NEET UG Day 37 Free Practice Test

NEET UG Day 37 Free Practice Test: Day 37 is here, and it’s time to challenge yourself with another exciting NEET UG Free Practice Test! Every single day of practice brings you one step closer to your dream medical college. These questions are carefully designed to match the NEET exam pattern and help you sharpen your concepts in Physics, Chemistry, and Biology. So grab your notebook, set a timer, and start practicing. Remember, consistent practice today will turn into confidence on exam day!

PHYSICS

Q.1. The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is

(1)1: √2

(2) 2 : 1

(3)√2 : 1

(4) 4 : 1

Q.2. A biconvex lens has radii of curvature of 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is

(1) Infinity

(2) +2 D

(3) +20 D

(4) +5 D

Q.3. An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 m/s. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : (g = 10 m/s^2)

(1) 23500

(2) 23000

(3) 20000

(4) 34500

Q.4. The angular speed of a flywheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/s2 is

(1) 104π

(2) 2π

(3) 4π

(4) 12π

Q.5. The peak voltage of the ac source is equal to:

(1)1/ √2times the rms value of the ac source

(2) The value of voltage supplied to the circuit

(3) The rms value of the ac source

(4) √2 times the rms value of the ac source

CHEMISTRY

Q.6. If a soap bubble expands, the pressure inside the bubble

(1) Is equal to the atmospheric pressure

(2) Decreases

(3) Increases

(4) Remains the same

Q.7. Let T1 and T2 be the energy of an electron in the first and second excited states of hydrogen atoms, respectively. According to the Bohr’s model of an atom, the ratio T1 : T2 is:

(1) 9 : 4

(2) 1 : 4

(3) 4 : 1

(4) 4 : 9

Q.8. Which statement regarding polymers is not correct?

(1) Thermosetting polymers are reusable

(2) Elastomers have polymer chains held together by weak intermolecular forces

(3) Fibers possess high tensile strength

(4) Thermoplastic polymers are capable of repeatedly softening and hardening on heating and cooling
respectively

Q.9. The IUPAC name of an element with atomic number 119 is

(1) ununoctium

(2) ununennium

(3) unnilennium

(4) unununnium

Q.10. Statement I: In the coagulation of a negative sol, the flocculating power of the three given ions is in the order Al3+ > Ba2+ > Na+
Statement II:
In the coagulation of a positive sol, the flocculating power of the three given salts is in the order
NaCl > Na2SO4 > Na3PO4. In the light of the above statements, choose the most appropriate answer from the options given below:

(1) Statement I is incorrect but Statement II is correct.

(2) Both Statement I and Statement II are correct.

(3) Both Statement I and Statement II are incorrect.

(4) Statement I is correct but Statement II is incorrect.

Read Also: NEET UG 2026 Application Deadline Extended

BIOLOGY

Q.11. DNA polymorphism forms the basis of :

(1) Translation

(2) Genetic mapping

(3) DNA finger printing

(4) Both genetic mapping and DNA finger printing

Q.12. In old trees the greater part of secondary xylem is dark brown and resistant to insect attack due to :

(a) secretion of secondary metabolities and their deposition in the lumen of vessels.

(b) deposition of organic compounds like tannins and resins in the central layers of stem.

(c) deposition of suberin and aromatic substances in the outer layer of stem.

(d) deposition of tannins, gum, resin and aromatic substances in the peripheral layers of stem.

(e) presence of parenchyma cells, functionally active xylem elements and essential oils.

Choose the correct answer from the options given below:

(1) (b) and (d) Only

(2) (a) and (b) Only

(3) (c) and (d) Only

(4) (d) and (e) Only

Q.13. Read the following statements and choose the set of correct statements :

(a) Euchromatin is loosely packed chromatin

(b) Heterochromatin is transcriptionally active

(c) Histone octomer is wrapped by negatively charged DNA in nucleosome

(d) Histones are rich in lysine and arginine

(e) A typical nucleosome contains 400 bp of DNA helix

Choose the correct answer from the options given below :

(1) (a), (c), (e) Only

(2) (b), (d), (e) Only

(3) (a), (c), (d) Only

(4) (b), (e) Only

Q.14. Identify the incorrect statement related to Pollination :

