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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.
| 201. |
Imagine a light planet revolving around a massive star in a circular orbit of raidus r with a a period of revolution T. If the gravitational force of attraction between planet and the star is proportioanl to `r^(-5)//^(2)`, then find the relation between T and r.A. `R^(3//2)`B. `R^(3//5)`C. `R^(7//2)`D. `R^(1//4)` |
| Answer» Correct Answer - D | |
| 202. |
A staellite in a circular orbit of raidus R has a period of 4h another satellite with orbital radius 3R around the same planet will have a period (in h)A. 16B. 4C. `4sqrt(27)`D. `4sqrt(8)` |
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Answer» Correct Answer - C According to kepler s third law `T^(2) prop R^(3) rarr (R_(2))/(R_(1))^(3//2)` `therefore (T_(2))/(T_(1))=(3R)/(R )^(3//2)=sqrt(27)` `therefore T_(2)=sqrt(27)T_(1)=sqrt(27)xx=s4sqrt(27)h` |
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| 203. |
The critical velocity of a satellite at height h from the surface of the earth is 5 km/s. The same for a satellite around another planet of double the radius and 4 times the mass of the earth and double the height will beA. `10km//s`B. `7km//s`C. `2.50m//s`D. `3.57km//s` |
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Answer» Correct Answer - b `(V_(C_(2)))/(V_(C_(1)))=sqrt((M_(2))/(M_(1))xx(r_(1))/(r_(2)))=sqrt((M_(2))/(M_(1))xx((R_(1)+h_(1)))/((R_(2)+h_(2))))` `=sqrt((4xx(R_(1)+h_(1)))/((2R_(1)+2h_(1))))=sqrt((4(R_(1)+h_(1)))/(2(R_(1)+h_(1))))` `V_(C_(2))=sqrt2xx5` `=1.141xx5=7.070"km/s."` |
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| 204. |
The gravitational field in a region is given by `vecE = (3hati- 4hatj) N kg^(-1)`. Find out the work done (in joule) in displacing a particle by `1 m` along the line `4y = 3x + 9`. |
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Answer» For the line `4y=3x+9` `4dy+3dx,4dy-3dx=0` ..(i) For work in the region `dW=vecE.(dxhati+dyhatj)=(3hati-4hatj).(dxhati+dyhatj)` `=3dx-4dy` (from equation (i)) =0 |
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| 205. |
Two stars of masses `m_(1)` and `m_(2)` distance r apart, revolve about their centre of mass. The period of revolution is :A. `2pi sqrt((r^(3))/(2G (m_(1)+m_(2)))`B. `2pi sqrt((r^(3)(m_(1)+m_(2)))/(2G m_(1)m_(2)))`C. `2pi sqrt((2r^(3))/(G (m_(1)+m_(2)))`D. `2pi sqrt((r^(3))/(G (m_(1)+m_(2)))` |
| Answer» Correct Answer - D | |
| 206. |
Two stars of masses `m_(1)` and `m_(2)` distance r apart, revolve about their centre of mass. The period of revolution is :A. `2pisqrt((r^(3))/(2G(m_(1)+m_(2))))`B. `2pisqrt((r^(3)(m_(1)+m_(2)))/(2Gm_(1)m_(2)))`C. `2pisqrt((2r^(3))/(G(m_(1)+m_(2))))`D. `2pisqrt((r^(3))/(G(m_(1)+m_(2))))` |
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Answer» Correct Answer - D |
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| 207. |
In a particular double star system two stars of mass `3.22xx10^(30)` kg each revolve about their common centre of mass `1.12xx10^(11)` on away. (i). Calculate their common period of revolution, in years (ii). Suppose that a meteoroid (small solid particle in space) passes through this centre of mass moving at right angles to the orbital plane of the stars. What must its speed be if it is to escape from the gravitational field of the double star? |
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Answer» Correct Answer - (i). `T=4pisqrt((r^(3))/(Gm))` (ii). `v=sqrt((4Gm)/(r)),r=(d)/(2)` (i) Necessary centripetal force=gravitational force `impliesomega^(2)r=(GM^(2))/(4r^(2))impliesT=4pisqrt((r^(3))/(GM))` (iii). COME: KE+PE=0 `implies(1)/(2)mv^(2)-(2GMm)/(r)=0impliesv=sqrt((4GM)/(r)),(r=(d)/(2))` |
