138487
A body is projected vertically upwards from the surface of a planet of radius $R$ with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is :
1 $\mathrm{R} / 2$
2 $R / 3$
3 $R / 5$
4 $R / 4$
Explanation:
B Suppose that, Escape velocity $=\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\therefore \mathrm{v}=\frac{1}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}$ By conservation of energy $\mathrm{P}_{\mathrm{A}}+\mathrm{K}_{\mathrm{A}}=\mathrm{P}_{\mathrm{B}}+\mathrm{K}_{\mathrm{B}}$ $\frac{-\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}}{2 \mathrm{R}}=0+\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\left\{\because \mathrm{P}_{\mathrm{E}}=-\frac{\mathrm{GMm}}{\mathrm{R}}\right\}$ $\operatorname{GMm}\left(\frac{-1}{\mathrm{R}}+\frac{1}{4 \mathrm{R}}\right)=\operatorname{GMm}\left(\frac{-1}{\mathrm{R}+\mathrm{h}}\right)$ $\frac{-1}{R}+\frac{1}{4 R}=\frac{-1}{R+h}$ $\frac{1}{4 R}=-\frac{1}{R+h}+\frac{1}{R}$ $\frac{1}{4 R}=\frac{-R+R+h}{R(R+h)}$ $\frac{1}{4}=\frac{h}{R+h}$ $R+h=4 h$ $R=3 h$ $h=\frac{R}{3}$
Karnataka CET-2002
Gravitation
138488
The velocity acquired by a body is falling to the surface of earth from a point at a height equal to the radius of the earth is approximately
1 $4 \mathrm{~km} / \mathrm{s}$
2 $8 \mathrm{~km} / \mathrm{s}$
3 $\frac{4}{2} \mathrm{~km} / \mathrm{s}$
4 $8 \sqrt{2} \mathrm{~km} / \mathrm{s}$
Explanation:
B Suppose that, Increase in kinetic energy $=$ Decrease in potential energy $\therefore \frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{2 \mathrm{GMm}-\mathrm{GMm}}{2 \mathrm{R}}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{R}}$ $\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\left(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \Rightarrow \frac{\mathrm{GM}}{\mathrm{R}}=\mathrm{gR}\right)$ $\mathrm{v}=\sqrt{\mathrm{gR}}$ $\left(\because\right.$ radius of earth $=6.4 \times 10^{6} \mathrm{~m}$ ) $\mathrm{v}=\sqrt{10 \times 6.4 \times 10^{6}}$ $\mathrm{v}=8 \times 10^{3} \mathrm{~m} / \mathrm{s}$ $\mathrm{v}=8 \mathrm{~km} / \mathrm{s}$
J and K CET- 1997
Gravitation
138489
A coin of mass $10 \mathrm{~g}$ rolls along a horizontal table with a velocity of $6 \mathrm{~cm} / \mathrm{s}$. Its total kinetic energy is
1 $9 \mu \mathrm{J}$
2 $18 \mu \mathrm{J}$
3 $27 \mu \mathrm{J}$
4 $36 \mu \mathrm{J}$
Explanation:
C Given that, mass of coin $=10 \mathrm{~g}=10 \times 10^{-3}$ $\mathrm{kg}$ Velocity $=6 \mathrm{~cm} / \mathrm{s}$ Total kinetic energy $=$ Rotational kinetic energy + Translational kinetic energy $\mathrm{K}=\mathrm{K}_{\mathrm{T}}+\mathrm{K}_{\mathrm{R}}$ $\mathrm{K}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ The moment of inertia of coin (circle) $I=\frac{1}{2} \mathrm{MR}^{2}$ Putting the value of I in equation (i), we get $\mathrm{K} =\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2}\left(\frac{\mathrm{MR}^{2}}{2}\right)\left(\frac{\mathrm{v}}{\mathrm{R}}\right)^{2}$ $\mathrm{~K} =\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{4} \mathrm{MR}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}$ $=\frac{3}{4} \mathrm{Mv}^{2}$ $=\frac{3}{4} \times\left(10 \times 10^{-3}\right)\left(6 \times 10^{-2}\right)^{2}$ $=27 \times 10^{-6} \mathrm{~J}$ $=27 \mu \mathrm{J}$
J and K-CET-2012
Gravitation
138490
A body is projected in vertically upward direction from the surface of the earth of radius ' $R$ ' into the space with velocity ' $n V_{e}$ ' ( $n$ $ \lt 1)$. The maximum height from the surface of earth to which a body can reach is $\left(V_{e}=\right.$ escape velocity)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138487
