

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.
2551. |
Figure shows the orbit of a planet `P` round the sun `S`. `AB` and `CD` are the minor and major axed of the ellipse. If `t_(1)` is the time taken by the planet to travel along `ACB` and `t_(2)` the time along `BDA`, thenA. `t_(1)=t_(2)`B. `t_(1)gt t_(2)`C. `t_(1)lt t_(2)`D. nothing can be concluded |
Answer» Correct Answer - B Since arial velocity is constant so `t_(1) gt t_(2)`. |
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2552. |
A mass of `6xx10^24` kg (equal to the mass of the earth) is to be compressed in a sphere in such a way that the escape velocity from its surface is 3xx10^8ms^-2`. What should be the radius of the sphere? |
Answer» The mass of the sphere `=6xx10^24kg` `Ve=(2GM)/R` or `R=(2GM)/(Ve^2)` `=(2xx6.67xx10^-11xx6xx10^24)/((3xx10^8)^2)` =2xx40.02xx10^13)/(9xx10^16)` `=80.02/9xx10^-3m` `=8.89xx10^-3m` `=9mm.` |
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2553. |
The maximum and minimum distance of a comet form the sun are `8xx10^(12)m and 1.6xx10^(12)m`. If its velocity when nearest to the sun is `60m//s`, what will be its velocity in m/s when it is farthestA. 12B. 60C. 112D. 6 |
Answer» Correct Answer - A | |
2554. |
A mass of `6xx10^24` kg (equal to the mass of the earth) is to be compressed in a sphere in such a way that the escape velocity from its surface is `3xx10^8ms^-2`. What should be the radius of the sphere?A. 9 mmB. 8 mmC. 7 mmD. 6 mm |
Answer» Correct Answer - A `As,v_(e)=sqrt(((2GM)/(R))),R=((2GM)/(v_(e)^(2))),thereforeR=(2xx6.67xx10^(-11)xx6xx10^(24))/((3xx10^(8))^(2-))=9xx10^(-3)m=9mm` |
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2555. |
Geostationary satellite :-A. is situated at a great geight above the surface of earthB. moves in equatorial planeC. have time period of 24 hoursD. have time period of 24 hours and moves in equatorial plane |
Answer» Correct Answer - D | |
2556. |
Two identical satellite are at `R` and `7R` away from earth surface, the wrong statement is (`R`=radius of earth)A. Ratio of total energy of both is 5B. Ratio of kinetic energy of both is 4C. Ratio of potential energy of both 4D. Ratio of total energy of both is 4 |
Answer» Correct Answer - A | |
2557. |
Gravitational potential difference between a point on surface of planet and another point 10 m above is 4J/kg. Considering gravitational field to be uniform, how much work is done in moving a mass of 2kg from the surface to a point 5 m above the surface? |
Answer» Correct Answer - 4J `W_("ext")+W_(g)=4K = 0` `W_("ext")-m_(4) V = 0` Wext `= 2xx(4)/(2)=4J` |
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2558. |
Gravitational potential difference between a point on surface of planet and another point 10 m above is 4J/kg. Considering gravitational field to be uniform, how much work is done in moving a mass of 2kg from the surface to a point 5 m above the surface?A. 4 JB. 5 JC. 6 JD. 7 J |
Answer» Correct Answer - A Gravitational field `g=- (Delta V)/(Delta x)=- ((-4)/10)=4/10` J/kg-m Work done in moving a mass of 2 kg from the surface to a point 5 m above the surface, `W=mgh=(2 kg) (4/10 J/("kg - m")) (5 m)=4 J` |
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2559. |
The atmospheric pressure and height of barometer column is `10^(5)P_(e)` and 760 mm respectively on the earth surface. If the barometer is taken to the moon then column height will be-A. zeroB. 76 mmC. 126.6 mmD. 760 mm |
Answer» Correct Answer - A There is no atmosphere on the moon. |
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2560. |
Two identical satellite are at `R` and `7R` away from earth surface, the wrong statement is (`R`=radius of earth)A. Ratio of total energy of both is5B. ratio of kinetic energy of both is 4C. ratio of potential energy of both 4D. ratio of total energy of both is 4 and ratio of magnitude fo potential to kinetic energy is 2. |
Answer» Correct Answer - A `K.E.=+(1)/(2)(GM_(1)M_(2))/(r)` `r=2R` for the first and `r=8R` for the `II^(nd)` `(K.E_(1))/(K.E_(2))=((1)/(2R)(8R)/(1))=4:1` Similarly P.E. is `implies-(GM_(1)M_(2))/(R).(P.E_(1))/(P.E_(2))=4:1` Put the ratio of `(K.E)/(P.E)=2` |
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2561. |
Gravitational potential difference between a point on surface of planet and another point 10 m above is 4J/kg. Considering gravitational field to be uniform, how much work is donw in moving a mass of 2kg from the surface to a point 5 m above the surface?A. 4JB. 5JC. 6JD. 7J |
Answer» Correct Answer - A Gravitational field `g=-(DeltaV)/(Deltax)=-((-4)/(10))=(4)/(10)J//kg` m Work done in moving a mass of 2kg from the surface to a point 5m above the surface `W=mgh=(2kg)((4)/(10)(J)/(kgm))(5m)=4J` |
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2562. |
A particle is projected from point `A`, that is at a distance `4R` form the centre of the earth, with speed `V_(1)` in a direction making `30^(@)` with the line joining the centre of the earth and point `A`, as shown. Consider gravitational interaction only between thesetwo. (Use `(GM)/R=6.4xx10^(7) m^(2)//s^(2)`). The speed `V_(1)` if particle pasess grazing the surface of the earth is A. `(8000)/(sqrt(2))`B. `800`C. `800sqrt(2)`D. None of these |
Answer» Correct Answer - A Conserving angular momentum `:m(v_(1)cos60^(@))4R=mv_(2)Rimplies(v_(2))/(v_(1))=2` Conserving energy of the system `:(GMm)/(4R)+(1)/(2)mv_(1)^(2)=(GMm)/(R)+(1)/(2)mv_(2)^(2)` `implies(1)/(2)v_(2)^(2)-(1)/(2)v_(1)^(2)=(3)/(4)(GM)/(R)impliesv_(1)^(2)=(1)/(2)(GM)/(R)impliesv_(1)=(1)/(sqrt(2))sqrt(64xx10^(6))=(8000)/(sqrt(2))m//s` |
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2563. |
Assertion : It we double the circular radius of a satellite, then its potential energy, kinetic energy and total mechanical energy will become half. Reason : Orbital speed of a satellite. `upsilon prop (1)/(sqrt(r )` where, `r` is its radius of orbit.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. |
Answer» Correct Answer - B `V = - (GMm)/(R)`, `K = (GMm)/(2r)`, `E = - (GMm)/(2r)` and `V = sqrt((GM)/(R))` |
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2564. |
Assertion : If the radius of earth is decreased kepping its constant, effective value of g may increase or decrease at pole. Reason : Value of g on the surface of earth is given by `g= (Gm)/(R^(2))`.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. |
Answer» Correct Answer - D By changing the radius, moment of inertia will change. Hence, angular speed `omega` will change. But `omega` has no effect on the value of g on pole. |
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2565. |
For the planet - sun system identify the correct satatement.A. the angular momentum of the planet is conserved about any pointB. the total energy of the system is conservedC. the momentum of the planet is conservedD. All of the above |
Answer» Correct Answer - B Angular momentum is conserved only about centre of sun. |
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2566. |
Why Is the law of gravitation known as universal law of gravitation? |
Answer» The law of gravitation is applicable to all material objects in the universe. Hence it is known as the universal law of gravitation. |
