Gravitation MCQs with Answers – Part 1 (NEET Physics)

Explanation: Newton's universal law of gravitation states that every pair of masses attracts each other. For point masses, the magnitude of this force is directly proportional to the product \(m_1m_2\). It is inversely proportional to the square of the separation \(r\). Combining these proportionalities introduces the universal gravitational constant \(G\), giving \(F=Gm_1m_2/r^2\). The masses must therefore be multiplied, not added or subtracted. The inverse-square dependence is also essential; replacing \(r^2\) by \(r\) gives a different physical law. The force acts along the line joining the two masses and is attractive. Hence option B correctly represents the universal law.

Explanation: \(\textbf{Original relation:}\) \[ F=G\frac{m_1m_2}{r^2} \] \(\textbf{Changed quantities:}\) \[ m_1'=2m_1,\qquad m_2'=\frac{m_2}{2},\qquad r'=2r \] The product of the two masses remains unchanged: \[ m_1'm_2'=(2m_1)\left(\frac{m_2}{2}\right)=m_1m_2 \] The squared separation becomes \[ (r')^2=(2r)^2=4r^2 \] Therefore, \[ F'=G\frac{m_1m_2}{4r^2}=\frac{F}{4} \] The mass changes cancel each other, but the doubling of separation reduces the force by the inverse-square factor \(1/4\). \(\textbf{Final answer:}\) \[ \boxed{F'=\frac{F}{4}} \]

Explanation: For two point masses, \(F=Gm_1m_2/r^2\). Statement I changes only the separation from \(r\) to \(3r\). The new force is therefore \(Gm_1m_2/(3r)^2=F/9\), so Statement I is correct. Statement II changes only one mass, say \(m_1\to3m_1\). Since gravitational force is directly proportional to each mass separately, the new force becomes \(3F\). Thus Statement II is also correct. A frequent mistake is to apply a linear \(1/r\) dependence to the separation change, but gravitation follows an inverse-square law. Another mistake is to think that both masses must change for the force to change proportionally; changing either one mass is sufficient. Hence both statements are correct.

Explanation: \(\textbf{Represent the new separation:}\) Let \[ r'=kr \] \(\textbf{Compare the two forces:}\) The original force is \[ F=G\frac{m_1m_2}{r^2} \] After the changes, \[ F'=G\frac{(3m_1)(2m_2)}{(kr)^2} \] Therefore, \[ \frac{F'}{F}=\frac{6}{k^2} \] The required force condition is \[ \frac{F'}{F}=\frac{3}{2} \] Hence, \[ \frac{6}{k^2}=\frac{3}{2} \] Cross-multiplying, \[ 12=3k^2, \] so \[ k^2=4,\qquad k=2 \] Only the positive value is physically relevant because \(k\) represents a separation factor. \(\textbf{Final answer:}\) \[ \boxed{r'=2r} \]

Explanation: \(\textbf{Force in system P:}\) \[ F_P=G\frac{(M)(2M)}{(2R)^2} \] Thus, \[ F_P=\frac{2GM^2}{4R^2}=\frac{GM^2}{2R^2} \] \(\textbf{Force in system Q:}\) \[ F_Q=G\frac{(3M)(M)}{(3R)^2} \] Hence, \[ F_Q=\frac{3GM^2}{9R^2}=\frac{GM^2}{3R^2} \] \(\textbf{Comparison:}\) \[ \frac{F_P}{F_Q} =\frac{GM^2/(2R^2)}{GM^2/(3R^2)} =\frac{3}{2} \] The larger mass product in Q does not by itself determine the larger force because Q also has the larger separation. The inverse-square dependence must be included together with the mass product. \(\textbf{Final answer:}\) \[ \boxed{\frac{F_P}{F_Q}=\frac{3}{2}} \]

Explanation: \(\textbf{Decisive geometry:}\) For two external spherically symmetric bodies, each may be treated gravitationally as if its mass were concentrated at its centre. The distance in Newton's law is therefore the centre-to-centre separation, not the surface gap. Here the centre separation is \[ d=R+R+2R=4R \] The first \(R\) is the radius of the smaller body, the middle \(R\) is the surface gap, and \(2R\) is the radius of the larger body. \(\textbf{Apply the gravitational law:}\) \[ F=G\frac{(M)(3M)}{(4R)^2} \] Therefore, \[ F=\frac{3GM^2}{16R^2} \] Using only the surface separation \(R\) would overestimate the force because it would use a separation much smaller than the actual centre-to-centre distance. \(\textbf{Final answer:}\) \[ \boxed{F=\frac{3GM^2}{16R^2}} \]

