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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.
| 2301. |
The universal law of gravitational is the force law known also as theA. triangular lawB. square lawC. inverse square lawD. parallelogram law |
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Answer» Correct Answer - C Universal law of gravitation states that the gravitational force of attraction acting between two of universe is `F=(Gm_(1)m_(2))/(r^(2))` where G is gravitational constant and `r` is the distance between them. Hence, it is also known as inverse square law. |
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| 2302. |
A satellite close to the earth is in orbit above the equator with a period of rotation of `1.5` hours. If it is above a point `P` on the equtor at some time, it will be above `P` again after time ...........A. (ii),(iv)B. (ii),(iii)C. (i),(iii)D. (i),(iv) |
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Answer» Correct Answer - A Let `omega_(0)`, angular velocity of earth, `omega_(0)=(2pi)/(T_(0))=(2pi)/24` `omega`: angular velocity of satellite, `omega=(2pi)/T=(2pi)/1.5` For satellite, rotating west to east, `omega_(1)=omega-omega_(0)` Time periode of rotation relative to earth `t_(1)=(2pi)/(omega_(1))=(2pi)/((2pi)/1.5-(2pi)/24)=1.6 hr` For satellite, rotating east to west, `omega_(2)=omega+omega_(0)` Time periode of rotation relative to earth `t_(2)=(2pi)/(omega_(2))=(2pi)/((2pi)/1.5+(2pi)/24)=24/17 hr` |
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| 2303. |
Two identical thin rings each of radius R are coaxially placed at a distance R. If the rings have a uniform mass distribution and each has masses `2m` and `4m` respectively, then the work done in moving a mass `m` from centre of one ring to that of the other isA. zeroB. `(sqrt(2)Gm^(2))/(R)(1-sqrt(2))`C. `(Gm^(2))/(sqrt(2)R)(sqrt(2)-1)`D. `(Gm^(2))/(sqrt(2)R)` |
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Answer» Correct Answer - B Work done, `W=U_(2)-U_(1)=mv_(2)-mv_(1)` `=m[{-(Gm_(2))/(R)-(Gm_(1))/(sqrt(2)R)}-{-(Gm_(1))/(R)-(Gm_(2))/(sqrt(2)R)}]` `=(Gm(m_(1)-m_(2)))/(sqrt(2)R)(sqrt(2)-1)=(sqrt(2)Gm^(2))/(R)(1-sqrt(2))`. |
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| 2304. |
Two rings having masses `M` and `2M` respectively, having the same radius are placed coaxially as shown in the figure. If the mass distribution on both the rings is non-uniform, then the gravitational potential at point `P` isA. `-(GM)/R[1/(sqrt(2))+2/(sqrt(5))]`B. `-(GM)/R[1+2/2]`C. zeroD. cannot be determined from given information |
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Answer» Correct Answer - A As all the point on the periphery of either ring are at same distance from point `P`, the potential at point `P` due to whole ring can be calculate as `V=-(GM)//(sqrt(R^(2)+x^(2)))` where `x` is axial distance from the centre of the ring. This expression is independent of the fact whether the distribution of mass is uniform or non-uniform. So, `V` at `P` is `V=-(GM)/(sqrt(2)R)-(Gxx2M)/(sqrt(5)R)` `=-(GM)/R[1/(sqrt(2))+2/(sqrt(5))]`. |
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| 2305. |
The ratio of the radii of planets A and B is `k_(1)` and ratio of acceleration due to gravity on them is `k_(2)` . The ratio of escape velocities from them will beA. `k_(1) k_(2)`B. `sqrt(k_(1)k_(2))`C. `sqrt(k_(1)/k_(2))`D. `sqrt(k_(2)/k_(1))` |
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Answer» Correct Answer - B |
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| 2306. |
Pertaining to two planets, the ratio of escape velocities from respective surfaces is `1:2` , the ratio of the time period of the same simple pendulum at their respective surfaces is `2:1` (in same order). Then the ratio of their average densities isA. `1:1`B. `1:2`C. `1:4`D. `8:1` |
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Answer» Correct Answer - C Time period of simple pendulum, `T=2pi sqrt((l)/(g))or T prop (1)/(sqrt(g))` `(T_(1))/(T_(2))=sqrt((g_(2))/(g_(1)))=(2)/(1)rArr (g_(2))/(g_(1))=4`....(i) `v_(e)=sqrt(2gR)or v_(e)prop sqrt(gR)rArr (v_(e1))/(v_(e2))=sqrt((g_(1))/(g_(2)).(R_(1))/(R_(2)))=1:2` `:. sqrt((1)/(4)xx(R_(1))/(R_(2)))=(1)/(2)" "(because "from Eq. (i)" (g_(2))/(g_(1))=4)` `:. (R_(1))/(R_(2))=1` Further, `g=(Gm)/(R^(2))=(G((4)/(3)piR^(3)rho))/(R^(2))or g prop R rho` `:. (g_(1))/(g_(2))=(R_(1))/(R_(2)).(rho_(1))/(rho_(2))or((1)/(4))=(1)(rho_(1))/(rho_(2))or (rho_(1))/(rho_(2))=(1)/(4)`. |
