Gravitation

Answer: (a)

1. Each corner is at the same distance $r$ from the centre, so each pull is $\frac{GM_im}{r^2}$ along the unit vector to that corner, $\frac{(\pm1, \pm1)}{\sqrt{2}}$.
2. Take $M(-1,-1)$, $2M(1,-1)$, $3M(1,1)$, $4M(-1,1)$: $\sum M_i(x,y) = M(0, 4)$, so $F_1 \propto \frac{4}{\sqrt{2}}$.
3. Swap $3M$ and $4M$: $\sum M_i(x,y) = M(2, 4)$, so $F_2 \propto \frac{\sqrt{20}}{\sqrt{2}}$.
4. $\frac{F_1}{F_2} = \frac{4}{\sqrt{20}} = \frac{2}{\sqrt{5}} \Rightarrow \alpha = 2$
5. So the answer is (a) 2.

Answer: 8

1. Acceleration scales as $\frac{Gm}{a^2}$ and distance as $a$, so $T \propto \sqrt{\frac{a}{Gm/a^2}} = \sqrt{\frac{a^3}{Gm}}$.
2. $\frac{T'}{T} = \sqrt{\frac{(2a)^3}{2m} \cdot \frac{m}{a^3}} = \sqrt{4} = 2$
3. $T' = 2 \times 4 = 8$ s

Answer: (b)

1. $F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$
2. Removed sphere: mass $\frac{M}{27}$, centre at $R - \frac{R}{3} = \frac{2R}{3}$ from $O$, so $\frac{4R}{3}$ from $m$.
3. Its pull: $\frac{G(M/27)m}{(4R/3)^2} = \frac{GMm}{48R^2}$
4. $F_2 = \frac{GMm}{4R^2} - \frac{GMm}{48R^2} = \frac{11GMm}{48R^2}$
5. $F_1 : F_2 = 12 : 11$, so the answer is (b).

Answer: (a)

1. Let the side be $a$. Two adjacent masses each pull with $\frac{Gm^2}{a^2}$ at $90^\circ$: resultant $\frac{\sqrt{2}Gm^2}{a^2}$ along the diagonal.
2. Diagonal mass at $\sqrt{2}a$ pulls along the same diagonal: $\frac{Gm^2}{2a^2}$.
3. Total $F = \left(\sqrt{2} + \frac{1}{2}\right)\frac{Gm^2}{a^2} = \frac{2\sqrt{2}+1}{2}\cdot\frac{Gm^2}{a^2}$
4. Set equal to $\frac{2\sqrt{2}+1}{32}\cdot\frac{Gm^2}{L^2}$: $a^2 = 16L^2 \Rightarrow a = 4L$
5. So the answer is (a) $4L$.

Answer: (b)

1. Each mass $M = \frac{4}{3}\pi R^3\rho$; centres in contact are $2R$ apart.
2. $F = \frac{GM^2}{(2R)^2} = \frac{G}{4R^2}\left(\frac{4}{3}\pi R^3\rho\right)^2 \propto R^4\rho^2$
3. So the answer is (b) $R^4\rho^2$.

Answer: (d)

1. Semicircle of length $L$: radius $r = \frac{L}{\pi}$, linear density $\lambda = \frac{M}{L}$.
2. Components along the diameter cancel; along the symmetry axis $F = \int_0^\pi \frac{Gm\lambda r\,d\theta}{r^2}\sin\theta = \frac{2Gm\lambda}{r}$.
3. $F = \frac{2Gm(M/L)}{L/\pi} = \frac{2\pi GMm}{L^2}$
4. So the answer is (d).

Answer: (a)

1. Every small element of the ring has an equal element diametrically opposite it at the same distance from the centre.
2. Their pulls on $m$ are equal and opposite, so they cancel in pairs.
3. So the answer is (a) zero.

Answer: (d)

1. The two masses at the ends of the side pull $m_2$ equally in opposite directions, so they cancel.
2. Only the opposite vertex acts, at distance equal to the height $h = \frac{\sqrt{3}}{2} \cdot \frac{L}{3} = \frac{\sqrt{3}L}{6}$.
3. $h^2 = \frac{3L^2}{36} = \frac{L^2}{12}$
4. $F = \frac{Gm_1m_2}{h^2} = \frac{12Gm_1m_2}{L^2}$
5. So the answer is (d).

Answer: (b)

1. Mass of each sphere $M \propto R^3$ (same density); separation of centres $= 2R$.
2. $F = \frac{GM^2}{4R^2} \propto \frac{R^6}{R^2} = R^4$
3. So $x = 4$, answer (b).

