the earth is ...... (in `Km//s`)A. 3.7 km / sB. 11.2 km / sC. 22.4 km / sD. 43.2 km / s |
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Answer» Correct Answer - C |
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| 2294. |
The value of acceleration due to gravity at the surface of earthA. is maximum at the polesB. is maximum at the equatorC. remains constant everywhere on the surface of the earthD. is maximum at the international time line |
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Answer» Correct Answer - A Value of acceleration due to gravity is `g = (GM)/(R^(2))` At poles, R is minimum and at equator R is maximum, hence `g` is maximum at poles. |
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| 2295. |
Two metal spheres each of radius `r` are kept in contact with each other. If `d` is the density of the material of the sphere,then the gravitational force between those spheres is proportional toA. `d^(2)r^(6)`B. `d^(2)r^(4)`C. `(d^(2))/(r^(4))`D. `(r^(2))/(d^(2))` |
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Answer» Correct Answer - B `F=(Gm_(1)m_(2))/(R^(2))`, Here `m`=volume`xx`density |
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| 2296. |
Two satellites `S_(1)` and `S_(2)` are revolving round a planet in coplanar and concentric circular orbit of radii `R_(1)` and `R_(2)` in te same direction respectively. Their respective periods of revolution are 1 hr and 8 hr. the radius of the orbit of satellite `S_(1)` is equal to `10^(4)`km. Find the relative speed in kmph when they are closest.A. `(pi)/2xx10^(4)`B. `pixx10^(4)`C. `2pixx10^(4)`D. `4pixx10^(4)` |
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Answer» Correct Answer - B `T^(2)alphaR^(3), V_(0)=(2piR)/T`, Rel. velocity `=V_(01)-V_(02)` |
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| 2297. |
At a height H from the surface of earth, the total energy of a satellite is equal to the potential energy of a body of equal mass at a height 3R from the surface of the earth (R = radius of the earth). The value of H isA. `R`B. `(4R)/(3)`C. `3R`D. `(R)/(3)` |
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Answer» Correct Answer - A We know that the total energy of the satellite is `E=-(1)/(2)(GM_(e)m)/(R)` At a height H from the surface of the earth, the total energy of satellite, `E=-(1)/(2)(GM_(e)m)/(R+H)`....(i) and the potential energy, `U=-(GM_(e)m)/(R+3R)=-(GM_(e)m)/(4R)`....(ii) According to the question, `E=U-(1)/(2).(GM_(e)m)/(R+H)=-(GM_(e)m)/(4R)` `rArr (1)/(2R+2H)=(1)/(4R)rArr2H=4R-2R rArr H=R`. |
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| 2298. |
A body is projected up with a velocity equal to `3//4th` of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth is `R`)A. `10R//9`B. `9R//7`C. `9R//8`D. `10R//3` |
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Answer» Correct Answer - B `1/2mv_(e)^(2)=(mgh)/(1+h/R)` |
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| 2299. |
If the escape velocity on the earth is `11.2 km//s`, its value for a planet having double the radius and `8` times the mass of the earth is ...... (in `Km//s`)A. `11.2`B. `22.4`C. `5.6`D. `8` |
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Answer» Correct Answer - B `v_(e)=sqrt((2GM)/R)rArr (v_(1))/(v_(2))=sqrt((M_(1))/(M_(2))(R_(2))/(R_(1)))` |
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| 2300. |
At what height from the surface of the earth, the total energy of satellite is equal to its potential energy at a height `3R` from the surface of the earth (`R`=radius of earth)A. `4R`B. `3R`C. `2R`D. `R` |
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Answer» Correct Answer - D `T.E.=P.E., -(GMm)/(2(R+h))=(GMm)/((4R))` |
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