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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.
| 2101. |
A sphere of mass 25 kg attracts another sphere of mass 24kg with a force of 0.1 milligram weight. if distance between the centres of two sphere is 20 cm, what is the value of G ? |
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Answer» Here, `m_(1)=25kg, m_(2)=24kg` `F=0.1` milligram wt. `=10^(-7) kg wt. =9.8xx10^(-7)N` `d=20cm=(1)/(5)m, G=?` From `F=(Gm_(1)m_(2))/d^(2), G=(F xx d^(2))/(m_(1)m_(2))=(9.8xx10^(-7)xx((1)/(5))^(2))/(25xx24)=6.53xx10^(-11)Nm^(2)kg^(2)` |
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| 2102. |
Calculate the force of gravitation between the earth the sun, given that the mass of the earth `=6xx10^(24)`kg and mass of the sun `=2xx10^(30)`kg. The average distance between the two is `1.5xx10^(11)m`. |
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Answer» Here, `F=?,m_(1)=6xx10^(24)kg, m_(2)=2xx10^(30)kg` `r=1.5xx10^(11)m, G=6.7xx10-11Nm^(2)//kg^(2)` `F=(Gm_(1)m_(2))/r^(2)=(6.7xx10^(-11)xx(6xx10^(24))xx(2xx10^(30)))/((1.5xx10^(11))^(2))=3.57xx10^(22)N` |
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| 2103. |
The mass of the moon is ` (1)/(8)` of the earth but the gravitational pull is ` (1)/(6)` earth It is due to the fact that .A. moon is the satellite of the earthB. the radius of the earth is `8/6` of the moonC. the radius of the earth is `(sqrt(8//6))` of the moonD. the radius of the moon is `(6//8)` of the earth |
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Answer» Correct Answer - C `(gM)/(g_(E))=((G(M_(E)//8))/(R_(E)^(2)))/((GM_(E))/(R_(E)^(2)))=(R_(E)^(2))/(8R_(M)^(2))..........(i)` Where the subcipts M and E are for the moon and the earth respectively. Given, `(gM)/(g_(E))=1/6........(ii)` From (i) and (ii), we get `(R_(E)^(2))/(8R_(M)^(2))=1/6` or `R_(E)=sqrt(8/6)R_(M)` |
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| 2104. |
The mass of the moon is ` (1)/(8)` of the earth but the gravitational pull is ` (1)/(6)` earth It is due to the fact that .A. moon is the satellite of the earthB. the radius of the earth is (8/6) of the moonC. the radius of the earth is `(sqrt(8//6))` of the moonD. the radius of the moon is (6/8) of the earth |
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Answer» Correct Answer - C (c ) `(g_(M))/(g_(E))=((G(M_(E)//8))/(R_(M)^(2)))/((GM_(E))/(R_(E)^(2)))=(R_(E)^(2))/(8R_(M)^(2))`.......(i) where the subscript M and E are for the moon and the earth respectively. Given, `(g_(M))/(g_(E))=(1)/(6)`……(ii) From (i) and (ii), we get `(R_(E)^(2))/(8R_(M)^(2))=(1)/(6) or R_(E)=sqrt((8)/(6))R_(M)` |
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| 2105. |
Calculate the force of attraction between the earth and the sun, given that mass of earth is `6xx10^(24)`kg abd mass of sun `=2xx10^(30)`kg. the average distance between the two is `1.5xx10^(11)m.` |
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Answer» Here, `m_(1)=6xx10^(24)kg m_(2)=2xx10^(30)kg` `d=1.5xx10^(11)m F=?` As `F=(Gm_(1)m_(2))/d^(2)=(6.67xx10^(-11)xx6xx10^(24)xx2xx10^(30))/((1.5xx10^(11))^(2))=3.58xx10^(22)N` |
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| 2106. |
When a mango falls from a mango tree thenA. Only the earth attracts the mangoB. Only the mango attracts the earthC. Both mango and earth attract each otherD. Both repel each other |
| Answer» Correct Answer - C | |
| 2107. |
The mass of the moon is `1/81` of the earth but the gravitational pull is `1/6` of the earth. It is due to the fact thatA. the radius of earth is `(9)/(sqrt(6))` of the moonB. the radius of moon is `(81)/(6)` of the moonC. moon is the satellite of the earthD. None of the above |
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Answer» Correct Answer - A Gravitational pull depends upon the acceleration due to gravity on that planet `M_(,)=(1)/(81)M_(e),g_(m)=(1)/(6)g_(e)` By the relation, `g=(GM)/(R^(2))` `(R_(e))/(R_(m))=((M_(e))/(M_(m))xx(g_(m))/(g_(e)))^(1//2)(81xx(1)/(6))^(1//2):. R_(e)=(9)/(sqrt(6))R_(m)`. |
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| 2108. |