Explanation: \(\textbf{Dimensional part:}\) The dimensions of force are \[ [F]=[MLT^{-2}] \] Therefore, \[ [Q]=\frac{[F][r]^3}{[m_1][m_2]} \] Substituting the dimensions, \[ [Q]=\frac{(MLT^{-2})(L^3)}{M^2} =[M^{-1}L^4T^{-2}] \] \(\textbf{Dependence on separation:}\) For fixed masses, \[ F\propto\frac{1}{r^2} \] Hence, \[ Q=\frac{Fr^3}{m_1m_2}\propto\frac{r^3}{r^2}=r \] So \(Q\) is directly proportional to the separation. Notice that \(Q\) does not have the dimensions of \(G\); it has one additional power of length because an \(r^3\) factor was used instead of the \(r^2\) required to isolate \(G\). \(\textbf{Final answer:}\) \(Q\) has dimensions \([M^{-1}L^4T^{-2}]\) and varies directly as \(r\).

Explanation: \(\textbf{Infer }G\textbf{ from the measurement:}\) Newton's law gives \[ F=G\frac{m_1m_2}{r^2} \] Rearranging, \[ G=\frac{Fr^2}{m_1m_2} \] Substitute the data: \[ G=\frac{(4.00\times10^{-8})(0.20)^2}{(4.0)(6.0)} \] Since \[ (0.20)^2=0.040, \] we obtain \[ G=\frac{1.60\times10^{-9}}{24} \] Thus, \[ G\approx6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2} \] \(\textbf{Universality check:}\) \(G\) is the universal gravitational constant; it is not the local acceleration due to gravity \(g\). The local value of \(g\) differs between Earth and Moon, but the value of \(G\) does not. \(\textbf{Final answer:}\) The inferred constant is approximately \(6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}\), and the same value applies on the Moon.

Explanation: \(\textbf{Reference force:}\) Define \[ C=\frac{GM^2}{R^2} \] Then \[ F_0=G\frac{(M)(2M)}{R^2}=2C \] \(\textbf{Use case P to determine }x\textbf{:}\) \[ F_P=G\frac{(3M)(xM)}{(2R)^2} =\frac{3x}{4}C \] The table states \[ F_P=\frac{3F_0}{2}=\frac{3}{2}(2C)=3C \] Therefore, \[ \frac{3x}{4}C=3C, \] which gives \[ x=4 \] \(\textbf{Now analyse case Q:}\) \[ F_Q=G\frac{(4M)(M)}{(3R/2)^2} \] Since \[ \left(\frac{3R}{2}\right)^2=\frac{9R^2}{4}, \] we get \[ F_Q=\frac{16}{9}C \] Therefore, \[ \frac{F_Q}{F_0} =\frac{(16/9)C}{2C} =\frac{8}{9} \] \(\textbf{Final answer:}\) \[ \boxed{\frac{F_Q}{F_0}=\frac{8}{9}} \]

Explanation: \(\textbf{Find the physically correct separation:}\) For external spherically symmetric bodies, the separation in Newton's law is measured from centre to centre. Hence, \[ r_{\text{true}}=10+20+20=50\,\text{cm} \] The student instead uses \[ r_{\text{wrong}}=20\,\text{cm} \] \(\textbf{Compare the inferred constants:}\) The true relation is \[ G=\frac{F r_{\text{true}}^2}{m_1m_2} \] The student's calculation is \[ G_{\text{calc}}=\frac{F r_{\text{wrong}}^2}{m_1m_2} \] Taking the ratio, \[ \frac{G_{\text{calc}}}{G} =\left(\frac{r_{\text{wrong}}}{r_{\text{true}}}\right)^2 \] Therefore, \[ \frac{G_{\text{calc}}}{G} =\left(\frac{20}{50}\right)^2 =\left(\frac{2}{5}\right)^2 =\frac{4}{25} \] The error is squared because separation occurs as \(r^2\) in the rearranged expression for \(G\). \(\textbf{Final answer:}\) \[ \boxed{G_{\text{calc}}=\frac{4}{25}G} \]

Explanation: \(\textbf{Use the dimensions of }K\textbf{:}\) Since \(K\) has the dimensions of \(G\), \[ [K]=[M^{-1}L^3T^{-2}] \] The proposed law has dimensions \[ [F]=[K][M]^a[M]^b[L]^c \] Therefore, \[ [MLT^{-2}] =[M^{-1+a+b}L^{3+c}T^{-2}] \] Comparing the mass powers gives \[ -1+a+b=1, \] so \[ a+b=2 \] Comparing the length powers gives \[ 3+c=1, \] hence \[ c=-2 \] \(\textbf{Use interchange symmetry:}\) The physical law must be unchanged when the labels \(m_1\) and \(m_2\) are exchanged. Thus their exponents must be equal: \[ a=b \] Combining \(a=b\) with \(a+b=2\) gives \[ a=b=1 \] Hence the reconstructed dependence is \(F\propto m_1m_2/r^2\), which is Newton's gravitational law. \(\textbf{Final answer:}\) \[ \boxed{a=1,\qquad b=1,\qquad c=-2} \]

