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NEET Practice Questions (MCQs) with Answers & Solutions

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NEET 2024

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Match List I with List II:

List I	List II
A. RNA polymerase III	I. snRNPs
B. Termination of transcription	II. Promoter
C. Splicing of Exons	III. Rho factor
D. TATA box	IV. snRNAs, tRNA

Choose the correct answer from the options given below:

A. A-IV, B-III, C-I, D-II B. A-II, B-IV, C-I, D-III C. A-III, B-II, C-IV, D-I D. A-III, B-IV, C-I, D-II

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Explanation

RNA pol III makes tRNA, 5S rRNA, snRNAs; Rho terminates prokaryotic transcription; snRNPs splice introns; TATA is in eukaryotic promoter.

NEET 2024

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Choose the correct statement given below regarding juxta medullary nephron.

A. Juxta medullary nephrons outnumber the cortical nephrons. B. Juxta medullary nephrons are located in the columns of Bertini. C. Renal corpuscle of juxta medullary nephron lies in the outer portion of the renal medulla. D. Loop of Henle of juxta medullary nephron runs deep into medulla.

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Explanation

Juxtamedullary nephrons have long loops of Henle that extend deep into the medulla — the basis of the counter-current mechanism.

NEET 2024

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Given below are two statements:

Statement I: Mitochondria and chloroplasts are both double membrane bound organelles.

Statement II: Inner membrane of mitochondria is relatively less permeable, as compared to chloroplast.

In the light of the above statements, choose the most appropriate answer from the options given below:

A. Statement I is incorrect but Statement II is correct. B. Both Statement I and Statement II are correct. C. Both Statement I and Statement II are incorrect. D. Statement I is correct but Statement II is incorrect.

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Explanation

Both statements are taken directly from NCERT (cell-organelles chapter).

NEET 2024

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Regarding catalytic cycle of an enzyme action, select the correct sequential steps:

A. Substrate enzyme complex formation. B. Free enzyme ready to bind with another substrate. C. Release of products. D. Chemical bonds of the substrate broken. E. Substrate binding to active site.

Choose the correct answer from the options given below:

A. E, D, C, B, A B. E, A, D, C, B C. E, B, D, C, A D. B, A, C, D, E

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Explanation

Binding (E) → ES complex (A) → bond cleavage (D) → product release (C) → free enzyme (B).

NEET 2024

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As per ABO blood grouping system, the blood group of father is $B^+$, mother is $A^+$ and child is $O^+$. Their respective genotype can be:

A. $I^Bi\ /\ I^Ai\ /\ ii$ B. $I^BI^B\ /\ I^AI^A\ /\ ii$ C. $I^AI^B\ /\ ii\,I^A\ /\ I^B$ D. $I^Ai\ /\ I^Bi\ /\ I^Ai$ E. $iiI^B\ /\ iiI^A\ /\ I^Ai$

Choose the most appropriate answer from the options given below:

A. D & E only B. A only C. B only D. C & B only

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Explanation

For child = $ii$, each parent must contribute one $i$. Father ($B$) must be $I^B i$; mother ($A$) must be $I^A i$ — only A satisfies this.

NEET 2025

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Consider a water tank shown in the figure. It has one wall at $x = L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $\rho$, the liquid surface makes angle $\theta_0\ (\theta_0 \ll 1)$ with the $x$-axis at $x = L$. If $y(x)$ is the height of the surface, then the equation for $y(x)$ is: (take $\theta(x) = \sin\theta(x) = \tan\theta(x) = \frac{dy}{dx}$, $g$ is the acceleration due to gravity)

A. $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{S}\, x$ B. $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{S}\, y$ C. $\dfrac{d^2y}{dx^2} = \sqrt{\dfrac{\rho g}{S}}$ D. $\dfrac{dy}{dx} = \sqrt{\dfrac{\rho g}{S}}\, x$

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Explanation

Balancing the Laplace pressure of the curved surface, $S\,\dfrac{d^2y}{dx^2} = \rho g\,y$ for small slopes, i.e. $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{S}\,y$.

NEET 2025

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A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is:

A. 100 B. 125 C. 150 D. 250

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Explanation

$m = \dfrac{L}{f_o}\cdot\dfrac{D}{f_e} = \dfrac{40}{2}\times\dfrac{25}{4} = 20\times 6.25 = 125$.

NEET 2025

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An electron (mass $9\times10^{-31}$ kg and charge $1.6\times10^{-19}$ C) moving with speed $c/100$ ($c$ = speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9\times10^{-4}$ T perpendicular to its direction of motion. We wish to apply a uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then ($c = 3\times10^8\ \text{ms}^{-1}$):

A. $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27\times10^4\ \text{V m}^{-1}$ B. $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27\times10^2\ \text{V m}^{-1}$ C. $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27\times10^2\ \text{V m}^{-1}$ D. $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27\times10^4\ \text{V m}^{-1}$

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Explanation

For zero deflection (velocity selector) the electric force must balance the magnetic force: $qE = qvB$ with $\vec{E}\perp\vec{B}$. $E = vB = \dfrac{c}{100}\times B = 3\times10^6\times 9\times10^{-4} = 2.7\times10^3 = 27\times10^2\ \text{V m}^{-1}$.

NEET 2025

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There are two inclined surfaces of equal length ($L$) and same angle of inclination $45^\circ$ with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on the rough surface than on the smooth surface. The coefficient of kinetic friction $(\mu_k)$ between the object and the rough surface is close to:

A. 0.25 B. 0.40 C. 0.5 D. 0.75

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Explanation

For equal $L$: $a_{smooth}t_1^2 = a_{rough}t_2^2$ with $t_2 = 2t_1\Rightarrow a_{smooth} = 4a_{rough}$. $g\sin45^\circ = 4g(\sin45^\circ-\mu\cos45^\circ)\Rightarrow \mu = \tfrac{3}{4}\tan45^\circ = 0.75$.

NEET 2025

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The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If $F_A$ and $F_B$ are the forces applied by the breaks on cars A and B, respectively, then the ratio $F_A/F_B$ is:

A. $\dfrac{2}{3}$ B. $\dfrac{3}{2}$ C. $\dfrac{1}{3}$ D. $\dfrac{1}{2}$

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Explanation

Work–energy: $F = \dfrac{KE}{d}$. $F_A = \dfrac{100}{1000} = 0.1$, $F_B = \dfrac{225}{1500} = 0.15$; $\dfrac{F_A}{F_B} = \dfrac{0.1}{0.15} = \dfrac{2}{3}$.

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