Answer: (b)

1. $F = \frac{Gm_1m_2}{r^2} \Rightarrow G = \frac{Fr^2}{m_1m_2} = [M^{-1}L^3T^{-2}]$: (a)-(ii).
2. Potential energy is energy: $[ML^2T^{-2}]$: (b)-(iv).
3. Potential $V = U/m = [L^2T^{-2}]$: (c)-(i).
4. Intensity $E = F/m = [LT^{-2}]$: (d)-(iii).
5. So the answer is (b).

Answer: (c)

1. Initially $F = \frac{Gm \cdot m}{r^2}$.
2. After transfer the masses are $\frac{2m}{3}$ and $\frac{4m}{3}$, product $\frac{8m^2}{9}$.
3. $F' = \frac{G(8m^2/9)}{r^2} = \frac{8}{9}F$
4. So the answer is (c) $\frac{8}{9}F$.

Answer: (a)

1. Statement I: Newton's law of gravitation is universal, so it holds for any pair of bodies. True.
2. Statement II: $g_d = g\left(1 - \frac{d}{R}\right) = 0$ at $d = R$, so weight $mg_d = 0$ at the centre. True.
3. So the answer is (a) both statements are true.

Answer: (b)

1. B pulls $P$ from $13$ m: $F_B = \frac{G(100)^2}{169} \approx 59.2G$.
2. A and C are at $13\sqrt{2}$ m: each $\frac{G(100)^2}{338} \approx 29.6G$, at $45^\circ$ to $PB$.
3. Sideways parts cancel; along $PB$: $2 \times 29.6G \cos45^\circ \approx 41.8G$.
4. $F \approx 59.2G + 41.8G \approx 101G$
5. So the answer is (b) $100G$.

Answer: (c)

1. The sphere acts as a point mass $M$ at its centre, on the axis of the ring.
2. Field of a ring on its axis: $E = \frac{Gm x}{(R^2 + x^2)^{3/2}}$, with $x = \sqrt{8}R$.
3. $R^2 + x^2 = 9R^2 \Rightarrow (9R^2)^{3/2} = 27R^3$
4. $F = \frac{GMm\sqrt{8}R}{27R^3} = \frac{\sqrt{8}}{27}\cdot\frac{GMm}{R^2}$
5. So the answer is (c).

Answer: (c)

1. $F_1 = \frac{GMm}{(3R)^2} = \frac{GMm}{9R^2}$
2. Cavity of radius $\frac{R}{2}$ has mass $\frac{M}{8}$; its centre B is $\frac{R}{2}$ from O, so $3R - \frac{R}{2} = \frac{5R}{2}$ from $m$.
3. Pull of the cavity part: $\frac{G(M/8)m}{(5R/2)^2} = \frac{GMm}{50R^2}$
4. $F_2 = GMm\left(\frac{1}{9} - \frac{1}{50}\right)\frac{1}{R^2} = \frac{41GMm}{450R^2}$
5. $F_1 : F_2 = \frac{50}{450} : \frac{41}{450} = 50 : 41$, so the answer is (c).

Answer: (b)

1. Element $dx$ at $x$ has mass $(A + Bx^2)dx$ and pulls $m$ with $dF = \frac{Gm(A + Bx^2)dx}{x^2}$.
2. $F = Gm\int_a^{a+L}\left(\frac{A}{x^2} + B\right)dx$
3. $F = Gm\left[A\left(\frac{1}{a} - \frac{1}{a+L}\right) + BL\right]$
4. So the answer is (b).

Answer: (a)

1. Difference between near and far points: $\Delta F \approx \frac{GMm}{d^3}\cdot 2(2R_e)$, so $\Delta F \propto \frac{M}{d^3}$.
2. $\frac{\Delta F_1}{\Delta F_2} = \frac{M_m}{M_s}\left(\frac{d_s}{d_m}\right)^3 = \frac{8 \times 10^{22}}{2 \times 10^{30}}\left(\frac{150}{0.4}\right)^3$
3. $= 4 \times 10^{-8} \times 375^3 = 4 \times 10^{-8} \times 5.27 \times 10^{7} \approx 2.1$
4. So the answer is (a) 2.

Answer: (a)

1. Gravity acts between every pair of masses, so the two astronauts attract each other.
2. With no other force acting, this mutual attraction pulls them together.
3. So the answer is (a) they move towards each other.

