NEET Practice Questions (MCQs) with Answers & Solutions

The context mentions 'Downstream processing technologies to purify the protein/organic compound.' and also states 'The processes include separation and purification, which are' (followed by an incomplete sentence, but the intent is clear from the applications chapter).

The summary lists: 'recombinant DNA technology... involves the use of restriction endonucleases, DNA ligase, appropriate plasmid or viral vectors to isolate and ferry the foreign DNA into host organisms, expression of the foreign gene, purification of the gene product...'. CRISPR-Cas9 is not mentioned in this context.

According to the NCERT text, 'The electric dipole potential falls off, at large distance, as $1/r^2$, not as $1/r$, characteristic of the potential due to a single charge.' Therefore, a point charge's potential decreases as $1/r$ and a dipole's potential decreases as $1/r^2$ at large distances.

The NCERT text states: '(i) The potential due to a dipole depends not just on $r$ but also on the angle between the position vector $\vec{r}$ and the dipole moment vector $\vec{p}$.' Equation (2.14) $V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$ clearly shows this dependence.

As per the NCERT text, 'The potential in the equatorial plane ($\theta = \pi/2$) is zero.' This is because for $\theta = \pi/2$, $\cos\theta = 0$, making the potential $V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2} = 0$.

The NCERT text states that the electric potential of a dipole is given by $V = \frac{1}{4\pi\epsilon_0} \frac{p \cdot \hat{r}}{r^2}$ or $V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$ for $r \gg a$. This formula (Equation 2.14 and 2.15) holds for large distances.

The NCERT text defines electrostatic potential as 'the work done in bringing a unit positive charge (without acceleration) from infinity to that point.' (Page 48).

The NCERT text states: 'For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative.' (Page 49, Example 2.1 note).

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'.