A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc, and a particle of mass m is placed at the centre of the arc. The gravitational force on the particle by the wire is

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Q.
A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc, and a particle of mass m is placed at the centre of the arc. The gravitational force on the particle by the wire is _____. [2024]

1 0

2 $\frac{G M m π}{2 L^{2}}$

3 $\frac{2 G M m π}{L^{2}}$

4 $\frac{G M m π}{L^{2}}$

Ans.
(3)

$E_{g} = \frac{2 \cdot G \cdot λ}{R} = \frac{2 G M}{L \cdot R}$

$also, π R = L \Rightarrow R = \frac{L}{π}$

$∴ E_{g} = \frac{2 G M}{L \cdot \frac{L}{π}} = \frac{2 G M π}{L^{2}}$

$F = m E_{g} = \frac{2 G M m π}{L^{2}}$

Solution

Q.
A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc, and a particle of mass m is placed at the centre of the arc. The gravitational force on the particle by the wire is _____. [2024]

1 0

2 $\frac{G M m π}{2 L^{2}}$

3 $\frac{2 G M m π}{L^{2}}$

4 $\frac{G M m π}{L^{2}}$

Ans.
(3)

$E_{g} = \frac{2 \cdot G \cdot λ}{R} = \frac{2 G M}{L \cdot R}$

$also, π R = L \Rightarrow R = \frac{L}{π}$

$∴ E_{g} = \frac{2 G M}{L \cdot \frac{L}{π}} = \frac{2 G M π}{L^{2}}$

$F = m E_{g} = \frac{2 G M m π}{L^{2}}$

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Solution

Q. A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc, and a particle of mass m is placed at the centre of the arc. The gravitational force on the particle by the wire is _____. [2024]

1 0

2 GMmπ2L2

3 2GMmπL2

4 GMmπL2

Ans. (3) Eg=2·G·λR=2GML·R also, πR=L ⇒ R=Lπ ∴ Eg=2GML·Lπ=2GMπL2 F=mEg=2GMmπL2

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Q.
A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc, and a particle of mass m is placed at the centre of the arc. The gravitational force on the particle by the wire is _____. [2024]

2 $\frac{G M m π}{2 L^{2}}$

3 $\frac{2 G M m π}{L^{2}}$

4 $\frac{G M m π}{L^{2}}$

Ans.
(3)

$E_{g} = \frac{2 \cdot G \cdot λ}{R} = \frac{2 G M}{L \cdot R}$

$also, π R = L \Rightarrow R = \frac{L}{π}$

$∴ E_{g} = \frac{2 G M}{L \cdot \frac{L}{π}} = \frac{2 G M π}{L^{2}}$

$F = m E_{g} = \frac{2 G M m π}{L^{2}}$