(1) Moths and butterflies are the most dominant pollinating agents among insects

(2) Pollination by water is quite rare in flowering plants

(3) Pollination by wind is more common amongst abiotic pollination

(4) Flowers produce foul odours to attract flies and beetles to get pollinated

Q.15. Addition of more solutes in a given solution will :

(1) not affect the water potential at all

(2) raise its water potential

(3) lower its water potential

(4) make its water potential zero

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ANSWERS & EXPLANATIONS

Ans.1. (3). √2 : 1. For a thin uniform disc, the moment of inertia about the center normal to the plane is $\frac{1}{2}m{R}^2$, so radius of gyration $k_{\perp }=\frac{R}{\sqrt{2}}$. About the diameter, it is $\frac{1}{4}m{R}^2$, so $k_d=\frac{R}{2}$. The ratio $k_{\perp }:k_d=\sqrt{2}:1$.

Ans.2. (4). +5 D. Using the lens maker’s formula for a biconvex lens: $\frac{1}{f}=(n-1)$, with $R_1=+20$, $R_2=-20$, $n=1.5$. This gives $f=20$ cm or 0.2 m, so power $P=\frac{1}{f}=+5$.

Ans.3. (4) 34500. At constant speed, the motor overcomes weight $mg=2000\times 10=20000$N and friction 3000 N, so total force $F=23000$ N. Power $P=F\times v=23000\times 1.5=34500$ W.

Ans.4. (3) 4π. Angular acceleration $\alpha =\frac{{\omega }_2-{\omega }_1}{t}=\frac{64\pi }{16}=4\pi$rad/s²

Ans.5. (4) √2 times the rms value of the ac source
The peak voltage $V_p$ relates to RMS voltage $V_{rms}$ by $V_p=V_{rms}\sqrt{2}$ for sinusoidal AC. This is the standard definition.

Ans.6. (2) Decreases. The excess pressure inside a soap bubble is $\mathrm{\Delta }P=\frac{4T}{r}$, where $T$ is surface tension and $r$ is radius. As the bubble expands, $r$ increases, so $\mathrm{\Delta }P$ decreases, meaning pressure inside decreases relative to outside.

Ans.7. (1). 9:4.  First excited state $n=2$, $T_1\propto \frac{1}{4}$; second excited $n=3$n=3, $T_2\propto \frac{1}{9}$ ​. Ratio $T_1:T_2=\frac{1\mathrm{/}4}{1\mathrm{/}9}=9:4$

Ans.8. (1) Thermosetting polymers are reusable
Thermosetting polymers harden permanently on heating and cannot be softened or reshaped, so they are not reusable. Thermoplastics soften repeatedly, elastomers have weak forces, and fibers have high tensile strength.

Ans.9. (2) ununennium. IUPAC systematic name for Z=119

Ans.10. (4) Statement I is correct but Statement II is incorrect
For negative sol, flocculating power of cations follows Hardy-Schulze rule: higher charge better, Al³⁺ > Ba²⁺ > Na⁺. For positive sol, anions: PO₄³⁻ > SO₄²⁻ > Cl⁻, so Na₃PO₄ > Na₂SO₄ > NaCl, opposite to Statement II. Thus, I correct, II incorrect.

Ans. 11. (4) Both genetic mapping and DNA finger printing
DNA polymorphism refers to variations in DNA sequences among individuals, serving as genetic markers. These variations enable both genetic mapping, which locates genes on chromosomes, and DNA fingerprinting, which creates unique profiles for identification.

Ans.12. (2)

Ans.13. (3) (a), (c), (d) Only
Euchromatin is loosely packed and transcriptionally active; heterochromatin is tightly packed and less active (b incorrect). Histones, rich in positively charged lysine and arginine, form an octamer wrapped by negatively charged DNA

Ans.14. (1) Moths and butterflies are the most dominant pollinating agents among insects
This is incorrect; bees are the most dominant insect pollinators, while moths/butterflies (lepidopterans) are significant but secondary. Water pollination is rare, wind common in abiotic, and foul odors do attract flies/beetles 

Ans.15. (3) lower its water potential
Water potential (ψ) decreases with added solutes due to solute potential (${\psi }_s$ψs) becoming more negative: $\psi ={\psi }_s+{\psi }_p$. More solutes lower ${\psi }_s$without affecting pressure potential significantly

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