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| 208. |
A particle of mass `M` is at a distance a from surface of a thin spherical shell of equal mass and having radius `a`. A. Gravitational field and potential both are zero at centre of the shell.B. Gravitational field is zero not only inside the shell but at a point outside the shell also.C. Inside the shell, gravitational field alone is zero.D. Neither gravitational field nor gravitational potential is zero inside the shell. |
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Answer» Correct Answer - D Both field and the potential inside the shell in non zero. |
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| 209. |
If there were a smaller gravitational effect, which of the following forces do you think would alter in some respectA. viscous forceB. Archimedes upliftC. electrostatic forceD. magnetic force. |
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Answer» Correct Answer - B Gravitational field and the electrostatic field both are conservation in nature. |
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| 210. |
The dimensions of universal gravitational constant are :-A. `[L^(1)M^(0)T^(0)]`B. `[L^(2)M^(1)T^(0)]`C. `[L^(-1)M^(1)T^(-2)]`D. `[L^(3)M^(-1)T^(-2)]` |
| Answer» Correct Answer - d | |
| 211. |
Calculate the gravitational field intensity at the centre of the base of a hollow hemisphere of mass `M` and radius `R`. (Assume the base of hemisphere to be open) |
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Answer» We consider the shaded elemental ring of mass `dm=M/((2piR^(2)))2piRsintheta(Rd theta)` Field due to this ring at `O` `dE=(GdmRcostheta)/(R^(3))` (see formulae for field due to a ring) or `dE=(GM)/(R^(2)) sinthetacosthetad theta` Hence `E=int_(0)^(pi//2) dE=int_(R^(2))^(pi//2) (GM)/(R^(2))sinthetacosthetad theta` or `E=(GM)/(2R^(2))` |
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| 212. |
The minimum and maximum distances of a satellite from the centre of the earth are 2R and 4R respectively where R is the radius of earth and M is the mass of the earth find radius of curvature at the point of minimum distance. |
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Answer» Correct Answer - `(8R)/(3)` COAM: `mv_(1)(2R)=mv_(2)(2R)impliesv_(1)=2v_(2)` .(i) COME: `(-GMm)/(2R)+(1)/(2)mv_(1)^(2)=-(GMm)/(4R)+(1)/(2)mv_(2)^(2)` ..(ii) `impliesv_(1)=sqrt((2GM)/(3R))` `therefore` radius of curvature at perigee `=(v_(1)^(2))/(g_(1))` `impliesR_(P)=(2GM)/(3R)xx(4R^(2))/(GM)=(8R)/(3)` |
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| 213. |
Calculate the gravitational potential at the centre of base of a solid hemisphere of mass `M`, radius `R`.A. Gravitational potential at the centre of curvature of a thin uniform wire of mass M, bent into a semicircle of radius R, is also equal V.B. In part (A) if the same wire is bent into a quarter of a circle then also the gravitational potential at the centre of curvature will be V.C. In part (A) if the wire mass is non uniformly distributed along its length audit is bent into a semicircle radius R, gravitational potential at the centre is VD. None of these |
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Answer» Correct Answer - A::C Gravitational potential due to hemisphere at the centre is V because distance of each mass particle from the centre O is R. If the distance between the point and mass is changed potential will also change. |
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| 214. |
The magnitude of the gravitational field at distance `r_(1)` and `r_(2)` from the centre of a uniform sphere of radius `R` and mass `M` are `F_(1)` and `F_(2)` respectively. Then:A. `(F_(r))/(F_(2))=(r_(1))/(r_(2))` if `r_(1)ltR` and `r_(2)ltR`B. `(F_(1))/(F_(2))=(r_(2)^(2))/(r_(1)^(2))` if `r_(1)gtR` and `r_(2)gtR`C. `(F_(1))/(F_(2))=(r_(1)^(3))/(r_(2)^(3))` if `r_(1)ltR` and `r_(2)ltR`D. `(F_(1))/(F_(2))=(r_(1)^(2))/(r_(2)^(2))` if `r_(1)ltR` and `r_(2)ltR` |
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Answer» Correct Answer - A::B Gravitational field intensity `F=(GMr)/(R^(3))` Inside the sphere `(F_(1)propr_(1),F_(2)propr_(2))` `(F_(1))/(F_(2))=(r_(1))/(r_(2))` or `r_(1)ltR&r_(2)ltR` Gravitational field intensity `Iprop(1)/(r^(2))` (out side the sphere) `therefore(F_(1))/(F_(2))=(r_(2)^(2))/(r_(1)^(2))` if `r_(1)gtR` and `r_(2)gtR` |
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| 215. |
The magnitude of the gravitational field at distance `r_(1)` and `r_(2)` from the centre of a uniform sphere of radius `R` and mass `M` are `F_(1)` and `F_(2)` respectively. Then:A. `F_(1)/F_(2)=r_(1)/r_(2)` if `r_(1) lt R` and `r_(2) lt R`B. `F_(1)/F_(2)=r_(1)^(2)/r_(2)^(2)` if `r_(1) gt R` and `r_(2) gt R`C. `F_(1)/F_(2)=r_(1)/r_(2)` if `r_(1) gt R` and `r_(2) gt R`D. `F_(1)/F_(2)=r_(2)^(2)/r_(1)^(2)` if `r_(1) lt R` and `r_(2) lt R` |
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Answer» Correct Answer - A::B |
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| 216. |
The magnitude of the gravitational field at distance `r_(1)` and `r_(2)` from the centre of a uniform sphere of radius `R` and mass `M` are `F_(1)` and `F_(2)` respectively. Then:A. `(F_(1))/(F_(2))=(r_(1))/(r_(2))` if `r_(1) gt R` and `r_(2) lt R`B. `(F_(1))/(F_(2)) = (r_(2)^(2))/(r_(1)^(2))` if `r_(!) gt R` and `r_(2) gt R`C. `(F_(1))/(F_(2)) = (r_(1)^(3))/(r_(2)^(3))` if `r_(1) lt R` and `r_(2) lt R`D. `(F_(1))/(F_(2)) =(r_(1)^(2))/(r_(2)^(2))` if `eta lt R` and `r_(2) lt R` |
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Answer» Correct Answer - B For `vleR, F=(GMr)/(R^(3))` `(F_(1))/(F_(2))=(r_(1))/(r_(2))` for `r_(1) gt R` and `r_(2) lt R` For `rgeR, F=(GM)/(r^(2))` It implies `F prop1//r^(2)` `:. (F_(1))/(F_(2))=(r_(2)^(2))/(r_(1)^(2))` for `r_(1)gtR` and `r_(2)gtR` |
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| 217. |
if velocity given to an object from the surface of the earth is n times the escapes velocity then what will be the residual velocity at inifinity. |
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Answer» Let residual velocity be v then from energy conservation `(1)/(2)m(nv_(e))^(2)-(GMm)/(R)=(1)/(2)mv^(2)+0` `impliesv^(2)=n^(2)v_(e)^(2)-(2GM)/(R)=n^(2)v_(e)^(2)-v_(e)^(2)-v_(e)^(2)=(n^(2)-1)v_(e)^(2)impliesv=(sqrt(n^(2)-1))v_(e)` |
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| 218. |
Which one of the following statements regarding artificial satellite of the earth is incorrectA. The orbital velocity depends on the mass of the satelliteB. A minimum velocity of 8 km / sec is required by a satellite to orbit quite close to the earthC. The period of revolution is large if the radius of its orbit is largeD. The height of a geostationary satellite is about 36000 km from earth |
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Answer» Correct Answer - A |
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| 219. |
A ball is dropped from a spacecraft revolving around the earth at a height of `1200km`. What will happen to the ball ? .A. It will continue to move with velocity v along the original orbit of spacecraftB. It will move with the same speed tangentially to the spacecraftC. It will fall down to the earth graduallyD. It will go very far in the space |
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Answer» Correct Answer - A |
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| 220. |
Gravitational force can be called ………………. A) contact force B) non-contacting force C) force at a distance D) both B & C |
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Answer» D) both B & C |
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| 221. |
The maximum vertical distance through which a full dressed astronaut can jump on the earth is 0.5m. Estimate the maximum vertical distance through which he can jump on the motion, which has a mean density 2/3 rd that of the earth and radius one-quarter that of the earth.A. 1.5mB. 3mC. 6mD. 7.5m |
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Answer» Correct Answer - B On the moon, `g_(m) = (4)/(3)pi G(R//4) (2rho//3)` `= (1)/(6) ((4)/(3)pi GR rho) = (1)/(6)g` Work done in jumping `=m xx g xx 0.5 = m xx (g//6) h_(1)` `h_(1) = 0.5 xx 6 = 3.0m` |
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| 222. |
An object of mass 50 g has a volume of `20 cm^(3)` Calculate the density of the object. If the density of water `1//g cm^(2)`, state whether the object will float or sink in water. |
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Answer» We know that: Destiy `=("Mass of the object" )/("volume of the object")` Here Mass of the object =50 g And volum of the object = `20=cm ^(3)` Now putting these values of mass of and volume of the object in the above formula ,we get : Density of object `=(50 g)/(20 cm)^(3)` `=2.5 g//cm^(3)` Thus the density of object is `2.5 g//cm^(3)` Since the density of object `(2.5 g// cm ^(3))` is greater than the density of water `(1 g // cm^(3))` therefore , the object will sink in water. |
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| 223. |
The difference in the value of `g` at poles and at a sphere of latitude, `45^(@)` isA. `Romega^(2)`B. `(Romega^(2))/2`C. `(Romega^(2))/4`D. `(Romega^(2))/3` |
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Answer» Correct Answer - B `g_(phi)=g-Romega^(2)cos^(2)phi, g_("poles")=g` |
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| 224. |
The acceleration due to gravity on the moon is 1//6th of that on the earth. If a man can jump upto a height of 1 m on the surface of the earth, how high he can jump on the surface of the moon ?A. 3 mB. 4 mC. 6 mD. 8 m |
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Answer» Correct Answer - C `mgh_(e )=m((g)/(6))xx h_(m) " " therefore h_(m)=6 m` |
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| 225. |
The acceleration due to gravity on the surface of the moon `= (1)/(6)` the acceleration due to gravity on the surface of the earth. What will be the mass of a steel ball on the surface of the moon, if its mass on the surface of the earth is 6 kg ?A. 1 kgB. zeroC. 6 kgD. 36 kg |
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Answer» Correct Answer - C There is no change in mass. Change in g produces a change in weight. |
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| 226. |
The radius and acceleration due to gravity of the moon are `1/4` and `1/5` that of the earth, the ratio of the mass of the earth to mass of the moon isA. `1:80`B. `80:1`C. `1:20`D. `20:1` |
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Answer» Correct Answer - B `g=(GM)/(R^(2))rArrgalphaM/(R^(2))` |
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| 227. |
The difference in the value of `g` at poles and at a latitude is `3/4R omega^(2)`. The latitude angle isA. `60^(@)`B. `30^(@)`C. `45^(@)`D. `95^(@)` |
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Answer» Correct Answer - B `Romega^(2)cos^(2)lambda=3/4Romega^(2)` |
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| 228. |
If suddenly the gravitational force of attraction between Earth and a stellite revolving around it becomes zero, then the stellite willA. Continue to move in its orbit with same velocityB. move tangentially to the original orbit with the same velocityC. becomes stationary in its orbitD. move towards the earth |
