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Physics NEET Questions — Free MCQ Practice

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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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A projectile is projected with velocity ${kv}_{e}$ in vertically upward direction from the ground into the space. ( $v_{e}$ is escape velocity and k<1). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be : (R = radius of earth)

A.

$\frac{R}{k^{2} + 1}$

B.

$\frac{R}{k^{2} - 1}$

C.

$\frac{R}{1 - k^{2}}$

D.

$\frac{R}{k + 1}$

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Explanation

(c)Loss in Kinetic energy = Gain in Potential energy

$\frac{1}{2} m ({kv}_{e})^{2} = \frac{mgh}{1 + \frac{h}{R}} \Rightarrow \frac{1}{2} {mk}^{2} 2 gR = \frac{mgh}{1 + \frac{h}{R}} \Rightarrow h = \frac{{Rk}^{2}}{1 - k^{2}}$

Height of projectile from the earth's surface = h

Height from the centre $r = R + h = R + \frac{{Rk}^{2}}{1 - k^{2}}$

By solving $r = \frac{R}{1 - k^{2}}$

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A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is $R_{0}$ and mass of the earth M, the angular momentum about the centre of the earth is

A.

$m \sqrt{G M R_{0}}$

B.

$M \sqrt{G m R_{0}}$

C.

$m \sqrt{\frac{G M}{R_{0}}}$

D.

$M \sqrt{\frac{G M}{R_{0}}}$

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Explanation

(a) Angular momentum = Mass ×Orbital velocity ×Radius

$= m \times (\sqrt{\frac{G M}{R_{0}}}) \times R_{0} = m \sqrt{G M R_{0}}$

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If the distance between the earth and the sun becomes half its present value, the number of days in a year would have been

A.

64.5

B.

129

C.

182.5

D.

730

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Explanation

(b) According to Kepler's third law, the ratio of the squares of the periods of any two planets revolving about the sun is equal to the ratio of the cubes of their average distances from the sun i.e.

${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{r_{1}}{r_{2}})}^{3} = {[\frac{r_{1}}{\frac{1}{2} r_{1}}]}^{3} = 8 \Rightarrow \frac{T_{1}}{T_{2}} = 2 \sqrt{2} ∴ T_{2} = \frac{T_{1}}{2 \sqrt{2}} = \frac{365 days}{2 \sqrt{2}} = 129 days$

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According to Kepler, the period of revolution of a planet (T) and its mean distance from the sun (r) are related by the equation

A.

$T^{3} r^{3} = c o n s \tan t$

B.

$T^{2} r^{‐ 3} = c o n s \tan t$

C.

$T r^{3} = c o n s \tan t$

D.

$T^{2} r = c o n s \tan t$

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Explanation

(b) $\frac{T^{2}}{r^{3}} = c o n s \tan t \Rightarrow T^{2} r^{‐ 3} = c o n s \tan t$

Physics MCQs for NEET — Practice Questions with Answers

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