NEET Physics PYQ 2025 — Set 1 (Q1–Q10) — Solved (English + Hindi)

NEET Physics PYQ 2025 — Set 1 (Questions 1–10) — Solved (English + Hindi)

This post contains 10 NEET-style physics questions with short, exam-focused solutions. Each solution is provided in English followed by a Hindi (Hinglish) explanation in parentheses for bilingual clarity.

Series: All NEET PYQs • Set 2 (Q11–Q20) • Set 3 (Q21–Q30) • Set 4 (Q31–Q40) • Set 5 (Q41–Q50)

Q1. A block of mass \(m\) is placed on a smooth horizontal surface and connected to a spring of spring constant \(k\). It is pulled by distance \(A\) and released. The time period of oscillation is:

(Mechanics: SHM)

(a) \(2\pi\sqrt{\dfrac{m}{k}}\)

(b) \(2\pi\sqrt{\dfrac{k}{m}}\)

(c) \(\pi\sqrt{\dfrac{m}{k}}\)

(d) \(2\sqrt{\dfrac{m}{k}}\)

Answer — (a) \(2\pi\sqrt{\dfrac{m}{k}}\)

English: Time period of simple harmonic motion for mass–spring: \(T=2\pi\sqrt{\dfrac{m}{k}}\).

Hindi (Hinglish): Spring–mass ka SHM ka formula \(T=2\pi\sqrt{m/k}\) hai. (यानि विकल्प (a) सही है।)

Q2. A projectile is fired at an angle \(\theta\) with initial speed \(u\). If its maximum height is \(H\), the time to reach height \(H/2\) (on the way up) is:

(Projectile motion)

(a) \(\dfrac{u\sin\theta}{g}\)

(b) \(\dfrac{u\sin\theta}{2g}\)

(c) \(\dfrac{u\sin\theta}{g}\Big(1-\dfrac{1}{\sqrt{2}}\Big)\)

(d) \(\dfrac{u\sin\theta}{g}\Big(1+\dfrac{1}{\sqrt{2}}\Big)\)

Answer — (c)

English: \(H=\dfrac{u^2\sin^2\theta}{2g}\). Height \(H/2\) corresponds to vertical velocity \(v_y\) satisfying \(v_y^2=(u\sin\theta)^2-2g(H/2)=(u\sin\theta)^2-(u\sin\theta)^2/2=(u\sin\theta)^2/2\). So \(v_y=(u\sin\theta)/\sqrt{2}\). Time from launch: \(t=(u\sin\theta - v_y)/g = \dfrac{u\sin\theta}{g}(1-\dfrac{1}{\sqrt{2}})\).

Hindi (Hinglish): \(H= u^2\sin^2\theta/(2g)\). \(H/2\) par vertical speed \((u\sin\theta)/\sqrt{2}\) hoti hai. Time \(t=(u\sin\theta - (u\sin\theta)/\sqrt2)/g\). (विकल्प c)।

Q3. A wire of length \(L\) and resistance \(R\) is stretched to double its length (volume preserved). New resistance becomes:

(Electricity)

(a) \(2R\)

(b) \(4R\)

(c) \(R/2\)

(d) \(R/4\)

Answer — (b) \(4R\)

English: Resistance \(R=\rho \dfrac{L}{A}\). If length doubles \(L' = 2L\), and volume \(V = LA\) constant ⇒ cross-section \(A' = A/2\). So \(R' = \rho (2L)/(A/2)=4\rho L/A = 4R\).

Hindi (Hinglish): L double hua toh area aadha ho jayega (volume constant). Isse resistance 4 guna ho jata hai. (विकल्प b)।

Q4. A particle in uniform circular motion has centripetal acceleration:

(Kinematics / Circular motion)

(a) Away from center

(b) Tangential

(c) Towards center

(d) Zero

Answer — (c) Towards center

English: Centripetal acceleration is directed radially inward: \(a_c = \dfrac{v^2}{r}\) towards the center.

