Physics MCQs for NEET — Practice Questions with Answers

The NCERT text states: 'From Eq. (2.15), potential on the dipole axis ($\theta = 0, \pi$) is given by $V = \pm \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$ (Eq. 2.16).' Here, for $\theta=0$, $\cos\theta=1$, so $V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$. For $\theta=\pi$, $\cos\theta=-1$, so $V = -\frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$.

The NCERT text clearly states: 'Work done is independent of the path' (page 48) and 'The work done corresponding to the later will be zero' (page 49). This implies that work done in an electrostatic field depends only on the initial and final positions, as shown in Example 2.1 where 'work done will be path independent'.

The NCERT text states: 'By the superposition principle, the potential V at P due to the system of charges is the algebraic sum of the potentials due to individual charges.' (Page 51).

According to the NCERT text (Eq. 2.8), the potential at P due to the charge Q is $V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$.

The NCERT text explains: '(It is, however, axially symmetric about $\vec{p}$. That is, if you rotate the position vector $\vec{r}$ about $\vec{p}$, keeping $\theta$ fixed, the points corresponding to P on the cone so generated will have the same potential as at P.)' (Page 51).

The NCERT text, specifically in problem 7.17 and 7.19, explicitly states G = $6.67 \times 10^{-11} \ N \ m^2 \ kg^{-2}$. This indicates the units are Newtons times meters squared per kilogram squared.

Though not explicitly stated in the provided excerpts, the constant G is intrinsic to Newton's Law of Universal Gravitation, which underpins all the calculations and concepts discussed in the 'Gravitation' chapter. The chapter refers to the gravitational force formula $F = G M_E m / R_E^2$ repeatedly, indicating its origin.

The text states: 'The measurement of G by Cavendish’s experiment (or otherwise), combined with knowledge of g and R_E enables one to estimate $M_E$ from Eq. (7.12). This is the reason why there is a popular statement regarding Cavendish : “Cavendish weighed the earth”.' (Page 133, Section 7.5)

Problems 7.17 and 7.19 explicitly provide the value of G as $6.67 \times 10^{-11} \ N \ m^2 \ kg^{-2}$.

The gravitational force is directly proportional to the gravitational constant G, as indicated by the formula $F = G M_1 M_2 / r^2$. Therefore, an increase in G would lead to an increase in the gravitational force between any two masses.