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| 2307. |
Assertion : Areal velocity of a planet around of surface area and density is same for two planets, escape velocities will be same for both Reason : Areal velocity `= (L)/(2m)`, Where L is angular momentum of planet about centre of sun.A. If both Assertin and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are correct but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is falseD. If Assertion is false but Reason is true |
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Answer» Correct Answer - D `(dA)/(dt)=(L)/(2m)="constant"=(mvr sin theta)/(2m)=(vr sin theta)/(2)` i.e., `(dA)/(dt)` is independent of `m`. |
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| 2308. |
The radii of two planets are respectively `R_(1) and R_(2)` and their densities are respectively `rho_(1) and rho_(2)`.The ratio of the accelerations due to gravity at their surface isA. `(R_(2))/(R_(1))(rho_(1))/(rho_(2))`B. `(R_(1))/(R_(2))(rho_(2))/(rho_(1))`C. `(R_(1)rho_(1))/(R_(2)rho_(2))`D. `sqrt((R_(1)rho_(2))/(R_(2)rho_(1)))` |
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Answer» Correct Answer - C `g=(GM)/(R^(2))=G(4)/(3)(pi R^(3)rho)/(R^(2))=(4)/(3)pi GR rho` `therefore (g_(1))/(g_(2))=(R_(1)rho_(1))/(R_(2)rho_(2))` |
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| 2309. |
The radii of two planets are respectively `R_(1) and R_(2)` and their densities are respectively `rho_(1) and rho_(2)`.The ratio of the accelerations due to gravity at their surface isA. `g_(1) : g_(2)=rho_(1)/R_(1)^(2) : rho_(2)/R_(2)^(2)`B. `g_(1) : g_(2)=R_(1)R_(2) : rho_(1)rho_(2)`C. `g_(1) : g_(2)=R_(1)rho_(2) : R_(2) rho_(1)`D. `g_(1) : g_(2) =R_(1) rho_(1) : R_(2)rho_(2)` |
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Answer» Correct Answer - D |
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| 2310. |
A satellite close to the earth is in orbit above the equator with a period of rotation of `1.5` hours. If it is above a point `P` on the equtor at some time, it will be above `P` again after time ........... |
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Answer» Correct Answer - 1.6 hours if it is rotating from west to east `(24)/(17)` hours if it is rotating east to west Relative velocity when satellite revolving anticlockwise `(omega_(1)+omega_(2))t=2piimplies((4pi)/(3)+(2pi)/(24))t=2pi,t=(24)/(17)` if it moves in same direction `((4pi)/(3)-(2pi)/(24))t=2pi` `((30pi)/(24))t=2pi impliest=(24)/(15)=1.6hrs` |
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| 2311. |
A satellite of mass `m` is in an elliptical orbit around the earth. The speed of the satellite at its nearest position is `(6GM)//(5r)` where r is the perigee (nearest point) distance from the centre of the earth. It is desired to transfer the satellite to the circular orbit of radius equal to its apogee (farthest point) distance from the centre of the earth. The change in orbital speed required for this purpose isA. `0.35sqrt((GM_(e))/r)`B. `0.085sqrt((GM_(e))/r)`C. `sqrt((2GM_(3))/r)`D. zero |
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Answer» Correct Answer - B `E=m/2xx(6GM_(e))/(5r)-(GM_(e)m)/(2r)=-2/3(GM_(e)m)/r` which is the total energy of the earth satellite system. So, semi major axis of the elliptical orbit is `a =(5r)/4` Speed of the satellite at the apogee position is `v_(A)=(v_(P)xxr)/(2a-r)=2/3sqrt((6GM_(e))/(5r))` For orbit to change to a circle of radius `3r//2=(2a-r)` the rocket has to be fired when the satellite is at the apogee position. New orbital speed is ` v_(0)=sqrt((GM_(e))/(3r//w))=sqrt((2FGM_(e))/(3r))` Required change in the orbital speed is `/_v=v_(A)-v_(0)=0.085sqrt((GM_(e))/R)`. |
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| 2312. |
