NEET Practice Questions

Hint: Torque is given by, \(\vec{\tau }=\frac{d\vec{L}}{dt}\).

Step 1: Check the assertion.
It is correct that the angular momentum of an isolated system of particles is conserved.

Step 2: Check the reason.
On an isolated system, there is no force acting. So the net torque on an isolated system of particles is zero.
The torque is given by the rate of change of angular momentum. So the rate of change of angular momentum is also equal to zero in an isolated system.

Hint: The reaction of nitrogen and oxygen depends upon the temperature.
Both ${\mathrm{N}}_2<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{O}}_2$ are less reactive at ordinary conditions. The reaction between nitrogen and oxygen requires a high temperature. N₂ bond order is 3 and it is non reactive at room temperature. Thus, it will not reacts with O₂ 

The bond dissociation energy of N₂ and O₂  is as follows:

N₂ = 945 kJ/mol
O₂ = 498 kJ/mol

Hence, both assertion and reason are correct, and the reason is the correct explanation for the assertion.

Hint: \(F=mw^2R\)

Step: Find the time period of the revolution.
Given: The  central fictitious force \(F\) that is inversely proportional to \(R^3\) i.e., 
\(F\propto \frac{1}{R^3}\Rightarrow F= \frac{K}{R^3}\) 
\(\Rightarrow \frac{K}{R^3}= m\omega^2 R\Rightarrow \omega^2 = \frac{K}{mR^4}\)
\(\Rightarrow \left(\frac{2\pi}{T}\right)^2= \frac{K}{mR^4}~~~\left[\omega = \frac{2\pi}{T}\right]\)
\(\Rightarrow T^2 \propto R^4\)
\(\Rightarrow T\propto R^2\)
Therefore, the time period of the revolution is proportional to \(R^2.\)
Hence, option (1) is the correct answer.

Hint: The formula of equilibrium constant is, \(K_{equ} = \frac{Z^{2}}{X\times Y}\)

Step 1:

The concentration of X, Y and Z is as follows:

Concentration = \(=\frac{number \ of \ mole}{volume \ of \ solution}\)

Hence, volume of solution is 1 L, concentration is equal to number of moles. So concentration of X, Y and Z is equal to its mole.

               X   +  Y  =  2Z
t = 0        1      1.5     0.5
At eq.     1-x   1.5-x      0.5 + 2x

The equilibrium concentration of Z is 1. Calculate the value of x as follows:

[Z] = 0.5 + 2x

\(\frac{1-0.5}{2}\)

x = 0.25 M

The concentration of X and Y at equilibrium is as follows:

[X] = 1 – x = 0.75, [Y] = 1.5 – x = 1.25

Step 2:

\(K_{equ} = \frac{Z^{2}}{X\times Y}\)

$K_{eq.}<mspace\; width="1em"></mspace>=\frac{1}{{\displaystyle \frac{3}{4}}\times {\displaystyle \frac{5}{4}}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{16}{15}$

Page 103, XI NCERT
 

Tight junctions form a barrier between adjacent cells, preventing leakage of substances between them. Gap junctions, on the other hand, allow the rapid transfer of ions and small molecules between cells, facilitating communication and coordination within a tissue. Adhering junctions help to anchor cells to each other and to the extracellular matrix, but do not have a major role in preventing leakage or facilitating communication.

Hint: $\frac{{\mathrm{\Lambda }}_{{\mathrm{CH}}_3\mathrm{COOH}}}{\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}}=\mathrm{\alpha }$ 

Calculate the value  $\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}$ as follows:

$\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}=\mathrm{\Lambda }{°}_{{\mathrm{H}}^{+}}+\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3{\mathrm{COO}}^{–}}$
$\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}$ $=350$ $+$ $50$
$\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}$ $=$ $400$ $\mathrm{S}$ ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{–1}$

Calculate the value of $\alpha$ as follows:

$\frac{{\mathrm{\Lambda }}_{{\mathrm{CH}}_3\mathrm{COOH}}}{\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}}=\mathrm{\alpha }$

$\frac{20}{400}=\mathrm{\alpha }$
$\mathrm{\alpha }=0.05$

Step 2:

Calculate the value of the dissociation constant as follows:

${\mathrm{CH}}_3\mathrm{COOH}\leftrightarrow {\mathrm{CH}}_3{\mathrm{COO}}^{–}$ $+$ ${\mathrm{H}}^{+}$
$\mathrm{C}$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $–$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $–$
$\mathrm{C}–\mathrm{C\alpha }$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $\mathrm{C\alpha }$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $\mathrm{C\alpha }$

${\mathrm{K}}_{\mathrm{a}}=\frac{\mathrm{C\alpha }\times \mathrm{C\alpha }}{\mathrm{C}(1–\mathrm{\alpha })}$
${\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{C\alpha }}^2}{(1–\mathrm{\alpha })}$
${\mathrm{K}}_{\mathrm{a}}=\frac{0.007\times 0.05\times 0.05}{1–0.05}$
K_a   =  1.75×10^-5

HINT: Oxidation occurs at anode.

Explanation: 

For the given cell : ${\mathrm{Zn}}_{\left(\mathrm{s}\right)<mspace\; width="1em"></mspace>}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2{{\mathrm{Ag}}^{+}}_{\left(\mathrm{aq}\right)}<mspace\; width="1em"></mspace>\to {{\mathrm{Zn}}^{+2}}_{\left(\mathrm{aq}\right)}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{Ag}}_{\left(\mathrm{s}\right)}<mspace\; width="1em"></mspace>$

Zn electrode is negatively charged because at this electrode, Zn oxidizes to ${\mathrm{Zn}}^{+2}$ and the leaving electrons accumulate on this electrode. Ions are the carriers of current in the cell.

The reaction taking place at the zinc electrode is : ${\mathrm{Zn}}_{\left(\mathrm{s}\right)\hspace{0.17em}}\to <mspace\; width="1em"></mspace>{{\mathrm{Zn}}^{+2}}_{\left(\mathrm{aq}\right)}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2{\mathrm{e}}^{–}$