NEET Practice Questions

Hint: For an ideal transformer; \( I_pV_p=I_sV_s
\)

Step: Find the current in the primary circuit. 
The power in the secondary circuit is given by;
\(P=VI\)
\(P_s=I_sV_s=44~\text W\)
The current in the primary circuit is given by;
As there is no power loss we know that;
\( I_pV_p=I_sV_s
\)
\(\Rightarrow I_p=\frac{I_sV_s}{V_p}\)
\(\Rightarrow I_p=\frac{44}{220}=0.2~\text A\)
Hence, option (3) is the correct answer.

Hint: Isostructural species have the same hybridization and structure

Isostructural species have the same hybridization and same shape. 

1. CH₃^– and CH₃⁺ 

Carbon hybridization in CH₃^–  is sp³ and carbon hybridization in CH₃⁺ is sp² hybridized. Hence, they are not isostructural. 

2. CO₂ and NH₃

Carbon hybridization in CO₂ is sp and nitrogen hybridization in NH₃ is sp³ . Hence, they are not isostructural.

3. SnCI₄ and CIO₄^– 

Sn hybridization in SnCl₄ is sp³ and molecular geometry is tetrahedral. Cl hybridization in ClO₄^– is sp³ and molecular geometry is tetrahedral. Hence, both are isostructural.

4. AICI₃   and   SO₂ 

AlCl₃ hybridization of Al is sp² and geometry in trigonal planar but in SO₂ sulphur hybridization is sp² but shape is V shape. 

Thus, option third is the correct answer. 

Hint: Potential energy is maximum at the highest point.

Step 1: Find the time taken to reach the maximum height.
\(v=u+at\\ 0=10-10t\\ t=1~\text{s}\)
Therefore, the potential energy will be maximum at \(t=1~\text{s}\)$$

Step 2: Find height where kinetic energy is equal to potential energy
\(v^2=u^2-2gh \dots(1)\\ K.E=P.E\\ \frac{1}{2}mv^2=mgh\\ \text{From equation (1)}\\ u^2-2gh=2gh\\ \Rightarrow u^2=4gh\\ \Rightarrow h=\frac{u^2}{4g}\\ \Rightarrow h= 2.5~\text{m} ~~~~\left[u= 10~\text{m/s}, g = 10 ~\text{m/s}^2\right]\)
Therefore, kinetic energy is equal to the potential energy at a height of \(2.5~\text{m}\) from the ground.
Hence, option (3) is the correct answer.

Correct option is (3)
 

Hint: Product B is an acid.

In the first step, a methyl bromide reacts with KCN and a nucleophilic substitution reaction takes place. In this reaction, Br

is substituted by the cyanide group.

In the second step, ethane nitrile undergoes hydrolysis reaction and CN group is converted into COOH group.

In the final step, a reduction reaction takes place. Ethanoic acid is reduced to ethanol. The sequence of reactions are as follows:

$\underset{\mathrm{methyl}}{{\mathrm{CH}}_3\mathrm{Br}}$ $\mathrm{bromide}\underset{–\mathrm{KBr}}{\overset{\mathrm{KCN}}{\to }}\underset{\mathrm{ethane}}{{\mathrm{CH}}_3\mathrm{CN}}$ $\mathrm{nitrile}$ $\left(\mathrm{A}\right)\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }$ $\underset{\mathrm{ethanoic}}{{\mathrm{CH}}_3\mathrm{COOH}}$ $\mathrm{acid}$ $\left(\mathrm{B}\right)\underset{\mathrm{ether}}{\overset{{\mathrm{LiAlH}}_4}{\to }}$ $\underset{\mathrm{ethanol}}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}}$ $\left(\mathrm{C}\right)$

As ethanol is also called ethyl alcohol so option 4 is correct.

Given, 
Mass of the planet = ${\mathrm{M}}_{\mathrm{p}}$
The diameter of the planet = ${\mathrm{D}}_{\mathrm{p}}$
Mass of the freely falling particle = m 
Force, 
$\mathrm{F}=\frac{{\mathrm{GM}}_{\mathrm{p}}\mathrm{m}}{{{\mathrm{R}}_{\mathrm{p}}}^2}$ 
Here, $\mathrm{F}=\frac{{\mathrm{GM}}_{\mathrm{p}}\mathrm{m}}{{({\mathrm{D}}_{\mathrm{p}}/2)}^2}$ 
$\text{F=}\frac{4{\mathrm{GM}}_{\mathrm{p}}\mathrm{m}}{{\mathrm{D}}_{\mathrm{p}}^2}$
Acceleration due to gravity, 
$\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{4{\mathrm{GM}}_{\mathrm{p}}}{{\mathrm{D}}_{\mathrm{p}}^2}$

• PAGE 311, XI NCERT

Hint: The reaction is an example of Coupling reaction.

Explanation:  

When diazonium salt is reacted with aniline then an electrophilic aromatic substitution reaction takes place. In this reaction, azo dye is obtained as a product. The reaction is as follows: