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JEE Advanced 2018 Paper 2

Section: Physics

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Question : 16 of 54

Marks: +1, -0

A planet of mass M, has two natural satellites with masses

m_1

m_2

.The radii of their circular orbits are

R_1

R_2

respectively. Ignore the gravitational force between the satellites. Define

v_1,L_1,K_1

T_1

to be,respectively,the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite 1; and

v_2,L_2,K_2

T_2

to be the corresponding quantities of satellite 2.Given

\frac{m_1}{m_2}=2

\frac{R_1}{R_2}=\frac{1}{4}

, match the ratios in List-I to the numbers in List-II

List - I	List - II
P. $\frac{v_1}{v_2}$	1. $\frac{1}{8}$
Q. $\frac{L_1}{L_2}$	2. 1
R. $\frac{K_1}{K_2}$	3. 2
S. $\frac{T_1}{T_2}$	4. 8

	P	Q	R	S
A)	4	2	1	3
B)	3	2	4	1
C)	2	3	1	4
D)	2	3	4	1

[JEE Adv 2018 P2]

D
A
C
B

Solution:

\frac{m_1}{m_2}=2\;\;\;\;\frac{R_1}{R_2}=\frac{1}{4}

\frac{GMm_1}{R_1^2}=\frac{m_1V_1^2}{R_1}\;\;\;\;V_1^2=\frac{GM}{R_1},V_2^2=\frac{GM}{R_2}

\frac{V_1^2}{V_2^2}=\frac{R_2}{R_1}=4\;\;\;\;\;\;\;(P)\frac{V_1}{V_2}=2

(Q)

L=mvR\;\;\;\frac{L_1}{L_2}=\frac{m_1v_1R_1}{m_2v_2R_2}=2\times2\times\frac{1}{4}=1

R)

K=\frac{1}{2}mv^2\;\;\;\;\frac{K_1}{K_2}=\frac{m_1v_1^2}{m_2v_2^2}=2\times(2)^2=8

(S)

T=2\pi R\text{ "/" }V\;\frac{T_1}{T_2}=\frac{R_1}{v_1}\times\frac{v_2}{R_2}=\frac{R_1}{R_2}\times\frac{v_2}{v_{}}=\frac{1}{4}\times\frac{1}{2}=\frac{1}{8}

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