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Answer» Correct Answer - 3 Since the centripetal force will disappear hence the satellite will move tangentially to the original orbit with speed `V`. |
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| 229. |
Why will a sheet of paper fall slower than one that is crumpled into a ball? |
| Answer» A sheet of paper has a larger area. Due to its large area, when a sheet of paper is dropped from a height ,it experience more resistance from air ,its speed decreases and if fall at a slower rate .On the other hand , a sheet of paper crumpled sheet of paper is dropped from a height , it experiences less resistance from air ,its speed increases and if falls at a faster rate. | |
| 230. |
Suppose the law of gravitational attraction suddenly changes and becomes an inverse cube law i.e. `F prop 1//r^(3)`, but still remaining a central force. ThenA. Keplers law of areas still holdsB. Keplers law of period still holdsC. Keplers law of areas and period still holdD. Neither the law of areas, nor the law of period still holds |
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Answer» Correct Answer - D |
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| 231. |
Why will sheet of paper fall slower than one that is crumpled into a ball? |
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Answer» The sheet of paper falls slower than that is compelled into a ball because in first case the area of the sheet is more. So it experiences large opposing force due to air. while the sheet crumpled into a ball experience less opposing force due to small area. |
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| 232. |
As observed from the earth, the sun appears to move an approx. circular orbit. For the motion of another planet like mercury as observed from the earth, this wouldA. be similarly trueB. not be true because the force between the earth and mercury is not inverse square lawC. not be true because the major gravitational force of mercury is due to the sunD. not be true becase mercury is influence by force other than gravitational forces |
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Answer» Correct Answer - C As observed form the earth, the sun appears to move in an approx. circular orbit. The gravitational force of attraction between the earth and the sun always follows inverse square law. Due to relative motion between the earth and mercury, the orbit of mercury, as observed form the earth will not be approx. circular, since the major gravitational force on mercury is due to the sun. |
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| 233. |
As observed from the earth, the sun appears to move an approx. circular orbit. For the motion of another planet like mercury as observed from the earth, this wouldA. the similary ture.B. not be true because the force between earth and mercury is not inverse square law.C. not be true because the major gravitational force on mercury is due to sun.D. not be true because mercury is influenced by force other than gravitational forces. |
| Answer» Correct Answer - C | |
| 234. |
Mass of moon is `7.34xx10^(22)kg`. If the acceleration due gravity on the moon is `1.4m//s^(2)`, the radius of the moon is`(G=6.667xx10^(-11)Nm^(2)//kg^(2))`A. `0.56xx10^(4) m`B. `1.87xx10^(6) m`C. `1.92xx10^(6) m`D. `1.01xx10^(8) m` |
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Answer» Correct Answer - B |
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| 235. |
A stone drops from the edge of the roof. It passes a window 2 m high in `0*1 s`. How far is the roof above the top of the window? |
