If the weight of an object on earth's surface is 400 N, then weight of the same particle at a depth \(\frac{R}{2}\) from surface would be (R is radius of earth)
- 100 N
- 300 N
- 200 N
- 250 N
The Correct Option is C
Solution and Explanation
Step 1: Use the formula for weight at depth.- Weight on surface: W = mg = 400N.- At depth d = R/2,
Step 2: Substitute the values.
W' = W(1 - dR)
W' = 400(1 - R/2R)
W' = 400(1 - 12)
W' = 400 $\cdot$ 12 = 200N
Final Answer: Weight at half the earth’s radius = 200N
Top IIT JEE Physics Questions
Proton (P) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume mp = 1849me):
- IIT JEE - 2023
- Physics
- Stereoisomers
- The reversible expansion ob an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct?
- IIT JEE - 2012
- Physics
- Kinetic molecular theory of gases
Three infinitely long charge sheets are placed as shown in figure. The electric field at point P is
- IIT JEE - 2005
- Physics
- Gauss Law
- The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Let $ \lambda_1 $ be the de-Broglie wavelength of the proton and $ \lambda_2 $ be the wavelength of the photon. The ratio $ \frac{\lambda_1}{\lambda_2} $ is proportional to
- IIT JEE - 2004
- Physics
- Kinetic Energy
- The variation of induced emf (e) with time (t) in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as
- IIT JEE - 2004
- Physics
- Faradays laws of induction
Top IIT JEE Stereoisomers Questions
Top IIT JEE Questions
- The major product of the following reaction is
- IIT JEE - 2022
- Hydrocarbons
- Let $ \alpha ( a) \, and \, \beta (a) $ be the roots of the equation $ (\sqrt [3] {1 + a} - 1) x^2 - (\sqrt {1 + a} - 1) x + (\sqrt [ 6] {1 + a} - 1) = 0$, where a > - 1. Then, $ lim_{a \to 0^+ } $, $ \alpha (a) \, and \, lim_{a \to 0^+ } \beta $ (a) are
- IIT JEE - 2012
- Quadratic Equations
- Let $f:(-1,1)\rightarrow$ R be such that $f(cos 4 \theta)=\frac{2}{2-sec^2 \theta}$ for $\theta \in \Bigg(0,\frac{\pi}{4}\Bigg)\cup\Bigg(\frac{\pi}{4},\frac{\pi}{2}\Bigg).$ Then, the value(s) of $f\Bigg(\frac{1}{3}\Bigg)$ is are
- IIT JEE - 2012
- Functions
- The reversible expansion ob an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct?
- IIT JEE - 2012
- Kinetic molecular theory of gases
- The Henry?? law constant for the solubility of $N_2$ gas in water at $298\, K$ is $1.0 \times 10^5\, atm$. The mole fraction of $N_2$ in air is $0.8$. The number of moles of $N_2$ from air dissolved in $10$ moles of water at $298\, K$ and $5 \,atm$ pressure is
- IIT JEE - 2009
- Colligative Properties
Concepts Used:
Newtons Law of Gravitation
Gravitational Force
Gravitational force is a central force that depends only on the position of the test mass from the source mass and always acts along the line joining the centers of the two masses.
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,
- Directly proportional to the product of their masses i.e. F ∝ (M1M2) . . . . (1)
- Inversely proportional to the square of the distance between their center i.e. (F ∝ 1/r2) . . . . (2)
By combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2 [f(r)is a variable, Non-contact, and conservative force]