NEET Practice Questions

Hint: The net electric field is given by, \(E=\sqrt{{E_1}^2+{E_2}^2+2E_1E_2cos\theta }\).

Step 1: Find the net electric field.
\(\vec{E}=E_{0}[\hat{\imath} \cos (\omega t-k z)+\hat{\jmath} \cos (\omega t-k x)]\)
\(\vec{E_1}=E_{0}\hat{\imath} \cos (\omega t-k z)\)
\(\vec{E_2}=E_{0}\hat{\jmath} \cos (\omega t-k x)\)
The angle between E₁ and E₂ = 90^o
\(E=\sqrt{{E}^2+{E}^2}=\sqrt2 E\)

Hint: The capacitor and the upper resistor \(R\) are in a series combination.

Step: Find the phase difference between the current and the voltage across the capacitor.
The circuit consists of a resistor \((R)\) in series with a capacitor \((C)\) and another resistor \((R)\) in parallel below it.
This implies the current in the vertical branch (the \(R\) at \(X\)) is zero. Therefore, the voltage at the junction \(C\) must be zero (the same as the ground potential at \(B\)).
The given condition states that the RMS voltage across \(R\) is equal to the RMS voltage across \(C.\)
For an RC circuit, the voltage across \(R\) and \(C\) are in quadrature (\(90^\circ\) phase difference).
The voltage across \(R\): \(V_R = I R,\) in phase with the current. 
The voltage across \(C\): \(V_C = I / (\omega C),\) leading the current by \(90^\circ.\)

The current through the capacitor \(C\) leads the voltage across \(C\) by \(90^\circ\) in a purely capacitive branch. The phase difference between the current through the capacitor and the voltage across it is \(90^\circ.\)
Hence, option (3) is the correct answer.

 

Hint:  \(n(\lambda)=A+\frac{B}{\lambda^2}\)

Explanation: Dispersion of light is the splitting of white light into its constituent colors due to the surface’s refractive index and the light’s wavelength. If the light entering the prism is with a single color then the emergent beam also has different colors arranged in a definite order.
This happens because different wavelengths of light travel at different speeds in a medium, causing them to refract by different angles.
The primary cause of dispersion is the dependence of refractive index on the wavelength of light.
Therefore, the dispersion takes place due to the wavelength of light.
Hence, option (3) is the correct answer.
 

Hint:  \(KE_{max} =PE_{max}\)

Step:

Apply the conservation of mechanical energy.
The kinetic energy for a particle is given by;
\(K = K_0 \sin^2(\omega t)\) 
\(K_i+U_i = K_f+U_f\) [When \(K_i = 0, \Rightarrow U_i = U_{\text{max}}\) Similarly \(U_f =0 \Rightarrow K_f = K_{\text{max}}\)]
Therefore maximum value of total energy is given as;
\(TE=KE _{\max }=K_0\)
As the total energy of the simple harmonic motion is constant therefore we can say that the maximum value of potential energy will be;
 \(\Rightarrow U_{max} =K_{max}\)
\(\Rightarrow U_{max} =K_0\)
Hence, option (1) is the correct answer.
 

Hint: Identify the series and parallel combinations in the circuit.

Step: Find the equivalent resistance between \(A\) and \(D.\) 
The circuit can be redrawn as shown in the figure below.

As we can see, there will be no current in the resistor between \(B\) and \(C,\) it is short-circuited.
 The equivalent resistance between \(A\) and \(D\) is given by;
\(\frac{1}{R_{eq}}=\frac{1}{2}+\frac{1}{2}+1\)
\(\Rightarrow R_{eq}=\frac{1}{2}~\Omega \)
Hence, option (3) is the correct answer.
 

Page 53-54, XI NCERT

\(F+e^-\rightarrow F^- , \Delta H=-x\\ \\ Cl+e^- \rightarrow Cl^-, \Delta H = -y\\ \\ y>x \)

Page 68, XI NCERT

Stem tendrils are specialized structures that some plants use for climbing and support. They are thin, elongated, coiling structures that help the plant in climbing and attaching to a support structure. Grapevine, watermelon, and cucumber are examples of plants that commonly produce stem tendrils for climbing and support. However, Citrus plants do not produce stem tendrils as they do not require climbing or support structures for their growth and development.

Hint: In trigonal bipyramidal, all bond lengths are not equal.

Explanation:

Species	The hybridisation of the central atom	Bond length
${\mathrm{BF}}_4^{\mathrm{\Theta }}$	sp3 (Tetrahedral)	All bond lengths equal
${\mathrm{XeF}}_4$	sp3d2 (square planar)	All bond lengths equal
${\mathrm{SF}}_4$	sp3d(see-saw)	axial bond length > equatorial length
${\mathrm{SiF}}_4$	sp3 (Tetrahedral)	all bond lengths equal

In SiF₄, axial and equatorial bonds are not equal. The axial bonds are longer than equatorial bonds.

Hence, option third is the correct answer.