JEE Main 24 Feb 2021 Shift 1 Solved Paper

Given, m=1kg,R=1m
We know that,
F‌‌=‌

Gm1m2

r2

∵‌‌F1‌‌=‌

Gmm

(2R)2

=‌

Gm2

4R2

‌ and ‌‌‌F2‌‌=‌

Gmm

(√2R)2

=‌

Gm2

2R2

Net force on one particle,
F‌net ‌‌‌=F1+F2‌cos‌45∘+F2‌cos‌45∘
‌‌=F1+2F2‌cos‌45∘
‌‌=‌

Gm2

4R2

+2(‌

Gm2

2R2

)⋅‌

1

√2

‌‌=‌

Gm2

4R2

+‌

Gm2

√2R2

‌‌=‌

Gm2

R2

[‌

1

4

+‌

1

√2

]
As the gravitational force provides the necessary centripetal force, so
F‌net ‌=FC=‌

mv2

R

Here, FC= centripetal force.
⇒‌

Gm2

R2

[‌

1

4

+‌

1

√2

]‌‌=‌

mv2

R

⇒‌‌v‌‌=‌

1

2

√‌

Gm

R

(1+2√2)
⇒‌‌v‌‌=‌

1

2

√G(1+2√2)