If the distance between the sun and the earth is increased by four times then the attraction between the two willA. Remain constantB. Decrease by 89%C. Decrease by 91%D. Decrease by 93.75% |
| Answer» Correct Answer - D | |
| 2109. |
Which of the following graphs between the square of the time period and cube of the distance of the planet from the sun is correct ?A. B. C. D. |
| Answer» Correct Answer - B | |
| 2110. |
If the mass of moon is `(M)/(81)`, where M is the mass of earth, find the distance of the point where gravitational field due to earth and moon cancel each other, from the centre of moon. Given the distance between centres of earth and moon is 60 R where R is the radius of earthA. 6RB. 8RC. 2RD. 4R |
| Answer» Correct Answer - A | |
| 2111. |
Two bodies of mass M and 4M kept at a distance y apart. Where should a small particle of mass m be placed from M so that the net gravitational force on it is zeroA. `(y)/(5)`B. `(y)/(2)`C. `(y)/(4)`D. `(y)/(3)` |
| Answer» Correct Answer - D | |
| 2112. |
If gravitational forces between a planet and a satellite is proportional to `R^(-5//2)`. If R is the orbit radius. Then the period of revolution of satellites is proportional to `R^(n)`. Find n. |
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Answer» Correct Answer - `7//2` |
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| 2113. |
Suppose, the acceleration due to gravity at the earth's surface is 10m/s2 and at the surface of Mars it is 4.0m/s2. A 60kg passenger goes from the earth to the Mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part of figure (11-Q1) best represents the weight (net gravitational force) of the passenger as a function of time. (a) A . (b) B. (c) C. (d) D. |
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Answer» The correct answer is (c) C. EXPLANATION: Since the acceleration due to gravity varies inversely to the square of the distance hence the apparent weight (net gravitational force) of the passenger with respect to time will not be a straight line but a curve. In between the earth and the mars, there will be a point where the gravitational force due to both of the bodies will be equal and opposite, hence the apparent weight will zero but never negative. Out of the three curves in the figure, only curve C fulfills this condition. |
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| 2114. |
If the acceleration due to gravity at the surface of earth is g, the work done in slowly lifting a body of mass m from the earth's surface to a height R equal to the radius of the earth is(a) 1/2mgR(b) 2mgR (c) mgR(d) 1/4mgR. |
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Answer» (a) The radius of the earth is 1/2mgR. |
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| 2115. |
Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing W on the earth will weigh (a) W (b) 2W (c) W/2 (d) 21/3W at the planet. |
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Answer» (d) 21/3 W at the planet. EXPLANATION: Since the density is same the volume and hence the radius of the planet will be more than the radius of the earth. Let the radius of the earth be R and that of the planet be R'. If the density is ρ, then 4πR'³ρ/3 = 2*4πR³ρ/3 →R'³ =2R³ →R' = 2¹/³R Acceleration due to gravity on this planet = G*2M/R'² (M is mass of the earth) =2GM/(2¹/³R)² =2⁽¹⁻²/³⁾(GM/R²) =2¹/³g Hence the weight on the planet =2¹/³*mg =2¹/³W. |
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| 2116. |
Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing W on the earth will weighA. `W`B. `2W`C. `W//2`D. `2^(1//3) W` at the planet |
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Answer» Correct Answer - D `M_(p)=rho.4/3piR_(p)^(3), M_(e)=rho.4/3piR_(e)^(3)` `(g_(p))/(g_(e))=(GM_(p)//R_(p)^(2))/(GM_(e).R_(e)^(2))=(M_(p))/(M_(e)).(R_(e)^(2))/(R_(p)^(2))=((M_(p))/(M_(e)))((M_(e))/(M_(p)))^(2//3)` `=((M_(p))/(M_(e)))^(1//3)=(2)^(1//3)` `(W_(p))/(W_(e))=(mg_(p))/(mg_(e))=(2)^(1//3)` `W_(p)=(2)^(1//3)W_(e)=(2)^(1//3) W` |
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| 2117. |
Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing W on the earth will weighA. WB. 2WC. `W//2`D. `2^(1//3)W` |