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2567. |
A straight rod of length `L` extends from `x=a` to `x=L+a`. Find the gravitational force exerts on a point mass `m` at `x=0` is (if the linear density of rod`mu=A+Bx^(2)`)A. `Gm(a(1/(alpha)-(1)/(alpha+1))+bl)`B. `(Gm(a+bx^(2)))/(l^(2))`C. `Gm(a(1/(alpha)-(1)/(alpha+1))+bl)`D. `Gm(a(1/(alpha+1)-(1)/(alpha))+bl)` |
Answer» Correct Answer - A |
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2568. |
You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means? |
Answer» Gravitational forces are independent of the intervening medium. A body can not be shielded from the gravitational influence of nearby matter. |
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2569. |
Answer the following:You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (You can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why? |
Answer» (a) No (b) Yes Gravitational influence of matter on nearby objects cannot be screened by any means. This is because gravitational force unlike electrical forces is independent of the nature of the material medium. Also, it is independent of the status of other objects. If the size of the space station is large enough, then the astronaut will detect the change in Earth’s gravity (g). Tidal effect depends inversely upon the cube of the distance while, gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun’s pull. |
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2570. |
Find the gravitational potential energy of a body of mass 10 kg when it is at a height of 6400 km from the earth’s surface.[Given: a mass of the earth and radius of the earth.] |
Answer» Answer is -3.127 × 108 J |
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2571. |
Find the gravitational potential energy of a body of mass 200 kg on the earth’s surface. [M(earth) = 6 × 1024 kg, R(earth) = 6400 km] |
Answer» Answer is -1.251 × 1010 J |
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2572. |
The gravitational constant G is equal to 6.67 × 10-11 N m2/kg2 in vacuum. Its value in a dense matter of density 1010 g/cm3 will be(A) 6.67 × 10-1 N m2/kg2(B) 6.67 × 10-11 N m2/kg2(C) 6.67 × 10-10 N m2/kg2(D) 6.67 × 10-21 N m2/kg2 |
Answer» (B) 6.67 × 10-11 N m2/kg2 |
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2573. |
Distinguish between the following:universal gravitational constant and gravitational acceleration of the earth. |
Answer» Universal gravitational constant:
Gravitational acceleration of the earth:
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2574. |
Gravitational force between any two bodies is always ………… type only. A) attractive B) repulsiveC) both D) none of these |
Answer» Correct option is A) attractive |
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2575. |
Distinguish between the following:mass and weight |
Answer» (1) Mass:
Weight:
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2576. |
When a body falls freely then it experiences ………………. A) weight B) mass C) weightlessness D) none of these |
Answer» C) weightlessness |
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2577. |
On observing the given figure, tell the theory explained byA) Rutherford B) Newton C) Bohr D) None |
Answer» Correct option is B) Newton |
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2578. |
The bodies of masses 40 kg and 80 kg are at a distance of `0.15` m from each other . Two force of gravitation between the bodies is `1.0` mg wt. calculate the constant of gravitation . Take g = `10 ms^(-2)` . |
Answer» Correct Answer - `7.0 xx 10^(-11) Nm^(2) kg^(-2)` |
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2579. |
