01. Acceleration due to Gravity
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Gravitation

138361 What is the percentage decrease in the weight of a body when it is taken to a height of $32 \mathrm{~km}$ from the surface of earth ?

1 $0.5 \%$
2 $2 \%$
3 $1.5 \%$
4 $1 \%$
MHT-CET 2020
Gravitation

138362 When the value of acceleration due to gravity ' $g$ ' becomes $\left(\frac{g}{3}\right)$ above the earth's surface at height ' $h$ ' then relation between ' $h$ ' and ' $R$ ' is $(R=$ radius of the earth)

1 $\mathrm{h}=\mathrm{R}(\sqrt{3}-1)$
2 $\mathrm{h}=(\sqrt{2}-1)$
3 $\mathrm{h}=2 \mathrm{R}$
4 $\mathrm{h}=\mathrm{R}$
MHT-CET 2020
Gravitation

138363 The acceleration due to gravity on moon is $\left(\frac{1}{6}\right)^{\text {th }}$ times the acceleration due to gravity on earth. If the ratio of the density of earth ' $\rho_{\mathrm{e}}$ ' to the density of moon ' $\rho_{m}$ ' is $\frac{5}{3}$, then the radius of moon ' $R_{m}$ ' in terms of the radius of earth ' $\mathbf{R}_{\mathbf{e}}$ ' is

1 $\left(\frac{3}{18}\right) \mathrm{R}_{\mathrm{e}}$
2 $\left(\frac{1}{2 \sqrt{3}}\right) \mathrm{R}_{\mathrm{e}}$
3 $\left(\frac{5}{18}\right) \mathrm{R}_{\mathrm{e}}$
4 $\left(\frac{7}{6}\right) \mathrm{R}_{\mathrm{e}}$
MHT-CET 2020
Gravitation

138365 For the weight of body of mass $5 \mathrm{~kg}$ to be zero on equator of the earth, angular velocity of the earth must be-
(The radius of earth $=6400 \mathrm{~km}$, acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^{2}$ )

1 $\frac{1}{800} \mathrm{rad} / \mathrm{s}$
2 $\frac{1}{400} \mathrm{rad} / \mathrm{s}$
3 $\frac{1}{80} \mathrm{rad} / \mathrm{s}$
4 $\frac{1}{1600} \mathrm{rad} / \mathrm{s}$
MHT-CET 2020
Gravitation

138366 A hole is drilled half way to the centre of the earth. A body weighs $300 \mathrm{~N}$ on the surface of the earth. How much will it weight at the bottom of the hole?

1 $150 \mathrm{~N}$
2 $250 \mathrm{~N}$
3 $200 \mathrm{~N}$
4 $120 \mathrm{~N}$
MHT-CET 2019
NEET DIGITAL LIBRARY
Gravitation

138361 What is the percentage decrease in the weight of a body when it is taken to a height of $32 \mathrm{~km}$ from the surface of earth ?

1 $0.5 \%$
2 $2 \%$
3 $1.5 \%$
4 $1 \%$
MHT-CET 2020
Gravitation

138362 When the value of acceleration due to gravity ' $g$ ' becomes $\left(\frac{g}{3}\right)$ above the earth's surface at height ' $h$ ' then relation between ' $h$ ' and ' $R$ ' is $(R=$ radius of the earth)

1 $\mathrm{h}=\mathrm{R}(\sqrt{3}-1)$
2 $\mathrm{h}=(\sqrt{2}-1)$
3 $\mathrm{h}=2 \mathrm{R}$
4 $\mathrm{h}=\mathrm{R}$
MHT-CET 2020
Gravitation

138363 The acceleration due to gravity on moon is $\left(\frac{1}{6}\right)^{\text {th }}$ times the acceleration due to gravity on earth. If the ratio of the density of earth ' $\rho_{\mathrm{e}}$ ' to the density of moon ' $\rho_{m}$ ' is $\frac{5}{3}$, then the radius of moon ' $R_{m}$ ' in terms of the radius of earth ' $\mathbf{R}_{\mathbf{e}}$ ' is

1 $\left(\frac{3}{18}\right) \mathrm{R}_{\mathrm{e}}$
2 $\left(\frac{1}{2 \sqrt{3}}\right) \mathrm{R}_{\mathrm{e}}$
3 $\left(\frac{5}{18}\right) \mathrm{R}_{\mathrm{e}}$
4 $\left(\frac{7}{6}\right) \mathrm{R}_{\mathrm{e}}$
MHT-CET 2020
Gravitation