Two rings having masses `M` and `2M` respectively, having the same radius are placed coaxially as shown in the figure. If the mass distribution on both the rings is non-uniform, then the gravitational potential at point `P` isA. `-(GM)/R[1/sqrt2+2/sqrt(5)]`B. `-(GM)/R[1+2/2]`C. zeroD. cannot be determined from the given information |
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Answer» Correct Answer - A As all the points on the periphery of eigher ring are the same distance from point `P`, the potential at point `P` due to the whole ring can be calculated as `V=-(GM)//(sqrt(R^(2)+x^(2)))` where `x` is the axis distance from the centre of the ring. This expression is independent of the fact whether the distribution of mass is uniform or non uniform. So at `P,V=-(GM)/(sqrt(2)R)-(Gxx2M)/(sqrt(5)R)=-(GM)/R[1/(sqrt(2))+2/(sqrt(5))]` |
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| 2313. |
The escape velocities on the two planets of denities `rho_(1) and rho_(2)` and having same radius are `v_(1) and v_(2)` respectively. ThenA. `(v_(1))/(v_(2))=(rho_(1))/(rho_(2))`B. `(v_(2))/(v_(1))=(rho_(1))/(rho_(2))`C. `(v_(1))/(v_(2))=((rho_(1))/(rho_(2)))^(2)`D. `(v_(1))/(v_(2))=sqrt((rho_(1))/(rho_(2)))` |
| Answer» Correct Answer - d | |
| 2314. |
A satellite is seen after every `8` hours over the equator at a place on the earth when its sense of rotation is opposite to the earth. The time interval after which it can be seen at the same place when the sense of rotation of earth and satellite is same will be:A. `8h`B. `12h`C. `24h`D. `6h` |
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Answer» Correct Answer - C Given `g=(2pi)/(omega_(1)+omega_(2))=(2pi)/((2pi)/(T_(1))+(2pi)/(T_(2)))` `implies T_(1)=24h` for the earth `impliesT_(2)=12h` (`T_(2)` being the time period of the satellite, it will remain the same as the distance from the centre of the earth remains constant) `implies T=(2pi)/(omega_(2)+omega_(1))=(2pi)/((2pi)/(T_(1))-(2pi)/(T_(2))=24h` |
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| 2315. |
Assertion: Two different planets have same escape velocity. Reason: Value of escape velocity is a universal constant.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 - D As escape velocity `=sqrt((2GM)/R)`, so its value depends on mass of planet and radius of the planet. The two different planets have same escape velocities, when these quantities (mass and radius) are equal. |
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| 2316. |
A planet revolves about the sun in elliptical orbit. The arial velocity `((dA)/(dt))` of the planet is `4.0xx10^(16) m^(2)//s`. The least distance between planet and the sun is `2xx10^(12) m`. Then the maximum speed of the planet in `km//s` is -A. 10B. 20C. 40D. None of these |
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Answer» Correct Answer - C `(dA)/(dt) = (r^(2)omega)/(2)` is constant `therefore (dA)/(dt) = (r_(max)^(2)omega_(min))/(2)=(r_(min)^(2)omega_(max))/(2)` `rArr omega_(min) = (2dA//dt)/(r_(max)^(2))` `V_(max)=omega_(min)r_(min)=(2dA//dt)/(r_(min))=40k m//s` |
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| 2317. |
A planet revolves about the sun in elliptical orbit. The arial velocity `((dA)/(dt))` of the planet is `4.0xx10^(16) m^(2)//s`. The least distance between planet and the sun is `2xx10^(12) m`. Then the maximum speed of the planet in `km//s` is -A. 10B. 20C. 40D. 80 |
| Answer» Correct Answer - C | |
| 2318. |
The gravitational potential in a region is given by `V=20Nkg^-1(x+y)`. A. Show that the equation is dimensionally correct B. Find the gravitational field at the point `(x,y)` Leave your answer in terms of the unit vector `veci,vecj,veck`. C. Calculate the magnitude of the gravitational force on a particle of mass 500 g placed at the origin. |
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Answer» (a) `V=20(x+y)` `E_(x)=(delV)/(delx)=-20` `E_(y)=-(delV)/(dely)=-20` `vecE=E_(x)hati+E_(y)hatj` `=-20hati-20hatj` (b) `|vecE|=E=20sqrt(2)N//kg` `F=mE=0.5xx20sqrt(2)=10sqrt(2)N` |
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| 2319. |
A planet revolves about the sun in elliptical orbit. The arial velocity `((dA)/(dt))` of the planet is `4.0xx10^(16) m^(2)//s`. The least distance between planet and the sun is `2xx10^(12) m`. Then the maximum speed of the planet in `km//s` is -A. `10`B. `20`C. `40`D. None of these |