Explanation: \(\textbf{Infer the separation at Q:}\) For fixed masses, \[ F\propto\frac{1}{r^2} \] Therefore, \[ \frac{F_Q}{F_P} =\frac{r_P^2}{r_Q^2} \] Using \(F_Q=F_0/9\) and \(F_P=F_0\), \[ \frac{1}{9}=\frac{r_0^2}{r_Q^2} \] Hence, \[ r_Q=3r_0 \] \(\textbf{Locate R from the stated condition:}\) Its separation is the arithmetic mean: \[ r_R=\frac{r_0+3r_0}{2}=2r_0 \] Thus, \[ F_R=F_0\left(\frac{r_0}{2r_0}\right)^2=\frac{F_0}{4} \] Now compare R with Q: \[ \frac{F_R}{F_Q} =\frac{F_0/4}{F_0/9} =\frac{9}{4} \] The arithmetic mean is taken between the separations, not between the force values, so the inverse-square relation must be reapplied after locating R. \(\textbf{Final answer:}\) \[ \boxed{\frac{F_R}{F_Q}=\frac{9}{4}} \]

Explanation: \(\textbf{Mass scaling:}\) For a homogeneous sphere of fixed density, \[ m\propto R^3 \] When every radius is doubled, each sphere's mass becomes \[ m'=2^3m=8m \] Thus the product of the two masses increases by \[ 8\times8=64 \] \(\textbf{Separation scaling:}\) The original centre-to-centre separation is the sum of the two radii and the surface gap: \[ r=R+2R+R=4R \] After every linear dimension is doubled, \[ r'=2r \] Therefore the inverse-square part changes by \[ \frac{r^2}{(r')^2}=\frac{1}{4} \] \(\textbf{Combine both effects:}\) \[ \frac{F'}{F} =64\times\frac14 =16 \] Although the separation becomes larger, the cubic increase of each mass dominates the inverse-square decrease caused by the increased separation. \(\textbf{Final answer:}\) \[ \boxed{F'=16F} \]

Explanation: \(\textbf{Initial centre separations:}\) For pair P, \[ r_P=R+2R+R=4R \] For pair Q, \[ r_Q=2R+R+R=4R \] \(\textbf{Use the initial equality of forces:}\) Pair P has \[ F_P=G\frac{(M)(2M)}{(4R)^2} =\frac{1}{8}\frac{GM^2}{R^2} \] Pair Q has \[ F_Q=G\frac{(3M)(xM)}{(4R)^2} =\frac{3x}{16}\frac{GM^2}{R^2} \] Since \(F_P=F_Q\), \[ \frac18=\frac{3x}{16} \] Hence, \[ x=\frac23 \] \(\textbf{Pair P after contact:}\) Its centre separation becomes \[ r_P'=R+R=2R \] Therefore, \[ F_P'=G\frac{(M)(2M)}{(2R)^2} =\frac12\frac{GM^2}{R^2} \] \(\textbf{Pair Q after contact:}\) Its centre separation becomes \[ r_Q'=2R+R=3R \] Thus, \[ F_Q'=G\frac{(3M)(2M/3)}{(3R)^2} =\frac{2}{9}\frac{GM^2}{R^2} \] Finally, \[ \frac{F_P'}{F_Q'} =\frac{1/2}{2/9} =\frac94 \] This result requires using the initial force equality to determine the unknown mass before analysing the changed geometry. \(\textbf{Final answer:}\) \[ \boxed{\frac{F_P'}{F_Q'}=\frac94} \]

Explanation: \(\textbf{Dimensions of the gravitational constant:}\) \[ [G]=[M^{-1}L^3T^{-2}] \] Let one new unit of length, mass and time be \[ L_u=10\,\text{m},\qquad M_u=2\,\text{kg},\qquad T_u=5\,\text{s} \] One unit of \(G\) in the new system therefore corresponds to \[ \frac{L_u^3}{M_uT_u^2} =\frac{10^3}{(2)(5^2)} \] in SI units. Hence, \[ \frac{10^3}{2\times25} =\frac{1000}{50} =20 \] So one new-system unit of \(G\) is \(20\) SI units of \(G\)'s dimensions. If the numerical SI value is denoted by \(G_{\text{SI}}\), the correct numerical value in the new system is therefore \[ G_u=\frac{G_{\text{SI}}}{20} \] \(\textbf{Consequence of the student's mistake:}\) When numerical masses and distances are already expressed in the new units, the force formula must use \(G_u\). The student instead inserts \(G_{\text{SI}}=20G_u\). Since \(G\) enters the force law linearly, \[ F_{\text{student}}=20F_{\text{correct}} \] in numerical value. This is a unit-representation error, not a physical change in the universal constant. \(\textbf{Final answer:}\) The calculated numerical force is \(20\) times too large.

