Physics MCQs for NEET — Practice Questions with Answers

The NCERT text states, 'Since VC and VL are always along the same line and in opposite directions, they can be combined into a single phasor (VC + VL) which has a magnitude $|v_{Cm} – v_{Lm}|$.' This corresponds to $|V_C - V_L|$ for the RMS or peak voltage magnitudes.

The NCERT text highlights this benefit: 'The analysis of an ac circuit is facilitated by the use of a phasor diagram.' And further explains, 'The rotating vectors that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know as the law of vector addition.'

The provided text in problem 7.17 and 7.19 explicitly states 'G = $6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$'. The other options represent acceleration due to gravity, mass of Earth, and radius of Earth respectively.

The text on page 133, under section 7.5, 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”.'

Points to Ponder, point 6, and page 135 state: 'Relative to infinity (i.e. if we presume that the potential energy of the object at infinity is zero), the gravitational potential energy of an object is negative.' and 'V = - $G m_1 m_2 / r$ (if we choose V = 0 as $r \rightarrow \infty$ )'. Since G, $m_1$, $m_2$, and r are positive, the expression - $G m_1 m_2 / r$ will always be negative.

Points to Ponder, point 8, states: 'Although the gravitational force between two particles is central, the force between two finite rigid bodies is not necessarily along the line joining their centre of mass. For a spherically symmetric body however the force on a particle external to the body is as if the mass is concentrated at the centre and this force is therefore central.'

The text on page 135 and Points to Ponder, point 5, explicitly states: 'The gravitational potential energy associated with two particles of masses $m_1$ and $m_2$ separated by distance by a distance r is given by $V = - G m_1 m_2 / r$ (if we choose V = 0 as $r \rightarrow \infty$ )'.

Equation (7.12) on page 133 is given as $g = G M_E / R_E^2$. In this equation, G is the universal gravitational constant, M_E is the mass of the Earth, and R_E is the radius of the Earth. F represents force, not a constant in this context.

Points to Ponder, point 9, and Exercise 7.1 (a) explicitly state: 'Gravitational shielding is not possible.'

Problems 7.17 and 7.19 consistently mention 'G = $6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$'. This translates to $N \text{ m}^2 / \text{ kg}^2$.