Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
A planet $(P_1)$ is moving around the star of mass $2M$ in the orbit of radius $R$. Another planet $(P_2)$ is moving around another star of mass $4M$ in a orbit of radius $2R$. Ratio of time periods of revolution of $P_2$ and $P_1$ is ________.
(1) $\frac{1}{2}$
(2) $2$
(3) $4$
(4) $\frac{1}{4}$
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Answer: Option (2)
Key idea: Kepler's Third Law of Planetary Motion
Apply Kepler's Third Law: $T^2 = \frac{4\pi^2}{GM} R^3$, where $M$ is the mass of the central star
If a body of mass 1 kg falls on the earth from infinity, it attains velocity $(v)$ and kinetic energy $(k)$ on reaching the surface of earth. The values of $v$ and $k$ respectively are __________.
(Take radius of earth to be 6400 km and $g = 9.8$ m/s$^2$)
(1) 11.2 km/s; $6.27 \times 10^7$ J
(2) 11.2 km/s; $12.54 \times 10^7$ J
(3) 8.8 km/s; $6.27 \times 10^7$ J
(4) 8.8 km/s; $12.54 \times 10^7$ J
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Answer: Option (1)
Key idea: Escape velocity and gravitational potential energy; Energy conservation for a body falling from infinity
Apply conservation of energy: Initial energy at infinity (= 0) equals final energy at Earth's surface
At surface: $0 = -\frac{GMm}{R} + \frac{1}{2}mv^2$, which gives $\frac{1}{2}mv^2 = \frac{GMm}{R}$
Use $g = \frac{GM}{R^2}$ to get $GM = gR^2$, so $v^2 = 2gR$
The height in terms of radius of the earth $(R)$, at which the acceleration due to gravity becomes $\frac{g}{9}$, where $g$ is acceleration due to gravity on earth's surface, is ________.
(1) $\sqrt{3}R$
(2) $2\sqrt{2}R$
(3) $2R$
(4) $\frac{4}{9}R$
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Answer: Option (3)
Key idea: Variation of acceleration due to gravity with height
Write the formula for gravity at Earth's surface: $g = \frac{GM}{R^2}$
Write the formula for gravity at height $h$: $g' = \frac{GM}{(R+h)^2}$
Given that $g' = \frac{g}{9}$, substitute: $\frac{GM}{(R+h)^2} = \frac{1}{9} \cdot \frac{GM}{R^2}$
When one moves from a point 16 km below the earth's surface to a point 16 km above the earth's surface. The change in g is approximately α %. The value of α is ______. (Take radius of the earth = 6400 km.)
(1) 0.12
(2) 0.25
(3) 0.50
(4) 0.75
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Answer: Option (2)
Key idea: Variation of acceleration due to gravity with height and depth
At depth $d = 16$ km below surface: $g_d = g\left(1 - \frac{d}{R}\right) = g\left(1 - \frac{16}{6400}\right)$
At height $h = 16$ km above surface: $g_h = g\left(1 - \frac{2h}{R}\right) = g\left(1 - \frac{32}{6400}\right)$
Change in $g$: $\Delta g = g_h - g_d = g\left[\left(1 - \frac{32}{6400}\right) - \left(1 - \frac{16}{6400}\right)\right]$
Simplify: $\Delta g = g\left[-\frac{32}{6400} + \frac{16}{6400}\right] = -\frac{g}{400}$
A body of mass $m$ is taken from the surface of earth to a height equal to twice the radius of earth $(R_e)$. The increase in potential energy will be __________. ($g$ is acceleration due to gravity at the surface of earth)
(1) $\frac{1}{2}mgR_e$
(2) $\frac{3}{4}mgR_e$
(3) $\frac{1}{4}mgR_e$
(4) $\frac{2}{3}mgR_e$
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Answer: Option (4)
Key idea: Gravitational Potential Energy and its variation with height
Identify positions: Initial at surface ($r_1 = R_e$), Final at height $2R_e$ ($r_2 = 3R_e$)
Use gravitational PE formula: $U(r) = -\frac{GMm}{r}$ where $GM = gR_e^2$