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Answer» Correct Answer - `19.4 m` Let h be the height of the roof above the top of the window and t be the time taken by the stone to reach the top of the window. `u=0,a=9.8m//s^(2)` from `h=ut+(1)/(2)g t^(2), h=0+(1)/(2)xx9.8(t+0.1)^(2)=4.9(t^(2)+0.01+0.2t)` and `h+2=0+(1)/(2)xx9.8(t+0.1)^(2)=4.9(t^(2)+0.01+0.2t)` using eqn. `4.9t^(2)=4.9t^(2)+4.9xx0.01=1.951` or `0.98t=2-4.9x0.01=1.951` or `t=1.951/0.98=1.999s` As `h=4.9t^(2), h=4.9(1.99)^(2)=19.4m`. |
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| 236. |
A stone is released from the top of tower of height 19.6 m. Calculate the final velocity just before touching the ground. |
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Answer» h = 19.6 u = 0 a = + g = 9.8 m/s2 v2 = 2gh v = \(\sqrt{ 2gh} = \sqrt{2\,\times\,9.8\,\times\,19.6} = 19.6 \,m/s\) |
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| 237. |
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground. |
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Answer» h = 19.6 m. Initial velocity u = 0 (∵ It starts from rest) From equation of motion. v2 = u2 + 2gh v2 = 0 + 2 x 9.8 x 19.6 = 19.6 x 19.6 ∴ v = \(\sqrt{19.6\times 19.6}\) = 19.6 ms-1 Final velocity of stone before touching the ground = 19.6 ms-1. |
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| 238. |
Assuming that acceleration remains constant `(5.34xx10^(-9)m//s^(2))`, How long will Mahendra take to move 1 cm towards Virat if he starts from rest? |
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Answer» Given Acceleration `(a)=6.34xx10^(-9)m//s^(2)` `=5xx10^(-9)m//s^(2)` Displacement `(s)=1cm=1/100m` Initial velocity `(u)=0ml//s` To find: time `(t)=?` Formula`s=ut+1/2at^(2)` Solution: `1/100=0xxt+1/2xx5xx10^(-9)xxt^(2)` `1/100=2.5xx10^(-9)xxt^(2)` `1/(100xx2.5xx10^(-9))=t^(2)` `(1xx10^(9))/250=t^(2)` `(1000xx10^(6))/250=t^(2)` `4xx10^(6)=t^(2)` `t=sqrt(4xx10^(6)s)` `t=2xx10^(3)s` Mahendra will take `2xx10^(3)s` to move towards Virat. |
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| 239. |
A ball thrown vertically upward reached a maximum height of 20m (i) What was the velocity of the stone at the instant of throwing up? (ii) How much time did the ball take to reach the height 20 m? |
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Answer» (i) v2 = u2 + 2as u2 = v2 - 2as = 02 - 2 x 10 x 20 = 400 u = 20 m/s (ii) v = u + t t = \(\frac{V-U}{a}\) = \(\frac{0-20}{-10}\) = 2 s \(v^2=u^2+2as\) |
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| 240. |
A stone is dropped from the edge of a roof. (a) How long does it take to fall 4.9m? (b) How fast does it move at the end of that fall? (c) How fast does it move at the end of 7.9m? (d) What is its acceleration after 1s and after 2 s? |
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Answer» (a) As the stone is dropped, its initial velocity,`u=0`, ` h=4.9m, a=g=9.8 m//s^(2)`, time `t=?` From `h=ut+(1)/(2)g t^(2), 4.9=0+(1)/(2)xx9.8t^(2)=4.9t^(2) or t^(2)=4.9/4.9=1,t=sqrt1=1s` (b) Finaly velocity,` v=?` at `t=1s` From `v=u+g t, v=0+9.8xx1=9.8 m//s` (c) Let v be Final velocity when`h=7.9m`. From `v^(2) -u^(2) = 2 gh, v^(2) - 0 = 2 (9.8) 7.9` or `v=sqrt(2xx9.8xx7.9)=sqrt(154.84)=12.4 m//s` (d) The accceleration of a freely failling body remains the same at all times, i.e,. `a=g=9.8 m//s^(2)`, after 1 s and after 2 s. |
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| 241. |
Starting from rest, due to the gravitational force of the Earth i.e. 733N, What is the speed of Mahendra after 1 second? If his mass is 75 kg. |