Answer: (a)

1. Removed sphere: mass $\frac{M}{8}$; it sits $3R$ from O. The cavity centre is $\frac{R}{2}$ from O, so $3R - \frac{R}{2} = \frac{5R}{2}$ from the removed sphere.
2. Remaining part = full sphere $M$ minus cavity sphere $\frac{M}{8}$.
3. $F = \frac{GM(M/8)}{(3R)^2} - \frac{G(M/8)(M/8)}{(5R/2)^2} = \frac{GM^2}{72R^2} - \frac{GM^2}{400R^2}$
4. $F = \frac{(50 - 9)GM^2}{3600R^2} = \frac{41GM^2}{3600R^2}$
5. So the answer is (a).

Answer: (a)

1. Moon's pull gives the earth an acceleration $a = \frac{GM_m}{r^2}$, toward the sun at eclipse and away from it on the other side.
2. $a = \frac{6.67 \times 10^{-11} \times 7.36 \times 10^{22}}{(3.8 \times 10^{8})^2} \approx 3.4 \times 10^{-5}$ m/s$^2$
3. Change $= 2a \approx 6.7 \times 10^{-5}$ m/s$^2$
4. So the answer is (a) $6.73 \times 10^{-5}$ m/s$^2$.

Answer: (a)

1. (1) True: C.G. is the point where the total weight of the body acts.
2. (2) True: for an earth of infinite radius $g$ is uniform over the body, and then C.M. and C.G. coincide.
3. (3) False: treating the whole mass as concentrated at one point is exact only for spherically symmetric bodies.
4. (4) False: radius of gyration $K$ is defined by $I = MK^2$, not as the perpendicular distance of the C.G. from the axis.
5. So the correct pair is (1) and (2), option (b).

Answer: (b)

1. No external force, so momentum is conserved: $50v = 0.5 \times 2 \Rightarrow v = 0.02$ m/s upward for the man.
2. No gravity, so the stone moves uniformly: $t = \frac{10}{2} = 5$ s.
3. Man rises $0.02 \times 5 = 0.1$ m, so his height $= 10 + 0.1 = 10.1$ m.
4. So the answer is (b) 10.1 m.

Answer: (b)

1. $F = \frac{GMm}{r^2}$ does not depend on the medium between the masses ($G$ is a universal constant).
2. So filling the space with a liquid leaves the force unchanged.
3. So the answer is (b) $F$.

Answer: (c)

1. Only internal forces act, so the centre of mass stays fixed.
2. They collide when the centres are $R + 2R = 3R$ apart, so together they move $12R - 3R = 9R$.
3. Distances are inversely proportional to masses: $\frac{x_1}{x_2} = \frac{5M}{M} = 5$.
4. $x_1 = \frac{5}{6} \times 9R = 7.5R$
5. So the answer is (c) $7.5R$.

Answer: (a)

1. $g = \frac{GM}{R^2} = \frac{4}{3}\pi G\rho R$, so $g \propto \rho R$.
2. $\frac{g_p}{g_e} = 3 \times 2.5 = 7.5$
3. So the answer is (a) 7.5.

Answer: (b)

1. $g_h = \frac{g}{(1 + h/R)^2}$; for $h = 2R$, $g_h = \frac{g}{9}$.
2. $T = 2\pi\sqrt{\frac{L}{g}} \Rightarrow T \propto \frac{1}{\sqrt{g}}$.
3. $\frac{T_2}{T_1} = \sqrt{\frac{g}{g/9}} = 3$
4. So the answer is (b) $T_1 : T_2 = 1 : 3$.

Answer: (c)

1. At $h = R$: $g_h = \frac{g}{(1+1)^2} = \frac{g}{4}$.
2. $T \propto \frac{1}{\sqrt{g}} \Rightarrow T' = T\sqrt{4} = 2T$
3. So the answer is (c) $x = 2$.

Answer: (c)

1. $\left(\frac{R}{R+h}\right)^2 = \frac{1}{9}$
2. $R + h = 3R \Rightarrow h = 2R$
3. So the answer is (c) $2R$.

Answer: (a)

1. (a) True: $g_h = \frac{gR^2}{(R+h)^2} < g$ and $g_d = g\left(1-\frac{d}{R}\right) < g$.
2. (b) False: for small heights $g_h \approx g\left(1-\frac{2h}{R}\right)$ but $g_d = g\left(1-\frac{d}{R}\right)$, so they differ when $h = d$.
3. (c) False: $g_h = \frac{g}{(1.1)^2} \approx 0.83g$, a decrease of about $17\%$, not $10\%$.
4. (d) False: $g' = g - \omega^2 R\cos^2\lambda$; at the poles $\lambda = 90^\circ$, so rotation causes no reduction there.
5. So the answer is (a).

Answer: (c)

1. Statement I: $g$ is maximum at the surface; it falls going up ($\propto 1/r^2$) and going down ($\propto r$). Correct.
2. Statement II: $g_h \approx g\left(1-\frac{2h}{R}\right)$, $g_d = g\left(1-\frac{d}{R}\right)$, so for $h = d$ they are not equal. Incorrect.
3. So the answer is (c).

Answer: (b)

1. For $h \ll R$: $g_h \approx g\left(1 - \frac{2h}{R}\right)$, so fractional decrease $= \frac{2h}{R}$.
2. $\frac{2 \times 48}{6400} \times 100\% = 1.5\%$
3. So the answer is (b) $1.5\%$.

Answer: (a)

1. Seconds pendulum: $T = 2\pi\sqrt{l/g}$ fixed, so $l \propto g$.
2. $\frac{l_h}{l_s} = \frac{g_h}{g} = \frac{R^2}{(R+h)^2} \Rightarrow \frac{\sqrt{l_h}}{\sqrt{l_s}} = \frac{R}{R+h}$
3. $R\sqrt{l_s} = (R+h)\sqrt{l_h} \Rightarrow R\left(\sqrt{l_s} - \sqrt{l_h}\right) = h\sqrt{l_h}$
4. So the answer is (a) $R = \frac{h\sqrt{l_h}}{\sqrt{l_s} - \sqrt{l_h}}$.

Answer: (d)

1. Capillary rise $h = \frac{2T\cos\theta}{r\rho g}$, so $h \propto \frac{1}{g}$
2. At depth $d$: $g_d = g\left(1 - \frac{d}{R}\right) = g\,\frac{R-d}{R}$
3. $\frac{h_2}{h_1} = \frac{g}{g_d} = \frac{R}{R-d}$
4. So the answer is (d) $\frac{R}{R-d}$.

Answer: (d)

1. Small height: $g_h = g\left(1 - \frac{2h}{R}\right)$; depth: $g_d = g\left(1 - \frac{d}{R}\right)$
2. Equal values: $\frac{2h}{R} = \frac{d}{R} \Rightarrow h = \frac{d}{2}$
3. $h = \frac{90}{2} = 45$ km
4. So the answer is (d) 45 km.

Answer: (b)

1. 16 km below: $g_1 = g\left(1 - \frac{16}{6400}\right) = g(1 - 0.0025)$
2. 16 km above: $g_2 = g\left(1 - \frac{2 \times 16}{6400}\right) = g(1 - 0.005)$
3. Change $= g_1 - g_2 = 0.0025g$, i.e. $0.25\%$
4. So $\alpha = 0.25$, answer (b).

Answer: (b)

1. At the equator: $g' = g - \omega^2 R$
2. $g - \omega^2 R = \frac{3}{5}g \Rightarrow \omega^2 = \frac{2g}{5R}$
3. $\omega = \sqrt{\frac{2g}{5R}}$ (angular speed)
4. So the answer is (b).

Answer: (b)

1. At the equator: $g' = g - \omega^2 R$
2. $g - \omega^2 R = \frac{3}{5}g \Rightarrow \omega^2 = \frac{2g}{5R}$
3. $\omega = \left(\frac{2g}{5R}\right)^{1/2}$
4. So the answer is (b).

Answer: (b)

1. $g = \frac{4}{3}\pi G\rho r$, so $g \propto \rho r$.
2. $g_A : g_B = r_1\rho_1 : r_2\rho_2$
3. So the answer is (b).

Answer: 9

1. $g = \frac{GM}{R^2}$; with $M$ fixed, $g \propto \frac{1}{R^2}$.
2. Diameter one third $\Rightarrow R' = \frac{R}{3} \Rightarrow g' = 9g$
3. So the answer is 9.

Answer: (a)

1. $W_h = \frac{W}{(1 + h/R)^2}$ with $h = \frac{R}{2}$.
2. $W_h = \frac{45}{(3/2)^2} = 45 \times \frac{4}{9}$
3. $W_h = 20\ \text{N}$
4. So the answer is (a) 20 N.

Answer: (d)

1. At $h = 2R$: $g_h = \frac{g}{(1+2)^2} = \frac{g}{9}$.
2. $T \propto \frac{1}{\sqrt{g}} \Rightarrow T' = T\sqrt{9} = 3T$
3. So the answer is (d) $x = 3$.

Answer: (d)

1. $W_h = \frac{W}{(1 + h/R)^2}$ with $h = \frac{R}{3}$.
2. $W_h = \frac{48}{(4/3)^2} = 48 \times \frac{9}{16}$
3. $W_h = 27\ \text{N}$
4. So the answer is (d) 27 N.