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Answer» Correct Answer - c `2m(dA)/(dt)=mvr` `2xx4xx10^(16)=v_(max)xx2xx10^(12)` `v_(max)=40 km//s` |
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| 2320. |
Satellites `A` and `B` are orbiting around the earth in orbits of ratio `R` and `4R` respectively. The ratio of their areal velocities is-A. `1:2`B. `1:4`C. `1:8`D. `1:16` |
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Answer» Correct Answer - a `L=mvrimpliesL=msqrt((GM)/r)r` `L=msqrt(GMr).....(1)` `L=2m(dA)/(dt)....(2)` From `(1)` and `(2)` `(dA)/(dt) propsqrt(r)implies(dA//dt)_(1)//(dA//dt)_(2)=sqrt(4/1)=2/1` |
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| 2321. |
The change in the value of acceleration of earth toward sun, when the moon coomes from the position of solar eclipse to the position on the other side of earth in line with sun is : (mass of moon `=7.36xx10^(22)` kg, orbital radius of moon `=3.8xx10^(8)` m.)A. `6.73xx10^(-2) m//s^(2)`B. `6.73xx10^(-3) m//s^(2)`C. `6.73xx10^(-4) m//s^(2)`D. `6.73xx10^(5) m//s^(2)` |
| Answer» Correct Answer - D | |
| 2322. |
Two astronauts are floating in gravitational free space after having lost contanct with their spaceship. The two will:A. Move towards each other.B. Move away from each other.C. Will become stationaryD. Keep floating at the same distance between them. |
| Answer» Correct Answer - D | |
| 2323. |
Two astronauts are floating in gravitational free space after having lost contanct with their spaceship. The two will:A. Move towards each otherB. Move away from each otherC. Will becomes stationaryD. Keep floating at the same distance between them |
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Answer» Correct Answer - A In the space, the external gravity is absent, but there will be a very small gravitational force between the astronauts, due to which both will move towards earth other with a very small acceleration. So, the best correct answer should be (a). |
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| 2324. |
Two astronauts are floating in gravitational free space after having lost contanct with their spaceship. The two will:A. keep floating at the same distance between themB. move towards each otherC. move away from each otherD. will beome stationary |
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Answer» Correct Answer - B |
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| 2325. |
Describe how an artificial satellite using two stage rocket is launched in an orbit around the Earth. |
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Answer» 1. Launching of a satellite in an orbit around the Earth cannot take place by use of single stage rocket. It requires minimum two stage rocket. 2. With the help of first stage of rocket, satellite can be taken to a desired height above the surface of the Earth. 3. Then the launcher is rotated in horizontal direction i.e.. through 900 using remote control and the first stage of the rocket is detached. 4. With the help of second stage of rocket, a specific horizontal velocity (vh) is given to satellite so that it can revolve in a circular path around the Earth. 5. The satellite follows different paths depending upon the horizontal velocity provided to it. |
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| 2326. |
Distinguish between inertial mass and gravitational mass. |
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Answer» Inertial mass of a body is a measure of its inertia. It is determined using Newton’s second law of motion. Gravitational mass is a measure of the gravitational pull acting on it. It is determined using Newton’s law of gravitation. |
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| 2327. |
Define the term Gravitational mass. |
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Answer» The mass of a body which determines the gravitational pull due to earth acting upon it is called gravitational mass. On the surface of the earth. Inertial mass of a body is equivalent to its gravitational mass. |
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| 2328. |
