JEE Physics Units Kinematics and Gravitation MCQs Set 01

Answer: (b) 8%

Explanation:
1. \(K = \tfrac{1}{2}mv^2\), so \(\dfrac{\Delta K}{K} = \dfrac{\Delta m}{m} + 2\dfrac{\Delta v}{v}\).
2. \(= 2\% + 2(3\%) = 8\%\).

Teacher's Note:
1. The square of speed doubles its percentage error.
2. Add, never subtract, errors for the maximum value.

Answer: (c) \(M^{-1}L^{0}T^{0}\)

Explanation:
1. The argument of a sine is dimensionless, so \(aEt\) and \(bVx\) are both dimensionless.
2. \([a] = \dfrac{1}{[E][t]} = \dfrac{1}{ML^2T^{-2}\cdot T} = M^{-1}L^{-2}T\).
3. \([b] = \dfrac{1}{[V][x]} = \dfrac{1}{LT^{-1}\cdot L} = L^{-2}T\).
4. \(\dfrac{[a]}{[b]} = M^{-1}L^{0}T^{0}\).

Teacher's Note:
1. Terms inside a trigonometric function must be dimensionless.
2. Use this to get the dimension of each constant separately and then divide.

Answer: (b) \(c^{1/2}h^{1/2}G^{-1/2}\)

Explanation:
1. Let \(M = c^{a}h^{b}G^{g}\) with \([c] = LT^{-1}\), \([h] = ML^2T^{-1}\), \([G] = M^{-1}L^3T^{-2}\).
2. Mass: \(b - g = 1\). Length: \(a + 2b + 3g = 0\). Time: \(-a - b - 2g = 0\).
3. Adding the last two gives \(b + g = 0\), so \(g = -b\) and \(2b = 1\).
4. Thus \(b = \tfrac12\), \(g = -\tfrac12\), \(a = \tfrac12\).
5. \(M = c^{1/2}h^{1/2}G^{-1/2}\).

Teacher's Note:
1. Write each fundamental constant in M, L, T and solve the linear system for the exponents.
2. Check: \(\sqrt{ch/G}\) is the Planck mass.

Answer: (b) \(\dfrac{a}{n+1}\)

Explanation:
1. \((n+1)\) vernier divisions equal \(n\) main scale divisions, so one vernier division = \(\dfrac{na}{n+1}\).
2. Least count = 1 MSD - 1 VSD = \(a - \dfrac{na}{n+1} = \dfrac{a}{n+1}\).

Teacher's Note:
1. Least count is the difference between one main scale division and one vernier division.
2. Equivalent form: smallest MSD divided by the number of vernier divisions.

Answer: (c) \(5\pi R\), \(2R\)

Explanation:
1. In 100 s he completes \(100/40 = 2.5\) rounds.
2. Distance = \(2.5 \times 2\pi R = 5\pi R\).
3. After two and a half rounds he is at the point diametrically opposite the start, so the displacement is \(2R\).

Teacher's Note:
1. Distance counts the whole path, displacement is only the straight line from start to end.
2. Half a round always gives displacement equal to the diameter.

Answer: (d) Acceleration will not be zero at any time

Explanation:
1. \(v = \dfrac{ds}{dt} = 6t^2 + 6t + 2\).
2. \(a = \dfrac{dv}{dt} = 12t + 6\).
3. \(a = 0\) gives \(t = -\tfrac12\) s, which is negative and therefore not a physical time.
4. So acceleration is not zero for any \(t \ge 0\).

Teacher's Note:
1. Differentiate twice and check whether the root lies in the allowed range of time.
2. A negative time means the event would have occurred before the motion started.

Answer: (a) \(\dfrac{2}{\sqrt{g}}\)

Explanation:
1. Compare with \(y = x\tan\theta - \dfrac{g x^2}{2u^2\cos^2\theta}\).
2. \(\tan\theta = 1\), so \(\theta = 45^\circ\).
3. \(\dfrac{g}{2u^2\cos^2\theta} = \dfrac12\) gives \(u^2 = \dfrac{g}{\cos^2 45^\circ} = 2g\).
4. \(T = \dfrac{2u\sin\theta}{g} = \dfrac{2\sqrt{2g}\cdot\frac{1}{\sqrt2}}{g} = \dfrac{2}{\sqrt{g}}\).

Teacher's Note:
1. Match the coefficients of x and \(x^2\) with the standard trajectory equation.
2. Once \(u\) and \(\theta\) are known, use \(T = 2u\sin\theta/g\).

Answer: (c) \(v = \dfrac{1}{2}\sqrt{\dfrac{Gm}{R}}\)

Explanation:
1. The two particles are at opposite ends of a diameter, so their separation is \(2R\).
2. The gravitational force provides the centripetal force: \(\dfrac{Gm^2}{(2R)^2} = \dfrac{mv^2}{R}\).
3. \(v^2 = \dfrac{Gm}{4R}\), so \(v = \dfrac{1}{2}\sqrt{\dfrac{Gm}{R}}\).

Teacher's Note:
1. Each particle moves in a circle of radius R, but the force depends on the full separation 2R.
2. Mixing up these two distances is the most common mistake.

Answer: (d) Graph (4)

Explanation:
1. Inside a thin uniform shell the gravitational field is zero, so \(E = 0\) for \(r < R\).
2. At \(r = R\) the field jumps to its maximum value \(GM/R^2\).
3. Outside, \(E = GM/r^2\), which falls off as \(1/r^2\).
4. Graph (4) shows exactly this: zero up to R, a jump at R and then a decaying curve.

Teacher's Note:
1. A shell has no field inside, unlike a solid sphere whose field rises linearly inside.
2. The field is discontinuous at the surface of a thin shell.

Answer: (d) \(\dfrac{1}{2}\sqrt{\dfrac{GM}{R}\left(1 + 2\sqrt{2}\right)}\)

Explanation:
1. The four particles sit at the corners of a square of diagonal \(2R\), so the side is \(\sqrt{2}R\).
2. Force from each of the two adjacent particles: \(\dfrac{GM^2}{2R^2}\). Their components towards the centre add to \(2 \times \dfrac{GM^2}{2R^2}\cos 45^\circ = \dfrac{GM^2}{\sqrt{2}R^2}\).
3. Force from the opposite particle (distance \(2R\)): \(\dfrac{GM^2}{4R^2}\).
4. Net inward force \(= \dfrac{GM^2}{4R^2}(2\sqrt{2} + 1)\).
5. Setting this equal to \(\dfrac{Mv^2}{R}\) gives \(v = \dfrac{1}{2}\sqrt{\dfrac{GM}{R}(1 + 2\sqrt{2})}\).

Teacher's Note:
1. Resolve the forces from neighbouring masses along the line to the centre, then add the force from the opposite mass.
2. The result is a standard one worth remembering.

Answer: (c) \(dR\)

Explanation:
1. \(g = \dfrac{GM}{R^2}\) with \(M = \dfrac{4}{3}\pi R^3 d\).
2. So \(g = \dfrac{4}{3}\pi G d R\), which is proportional to \(dR\).

Teacher's Note:
1. Expressing mass through density turns \(R^2\) in the denominator into a net power of R.
2. For planets of the same density, g is proportional to radius.

Answer: (c) 17

Explanation:
1. At the equator \(g_{\text{eff}} = g - \omega'^2 R\).
2. For this to be zero, \(\omega'^2 R = g\), so \(\omega' = \sqrt{g/R} = \sqrt{9.8/6.4\times10^{6}} = 1.24\times10^{-3}\) rad/s.
3. \(x = \dfrac{\omega'}{\omega} = \dfrac{1.24\times10^{-3}}{7.27\times10^{-5}} \approx 17\).

Teacher's Note:
1. Effective gravity at the equator falls by \(\omega^2 R\) due to rotation.
2. The earth would need to spin about 17 times faster (day of about 84 minutes) for objects to float.

Answer: (c) \(\dfrac{5}{6}mgh\)

Explanation:
1. \(\Delta U = \dfrac{GMm}{R} - \dfrac{GMm}{R+h} = \dfrac{GMm\,h}{R(R+h)} = mg\,\dfrac{R h}{R+h}\).
2. With \(h = R/5\): \(\dfrac{Rh}{R+h} = \dfrac{R \cdot R/5}{6R/5} = \dfrac{R}{6}\).
3. \(\Delta U = \dfrac{mgR}{6}\). Since \(R = 5h\), \(\Delta U = \dfrac{5}{6}mgh\).

Teacher's Note:
1. The formula mgh is accurate only for heights that are tiny compared to R.
2. For large heights use the difference of \(-GMm/r\) at the two positions.

Answer: (d) \(\dfrac{GMm}{6R}\)

Explanation:
1. Required energy = change in potential energy \(= -\dfrac{GMm}{3R} - \left(-\dfrac{GMm}{2R}\right)\).
2. \(= GMm\left(\dfrac{1}{2R} - \dfrac{1}{3R}\right) = \dfrac{GMm}{6R}\).

Teacher's Note:
1. Moving outward needs positive energy because potential energy rises towards zero.
2. Always subtract the initial potential energy from the final one.

Answer: (a) \(\sqrt{2}\,v\)

Explanation:
1. \(v_e = \sqrt{\dfrac{2GM}{R}} = R\sqrt{\dfrac{8\pi G\rho}{3}}\), so \(v_e \propto R\sqrt{\rho}\).
2. New value \(= v \times 2 \times \sqrt{\tfrac{1}{2}} = \sqrt{2}\,v\).

Teacher's Note:
1. Write escape velocity in terms of radius and density when both change.
2. Double radius gives a factor 2, half density gives \(1/\sqrt{2}\).

Answer: (d) 1 : 1

Explanation:
1. The two forces form an action-reaction pair under Newton's third law.
2. They are equal in magnitude and opposite in direction, so \(F_1 : F_2 = 1 : 1\).

Teacher's Note:
1. The gravitational force between two bodies is the same on each, only the accelerations differ.
2. The lighter body accelerates more.