Physics MCQs for NEET — Practice Questions with Answers

As per the NCERT text, '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 NCERT text explains, 'The force on m due to the outer shell of thickness d is zero because the result quoted in the previous section. As far as the smaller sphere of radius $(R_E - d)$ is concerned, the point mass is outside it and hence according to the result quoted earlier, the for ce due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre.'

The force on mass m at P (distance r from center) inside a homogeneous solid sphere is given by $F = G \frac{M_r m}{r^2}$. Since density is uniform, $M_r = \frac{4}{3}\pi r^3 \rho$. Also, $M_E = \frac{4}{3}\pi R_E^3 \rho$. So, $M_r = M_E \frac{r^3}{R_E^3}$. Substituting this into the force equation gives $F = G \frac{M_E m r}{R_E^3}$. Therefore, F is proportional to r.

The NCERT states, 'We have to add up these forces vectorially for all the point masses in the extended object to get the total force.' Also, 'From the principle of superposition each force acts independently and uninfluenced by the other bodies. The resultant force $F_R$ is then found by vector addition'.

The NCERT clearly states, 'Gravitational shielding is not possible.' Options 1, 2, and 3 are correct descriptions based on the provided text.

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.