A mass falls from a height 'h' and its time of fall $'l'$ is recorded in terms of time period T of a simple pendulum. On the surface of earth it is found that I = 2T. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as t' and T'.
- t ' = $\sqrt{ 2}$ T '
- t ' > 2 T'
- t ' < 2 T'
- t ' = 2 T'
The Correct Option is D
Solution and Explanation
Time period of pendulum $ = 2\pi \sqrt{\frac{1}{g}} \propto \frac{1}{\sqrt{g}}$
Ratio of time of flight & time period of pendulum is independent of g. Hence $t' = 2T$
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Match the LIST-I with LIST-II
\[ \begin{array}{|l|l|} \hline \text{LIST-I} & \text{LIST-II} \\ \hline \text{A. Gravitational constant} & \text{I. } [LT^{-2}] \\ \hline \text{B. Gravitational potential energy} & \text{II. } [L^2T^{-2}] \\ \hline \text{C. Gravitational potential} & \text{III. } [ML^2T^{-2}] \\ \hline \text{D. Acceleration due to gravity} & \text{IV. } [M^{-1}L^3T^{-2}] \\ \hline \end{array} \]
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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].