Distance between the centres of two stars is `10a`. The masses of these stars are `M` and `16 M` and their radii `a` and `2a` respectively. A body of mass `m` is fired straight from the surface of the larger star towards the surface of the smaller star. What should be its minimum initial speed to reach the surface of the smaller star? Obtain the expression in terms of `G`, `M` and `a`. |
Answer» Correct Answer - 3 As `v_(e)=sqrt((2GM)/R)` `R=(2GM)/(v_(e)^(2))=(2xx6.67xx10^(-11)xx6xx10^(24))/((3xx10^(8))^(2))~~8.89xx10^(-3)m` `:. R~~9 mm` |
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2580. |
The density of the core a planet is `rho_(1)` and that of the outer shell is `rho_(2)`. The radii of the core and that of the planet are `R` and `2R` respectively. The acceleration due to gravity at the surface of the planet is same as at a depth `R`. Find the ratio of `(rho_(1))/(rho_(2))` |
Answer» Correct Answer - `[(rho_(1))/(rho_(2)) = 7/3]` |
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2581. |
The density of the core a planet is `rho_(1)` and that of the outer shell is `rho_(2)`. The radii of the core and that of the planet are `R` and `2R` respectively. The acceleration due to gravity at the surface of the planet is same as at a depth `R`. Find the ratio of `(rho_(1))/(rho_(2))` A. `3//4`B. `5//3`C. `7//3`D. `3//5` |
Answer» Correct Answer - C Let `m_(1)` is mass of core and `m_(2)` outer portion `m_(1)=4/3piR^(3) rho_(1), m_(2)=4/3pi[(2R)^(3)-R^(3)] rho_(2)` Given that `(Gm_(1))/(R^(2))=(G(m_(1)+m_(2)))/((2R)^(2)) implies (rho_(1))/(rho_(2))=7/3` |
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2582. |
The density of the core of planet is `rho_(1)` and that of the outer shell is `rho_(2)`. The radius of the core that of the planet are R and 2R respectively. Gravitastional intensity at the surface of the planet is same as at a depth R. Find the ratio `(rho_(1))/(rho_(2))` , assuming them to be uniform independently. |
Answer» Correct Answer - `7//3` `(GM_("core"))/(R^(2))=(GM_("Total"))/(4R^(2)), (G(4)/(3)piR^(3)rho)/(R^(2))=(G(4)/(3)piR^(3)[rho_(1)+7rho_(2)])/(4R^(2))` `4rho_(1)=rho_(1)+7rho_(2) rArr (rho_(1))/(rho_(2))=(7)/(3)` |
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2583. |
What will be acceleration due to gravity on the surface of the moon if its radius were `(1//4)^(th)` the radius of earth and its mass `(1//80)^(th)` the mass of earth? What will be the escape velocity on the surface of moon if it is `11.2 km//s` on the surface of the earth? (given that `g=9.8 m//s^(2)`) |
Answer» `R_(m)=(R_(e))/4,M_(m)=(M_(e))/80` `g_(e)=(GM_(e))/(R_(e)^(2)),g_(m)=(GM_(m))/(R_(m)^(2))` `(g_(m))/(g_(e))=(M_(m))/(M_(e)).((R_(e))/(R_(m)))^(2)=1/80xx(4)^(2)=1/5` `g_(m)=(g_(e))/5=9.8/5=1.96m//s^(2)` `(v_(e))_(e)=sqrt((2GM_(e))/(R_(e))), (v_(e))_(m)=sqrt((2GM_(m))/(R_(m)))=sqrt((2GM_(e))/(80R_(e)//4))` `=1/(sqrt(20)) v_(e)` `(v_(e))_(m)=11.2/(sqrt(20))=2.5 km//s` |
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2584. |
The density of the core a planet is `rho_(1)` and that of the outer shell is `rho_(2)`. The radii of the core and that of the planet are `R` and `2R` respectively. The acceleration due to gravity at the surface of the planet is same as at a depth `R`. Find the ratio of `(rho_(1))/(rho_(2))` A. `3/7`B. `9/4`C. `7/3`D. `3/8` |
Answer» Correct Answer - C Let `m_(1)` be the mass of the core and `m_(2)` be the mass of outer shell. `g_(A)=g_(B)` (given) Then `(Gm_(1))/(R^(2))=(G(m_(1)+m_(2)))/((2R)^(2))` `:. 4m_(1)=(m_(1)+m_(2))` or `4{4/3piR^(2)rho_(1)}=4/3piR^(3)rho_(1)+{4/3pi(2R)^(3)-4/3piR^(3}rho_(2)` `:. 4rho_(1)=rho_(1)+7rho_(2)rArr (rho_(1))/(rho_(2))=7/3` |
