India's Mangalyan was sent to… - JEE Main 2014 Physics PYQ

JEE Main 2014 — Physics Mechanics

2014mcqmedium

India's Mangalyan was sent to the Mars by launching it into a transfer orbit EOM around the sun. It leaves the earth at $E$ and meets Mars at $M$ . If the semi-major axis of Earth's orbit is ${\text{a}}_{\text{e}} = \text{1.5} \times 1 {0}^{ 1 1 } \text{m}$ , that of Mar's orbit ${\text{a}}_{\text{m}} = \text{2.28} \times 1 {0}^{ 1 1 } \text{m}$ , taking Kepler's laws, give the estimate of time for Mangalyan to reach Mars from Earth.

JEE Main 2014 Physics, Gravitation — question figure

Official previous-year question

Held on 9 Apr 2014 · Verified 6 Jul 2026.

Options

A
$220$ days.
B
$500$ days.
C
$260$ days.
D
$320$ days.

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Solution

Semi major axis of mangalyan,

${\text{a}}_{\text{Mn}} = \frac{ {\text{a}}_{\text{e}} + {\text{a}}_{\text{m}} }{ 2 }$

$= \text{1.89} \times 1 {0}^{ 1 1 } \text{m}$

From Kepler's law,

${\text{T}}^{2} \propto {\text{R}}^{3}$

$\therefore { ( \frac{ {\text{T}}_{\text{Mn}} }{ \text{Te} } ) }^{2} = { ( \frac{ {\text{R}}_{\text{M}}}{{\text{R}}_{\text{e}} } ) }^{3}$

$\therefore {\text{T}}_{\text{Mn}}\text{≈}518\text{ days}$

$\therefore$ Required time $= \frac{ 5 1 8 }{ 2 } = 2 6 0 \text{ days}$ .

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Solution

Semi major axis of mangalyan,

${\text{a}}_{\text{Mn}} = \frac{ {\text{a}}_{\text{e}} + {\text{a}}_{\text{m}} }{ 2 }$

$= \text{1.89} \times 1 {0}^{ 1 1 } \text{m}$

From Kepler's law,

${\text{T}}^{2} \propto {\text{R}}^{3}$

$\therefore { ( \frac{ {\text{T}}_{\text{Mn}} }{ \text{Te} } ) }^{2} = { ( \frac{ {\text{R}}_{\text{M}}}{{\text{R}}_{\text{e}} } ) }^{3}$

$\therefore {\text{T}}_{\text{Mn}}\text{≈}518\text{ days}$

$\therefore$ Required time $= \frac{ 5 1 8 }{ 2 } = 2 6 0 \text{ days}$ .

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics