NEET Practice Questions (MCQs) with Answers & Solutions

The NCERT states, 'For the shells of radius greater than r, the point P lies inside. Hence according to result stated in the last section, they exert no gravitational force on mass m kept at P. The shells with radius $\le r$ make up a sphere of radius r for which the point P lies on the sur face. This smaller sphere therefore exerts a force on a mass m at P as if its mass Mr is concentrated at the centre.'

For a point inside a homogeneous solid sphere, the force is due to the mass $M_r$ of the sphere of radius r. If the sphere has uniform density $\rho = \frac{M}{(4/3)\pi R^3}$, then $M_r = \frac{4}{3}\pi r^3 \rho = \frac{4}{3}\pi r^3 \frac{M}{(4/3)\pi R^3} = M \frac{r^3}{R^3}$. The force is then $F = G \frac{M_r m}{r^2} = G \frac{(M \frac{r^3}{R^3}) m}{r^2} = G \frac{M m r}{R^3}$. This matches the derivation in the NCERT for a point inside the Earth.

According to the NCERT text, 'Though voltage and current in ac circuit are represented by phasors – rotating vectors, they are not vectors themselves. They are scalar quantities.' The representation by phasors is a mathematical convenience to add these quantities using vector addition rules for their amplitudes and phases.

The NCERT text states, 'From Fig. 7.4(a) we see that phasors V and I for the case of a resistor are in the same direction. This is so for all times. This means that the phase angle between the voltage and the current is zero.'

While not explicitly stated in the provided snippets for a purely inductive circuit, the text mentions for an RLC circuit that 'VL is $\pi/2$ ahead of I'. This implies that in a purely inductive circuit, the voltage across the inductor leads the current by $\pi/2$ or 90 degrees.

The NCERT text states, 'The current through the capacitor is $\pi/2$ ahead of the applied voltage.'

According to the NCERT text, 'From previous section, we know that VR is parallel to I, VC is $\pi/2$ behind I and VL is $\pi/2$ ahead of I.'

The NCERT text states, 'The phasor relation whose vertical component gives the above equation is $\text{VL} + \text{VR} + \text{VC} = \text{V}$'.

The NCERT text explains, 'If XC > XL, $\phi$ is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage.'

From the impedance diagram (Fig. 7.12) and the formula for impedance $Z = \sqrt{R^2 + (X_C - X_L)^2}$, the vertical side of the right-triangle is $(X_C - X_L)$ or $(X_L - X_C)$ depending on which reactance is larger. The magnitude is $|X_C - X_L|$.