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Physics MCQs for NEET — Practice Questions with Answers

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A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a satellite orbiting a few hundred kilometres above the earth’s surface ( $R_{earth}$ = 6400 km) will approximately be -

1/2 h

1 h

2 h

4 h

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Explanation

(c) $\frac{T_{2}}{T_{1}} = {(\frac{r_{2}}{r_{1}})}^{3 / 2} \Rightarrow T_{2} = 24 {(\frac{6400}{36000})}^{3 / 2} ≅ 2 hour$

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A planet revolves around sun whose mean distance is 1.588 times the mean distance between earth and sun. The revolution time of planet will be

1.25 years

1.59 years

0.89 years

2 years

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Explanation

$\frac{T_{p l a n e t}}{T_{e a r t h}} = {[\frac{r_{p l a n e t}}{r_{e a r t h}}]}^{3 / 2} = {(1.588)}^{3 / 2} = 2 ∴ T_{p l a n e t} = 2 y e a r s$

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The period of revolution of planet A around the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun ?

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Explanation

(c) $\frac{T_{A}}{T_{B}} = {(\frac{r_{A}}{r_{B}})}^{3 / 2} \Rightarrow 8 = {(\frac{r_{A}}{r_{B}})}^{3 / 2} \Rightarrow r_{A} = {(8)}^{2 / 3} r_{B} = 4 r_{B}$

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If the radius of earth's orbit is made 1/4 th, the duration of an year will become -

8 times

4 times

1/8 times

1/4 times

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Explanation

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If mass of a satellite is doubled and time period remain constant, the ratio of orbit in the two cases will be

1 : 2

1 : 1

1 : 3

None of these

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Explanation

(b) Mass of satellite does not affects on orbital radius.

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The earth revolves round the sun in one year. If the distance between them becomes double, the new period of revolution will be

1/2 year

$2 \sqrt{2}$ years

4 years

8 years

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Explanation

(b) $\frac{T_{2}}{T_{1}} = {(\frac{r_{2}}{r_{1}})}^{3 / 2} = {(2)}^{3 / 2} = 2 \sqrt{2} \Rightarrow T_{2} = 2 \sqrt{2} y e a r s$

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The maximum and minimum distances of a comet from the sun are $8 \times 10^{12}$ m and $1.6 \times 10^{12}$ m. If its velocity when nearest to the sun is 60 m/s, what will be its velocity in m/s when it is farthest -

112

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Explanation

(a) By conservation of angular momentum mvr = constant

$V_{m i n} \times V_{m a x} = V_{m a x} \times V_{m i n} ∴ V_{m i n} \frac{60 \times 1.6 \times 10^{12}}{8 \times 10^{12}} = \frac{60}{5} = 12 m / s$

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A body revolved around the sun 27 times faster then the earth . What is the ratio of their radii ?

1/3

1/9

1/27

1/4

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Explanation

(b)

$ω_{b o d y} = 27 ω_{e a r t h} T^{2} \propto r^{3} \Rightarrow ω \propto \frac{1}{r^{3 / 2}} \Rightarrow \frac{1}{r^{3 / 2}} ∴ r \propto \frac{1}{ω^{2 / 3}} \Rightarrow \frac{r_{b o d y}}{r_{e a r t h}} = {[\frac{ω_{e a r t h}}{ω_{b o d y}}]}^{2 / 3} = {(\frac{1}{27})}^{2 / 3} = \frac{1}{9}$

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The period of moon’s rotation around the earth is nearly 29 days. If moon’s mass were 2 fold, its present value and all other things remained unchanged, the period of moon’s rotation would be nearly -

$29 \sqrt{2} d a y s$

$29 / \sqrt{2} d a y s$

$29 \times 2 d a y s$

29 days

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Explanation

(d) Time period does not depends upon the mass of satellite.

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Two planets at mean distances $d_{1}$ and $d_{2}$ from the sun and their frequencies are n₁ and n₂ respectively, then;

${n_{1}}^{2} {d_{1}}^{2} = n_{2} {d_{2}}^{2}$

${n_{2}}^{2} {d_{2}}^{3} = {n_{1}}^{2} {d_{1}}^{3}$

$n_{1} {d_{1}}^{2} = n_{2} {d_{2}}^{2}$

${n_{1}}^{2} d_{1} = {n_{2}}^{2} d_{2}$

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Explanation

(b)

$\frac{T^{2}}{R^{3}} = \frac{T^{2}}{d^{3}} = \frac{1}{n^{2} d^{3}} = c o n s \tan t ∴ {n_{1}}^{2} {d_{1}}^{3} = {n_{2}}^{2} {d_{2}}^{2} [w h e r e n = f r e q u e n c y]$

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