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Answer» Correct Answer - D |
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| 2118. |
Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing Won the earth will weigh(a) W(b) 2 W(c) W/2(d) 21/3 W at the planet. |
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Answer» (d) 21/3 W at the planet. EXPLANATION: Since the density is same the volume and hence the radius of the planet will be more than the radius of the earth. Let the radius of the earth be R and that of the planet be R'. If the density is ρ, then 4πR'³ρ/3 = 2*4πR³ρ/3 →R'³ =2R³ →R' = 2¹/³R Acceleration due to gravity on this planet = G*2M/R'² (M is mass of the earth) =2GM/(2¹/³R)² =2⁽¹⁻²/³⁾(GM/R²) =2¹/³g Hence the weight on the planet =2¹/³*mg =2¹/³W. |
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| 2119. |
Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing W on the earth will weighA. 2 wB. 3 wC. 3/wD. `2^(1//3)w` |
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Answer» Correct Answer - d `V_(P)=2V_(e)` `(4pi)/(3)xxR_(p)^(3)=2(4pi)/(3)xxR_(e)^(3)` `therefore" "R_(p)=root3(2)R_(e)` `F_(p)=(GM_(p)m)/(R_(p)^(2))` |
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| 2120. |
The earth `(mass = 10^(24) kg)` revolves round the Sun with an angular velocity `2 xx 10^(-7) rad s^(-1)` in a circular orbit of radius `1.5 xx 10^(8)km`. Find the force exerted by the Sun on the earth (in `xx 10^(21)N)`. |
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Answer» Correct Answer - 6 `F=mromega^(2)=10^(24)xx1.5x10^(8)xx10^(3)xx(2xx10^(-7))^(2)=6xx10^(21)N` |
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| 2121. |
Binding energy of moon and earth is :-A. `(GM_(e)M_(m))/r_(em)`B. `(GM_(e)M_(m))/(2r_(em))`C. `-(GM_(e)M_(m))/r_(em)`D. `-(GM_(e)M_(m))/(2r_(em))` |
| Answer» Correct Answer - B | |
| 2122. |
The earth (mass `=6xx10^(24)kg`) revolves round the sun with an angular velocity of `2xx10^(-7)"rad/s"` in a circular orbit of radius `1.5xx10^(8)km.` The gravitational force exerted by the sun on the earth, in newtons, isA. Zero NB. `18xx10^(25)N`C. `27xx10^(39)N`D. `36xx10^(21)N` |
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Answer» Correct Answer - D The force exerted by the sun on the earth is the gravitational force given by `F=(GMm)/(R^(2))=(mv^(2))/(r )= mr omega^(2)` `therefore F = mr omega^(2)` `= 6xx10^(24)xx1.5xx10^(8)xx10^(3)xx2xx10^(-7)xx2xx10^(-7)` `therefore F = 36xx10^(21)N` |
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| 2123. |
The earth (mass `=6xx10^(24)kg`) revolves round the sun with an angular velocity of `2xx10^(-7)"rad/s"` in a circular orbit of radius `1.5xx10^(8)km.` The gravitational force exerted by the sun on the earth, in newtons, isA. zeroB. `18xx10^(25)`C. `36xx10^(21)`D. `3.6xx10^(18)` |
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Answer» Correct Answer - c `F=mr omega^(2)` |
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| 2124. |
The earth (mass `=6xx10^(24)kg`) revolves round the sun with an angular velocity of `2xx10^(-7)"rad/s"` in a circular orbit of radius `1.5xx10^(8)km.` The gravitational force exerted by the sun on the earth, in newtons, isA. zeroB. `18xx10^(25)`C. `27xx10^(39)`D. `36xx10^(21)` |
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Answer» Correct Answer - C `F=mR omega^(2)=6xx10^(24)xx1.5xx10^(11)(2xx10^(-7))^(2)` `=36xx10^(21)N` |
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| 2125. |
The average radii of orbits of mercury and earth around the sun are `6xx10^(7)` km and `1.5xx10^(8)` km respectively. The ratio of their orbital speeds will be :-A. `sqrt(5) : sqrt(2)`B. `sqrt(2) : sqrt(5)`C. `2.5 :1`D. `1 : 25` |
| Answer» Correct Answer - A | |
| 2126. |
It is not possible to keep a geo-stationary satellite over Delhi. Since DelhiA. is not present in `A.P`B. is capital of indiaC. is not in the equatorial plane of the earthD. is near Agra. |
| Answer» Correct Answer - C | |
| 2127. |
The orbit of geo-stationary satellite is circular, the time period of satellite dependsd onA. mass of the EarthB. radius of the orbitC. height of the satellite from the surface of EarthD. all the above |
