JEE Physics Gravitation MCQs Set 05

Answer: (b) at the equator

Explanation:
1. A 1 kg weight means a fixed pull mg
2. g is smallest at the equator
3. So the mass of sugar is largest there

Teacher's Note:
1. The weight is fixed, so mass is inversely proportional to g.
2. g is minimum at the equator.

Answer: (a) increases

Explanation:
1. The earth is flattened at the poles
2. Centrifugal reduction also vanishes at the poles
3. So g increases

Teacher's Note:
1. Both radius and rotation effects make g larger at the poles.

Answer: (a) increase by 2%

Explanation:
1. \( g \propto R^{-2} \)
2. \( \frac{\Delta g}{g} = -2\frac{\Delta R}{R} = +2\% \)

Teacher's Note:
1. Use the percentage-error rule for powers.

Answer: (a) decrease by 0.5%

Explanation:
1. At height: \( \frac{\Delta g}{g} = \frac{2h}{R} = 1\% \), so \( \frac hR = 0.5\% \)
2. At depth: \( \frac{\Delta g}{g} = \frac hR = 0.5\% \) decrease

Teacher's Note:
1. Depth reduces g half as fast as height for small values.

Answer: (b) 0.99 g

Explanation:
1. \( \frac{\Delta g}{g} = \frac{2h}{R} = \frac{2\times32}{6400} = 0.01 \)
2. \( g' = 0.99g \)

Teacher's Note:
1. Use the small-height formula as h much less than R.

Answer: (a) 1 : 1

Explanation:
1. These are an action-reaction pair
2. They are equal in magnitude

Teacher's Note:
1. Newton's third law applies to gravitation.

Answer: (a) g' = 3g

Explanation:
1. \( g = \frac43\pi G\rho R \)
2. \( g \propto R \) at fixed density
3. \( g' = 3g \)

Teacher's Note:
1. At equal density, g is proportional to R.

Answer: (b) d = 2h

Explanation:
1. Height: \( \frac{2h}{R} \)
2. Depth: \( \frac{d}{R} \)
3. Equating gives \( d = 2h \)

Teacher's Note:
1. Compare the two fractional changes.

Answer: (b) decreases

Explanation:
1. \( g_{eff} = g - \omega^2R \)
2. A larger \( \omega \) lowers weight

Teacher's Note:
1. Only the equator feels the maximum centrifugal effect.

Answer: (c) \( \frac{4W}{9} \)

Explanation:
1. \( W' = W\left(\frac{R}{R+R/2}\right)^2 = W\left(\frac23\right)^2 = \frac{4W}{9} \)

Teacher's Note:
1. The distance from the centre becomes 1.5 R.

Answer: (a) \( 1.25\times10^{-3} \) rad/s

Explanation:
1. Set \( g - \omega^2R = 0 \)
2. \( \omega = \sqrt{g/R} = \sqrt{\frac{10}{6.4\times10^6}} \)
3. \( \omega = 1.25\times10^{-3} \) rad/s

Teacher's Note:
1. The body then needs the whole of g as centripetal acceleration.

Answer: (d) acceleration

Explanation:
1. Free-fall acceleration g does not depend on mass
2. KE and momentum depend on mass

Teacher's Note:
1. Neglect air resistance.
2. Speed is the same, but energy and momentum differ.

Answer: (a) 2 J

Explanation:
1. Field \( = \frac{2}{20} = 0.1 \) J/kg per metre
2. Potential rise over 4 m \( = 0.4 \) J/kg
3. \( W = 5\times0.4 = 2 \) J

Teacher's Note:
1. Uniform field means potential changes linearly with height.

Answer: (c) \( -GmM_m\left(\frac{81}{R}+\frac1r\right) \)

Explanation:
1. \( U = -\frac{GM_em}{R} - \frac{GM_mm}{r} \)
2. Substitute \( M_e = 81M_m \)
3. \( U = -GmM_m\left(\frac{81}{R}+\frac1r\right) \)

Teacher's Note:
1. Potential energies of the two sources add as scalars.

Answer: (a) infinity

Explanation:
1. \( U = -\frac{GMm}{r} \) is negative and rises to zero
2. The maximum value, zero, occurs at infinity

Teacher's Note:
1. Potential energy is zero at infinity by convention.

Answer: (b) negative

Explanation:
1. Total energy is negative for a bound body
2. Escape needs total energy of at least zero

Teacher's Note:
1. Below escape velocity, the missile stays bound.

Answer: (a) \( \sqrt{\frac{GM}{R}} \)

Explanation:
1. \( \frac12v^2 = GM\left(\frac1R-\frac1{2R}\right) = \frac{GM}{2R} \)
2. \( v = \sqrt{\frac{GM}{R}} \)

Teacher's Note:
1. Use energy conservation with final speed zero at r = 2R.

Answer: (b) \( \frac{U}{r} \)

Explanation:
1. \( U = -\frac{GMm}{r} \)
2. \( F = \frac{GMm}{r^2} = \frac{
3. U
4. }{r} \)

Teacher's Note:
1. Force is the negative gradient of potential energy.

Answer: (d) \( \sqrt{2Rg} \)

Explanation:
1. \( \frac12v^2 = \frac{GM}{R} = gR \)
2. \( v = \sqrt{2gR} \)

Teacher's Note:
1. This equals the escape velocity.

Answer: (a) \( \frac{GMmh}{R(R+h)} \)

Explanation:
1. \( \Delta U = GMm\left(\frac1R-\frac1{R+h}\right) = \frac{GMmh}{R(R+h)} \)

Teacher's Note:
1. Subtract final from initial potential energy.

Answer: (d) \( -\frac{4GM}{r} \)

Explanation:
1. Each sphere is at \( r/2 \)
2. \( V = 2\times\left(-\frac{GM}{r/2}\right) = -\frac{4GM}{r} \)

Teacher's Note:
1. Scalar addition of potentials.

Answer: (d) \( 2E_0 \)

Explanation:
1. \( E_0 = -\frac{GMm}{2r},\ U = -\frac{GMm}{r} \)
2. \( U = 2E_0 \)

Teacher's Note:
1. Potential energy is twice the (negative) total energy.