Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The kinetic energy needed to project a body of mass m from earth surface to infinity is (1)/(2)mgR$\frac{1}{2}mgR$, where R is the radius of earth.
Reason R: The maximum potential energy of a body is zero when it is projected to infinity from earth surface.
In the light of the above statements, choose the correct answer from the option given below
A.A is False but R is true
B.Both A and R are true and R is the correct explanation of A
C.A is true but R is false
D.Both A and R are true but R is NOT the correct explanation of A
At r → ∞$r \to \infty$, Umax = 0$U_{\text{max}} = 0$.
Core Logic
Assertion Check: The minimum work required to project an object from Earth's surface (r = R$r = R$) to infinity (r → ∞$r \to \infty$) equals the change in gravitational potential energy:
Therefore, the required kinetic energy is mgR$mgR$. The statement asserts it is (1)/(2)mgR$\frac{1}{2}mgR$ (which corresponds to orbital kinetic energy near Earth's surface). Hence, Assertion A is false.
Reason Check: Since the gravitational force is attractive, potential energy is negative everywhere in the field and reaches its maximum value asymptotically at infinity (Umax = 0$U_{\text{max}} = 0$). Hence, Reason R is true.
Pattern Recognition
Escape energy from the surface is mgR$mgR$, while circular orbital kinetic energy near the surface is (1)/(2)mgR$\frac{1}{2}mgR$. Because gravitational potential energy is defined with zero at infinity, every bound state has U < 0$U < 0$, making zero the absolute maximum potential energy.
Evaluation Rubric / Model Answer
Option A: A is false but R is true
Chapter Mix
Class 11 Physics: Gravitation
More Gravitation Previous-Year Questions — Page 4
Q44jee_main_2024_29_january_eveningKepler's Laws of Planetary Motion
A planet takes 200 days$200\text{ days}$ to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution?
A.25$25$
B.50$50$
C.100$100$
D.20$20$
Solution
Related Formula
According to Kepler's Third Law (Law of Periods):
T² ∝ r³$T^2 \propto r^3$
where:
T$T$ is the time period of revolution.
r$r$ is the orbital radius of the planet.
Core Logic
Using the proportionality relationship for two states:
If orbital distance scales by x$x$, the period scales by x3/2$x^{3/2}$. Here, distance scales by (1)/(4)$\frac{1}{4}$, so the period scales by ((1)/(4))3/2 = (1)/(8)$\left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$. Thus, 200 × (1)/(8) = 25 days$200 \times \frac{1}{8} = 25\text{ days}$.
Chapter Mix
Class 11 Physics: Gravitation
Q35jee_main_2024_27_jan_morningAcceleration due to Gravity
Inverse square dependence means halving the distance scale amplifies the surface field metric by a factor of 2² = 4$2^2 = 4$ matching constant mass bounds.
Chapter Mix
Class 11 Physics: Gravitation
Qjee_main_2024_29_jan_morningAcceleration due to Gravity
At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as R.)
A.√(5) R - R$\sqrt{5} \mathrm{R} - \mathrm{R}$
B.√(3) R - R2$\frac{\sqrt{3} \mathrm{R} - \mathrm{R}}{2}$
C.(R)/(2)$\frac{R}{2}$
D.√(5) R - R2$\frac{\sqrt{5} \mathrm{R} - \mathrm{R}}{2}$
Solution
Related Formula
Acceleration due to gravity at a height h$h$ above the Earth's surface:
Therefore, the required distance is √(5)R - R2$\frac{\sqrt{5}R - R}{2}$.
Pattern Recognition
Do not use the linear approximation formula gh ≈ g(1 - (2h)/(R))$g_h \approx g(1 - \frac{2h}{R})$ unless the problem explicitly states h ll R$h \ll R$. Equating the approximated form to depth gives hheight = (1)/(2) hdepth$h_{\text{height}} = \frac{1}{2} h_{\text{depth}}$, which fails when looking for a single unified distance value h$h$.
Escape velocity of a body from earth is 11.2 km/s$11.2 \,\mathrm{km/s}$. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is:
A.11.2 ~km / s$11.2 \mathrm{~km / s}$
B.8.4 ~km / s$8.4 \mathrm{~km / s}$
C.4.2 ~km / s$4.2 \mathrm{~km / s}$
D.7.9 ~km / s$7.9 \mathrm{~km / s}$
Solution
Related Formula
Vₑ = √((2GM)/(R))$$V_e = \sqrt{\frac{2GM}{R}}$$
Core Logic
For Earth: Vₑ = √((2GME)/(RE)) = 11.2 ~km/s$V_e = \sqrt{\frac{2GM_E}{R_E}} = 11.2 \mathrm{~km/s}$
For the planet:
RP = RE3$\mathrm{R}_{\mathrm{P}} = \frac{\mathrm{R}_{\mathrm{E}}}{3}$ and MP = ME6$\mathrm{M}_{\mathrm{P}} = \frac{\mathrm{M}_{\mathrm{E}}}{6}$
We can express the escape velocity of the planet Vₚ$V_p$ as a ratio of the Earth's escape velocity.
Any scaling of a planet's mass by factor α$\alpha$ and radius by factor β$\beta$ scales the escape velocity by a factor of √(α / β)$\sqrt{\alpha / \beta}$.
Chapter Mix
Class 11 Physics: Gravitation
Q42jee_main_2024_30_jan_morningGravitational Potential and Field
The gravitational potential at a point above the surface of earth is -5.12 × 10⁷ ~J / kg$-5.12 \times 10^{7} \mathrm{~J / kg}$ and the acceleration due to gravity at that point is 6.4 ~m/s²$6.4 \mathrm{~m/s^2}$. Assume that the mean radius of earth to be 6400 ~km$6400 \mathrm{~km}$. The height of this point above the earth's surface is:
The gravitational potential (V$V$) and acceleration due to gravity (g'$g'$) at a distance r = RE + h$r = R_E + h$ from the center of the earth can be related by dividing their magnitudes: |V| / g' = r$|V| / g' = r$.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