| Answer» Correct Answer - D | |
| 2128. |
A body is dropped by a satellite in its geo-stationary orbit :A. it will burn on entering in to the atmosphereB. it will remain in the same place with respect to the earthC. it will reach the earth is 24 hoursD. it will perform uncertain motion |
| Answer» Correct Answer - B | |
| 2129. |
It is possible to keep a geo-stationary satellite in an orbit so that it always remains overA. New DelhiB. PuneC. NewyorkD. Any place on the equator |
| Answer» Correct Answer - d | |
| 2130. |
If a satellite performes elliptical path, thenA. `v_(h) lt v_(c)`B. `v_(c)=v_(h)`C. `v_(c) lt v_(h) lt v_(e)`D. `v_(h) gt v_(e)` |
| Answer» Correct Answer - c | |
| 2131. |
A satellite of mass m moves along an elliptical path arouned the earth. The areal velocity of the satellite is proportional toA. mB. `m^(-1)`C. `m^(0)`D. `m^(1//2)` |
| Answer» Correct Answer - C | |
| 2132. |
Which of the following is correct?A. an astronaut in going from earth to moon will experience weightlessness once.B. When a thin unform spherical shell gradually shrinks maintaining its shape, the gravitational potential at its centre decreases.C. In the case fo spherical shell, the piot of `V` versus `r` is continuous.D. In the case spherical shell, the plot of gravitational field intensity, I versus `r`, is continuous |
| Answer» Correct Answer - A::B::C | |
| 2133. |
When do you say that a body is freely falling? |
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Answer» A body is said to be free-fall body when only one gravitational force acts on that body. |
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| 2134. |
The centre of mass of an extended body on the surface of the earth and its centre of gravityA. are always at the same point for any size of the bodyB. are always at the same point only for spherical bodiesC. can never be at the same pointD. in close to each other for objects, say of sizes less than `100 m` |
| Answer» Correct Answer - D | |
| 2135. |
Define weight. |
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Answer» Weight of a body is the force of attraction on the body due to earth. W = mg |
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| 2136. |
Give one example each of central force and non-central force. |
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Answer» Central force; gravitational force of a point mass, electrostatic force due to a point charge. Non-central force: spin-dependent nuclear forces, magnetic force between two current carrying loops. |
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| 2137. |
The centre of mass of an extended body on the surface of the earth and its centre of gravity (a) are always at the same point for any size of the body. (b) are always at the same point only for spherical bodies. (c) can never be at the same point. (d) is close to each other for objects, say of sizes less than 100 m. (e) both can change if the object is taken deep inside the earth. |
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Answer» (d) is close to each other for objects, say of sizes less than 100 m. |
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| 2138. |
Molecules in air in the atmosphere are attracted by gravitational force of the earth. Explain why all of them do not fall into the earth just like an apple falling from a tree. |
| Answer» Molecules experience the vertically downward force due to gravity just like an apple falling from a tree. Due to thermal motion, which is random, their velocity is not in the vertical direction. The downward force of gravity causes the density of air in the atmosphere close to earth higher than the density as we go up. | |
| 2139. |
What is centre of gravity? |
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Answer» The point where total weight appears to act is called centre of gravity. |
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| 2140. |
A car falls off a ledge and drops to the ground in `0.5 s`. let` g=10 m//s^(2)` (for simplifyng the calculation). (i) what is its speed on striking the ground ? what is its average speed during 0.5s? (iii) How high is the ledge from the ground ? |
