There is a shell of mass $M$ and density of the shell is uniform. The work done to take a point mass from point $A$ to $B$ is $[A B=r]$
- $\frac{GmM}{r}$
- $\frac{GmM}{R}$
- $ - \frac{GmM}{r}$
- zero
The Correct Option is D
Solution and Explanation
Top Questions on Gravitation
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(R) The mass of the pendulum remains unchanged at Earth and the other planet.
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A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is:
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The height from Earth's surface at which acceleration due to gravity becomes \(\frac{g}{4}\) is \(\_\_\)? (Where \(g\) is the acceleration due to gravity on the surface of the Earth and \(R\) is the radius of the Earth.)
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Questions Asked in BITSAT exam
- Let \( ABC \) be a triangle and \( \vec{a}, \vec{b}, \vec{c} \) be the position vectors of \( A, B, C \) respectively. Let \( D \) divide \( BC \) in the ratio \( 3:1 \) internally and \( E \) divide \( AD \) in the ratio \( 4:1 \) internally. Let \( BE \) meet \( AC \) in \( F \). If \( E \) divides \( BF \) in the ratio \( 3:2 \) internally then the position vector of \( F \) is:
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- Let \( \mathbf{a} = \hat{i} - \hat{k}, \mathbf{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}, \mathbf{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} \). Then, \( [\mathbf{a} \, \mathbf{b} \, \mathbf{c}] \) depends on:}
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].