Why does ice float in water ? |
| Answer» The density of ice is less than that of water, so ice floats in water. | |
| 2329. |
What force acting on an area of 0.5 m2 will produce a pressure of 500 Pa ? |
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Answer» Pressure = Force/Area Force = Area x pressure = 0.5 x 500 = 250 N |
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| 2330. |
An object of weight 200 N is oating in a liquid. What is the magnitude of buoyant force acting on it ? |
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Answer» Correct Answer - 200 N |
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| 2331. |
What name is given to ‘thrust per unit area’ ? |
| Answer» Pressure is ‘thrust per unit area’. | |
| 2332. |
What name is given to trust per unit area ? |
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Answer» Correct Answer - Pressure |
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| 2333. |
lire relative density of mercury is 13.6. What does this statement mean ? |
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Answer» The relative density of mercury is 13.6, this means that mercury is 13.6 times as heavy as an equal volume of water. |
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| 2334. |
Name the principle which gives the magnitude of buoyant force acting on an object immersed in a liquid. |
| Answer» Archimedes’ Principle. | |
| 2335. |
Name the scientist who gave he magnitude of buoyant force acting on a solid object immersed in a liquid. |
| Answer» Archimedes gave the magnitude of buoyant force acting on a solid object immersed in a liquid. | |
| 2336. |
Why does a block of wood held under water rise to the surface when released ? |
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Answer» Because the weight of the block of wood is less than the weight of an equal volume of water. So when it is completely submerged in water, the upward buoyant force on it is greater than the downward gravitational force on it. Hence, the lock rises to the surface. |
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| 2337. |
Fill in the following blanks with suitable words : (a) Force acting on a unit area is called………………….. (b) It is the……………. force which makes objects appear lighter in water. (c) A heavy ship floats in water because its……………………. density is less than that of water. (d) In fluids (liquids and gases), pressure acts in……………………….. directions, and pressure………………… as the depth increases. (e) In order to sink in a fluid, the density of an object must be……………………… than the………………… of the fluid. (f) Snow shoes work by spreading out a person’s……………………… over a much bigger……………………….. (g) If the area of a snow shoe is five times……………………. than the area of an ordinary shoe, then the pressure of a snow shoe on the snow is five times…………………… |
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Answer» (a) pressure (b) buoyant (c) average (d) all; increases (e) less; density (f) weight; area (g) bigger; smaller |
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| 2338. |
What is the mass of `5m^(3)` of cement of density `3000 kg //m^(3)` ? |
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Answer» Correct Answer - 15000 kg |
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| 2339. |
The density of a body is 800 kg/m3. Will it sink or float when dipped in a bucket of water ? (Density’ of water = 1000 kg/m3 ). |
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Answer» The body will float when dipped in a bucket of water as its density is less than that of water. |
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| 2340. |
Wjhat is the density of a substance of mass 100 g and volume `10 cm^(3)` |
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Answer» Correct Answer - `10g//cm^(3)` |
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| 2341. |
The density of a body is `800 kg // m^(3)`.Will it sink of float when dipped in a bucket of water ? (Density of water `=1000 kg//m^(3)`) |
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Answer» Correct Answer - Float in water |
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| 2342. |