Explanation: \(\textbf{Original gravitational force:}\) \[ F=G\frac{(m)(4m)}{r^2} =\frac{4Gm^2}{r^2} \] \(\textbf{Force after the changes:}\) The new masses are \(2m\) and \(4m\), while the new separation is \(r/2\). Thus, \[ F'=G\frac{(2m)(4m)}{(r/2)^2} \] Since \[ \left(\frac r2\right)^2=\frac{r^2}{4}, \] we obtain \[ F'=\frac{8Gm^2}{r^2/4} =\frac{32Gm^2}{r^2} \] Comparing with the original force, \[ \frac{F'}{F} =\frac{32}{4} =8 \] Therefore, \[ F'=8F \] \(\textbf{Acceleration comparison:}\) The gravitational forces on the two bodies have equal magnitudes and opposite directions, even though their masses are unequal. For the smaller body, \[ a_s=\frac{F'}{2m} \] For the larger body, \[ a_L=\frac{F'}{4m} \] Hence, \[ \frac{a_s}{a_L} =\frac{F'/(2m)}{F'/(4m)} =2 \] Because gravitational interaction is attractive, both acceleration vectors point towards the other body. Equal interaction forces therefore do not imply equal accelerations when the masses differ. \(\textbf{Final answer:}\) \[ F'=8F,\qquad \frac{a_s}{a_L}=2, \] with both accelerations directed towards each other.

Explanation: The mutual attraction between the two laboratory masses is governed by \[ F=G\frac{m_1m_2}{r^2} \] The masses and their centre-to-centre separation are unchanged when the apparatus is moved from P to Q. The universal gravitational constant \(G\) also remains unchanged. Therefore the mutual gravitational interaction of the two laboratory masses is unchanged: \[ F_Q=F_P \] Weight is a different quantity and is given by \[ W=mg \] Since the local value of \(g\) at Q is one-fourth of that at P, \[ \frac{W_Q}{W_P}=\frac{g_0/4}{g_0}=\frac14 \] Thus a change in local \(g\) must not be confused with a change in the universal constant \(G\). \(\textbf{Final answer:}\) \[ F_Q=F_P,\qquad W_Q=\frac{W_P}{4} \]

Explanation: \(\textbf{True gravitational relation:}\) For the actual experimental arrangement, \[ F=G\frac{(M)(2M)}{r^2} =\frac{2GM^2}{r^2} \] \(\textbf{Student's inferred constant:}\) The student uses \[ G_{\text{calc}}=\frac{F(3r/2)^2}{(M)(3M)} \] Now, \[ \left(\frac{3r}{2}\right)^2=\frac{9r^2}{4} \] Therefore, \[ G_{\text{calc}} =\frac{F(9r^2/4)}{3M^2} =\frac{3Fr^2}{4M^2} \] Substitute the correctly measured force: \[ G_{\text{calc}} =\frac{3}{4M^2}\left(\frac{2GM^2}{r^2}\right)r^2 \] Hence, \[ G_{\text{calc}}=\frac{3G}{2} \] The overestimated separation increases the inferred value through a square factor, while the overestimated mass partly compensates for that error. \(\textbf{Final answer:}\) \[ \boxed{G_{\text{calc}}=\frac{3G}{2}} \]

Explanation: Statement I is correct because Newton's inverse-square expression is written directly for point masses. Statement II is also correct for an external point when the source body is spherically symmetric. Spherical symmetry causes the contributions from different parts of the body to combine exactly as though the total mass were concentrated at the centre. Statement III is incorrect because an arbitrary extended nonspherical body need not produce exactly the same nearby gravitational field as a point mass placed at its centre of mass. The centre-of-mass replacement is therefore not a universal close-range rule for all finite bodies. Statement IV is correct because each non-overlapping spherically symmetric body is external to the other. Each can consequently be represented by an effective point mass at its own centre for their mutual gravitational interaction. The relevant \(r\) is then the centre-to-centre distance. Using merely a surface gap in such a situation would generally give an incorrect force. Therefore Statements I, II and IV are correct.

Explanation: The crucial condition is spherical symmetry, not uniform density throughout the body. For an external point, a spherically symmetric mass distribution produces the same gravitational effect as its total mass concentrated at the centre. The test mass lies at \(r\gt R\), so it remains outside the entire body after the redistribution. The total source mass is still \(M\). The centre of spherical symmetry is also unchanged. Therefore the external force remains \[ F=G\frac{Mm}{r^2}. \] Moving material between different radial layers can change the gravitational field at interior points. It does not alter the external inverse-square force when total mass and spherical symmetry are preserved. Hence detailed knowledge of the radial density profile is unnecessary for this external-force calculation.