Hindi (Hinglish): Circular motion mein centripetal acceleration centre ki taraf hota hai (radially inward). (विकल्प c)।

Q5. The electric field at distance \(r\) from a long charged wire (linear charge density \(\lambda\)) is:

(Electrostatics)

(a) \(\dfrac{\lambda}{4\pi\epsilon_0 r^2}\)

(b) \(\dfrac{\lambda}{2\pi\epsilon_0 r}\)

(c) \(\dfrac{\lambda r}{4\pi\epsilon_0}\)

(d) \(\dfrac{2\lambda}{\epsilon_0 r}\)

Answer — (b) \(\dfrac{\lambda}{2\pi\epsilon_0 r}\)

English: By Gauss's law for a long line charge: \(E \cdot 2\pi r L = \dfrac{\lambda L}{\epsilon_0}\) ⇒ \(E=\dfrac{\lambda}{2\pi\epsilon_0 r}\).

Hindi (Hinglish): Gauss law lagane par field \(E=\lambda/(2\pi\epsilon_0 r)\) milta hai. (विकल्प b)।

Q6. Dimensional formula of Planck's constant \(h\) is:

(Modern Physics / Dimensions)

(a) \(ML^2T^{-1}\)

(b) \(MLT^{-2}\)

(c) \(M^2L^2T^{-2}\)

(d) \(ML^2T^{-2}\)

Answer — (a) \(ML^2T^{-1}\)

English: Energy × time has dimensions \([E][T] = (ML^2T^{-2})(T) = ML^2T^{-1}\). Planck constant has same dimensions because \(E=h\nu\).

Hindi (Hinglish): \(h\) ki dimension energy×time ke barabar hoti hai: \(ML^2T^{-1}\). (विकल्प a)।

Q7. Two capacitors \(C_1\) and \(C_2\) are connected in series. Equivalent capacitance is:

(Electrostatics)

(a) \(C_1 + C_2\)

(b) \(\dfrac{C_1 + C_2}{2}\)

(c) \(\dfrac{C_1 C_2}{C_1 + C_2}\)

(d) \(\dfrac{1}{C_1} + \dfrac{1}{C_2}\)

Answer — (c) \(\dfrac{C_1 C_2}{C_1 + C_2}\)

English: For series, reciprocals add: \(\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\) ⇒ \(C_{eq}=\dfrac{C_1C_2}{C_1+C_2}\).

Hindi (Hinglish): Series mein \(1/C\) add hota hai → equivalent \(C_1C_2/(C_1+C_2)\). (विकल्प c)।

Q8. A lens forms a diminished, erect image. The lens must be:

(Optics)

(a) Convex, object beyond 2F

(b) Concave, object anywhere

(c) Convex, object at 2F

(d) Plane lens

Answer — (b) Concave, object anywhere

English: A concave (diverging) lens always forms a virtual, erect, diminished image for any real object position. Convex lens can form erect diminished image only for virtual objects (not usual).

Hindi (Hinglish): Diminished + erect image hamesha concave lens deta hai (for real object positions). (विकल्प b)।

Q9. Work done in an isothermal process for an ideal gas is:

(Thermodynamics)

(a) \(W = nRT \ln\dfrac{V_2}{V_1}\)

(b) \(W = nR(T_2 - T_1)\)

(c) \(W = p(V_2 - V_1)\)

(d) Zero

Answer — (a) \(W = nRT \ln\dfrac{V_2}{V_1}\)

English: For isothermal process \(T=\) constant ⇒ \(W=\int p\,dV = nRT\int \dfrac{dV}{V} = nRT\ln(V_2/V_1)\).

Hindi (Hinglish): Isothermal mein internal energy change zero, aur work \(nRT\ln(V_2/V_1)\) hota hai. (विकल्प a)।

Q10. A transformer works on the principle of:

(Electromagnetism)

(a) Electrostatics

(b) Mutual induction

(c) Photoelectric effect

(d) Joule heating

Answer — (b) Mutual induction

English: Transformer transfers energy between coils by changing magnetic flux → Faraday's law & mutual induction.

Hindi (Hinglish): Transformer kaam mutual induction se hota hai — ek coil ka changing flux doosre coil mein emf induce karta hai. (विकल्प b)।

Want the next set? Use the links above or click: Set 2 (Q11–Q20).

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