A body is projected vertically upwards from the surface of a planet of radius $R$ with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is :
1 $\mathrm{R} / 2$
2 $R / 3$
3 $R / 5$
4 $R / 4$
Explanation:
B Suppose that, Escape velocity $=\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\therefore \mathrm{v}=\frac{1}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}$ By conservation of energy $\mathrm{P}_{\mathrm{A}}+\mathrm{K}_{\mathrm{A}}=\mathrm{P}_{\mathrm{B}}+\mathrm{K}_{\mathrm{B}}$ $\frac{-\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}}{2 \mathrm{R}}=0+\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\left\{\because \mathrm{P}_{\mathrm{E}}=-\frac{\mathrm{GMm}}{\mathrm{R}}\right\}$ $\operatorname{GMm}\left(\frac{-1}{\mathrm{R}}+\frac{1}{4 \mathrm{R}}\right)=\operatorname{GMm}\left(\frac{-1}{\mathrm{R}+\mathrm{h}}\right)$ $\frac{-1}{R}+\frac{1}{4 R}=\frac{-1}{R+h}$ $\frac{1}{4 R}=-\frac{1}{R+h}+\frac{1}{R}$ $\frac{1}{4 R}=\frac{-R+R+h}{R(R+h)}$ $\frac{1}{4}=\frac{h}{R+h}$ $R+h=4 h$ $R=3 h$ $h=\frac{R}{3}$
Karnataka CET-2002
Gravitation
138488
The velocity acquired by a body is falling to the surface of earth from a point at a height equal to the radius of the earth is approximately
1 $4 \mathrm{~km} / \mathrm{s}$
2 $8 \mathrm{~km} / \mathrm{s}$
3 $\frac{4}{2} \mathrm{~km} / \mathrm{s}$
4 $8 \sqrt{2} \mathrm{~km} / \mathrm{s}$
Explanation:
B Suppose that, Increase in kinetic energy $=$ Decrease in potential energy $\therefore \frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{2 \mathrm{GMm}-\mathrm{GMm}}{2 \mathrm{R}}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{R}}$ $\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\left(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \Rightarrow \frac{\mathrm{GM}}{\mathrm{R}}=\mathrm{gR}\right)$ $\mathrm{v}=\sqrt{\mathrm{gR}}$ $\left(\because\right.$ radius of earth $=6.4 \times 10^{6} \mathrm{~m}$ ) $\mathrm{v}=\sqrt{10 \times 6.4 \times 10^{6}}$ $\mathrm{v}=8 \times 10^{3} \mathrm{~m} / \mathrm{s}$ $\mathrm{v}=8 \mathrm{~km} / \mathrm{s}$
J and K CET- 1997
Gravitation
138489
A coin of mass $10 \mathrm{~g}$ rolls along a horizontal table with a velocity of $6 \mathrm{~cm} / \mathrm{s}$. Its total kinetic energy is
1 $9 \mu \mathrm{J}$
2 $18 \mu \mathrm{J}$
3 $27 \mu \mathrm{J}$
4 $36 \mu \mathrm{J}$
Explanation:
C Given that, mass of coin $=10 \mathrm{~g}=10 \times 10^{-3}$ $\mathrm{kg}$ Velocity $=6 \mathrm{~cm} / \mathrm{s}$ Total kinetic energy $=$ Rotational kinetic energy + Translational kinetic energy $\mathrm{K}=\mathrm{K}_{\mathrm{T}}+\mathrm{K}_{\mathrm{R}}$ $\mathrm{K}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ The moment of inertia of coin (circle) $I=\frac{1}{2} \mathrm{MR}^{2}$ Putting the value of I in equation (i), we get $\mathrm{K} =\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2}\left(\frac{\mathrm{MR}^{2}}{2}\right)\left(\frac{\mathrm{v}}{\mathrm{R}}\right)^{2}$ $\mathrm{~K} =\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{4} \mathrm{MR}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}$ $=\frac{3}{4} \mathrm{Mv}^{2}$ $=\frac{3}{4} \times\left(10 \times 10^{-3}\right)\left(6 \times 10^{-2}\right)^{2}$ $=27 \times 10^{-6} \mathrm{~J}$ $=27 \mu \mathrm{J}$
J and K-CET-2012
Gravitation
138490
A body is projected in vertically upward direction from the surface of the earth of radius ' $R$ ' into the space with velocity ' $n V_{e}$ ' ( $n$ $ \lt 1)$. The maximum height from the surface of earth to which a body can reach is $\left(V_{e}=\right.$ escape velocity)
138487
A body is projected vertically upwards from the surface of a planet of radius $R$ with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is :
1 $\mathrm{R} / 2$
2 $R / 3$
3 $R / 5$
4 $R / 4$
Explanation:
B Suppose that, Escape velocity $=\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\therefore \mathrm{v}=\frac{1}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}$ By conservation of energy $\mathrm{P}_{\mathrm{A}}+\mathrm{K}_{\mathrm{A}}=\mathrm{P}_{\mathrm{B}}+\mathrm{K}_{\mathrm{B}}$ $\frac{-\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}}{2 \mathrm{R}}=0+\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\left\{\because \mathrm{P}_{\mathrm{E}}=-\frac{\mathrm{GMm}}{\mathrm{R}}\right\}$ $\operatorname{GMm}\left(\frac{-1}{\mathrm{R}}+\frac{1}{4 \mathrm{R}}\right)=\operatorname{GMm}\left(\frac{-1}{\mathrm{R}+\mathrm{h}}\right)$ $\frac{-1}{R}+\frac{1}{4 R}=\frac{-1}{R+h}$ $\frac{1}{4 R}=-\frac{1}{R+h}+\frac{1}{R}$ $\frac{1}{4 R}=\frac{-R+R+h}{R(R+h)}$ $\frac{1}{4}=\frac{h}{R+h}$ $R+h=4 h$ $R=3 h$ $h=\frac{R}{3}$
Karnataka CET-2002
Gravitation
138488
The velocity acquired by a body is falling to the surface of earth from a point at a height equal to the radius of the earth is approximately
1 $4 \mathrm{~km} / \mathrm{s}$
2 $8 \mathrm{~km} / \mathrm{s}$
3 $\frac{4}{2} \mathrm{~km} / \mathrm{s}$
4 $8 \sqrt{2} \mathrm{~km} / \mathrm{s}$
Explanation:
B Suppose that, Increase in kinetic energy $=$ Decrease in potential energy $\therefore \frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{2 \mathrm{GMm}-\mathrm{GMm}}{2 \mathrm{R}}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{R}}$ $\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\left(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \Rightarrow \frac{\mathrm{GM}}{\mathrm{R}}=\mathrm{gR}\right)$ $\mathrm{v}=\sqrt{\mathrm{gR}}$ $\left(\because\right.$ radius of earth $=6.4 \times 10^{6} \mathrm{~m}$ ) $\mathrm{v}=\sqrt{10 \times 6.4 \times 10^{6}}$ $\mathrm{v}=8 \times 10^{3} \mathrm{~m} / \mathrm{s}$ $\mathrm{v}=8 \mathrm{~km} / \mathrm{s}$
J and K CET- 1997
Gravitation
138489
A coin of mass $10 \mathrm{~g}$ rolls along a horizontal table with a velocity of $6 \mathrm{~cm} / \mathrm{s}$. Its total kinetic energy is
1 $9 \mu \mathrm{J}$
2 $18 \mu \mathrm{J}$
3 $27 \mu \mathrm{J}$
4 $36 \mu \mathrm{J}$
Explanation:
C Given that, mass of coin $=10 \mathrm{~g}=10 \times 10^{-3}$ $\mathrm{kg}$ Velocity $=6 \mathrm{~cm} / \mathrm{s}$ Total kinetic energy $=$ Rotational kinetic energy + Translational kinetic energy $\mathrm{K}=\mathrm{K}_{\mathrm{T}}+\mathrm{K}_{\mathrm{R}}$ $\mathrm{K}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ The moment of inertia of coin (circle) $I=\frac{1}{2} \mathrm{MR}^{2}$ Putting the value of I in equation (i), we get $\mathrm{K} =\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2}\left(\frac{\mathrm{MR}^{2}}{2}\right)\left(\frac{\mathrm{v}}{\mathrm{R}}\right)^{2}$ $\mathrm{~K} =\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{4} \mathrm{MR}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}$ $=\frac{3}{4} \mathrm{Mv}^{2}$ $=\frac{3}{4} \times\left(10 \times 10^{-3}\right)\left(6 \times 10^{-2}\right)^{2}$ $=27 \times 10^{-6} \mathrm{~J}$ $=27 \mu \mathrm{J}$
J and K-CET-2012
Gravitation
138490
A body is projected in vertically upward direction from the surface of the earth of radius ' $R$ ' into the space with velocity ' $n V_{e}$ ' ( $n$ $ \lt 1)$. The maximum height from the surface of earth to which a body can reach is $\left(V_{e}=\right.$ escape velocity)
138487
A body is projected vertically upwards from the surface of a planet of radius $R$ with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is :
1 $\mathrm{R} / 2$
2 $R / 3$
3 $R / 5$
4 $R / 4$
Explanation:
B Suppose that, Escape velocity $=\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\therefore \mathrm{v}=\frac{1}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}$ By conservation of energy $\mathrm{P}_{\mathrm{A}}+\mathrm{K}_{\mathrm{A}}=\mathrm{P}_{\mathrm{B}}+\mathrm{K}_{\mathrm{B}}$ $\frac{-\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}}{2 \mathrm{R}}=0+\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\left\{\because \mathrm{P}_{\mathrm{E}}=-\frac{\mathrm{GMm}}{\mathrm{R}}\right\}$ $\operatorname{GMm}\left(\frac{-1}{\mathrm{R}}+\frac{1}{4 \mathrm{R}}\right)=\operatorname{GMm}\left(\frac{-1}{\mathrm{R}+\mathrm{h}}\right)$ $\frac{-1}{R}+\frac{1}{4 R}=\frac{-1}{R+h}$ $\frac{1}{4 R}=-\frac{1}{R+h}+\frac{1}{R}$ $\frac{1}{4 R}=\frac{-R+R+h}{R(R+h)}$ $\frac{1}{4}=\frac{h}{R+h}$ $R+h=4 h$ $R=3 h$ $h=\frac{R}{3}$
Karnataka CET-2002
Gravitation
138488
The velocity acquired by a body is falling to the surface of earth from a point at a height equal to the radius of the earth is approximately
1 $4 \mathrm{~km} / \mathrm{s}$
2 $8 \mathrm{~km} / \mathrm{s}$
3 $\frac{4}{2} \mathrm{~km} / \mathrm{s}$
4 $8 \sqrt{2} \mathrm{~km} / \mathrm{s}$
Explanation:
B Suppose that, Increase in kinetic energy $=$ Decrease in potential energy $\therefore \frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{2 \mathrm{GMm}-\mathrm{GMm}}{2 \mathrm{R}}$ $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{R}}$ $\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\left(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \Rightarrow \frac{\mathrm{GM}}{\mathrm{R}}=\mathrm{gR}\right)$ $\mathrm{v}=\sqrt{\mathrm{gR}}$ $\left(\because\right.$ radius of earth $=6.4 \times 10^{6} \mathrm{~m}$ ) $\mathrm{v}=\sqrt{10 \times 6.4 \times 10^{6}}$ $\mathrm{v}=8 \times 10^{3} \mathrm{~m} / \mathrm{s}$ $\mathrm{v}=8 \mathrm{~km} / \mathrm{s}$
J and K CET- 1997
Gravitation
138489
A coin of mass $10 \mathrm{~g}$ rolls along a horizontal table with a velocity of $6 \mathrm{~cm} / \mathrm{s}$. Its total kinetic energy is
1 $9 \mu \mathrm{J}$
2 $18 \mu \mathrm{J}$
3 $27 \mu \mathrm{J}$
4 $36 \mu \mathrm{J}$
Explanation:
C Given that, mass of coin $=10 \mathrm{~g}=10 \times 10^{-3}$ $\mathrm{kg}$ Velocity $=6 \mathrm{~cm} / \mathrm{s}$ Total kinetic energy $=$ Rotational kinetic energy + Translational kinetic energy $\mathrm{K}=\mathrm{K}_{\mathrm{T}}+\mathrm{K}_{\mathrm{R}}$ $\mathrm{K}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ The moment of inertia of coin (circle) $I=\frac{1}{2} \mathrm{MR}^{2}$ Putting the value of I in equation (i), we get $\mathrm{K} =\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2}\left(\frac{\mathrm{MR}^{2}}{2}\right)\left(\frac{\mathrm{v}}{\mathrm{R}}\right)^{2}$ $\mathrm{~K} =\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{4} \mathrm{MR}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}$ $=\frac{3}{4} \mathrm{Mv}^{2}$ $=\frac{3}{4} \times\left(10 \times 10^{-3}\right)\left(6 \times 10^{-2}\right)^{2}$ $=27 \times 10^{-6} \mathrm{~J}$ $=27 \mu \mathrm{J}$
J and K-CET-2012
Gravitation
138490
A body is projected in vertically upward direction from the surface of the earth of radius ' $R$ ' into the space with velocity ' $n V_{e}$ ' ( $n$ $ \lt 1)$. The maximum height from the surface of earth to which a body can reach is $\left(V_{e}=\right.$ escape velocity)
