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Gravitation

JEE Physics · 166 questions · Page 8 of 17 · Click an option or "Show Solution" to reveal answer

Q71
If RE be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is : (Given : r < RE)
A 1−rRE−r2RE2−r3RE31 - {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}}
B 1+rRE+r2RE2+r3RE31 + {r \over {{R_E}}} + {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}}
C 1+rRE−r2RE2+r3RE31 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}}
D 1+rRE−r2RE2−r3RE31 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}}
Correct Answer
Option D
Solution
gup=g(1+rR)2{g_{up}} = {g \over {{{\left( {1 + {r \over R}} \right)}^2}}}
gdown=g(1−rR){g_{down}} = g\left( {1 - {r \over R}} \right)
gdowngup=(1−rR)(1+rR)2{{{g_{down}}} \over {{g_{up}}}} = \left( {1 - {r \over R}} \right){\left( {1 + {r \over R}} \right)^2}
=(1−rR)(1+2rR+r2R2)= \left( {1 - {r \over R}} \right)\left( {1 + {{2r} \over R} + {{{r^2}} \over {{R^2}}}} \right)
=1+rR−r2R2−r3R3= 1 + {r \over R} - {{{r^2}} \over {{R^2}}} - {{{r^3}} \over {{R^3}}}
Q72
The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three times its previous value, then approximate new time period of the satellite will be
A 40 hours
B 36 hours
C 30 hours
D 25 hours
Correct Answer
Option B
Solution
T22=(R2R1)3T12T_2^2 = {\left( {{{{R_2}} \over {{R_1}}}} \right)^3}T_1^2
⇒T2=(3)3/2×7≈5.2×7\Rightarrow {T_2} = {(3)^{3/2}} \times 7 \approx 5.2 \times 7
T2≅36{T_2} \cong 36

hrs

Q73
Water falls from a 40 m high dam at the rate of 9 ×\times 104 kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of 100 W lamps, that can be lit, is : (Take g = 10 ms−-2)
A 25
B 50
C 100
D 18
Correct Answer
Option B
Solution

Total gravitational PE of water per second

=mghT= {{mgh} \over T}
=9×104×10×403600=104= {{9 \times {{10}^4} \times 10 \times 40} \over {3600}} = {10^4}

J/sec 50% of this energy can be converted into electrical energy so total electrical energy

=1042=5000= {{{{10}^4}} \over 2} = 5000

W So total bulbs lit can be

=5000 W100 W=50= {{5000\,W} \over {100\,W}} = 50

bulbs

Q74
Two planets A and B of equal mass are having their period of revolutions TA and TB such that TA = 2TB. These planets are revolving in the circular orbits of radii rA and rB respectively. Which out of the following would be the correct relationship of their orbits?
A 2rA2=rB32r_A^2 = r_B^3
B rA3=2rB3r_A^3 = 2r_B^3
C rA3=4rB3r_A^3 = 4r_B^3
D TA2−TB2=π2GM(rB3−4rA3)T_A^2 - T_B^2 = {{{\pi ^2}} \over {GM}}\left( {r_B^3 - 4r_A^3} \right)
Correct Answer
Option C
Solution
TA=2TB{T_A} = 2{T_B}

Now

TA2∝rA3T_A^2 \propto r_A^3
⇒(rArB)3=(TATB)2\Rightarrow {\left( {{{{r_A}} \over {{r_B}}}} \right)^3} = {\left( {{{{T_A}} \over {{T_B}}}} \right)^2}
⇒rA3=4rB3\Rightarrow r_A^3 = 4r_B^3
Q75
The distance of the Sun from earth is 1.5 ×\times 1011 m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be :
A 2.45 ×\times 1010 m
B 1.45 ×\times 1010 m
C 1.45 ×\times 109 m
D 0.14 ×\times 109 m
Correct Answer
Option C
Solution

Diameter = r ×\times δ\delta

=1.5×1011×(2000)×(13600)×(π180)= 1.5 \times {10^{11}} \times (2000) \times \left( {{1 \over {3600}}} \right) \times \left( {{\pi \over {180}}} \right)
=1.45×109= 1.45 \times {10^9}

m

Q76
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : If we move from poles to equator, the direction of acceleration due to gravity of earth always points towards the center of earth without any variation in its magnitude. Reason R : At equator, the direction of acceleration due to the gravity is towards the center of earth. In the light of above statements, choose the correct answer from the options given below:
A Both A and R are true and R is the correct explanation of A.
B Both A and R are true but R is NOT the correct explanation of A.
C A is true but R is false.
D A is false but R is true.
Correct Answer
Option D
Solution

Assertion A is false.

While the direction of acceleration due to gravity does always point towards the center of the Earth, its magnitude actually varies from the poles to the equator.

