SATHEE: Gravitation Question 341

Gravitation Question 341

Question: Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and length $[ℓ]$ is

Options:

A) $[\frac{G m}{ℓ^{2}} a l o n g + x - a x i s]$

B) $[\frac{G m}{π ℓ^{2}} a l o n g + y - a x i s]$

C) $[\frac{2 π G m}{ℓ^{2}} a l o n g + x - a x i s]$

D) $[\frac{2 π G m}{ℓ^{2}} a l o n g + y - a x i s]$

Show Answer

Answer:

Correct Answer: D

Solution:

Let mass per unit light of wire, $[λ = \frac{M}{ℓ}]$

and $[π r = ℓ, r = \frac{ℓ}{π}]$ .

mass of element,

$[d m = λ r d θ$ then $d E = \frac{G d m}{r^{2}}]$

$[\int_{0}^{π} d E \int_{0}^{π} \frac{G λ r d θ}{r^{2}} (\hat{i} \cos θ + \hat{j} \sin θ)]$

$[E = \frac{G λ}{r} [\int_{0}^{π} \hat{i} \cos θ + \int_{0}^{π} \hat{j} \sin θ]]$

$[= \frac{2 G λ}{r} \hat{j} = \frac{2 G M}{ℓ r} \hat{j} = \frac{2 G m π}{ℓ^{2}} \hat{j}]$ (along y-axis

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Gravitation Question 341

Question: Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and length [ℓ] is

Options:

Answer:

Solution:

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Question: Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and length $[ℓ]$ is