From an inertial frame of reference, explain the apparent weight for a person standing in a lift having zero acceleration. |
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Answer» 1. A passenger inside a lift experiences only two forces:
2. As these forces are oppositely directed, the net force in the downward direction will be F = ma – N. 3. Though the weight of a passenger is the gravitational force acting upon it, the person experiences his weight only due to the normal reaction force N exerted by the floor. 4. A lift has zero acceleration when the lift is at rest or is moving upwards or downwards with constant velocity. 5. Thus, a net force acting on the passenger inside the lift will be, F = 0 = mg – N ∴ mg = N Hence, in this case the passenger feels his normal weight mg. |
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| 2343. |
What happens to the apparent weight of the person inside the lift moving with net upward acceleration? |
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Answer» 1. The lift is said to be moving with net upward acceleration in two possible conditions:
2. As the net acceleration is upwards, the upward force must be greater. ∴ F = ma = N – mg ∴ N = mg + ma ∴ N > mg 3. Thus, for a passenger inside this lift, his apparent weight is more than his actual weight when the lift was not accelerated. |
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| 2344. |
When does a weighing machine will record zero for a passenger in a lift? |
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Answer» If the cables of the lift are cut, the downward acceleration of the lift, ad = g. In this case, we get, N = mg – mad = 0 |
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| 2345. |
A particle of mass an charge `q` is projected vertically upwards .A uniform electric field `vec(E)` is acted vertically downwards.The most appropriate graph between potential energy `U` (gravitation plus electrostatic) and height h(`lt lt` radius of earth) is :(assume `U` to be zero on surface of earth)A. B. C. D. |
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Answer» Correct Answer - A |
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| 2346. |
Why does a passenger feel lighter when the lift is about to stop at a higher floor during its upward motion? |
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Answer» 1. When the lift is about to stop at a higher floor during its upward motion it has a net downward acceleration. 2. As the net acceleration is downwards, the downward force must be greater. ∴ F = ma = mg – N ∴ N = mg – ma i.e., N < mg Hence, a passenger feels lighter when the lift is about to stop at a higher floor during its upward motion. |
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| 2347. |
The orbit velocity of a satellite at a height `R` above the surface of Earth is `v`. The escape velocity from the location isA. `sqrt(2)v`B. `2v`C. `4v`D. none of these |
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Answer» Correct Answer - A `v_(e)=sqrt((2GM)/r)` `=sqrt(2) sqrt((GM)/r)=sqrt(2)v` |
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| 2348. |
Mass `M=1` unit is divided into two parts `X` and `(1-X)`. For a given separation the value of `X` for which the gravitational force between them becomes maximum isA. `1//2`B. `3//5`C. `1`D. `2` |
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Answer» Correct Answer - A `F=(Gxxm(1-x)mx)/(R^(2))` is maximum when `x=1/2` |
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| 2349. |
Mass `M=1` unit is divided into two parts `X` and `(1-X)`. For a given separation the value of `X` for which the gravitational force between them becomes maximum isA. `1/2`B. `3/5`C. `1`D. `2` |
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Answer» Correct Answer - A |
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| 2350. |
Explain the reason of weightlessness inside a satellite.A. Zero gravityB. Centre of massC. Zero reaction force by satellite surfaceD. None |
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Answer» Correct Answer - C |
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