138365 For the weight of body of mass $5 \mathrm{~kg}$ to be zero on equator of the earth, angular velocity of the earth must be-
(The radius of earth $=6400 \mathrm{~km}$, acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^{2}$ )

1 $\frac{1}{800} \mathrm{rad} / \mathrm{s}$
2 $\frac{1}{400} \mathrm{rad} / \mathrm{s}$
3 $\frac{1}{80} \mathrm{rad} / \mathrm{s}$
4 $\frac{1}{1600} \mathrm{rad} / \mathrm{s}$
MHT-CET 2020
Gravitation

138366 A hole is drilled half way to the centre of the earth. A body weighs $300 \mathrm{~N}$ on the surface of the earth. How much will it weight at the bottom of the hole?

1 $150 \mathrm{~N}$
2 $250 \mathrm{~N}$
3 $200 \mathrm{~N}$
4 $120 \mathrm{~N}$
MHT-CET 2019
Join With Us for More Updates
Gravitation

138361 What is the percentage decrease in the weight of a body when it is taken to a height of $32 \mathrm{~km}$ from the surface of earth ?

1 $0.5 \%$
2 $2 \%$
3 $1.5 \%$
4 $1 \%$
MHT-CET 2020
Gravitation

138362 When the value of acceleration due to gravity ' $g$ ' becomes $\left(\frac{g}{3}\right)$ above the earth's surface at height ' $h$ ' then relation between ' $h$ ' and ' $R$ ' is $(R=$ radius of the earth)

1 $\mathrm{h}=\mathrm{R}(\sqrt{3}-1)$
2 $\mathrm{h}=(\sqrt{2}-1)$
3 $\mathrm{h}=2 \mathrm{R}$
4 $\mathrm{h}=\mathrm{R}$
MHT-CET 2020
Gravitation

138363 The acceleration due to gravity on moon is $\left(\frac{1}{6}\right)^{\text {th }}$ times the acceleration due to gravity on earth. If the ratio of the density of earth ' $\rho_{\mathrm{e}}$ ' to the density of moon ' $\rho_{m}$ ' is $\frac{5}{3}$, then the radius of moon ' $R_{m}$ ' in terms of the radius of earth ' $\mathbf{R}_{\mathbf{e}}$ ' is

1 $\left(\frac{3}{18}\right) \mathrm{R}_{\mathrm{e}}$
2 $\left(\frac{1}{2 \sqrt{3}}\right) \mathrm{R}_{\mathrm{e}}$
3 $\left(\frac{5}{18}\right) \mathrm{R}_{\mathrm{e}}$
4 $\left(\frac{7}{6}\right) \mathrm{R}_{\mathrm{e}}$
MHT-CET 2020
Gravitation

138365 For the weight of body of mass $5 \mathrm{~kg}$ to be zero on equator of the earth, angular velocity of the earth must be-
(The radius of earth $=6400 \mathrm{~km}$, acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^{2}$ )

1 $\frac{1}{800} \mathrm{rad} / \mathrm{s}$
2 $\frac{1}{400} \mathrm{rad} / \mathrm{s}$
3 $\frac{1}{80} \mathrm{rad} / \mathrm{s}$
4 $\frac{1}{1600} \mathrm{rad} / \mathrm{s}$
MHT-CET 2020
Gravitation

138366 A hole is drilled half way to the centre of the earth. A body weighs $300 \mathrm{~N}$ on the surface of the earth. How much will it weight at the bottom of the hole?

1 $150 \mathrm{~N}$
2 $250 \mathrm{~N}$
3 $200 \mathrm{~N}$
4 $120 \mathrm{~N}$
MHT-CET 2019
For More Updates join with Us
Gravitation

138361 What is the percentage decrease in the weight of a body when it is taken to a height of $32 \mathrm{~km}$ from the surface of earth ?

1 $0.5 \%$
2 $2 \%$
3 $1.5 \%$
4 $1 \%$
MHT-CET 2020
Gravitation

138362 When the value of acceleration due to gravity ' $g$ ' becomes $\left(\frac{g}{3}\right)$ above the earth's surface at height ' $h$ ' then relation between ' $h$ ' and ' $R$ ' is $(R=$ radius of the earth)

1 $\mathrm{h}=\mathrm{R}(\sqrt{3}-1)$
2 $\mathrm{h}=(\sqrt{2}-1)$
3 $\mathrm{h}=2 \mathrm{R}$
4 $\mathrm{h}=\mathrm{R}$
MHT-CET 2020
Gravitation

138363 The acceleration due to gravity on moon is $\left(\frac{1}{6}\right)^{\text {th }}$ times the acceleration due to gravity on earth. If the ratio of the density of earth ' $\rho_{\mathrm{e}}$ ' to the density of moon ' $\rho_{m}$ ' is $\frac{5}{3}$, then the radius of moon ' $R_{m}$ ' in terms of the radius of earth ' $\mathbf{R}_{\mathbf{e}}$ ' is

1 $\left(\frac{3}{18}\right) \mathrm{R}_{\mathrm{e}}$
2 $\left(\frac{1}{2 \sqrt{3}}\right) \mathrm{R}_{\mathrm{e}}$
3 $\left(\frac{5}{18}\right) \mathrm{R}_{\mathrm{e}}$
4 $\left(\frac{7}{6}\right) \mathrm{R}_{\mathrm{e}}$
MHT-CET 2020
Gravitation

138365 For the weight of body of mass $5 \mathrm{~kg}$ to be zero on equator of the earth, angular velocity of the earth must be-
(The radius of earth $=6400 \mathrm{~km}$, acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^{2}$ )

1 $\frac{1}{800} \mathrm{rad} / \mathrm{s}$
2 $\frac{1}{400} \mathrm{rad} / \mathrm{s}$
3 $\frac{1}{80} \mathrm{rad} / \mathrm{s}$
4 $\frac{1}{1600} \mathrm{rad} / \mathrm{s}$
MHT-CET 2020
Gravitation

138366 A hole is drilled half way to the centre of the earth. A body weighs $300 \mathrm{~N}$ on the surface of the earth. How much will it weight at the bottom of the hole?

1 $150 \mathrm{~N}$
2 $250 \mathrm{~N}$
3 $200 \mathrm{~N}$
4 $120 \mathrm{~N}$
MHT-CET 2019
NEET DIGITAL LIBRARY
Gravitation

138361 What is the percentage decrease in the weight of a body when it is taken to a height of $32 \mathrm{~km}$ from the surface of earth ?

1 $0.5 \%$
2 $2 \%$
3 $1.5 \%$
4 $1 \%$
MHT-CET 2020
Gravitation

138362 When the value of acceleration due to gravity ' $g$ ' becomes $\left(\frac{g}{3}\right)$ above the earth's surface at height ' $h$ ' then relation between ' $h$ ' and ' $R$ ' is $(R=$ radius of the earth)

1 $\mathrm{h}=\mathrm{R}(\sqrt{3}-1)$
2 $\mathrm{h}=(\sqrt{2}-1)$
3 $\mathrm{h}=2 \mathrm{R}$
4 $\mathrm{h}=\mathrm{R}$
MHT-CET 2020
Gravitation

138363 The acceleration due to gravity on moon is $\left(\frac{1}{6}\right)^{\text {th }}$ times the acceleration due to gravity on earth. If the ratio of the density of earth ' $\rho_{\mathrm{e}}$ ' to the density of moon ' $\rho_{m}$ ' is $\frac{5}{3}$, then the radius of moon ' $R_{m}$ ' in terms of the radius of earth ' $\mathbf{R}_{\mathbf{e}}$ ' is

1 $\left(\frac{3}{18}\right) \mathrm{R}_{\mathrm{e}}$
2 $\left(\frac{1}{2 \sqrt{3}}\right) \mathrm{R}_{\mathrm{e}}$
3 $\left(\frac{5}{18}\right) \mathrm{R}_{\mathrm{e}}$
4 $\left(\frac{7}{6}\right) \mathrm{R}_{\mathrm{e}}$
MHT-CET 2020
Gravitation

138365 For the weight of body of mass $5 \mathrm{~kg}$ to be zero on equator of the earth, angular velocity of the earth must be-
(The radius of earth $=6400 \mathrm{~km}$, acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^{2}$ )

1 $\frac{1}{800} \mathrm{rad} / \mathrm{s}$
2 $\frac{1}{400} \mathrm{rad} / \mathrm{s}$
3 $\frac{1}{80} \mathrm{rad} / \mathrm{s}$
4 $\frac{1}{1600} \mathrm{rad} / \mathrm{s}$
MHT-CET 2020
Gravitation

138366 A hole is drilled half way to the centre of the earth. A body weighs $300 \mathrm{~N}$ on the surface of the earth. How much will it weight at the bottom of the hole?

1 $150 \mathrm{~N}$
2 $250 \mathrm{~N}$
3 $200 \mathrm{~N}$
4 $120 \mathrm{~N}$
MHT-CET 2019