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Answer» Correct Answer - (i) `5m//s` (ii)` 2.5 m//s` (iii)`1.25m` Here, `u = 0, t = 0.5 s, g = 10 m//s^(2)` (i) `v = u +g t = 0 +10 xx 0.5 = 5 m//s` (ii) Average spped `=(u +v)/(2) = (0+5)/(2) = 2/5 m//s` (iii) From `h = ut +(1)/(2) g t^(2), h = 0+(1)/(2) xx 10(0.5)^(2) = 1.25m` |
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| 2141. |
A car falls off an edge and drops to the ground in 0.5 s. i) What is its speed on striking the ground?ii) What is the average speed during the 0.5 s? iii) What is the height of the edge from the ground? |
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Answer» i) t = 0.5 s, u = 0, v = ? ,a = +g = 10 m/s2 . v = u +at = 0 + 10 x 0.5 = 5 m/s ii) Average speed = (u + v)/2 = (0 + 5 )/ 2 = 2.5 m/s. iii) Distance travelled s = 1/2 gt2 = 1/2 x 10 x 0.52 =1/2 x 10 x 0.25 = 1.25 m. |
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| 2142. |
When will a body be stable? |
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Answer» If we draw a line straight down from the centre of gravity of an object of any shape and it falls inside the base of the object, then the object will be stable. |
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| 2143. |
What is time period? |
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Answer» The time taken by a body executing uniform circular motion, to complete one revolution is called time period. |
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| 2144. |
When could we say that an object is stable? |
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Answer» If we draw a line straight down from the centre of gravity of an object of any shape and it falls inside the base of the object, then the object will stable. If the line through the centre of gravity falls outside the base then the object will be unstable. |
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| 2145. |
The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why? |
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Answer» Moon attracts with same gravitational forces as earth attracts moon. They form action reaction pair. |
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| 2146. |
Molecules in air in the atmosphere are attracted by gravitational force of the earth. Explain why all of them do not fall into earth just like an apple falling from a tree. |
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Answer» Air molecules in the atmosphere experience the vertically downward force due to gravity just like an apple falling from a tree. Due to thermal motion, air molecules move randomly, their velocity is not in the vertical direction. The downward force gravity causes the density of air in the atmosphere close to earth higher than the density as we go up. But in apple’s case, only vertical motion dominates because of heavier molecules than air molecules. |
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| 2147. |
A body of mass 1 kg is attracted by the Earth with a force which is equal to _______________A. `9.8N`B. `6.67xx10^(-1)`C. `1N`D. `9.8m//s` |
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Answer» Correct Answer - A `9.8N` |
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| 2148. |
The weight of body gradually decreases from ______________.A. equator of polesB. poles to equatorC. pole to poleD. height to surface |
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Answer» Correct Answer - B poles to equator |
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| 2149. |
A geo-stationary satellite has an orbital period ofA. 2 hB. 6 hC. 12 hD. 24 h |
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Answer» Correct Answer - d T = 24 h Period of Geostationary satellite is 24 h. |
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| 2150. |
The angular velocity of rotation of star (of mass M and radius R ) at which the matter start to escape from its equator will beA. `sqrt((2GM)/(M))`B. `sqrt((2GM)/(R^(3)))`C. `sqrt((2GM)/(R))`D. `sqrt(((2GM)^(2))/(R))` |
| Answer» Correct Answer - b | |