The time period of a pendulum increases as g decreases. At higher altitudes, pendulums run slower, not faster.
The time period of a pendulum is given by:
\[ T \propto \sqrt{\frac{1}{g}} \]
At Mount Everest, \( g \) is less than on Earth’s surface. A decrease in \( g \) increases \( T \), making the pendulum slower, not faster. Thus:
T∝\(\frac {1}{\sqrt g}\)
Time period of pendulum is inversely proportional to acceleration due to gravity.
So, the correct answer is (C): A is not correct but R is correct
If $ \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p $, then $ 96 \log_e p $ is equal to _______
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
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.
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
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].