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2585. |
Two planets revolve round the sun with frequencies `N_(1)` and `N_(2)` revolutions per year. If their average orbital radii be `R_(1)` and `R_(2)` respectively, then `R_(1)//R_(2)` is equal toA. `(N_(1)//N_(2))^(3//2)`B. `(N_(2)//N_(1))^(3//2)`C. `(N_(1)//N_(2))^(2//3)`D. `(N_(2)//N_(1))^(2//3)` |
Answer» Correct Answer - d `Tprop R^(3//2)" as "T_(1)=(1)/(N_(1))` `therefore" "(1)/(N_(1)) prop R_(1)^(3//2)" and ""(1)/(N_(2)) prop R_(2)^(3//2)` `(N_(2))/(N_(1))=((R_(1))/(R_(2)))^(3//2) or (R_(1))/(R_(2))=((N_(2))/(N_(1)))^(2//3)` |
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2586. |
Two satellites of masses of `m_(1)` and `m_(2)(m_(1)gtm_(2))` are revolving round the earth in circular orbits of radius `r_(1)` and `r_(2)(r_(1)gtr_(2))` respectively. Which of the following statements is true regarding their speeds `v_(1)` and `v_(2)`?A. `v_(1)=v_(2)`B. `v_(1) lt v_(2)`C. `v_(1)gt v_(2)`D. `(v_(1))/(r_(1))=(v_(2))/(r_(2))` |
Answer» Correct Answer - b `v=sqrt((GM)/(R+h))" "thereforev=Rsqrt((g)/(R+h))` |
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2587. |
Two satellites of masses of `m_(1)` and `m_(2)(m_(1)gtm_(2))` are revolving round the earth in circular orbits of radius `r_(1)` and `r_(2)(r_(1)gtr_(2))` respectively. Which of the following statements is true regarding their speeds `v_(1)` and `v_(2)`?A. `v_(1)=v_(2)`B. `v_(1) lt v_(2)`C. `v_(1) gt v_(2)`D. `v_(1)/r_(1)=v_(2)/r_(2)` |
Answer» Correct Answer - B |
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2588. |
Two satellites of masses of `m_(1)` and `m_(2)(m_(1)gtm_(2))` are revolving round the earth in circular orbits of radius `r_(1)` and `r_(2)(r_(1)gtr_(2))` respectively. Which of the following statements is true regarding their speeds `v_(1)` and `v_(2)`?A. `v_(1) = v_(2)`B. `v_(1) lt v_(2)`C. `v_(1 gt v_(2)`D. `(v_(1))/(r_(1))=(v_(2))/(r_(2))` |
Answer» Correct Answer - B `v=sqrt((GM)/r)` if `r_(1) gt r_(2)` then `v_(1) lt v_(2)` Orbital speed of satellite does not depend upon the mass of the satellite. |
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2589. |
Two satellites of masses of `m_(1)` and `m_(2)(m_(1)gtm_(2))` are revolving round the earth in circular orbits of radius `r_(1)` and `r_(2)(r_(1)gtr_(2))` respectively. Which of the following statements is true regarding their speeds `v_(1)` and `v_(2)`?A. `v_(1)=v_(2)`B. `v_(1)gtv_(2)`C. `v_(1)ltv_(2)`D. `(v_(1))/(r_(1))=(v_(2))/(v_(2))` |
Answer» Correct Answer - C `m_(1)v_(1)^(2)= m_(1)(GM)/(r_(1)^(2)), (m_(2)v_(2)^(2))/(r_(2))=m_(2)(GM)/(r_(2)^(2))` `(v_(1)^(2))/(v_(2)^(2))=(r_(2))/(r_(1)),r_(2)ltr_(1)impliesv_(1)^(2)ltv_(2)^(2)` Hence `v_(1)ltv_(2)` |
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2590. |
Time period of revolution of poalr satellite is aroungA. 6 minutesB. 100 minutesC. 8 hoursD. 24 hours |
Answer» Correct Answer - B | |
2591. |
Escape velocity on a planet is e v . If radius of the planet remains same and mass becomes 4 times, the escape velocity becomesA. `4 v_(e)`B. `2 v_(e)`C. `v_(e)`D. `1/2 v_(e)` |
Answer» Correct Answer - B |
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2592. |
A coke can of negligible mass is in the shape of a cylinder. Its volume is 500 ml and its base area is `(100)/(3)cm^(2)`. A person consumes nearly 25 ml of coke for every sip. After he consumes 12 sips of the drink, what is the height of the center of gravity of the can from its base when placed vertically? If the mass of the can is not negligible, how would this answer vary? |
