NEET Practice Questions

Step: Find the change in the voltage sensitivity.
The voltage sensitivity \((S_V)\) of a moving coil galvanometer is given by;
\(S_V=\frac{NAB}{KR}\)
where, \(N = \) Number of turns, \(B=\) Magnetic field, \(A=\) Area of the coil, \(K=\) Torsional constant, and \(R=\) Resistance of the coil.
The current sensitivity of the coil is given by;
\(S_I = \frac{\theta}{I}= \frac{NBA}{K}\)
The voltage sensitivity is related to current sensitivity by the expression as;
\(S_V = \frac{S_I}{R}\)
If the number of turns \(N\) increases by \(25\%\), it would increase the current sensitivity \((S_I)\) proportionally. However, the resistance \(R\) of the coil also increases proportionally to \(N\) (since more turns mean longer wire).
Since \(S_V = \frac{S_I}{R}\) and both \(S_I\)​ and \(R\) increase proportionally, the voltage sensitivity remains unchanged.
Therefore, the change in the voltage sensitivity is zero.
Hence, option (1) is the correct answer.
 

Step: Find the time period of the combination of the springs.
As both springs are connected in parallel then the effective spring constant is given by; 
\({k}_{eq}{=}{k}_{1}{+}{k}_{2}\)
The time period of the spring combination is given by;  
\({T}{=}{2}\mathit{\pi}\sqrt{\frac{m}{k_{eq}}}\)
\(\Rightarrow {T}{=}{2}\mathit{\pi}\sqrt{\frac{m}{{k}_{1}{+}{k}_{2}}}\)
Hence, option (3) is the correct answer.
 

Explanation:

(i) Nitriles can be hydrolyzed to carboxylic acids in acidic aqueous solutions. In the first stage, it forms the corresponding amide group, and in the second stage, it forms the corresponding carboxylic acid from the amide.

(ii) Carboxylic acids having α-hydrogen are halogenated at α-position on treatment with Br₂ in the presence of red phosphorous to give \(\alpha\)-halocarboxylic acids. This is the Hell-Volhard-Zelinsky reaction.

So, the final product will be a \(\alpha\)-bromocarboxylic acid. The complete reaction is as follows: 

So, option 2 is the correct choice. 

Explanation:

In case of alkyl halide, the leaving group order is: Iodide > Bromide > Chloride

Iodide is the best leaving group than other halogen atoms due to its larger size. Due to larger size, charge density decreases and it becomes stable. 

So, the order of S_N2 reaction is:  RI > RBr > RCl.  Because as the atoms get bigger from Cl to I , bonds get longer, so weaker and react faster. Length and strength dominate, not polarity of the carbon-halogen bond, which would give the opposite order.
 

In metal carbonyl, synergic bonding takes place. The metal-carbon bond in metal carbonyls possess both σ and π character. The M–C σ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal.
The M–C π bond is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding π* orbital of carbon monoxide.   

The diagram is as follows:

Thus, the assertion is correct but the reason is false.