This is due to the Earth's rotation and the fact that Earth is not a perfect sphere but an oblate spheroid, meaning it's slightly flattened at the poles and bulging at the equator.

Reason R is true.

At the equator, the direction of acceleration due to gravity is towards the center of the Earth.

Q77
The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be : (Given g = acceleration due to gravity at the surface of earth).
A g/2
B g/4
C g/3
D g/9
Correct Answer
Option D
Solution

The acceleration due to gravity (g) at a distance (r) from the center of a planet is given by:

g=GMr2g = \frac{G M}{r^2}

where: G is the gravitational constant, M is the mass of the planet, r is the distance from the center of the planet.

If the height h of a point P above the surface of the Earth is equal to the diameter of the Earth, then the distance r from the center of the Earth to the point P is 3 times the radius of the Earth.

Substituting this into the equation for g gives:

g′=g(R3R)2=g9g' = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}

where: g' is the acceleration due to gravity at the point P, R is the radius of the Earth.

Therefore, the value of acceleration due to gravity at point P is g/9.

Q78
The radii of two planets A and B are in the ratio 2 : 3. Their densities are 3ρ\rho and 5ρ\rho respectively. The ratio of their acceleration due to gravity is :
A 9 : 4
B 9 : 8
C 9 : 10
D 2 : 5
Correct Answer
Option D
Solution

Given,

rArB=23{{{r_A}} \over {{r_B}}} = {2 \over 3}
ρAρB=35{{{\rho _A}} \over {{\rho _B}}} = {3 \over 5}

We know, Acceleration due to gravity

g=GMr2=Gr2×43πr3×ρg = {{GM} \over {{r^2}}} = {G \over {{r^2}}} \times {4 \over 3}\pi {r^3} \times \rho
=4πGrρ3= {{4\pi Gr\rho } \over 3}

∴\therefore

g∝ρrg \propto \rho r

∴\therefore

gAgB=ρAρB×rArB{{{g_A}} \over {{g_B}}} = {{{\rho _A}} \over {{\rho _B}}} \times {{{r_A}} \over {{r_B}}}
=35×23= {3 \over 5} \times {2 \over 3}
=25= {2 \over 5}
Q79
Three identical particles A,B\mathrm{A}, \mathrm{B} and C\mathrm{C} of mass 100 kg100 \mathrm{~kg} each are placed in a straight line with AB=BC=13 m\mathrm{AB}=\mathrm{BC}=13 \mathrm{~m}. The gravitational force on a fourth particle P\mathrm{P} of the same mass is F\mathrm{F}, when placed at a distance 13 m13 \mathrm{~m} from the particle B\mathrm{B} on the perpendicular bisector of the line AC\mathrm{AC}. The value of F\mathrm{F} will be approximately :
A 21 G
B 100 G
C 59 G
D 42 G
Correct Answer
Option B
Solution

m = 100 kg

FAP=Gm2(132)2{F_{AP}} = {{G{m^2}} \over {{{\left( {13\sqrt 2 } \right)}^2}}}
FBP=Gm2132{F_{BP}} = {{G{m^2}} \over {{{13}^2}}}
FCP=Gm2(132)2{F_{CP}} = {{G{m^2}} \over {{{\left( {13\sqrt 2 } \right)}^2}}}
Fnet=FBP+FAPcos⁡45∘+FCPcos⁡45∘{F_{net}} = {F_{BP}} + {F_{AP}}\cos 45^\circ + {F_{CP}}\cos 45^\circ
=Gm2132(1+12)= {{G{m^2}} \over {{{13}^2}}}\left( {1 + {1 \over {\sqrt 2 }}} \right)
=G1002169(1+0.707)= {{G{{100}^2}} \over {169}}(1 + 0.707)
≃100 G\simeq 100\,G
Q80
The length of a seconds pendulum at a height h = 2R from earth surface will be: (Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = π\pi2 ms−-2)
A 29{2 \over 9} m
B 49{4 \over 9} m
C 89{8 \over 9} m
D 19{1 \over 9} m
Correct Answer
Option D
Solution
g=GM(R+h)2=GM9R2=g09g = {{GM} \over {{{(R + h)}^2}}} = {{GM} \over {9{R^2}}} = {{{g_0}} \over 9}
⇒T=2πlg=2πlg09\Rightarrow T = 2\pi \sqrt {{l \over g}} = 2\pi \sqrt {{l \over {{{{g_0}} \over 9}}}}
⇒2=2π9lg0\Rightarrow 2 = 2\pi \sqrt {{{9l} \over {{g_0}}}}
⇒l=g09π2=19 m\Rightarrow l = {{{g_0}} \over {9{\pi ^2}}} = {1 \over 9}\,m
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