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Answer» Given: Initial velocity `(u)=0m//s` Force `(F)=733N` Mass `(m)=75kg` Time `(t)=1s` To find: Final velocity `(v)=?` Fromulae`v=u+at, a=F/M` Solution: `a=733/75` `:.a=9.77m//s^(2)` `v=u+at` `0+9.77xx1` `=9.77m//s` The speedof Mahendra after 1 second is 9.77 m/s |
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| 242. |
A stone is thrown verticaly upward with an initial velocity of `40m//s`. Taking `g=10m//s^(2)`, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone? |
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Answer» Here, `u=40m//s, g=-10m//s,h=?,v=0` From `v^(2)-u^(2)=2gh, 0-(40)^(2)=2(-10)h or h=(40xx40)/20=80m` As final position of the stone coincides with its initial position, net displacement =0. Total distance coverd by the stone `=h+h=2h=2xx80m=160m`. |
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| 243. |
What is the nature of the distance-time graphs for uniform and non-uniform motion of an object ? |
| Answer» In uniform motion, distance time graph is a straight line with some slope. In non uniform motion, the distance time graph may be any curve. | |
| 244. |
Calculaste the gravitational force due to the Earth on Mahendra if mass of Earth is `6xx10^(24)kg` Radius is `6.4xx10^(6)m,g=9.77m//s^(2)` and mass of Mahendra is 75 kg. |
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Answer» Given: Mass of Earth `(M)=6xx10^(24)kg` Radius of Earth `(R)=6.4xx10^(6)m` Mass of object `(m)=75kg` Gravitational acceleration `(g)=9.77m//s^(2)` To find : Force `F=?` Forumula `F=(GMm)/(R^(2))` Solution `F=(6.67xx10^(-11)xx6xx10^(24)xx75)/(40.96xx10^(12))` `=(9.77xx75)/1` `=732.75N` `F=733N` The gravitational force is 733N |
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| 245. |
Calculate the force of gravity acting on your friend of mass 60kg.Given mass of earth = `6xx10^(24)` kg and radius of Earth=`6.4xx10^(6)m`. |
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Answer» Here,`m= 60kg, M=6xx10^(24)kg,R=6.4xx10^(6)m,F=?,G=6.67xx10^(-11)Nm^(2)//kg^(2)` Thus, `F=(GMm)/(R^(2))=(6.67xx10^(11)xx6xx10^(24)xx60)/((6.4xx10^(6))^(2))=58.62N` |
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| 246. |
A particle is thrown up verticaly with a velocity with a velocity of `50 m//s`. what will be its velocity at the highest point of the journey ? How high would the particle rise ? What time would it take it take to reach the highest point ? Take `g=10m//s^(2)`. |
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Answer» Here, intial velocity, `u=50m//s`, final velocity, `v=?` height coverd, `h=?` time taken, `t=?` `g=10m//s^(2)` At the highest point, final velocity `v=0`. From `v^(2)-u^(2)=2gh`, where `g=-10m//s^(2)` for upward journey, `0-(50)^(2)=(-10)h` or `h=-(2500)/(-20)=125m` `v=u+g t, 0=50+(-10)t` or `t=50//10=5s` |
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| 247. |
A train starting from a railway station and moving with uniform acceleration attains a speed of `40km//h` in 5 min. find its acceleration. |
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Answer» Here, `u=0,a=?,v=40 km//h=(40xx1000)/(60xx60)=(100)/(9)m//s` `t=5 min.=5xx60 s=300 s` from `v=u+at` `(100)/(9)=0+axx300` `a=(100)/(9xx300)=(1)/(27)m/s^(2)` |
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| 248. |
Artificial satellite moving around the earth is just like aA. projectileB. freely falling bodyC. body producted vertically upD. body at rest |
| Answer» Correct Answer - b | |
| 249. |
Is it necessary for the plane of the orbit of a satellite to pass through the centre of the earth?A. centre of earthB. south poleC. north poleD. can not be predicted |
| Answer» Correct Answer - a | |
| 250. |
The value of G was successfully determined for the first time in the labortaory byA. FaradayB. CavandishC. NewtonD. Sir Airy |
| Answer» Correct Answer - b | |