Answer» (i) Find the volume and base area of the coke can from the given data. Find the centre of gravity `(G_(1))`, when the coke can is filled completely with coke. Does this `G_(1)` lie at half of the height of the cylinder? Find the amount of volume of coke left after 12 sips. Find the height of the coke in the can by the formula, `h = ("volume")/("base area")` Then the centre of gravity `(G_(2))` will be equal to (h/2). If the mass of the can is not negligible, will this have its own centre of gravity? Will the position `G_(2)` shifts upwards? (ii) 3 cm |
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2593. |
A body weighs 1 kg on the surface of earth. What is its mass on moon ? |
Answer» Mass on moon=mass on earth=1 kg | |
2594. |
A person weighs 600 N earth. Find his mass (g = 10 ms-2). |
Answer» Given, weight on earth (W) = 600 Ng= 10 ms-2 Let the mass on earth be m. As we know, Weight = mass × acceleration due to gravity placeW = mg ⇒ m = \(\frac{W}{g}\) = \(\frac{100}{60}\) = 60 kg. |
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2595. |
Given mass of the moon is 1/81 of the mass of the earth and corresponding radius is 1/4 of the earth. If escape velocity on the earth surface is 11.2 km / s , the value of same on the surface of the moon isA. 0.14 km / sB. 0.5 km / sC. 2.5 km / sD. 5 km / s |
Answer» Correct Answer - C |
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2596. |
The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon. The ratio of the escape velocity on the surface of earth to that on the surface of moon will beA. 0.2B. 2.57C. 4.81D. 0.39 |
Answer» Correct Answer - C Esacpe velocity `v_(e )=sqrt(2GM)/(R )` `therefore (v_(e ))/(v_(m))=sqrt(M_(e )R_(m))/(M_(m)R_(e))=sqrt(81)/(3.5)=4.81` |
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2597. |
If he weighs 100 N on the moon, what is the acceleration due to gravity on the moon? |
Answer» Given, Weight on the moon (W’) = 100 NMass (m)=60 kg Let the acceleration due to gravity on the moon be g’. Weight = mass x acceleration due to gravity at that place ⇒ W' = m x g' ⇒ g' = \(\frac{W}{m}\) = \(\frac{100}{60}\) = 1.67 ms-2 |
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2598. |
Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem? |
Answer» (b), (c), and (d) Legs hold the entire mass of a body in standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen feet of an astronaut do not affect him/her in space. A swollen face is caused generally because of apparent weightlessness in space. Sense organs such as eyes, ears nose, and mouth constitute a person’s face. This symptom can affect an astronaut in space. Headaches are caused because of mental strain. It can affect the working of an astronaut in space. Space has different orientations. Therefore, orientational problem can affect an astronaut in space. |
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2599. |
A planet has twice the radius but the mean density is `1/4` th as compared to earth. What is the ratio of escape velocity from earth to that from the planetA. `3:1`B. `1:2`C. `1:1`D. `2:1` |
Answer» Correct Answer - c `(v_(rho_(2)))/(v_(rho_(1)))=(R_(2))/(R_(1))sqrt((rho_(2))/(rho_(1)))=2xxsqrt((1)/(4))=2xx(1)/(2)=1` `(v_(rho_(2)))/(v_(rho_(1)))=(1)/(1)` |
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2600. |
A planet has twice the radius but the mean density is `1/4` th as compared to earth. What is the ratio of escape velocity from earth to that from the planetA. `3 : 1`B. `1 : 2`C. `1 : 1`D. `2 : 1` |
Answer» Correct Answer - C |
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