If V is the gravitational potential on the surface of the earth, then what is its value at the centre of earth?
- zero
- 2V
- 3V
- $\frac{3}{2}V$
The Correct Option is D
Solution and Explanation
$V_{c}=-\frac{3}{2} \frac{G M}{R}$, on surface $V=-\frac{G M}{R}$
$\therefore V_{c}=\frac{3}{2} V$
Top Questions on Gravitation
- Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R):
(A) A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.
(R) The mass of the pendulum remains unchanged at Earth and the other planet.
In light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
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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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- Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R): \begin{itemize} \item[(A)] A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet. \item[(R)] The mass of the pendulum remains unchanged at Earth and the other planet. \end{itemize} In light of the above statements, choose the correct answer from the options given below:
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- "The height from Earth's surface at which acceleration due to gravity becomes \( g/4 \), where \( g \) is acceleration due to gravity on the surface of Earth and \( R \) is the radius of Earth?"
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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 KEAM exam
- In the measurement of length 6 $\mu$m is equal to $x$ pm. Then the value of $x$ is
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- An organic compound contains 37.5% C, 12.5% H and the rest oxygen. What is the empirical formula of the compound?
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- Let A, B, C denote the set of students in a college who play football, basketball, and cricket respectively. If \( n(A) = 60 \), \( n(B) = 55 \), \( n(C) = 70 \), \( n(A \cup B \cup C) = 100 \) and \( n(A \cap B \cap C) = 20 \), then the number of students who play exactly two of these sports is
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For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively:
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- Grignard’s reagents and their uses
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].