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NEET NEET Questions — Free MCQ Practice

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The distance of a geo-stationary satellite from the centre of the earth (Radius R = 6400 km) is nearest to -

A.

5 R

B.

7 R

C.

10 R

D.

18 R

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Explanation

(b) 6R from the surface of earth and 7R from the centre.

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In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be (g=10 ${ms}^{- 2}$ and radius of earth is 6400 kms)

A.

0 rad/sec

B.

$\frac{1}{800}$ rad/sec

C.

$\frac{1}{80}$ rad/sec

D.

$\frac{1}{8}$ rad/sec

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Explanation

(b) $g' = g - ω^{2} {Rcos}^{2} λ$

For weightlessness at equator $λ = 0$ and $g' = 0$

$∴ 0 = g - ω^{2} R \Rightarrow ω = \sqrt{\frac{g}{R}} = \frac{1}{800} \frac{rad}{s}$

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A simple pendulum has a time period $T_{1}$ when on the earth’s surface and $T_{2}$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of is $T_{2} / T_{1}$ is

A.

1

B.

$\sqrt{2}$

C.

4

D.

2

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Explanation

(d) If acceleration due to gravity is g at the surface of earth then at height R its value becomes $g' = g {(\frac{R}{R + h})}^{2} = \frac{g}{4}$

$T_{1} = 2 π \sqrt{\frac{l}{g}}$ and $T_{2} = 2 π \sqrt{\frac{l}{g / 4}}$

$∴ \frac{T_{2}}{T_{1}} = 2$

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A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will be

A.

mgR

B.

$\frac{1}{2}$ mgR

C.

2 mgR

D.

$\frac{1}{4}$ mgR

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Explanation

(b) $∆ U = \frac{mgh}{1 + \frac{h}{R}} = \frac{1}{2} mgR (∵ h = R)$

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Periodic time of a satellite revolving above Earth’s surface at a height equal to R, radius of Earth, is
[g is acceleration due to gravity at Earth’s surface]

A.

$2 π \sqrt{\frac{2 R}{g}}$

B.

$4 \sqrt{2} π \sqrt{\frac{R}{g}}$

C.

$2 π \sqrt{\frac{R}{g}}$

D.

$8 π \sqrt{\frac{R}{g}}$

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Explanation

(b) $T = 2 π \sqrt{\frac{{(R + h)}^{3}}{{gR}^{2}}} = 2 π \sqrt{\frac{{(2 R)}^{3}}{{gR}^{2}}} = 4 \sqrt{2 π} \sqrt{\frac{R}{g}}$

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An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy $E_{0}$ . Its potential energy is

A.

$- E_{0}$

B.

$1.5 E_{0}$

C.

$2 E_{0}$

D.

$E_{0}$

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Explanation

(c) Potential energy = 2 x (Total energy) = $2 E_{0}$

Because we know = $U = \frac{- GMm}{r}$ and $E_{0} = \frac{- GMm}{2 r}$

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Given radius of Earth ‘R’ and length of a day ‘T’ , the height of a geostationary satellite is [G–Gravitational Constant, M–Mass of Earth]

A.

${(\frac{4 π^{2} GM}{T^{2}})}^{\frac{1}{3}}$

B.

${(\frac{4 πGM}{R^{2}})}^{\frac{1}{3}} - R$

C.

${(\frac{{GMT}^{2}}{4 π^{2}})}^{\frac{1}{3}} - R$

D.

${(\frac{{GMT}^{2}}{4 π^{2}})}^{\frac{1}{3}} + R$

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Explanation

(c) $T = 2 π \sqrt{\frac{r^{3}}{GM}} \Rightarrow T^{2} = \frac{4 π^{2}}{G M} {(R + h)}^{3} \Rightarrow R + h = {[\frac{G M T^{2}}{4 π^{2}}]}^{\frac{1}{3}} \Rightarrow h = {[\frac{G M T^{2}}{4 π^{2}}]}^{\frac{1}{3}} - R$

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A rocket of mass M is launched vertically from the surface of the earth with an initial speed V. Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is

A.

$R / (\frac{gR}{2 V^{2}} - 1)$

B.

$R (\frac{gR}{2 V^{2}} - 1)$

C.

$R / (\frac{2 gR}{V^{2}} - 1)$

D.

$R (\frac{2 gR}{V^{2}} - 1)$

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Explanation

(c) $∆ K . E . = ∆ U$

$\Rightarrow \frac{1}{2} {MV}^{2} = {GM}_{e} M (\frac{1}{R} - \frac{1}{R + h})$ ...(i)

Also $g = \frac{{GM}_{e}}{R^{2}}$ ...(ii)

On solving (i) and (ii) $h = \frac{R}{(\frac{2 gR}{V^{2}} - 1)}$

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Two bodies of masses $m_{1}$ and $m_{2}$ are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is

A.

${[2 G \frac{(m_{1} - m_{2})}{r}]}^{1 / 2}$

B.

${[\frac{2 G}{r} (m_{1} + m_{2})]}^{1 / 2}$

C.

${[\frac{r}{2 G (m_{1} m_{2)}}]}^{1 / 2}$

D.

${[\frac{2 G}{r} m_{1} m_{2}]}^{1 / 2}$

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Explanation

(b) Let velocities of these masses at r distance from each other be $v_{1}$ and $v_{2}$ respectively. By conservation of momentum

$m_{1} v_{1} - m_{2} v_{2} = 0$

$\Rightarrow m_{1} v_{1} = m_{2} v_{2}$ ...(i)

By conservation of energy

change in P.E.= change in K.E.

$\frac{{Gm}_{1} m_{2}}{r} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

$\Rightarrow \frac{m_{1}^{2} v_{1}^{2}}{m_{1}} + \frac{m_{2}^{2} v_{2}^{2}}{m_{2}} = \frac{2 {Gm}_{1} m_{2}}{r}$ ...(ii)

On solving equation (i) and (ii)

$v_{1} = \sqrt{\frac{2 {Gm}_{2}^{2}}{r (m_{1} + m_{2})}}$ and $v_{2} = \sqrt{\frac{2 {Gm}_{1}^{2}}{r (m_{1} + m_{2)}}}$

$∴ ν_{app} = |v_{1}| + |v_{2}| = \sqrt{\frac{2 G}{r} (m_{1} + m_{2})}$

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The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be

A.

83 minutes

B.

$83 \times \sqrt{8}$ minutes

C.

664 minutes

D.

249 minutes

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Explanation

(c) For first satellite and $r_{1} = R a n d T_{1} = 83 m i n u t e s$

For second satellite $r_{2} = 4 R$

$T_{2} = T_{1} {(\frac{r_{2}}{r_{1}})}^{3 / 2} = T_{1} {(4)}^{3 / 2} = 8 T_{1} = 8 \times 83 = 664 m i n u t e s$

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