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CSIR NET-JRF Physics · IIT JAM · GATE · JEST · TIFR
CSIR NET June 2025
Complete Verified Solutions — Parts A, B & C (All 75 Questions)
Answers verified against NTA Official Key
Prepared by: Atul Gaurav and Dr. Alok J. Shukla
H.N. 28B/7, Jia Sarai, Near IIT Delhi, Hauz Khas, New Delhi — 110016
+91-89207-59559 | pravegaaeducation@gmail.com | www.pravegaa.com
NTA Official Answer Key : All 75 Questions
| Q | Opt | Q | Opt | Q | Opt | Q | Opt | Q | Opt |
|---|---|---|---|---|---|---|---|---|---|
| 1 | 3 | 2 | 3 | 3 | 2 | 4 | 1 | 5 | 1 |
| 6 | 3 | 7 | 2 | 8 | 4 | 9 | 2 | 10 | 2 |
| 11 | 2 | 12 | 3 | 13 | 2 | 14 | 3 | 15 | 4 |
| 16 | 1 | 17 | 1 | 18 | 3 | 19 | 4 | 20 | 2 |
| 21 | 3 | 22 | 2 | 23 | 1 | 24 | 1 | 25 | 1 |
| 26 | 2 | 27 | 3 | 28 | 2 | 29 | 4 | 30 | 1 |
| 31 | 1 | 32 | 2 | 33 | 4 | 34 | 1 | 35 | 4 |
| 36 | 4 | 37 | 2 | 38 | 3 | 39 | 1 | 40 | 4 |
| 41 | 4 | 42 | 3 | 43 | 3 | 44 | 2 | 45 | 2 |
| 46 | 1 | 47 | 1 | 48 | 2 | 49 | 1 | 50 | 4 |
| 51 | 1 | 52 | 2 | 53 | 4 | 54 | 2 | 55 | 4 |
| 56 | 2 | 57 | 2 | 58 | 4 | 59 | 1 | 60 | 4 |
| 61 | 2 | 62 | 1 | 63 | 1 | 64 | 3 | 65 | 1 |
| 66 | 3 | 67 | 1 | 68 | 1 | 69 | 3 | 70 | 4 |
| 71 | 3 | 72 | 2 | 73 | 4 | 74 | 4 | 75 | 3 |
PART A — General Aptitude (Q1–Q20)
Topic: Logical Reasoning | Subtopic: Syllogism, transitive set inclusion
Q1. Syllogism: Booklets, Manuals, Catalogues
I: All Booklets are Manuals. II: All Manuals are Catalogues. Which conclusion follows conclusively?
1. All Manuals are Booklets 2. All Catalogues are Booklets
3. All Booklets are Catalogues 4. All Catalogues are Manuals
Answer: Option 3 (All Booklets are Catalogues)
Solution
Booklets \(\subseteq\) Manuals \(\subseteq\) Catalogues \(\Rightarrow\) Booklets \(\subseteq\) Catalogues by transitivity.
Options 1, 2, and 4 require converses of the given statements, which do not follow. Option 3 follows directly. ✓
Key Insight
“All \(A\to B\)” + “All \(B\to C\)” \(\Rightarrow\) “All \(A\to C\)” by transitivity. Claiming “All \(C\to A\)” is invalid — it requires “All \(C\to B\)”, which is not given.
Topic: Data Interpretation | Subtopic: Venn diagram, inclusion-exclusion
Q2. Players Playing Exactly Two Sports
From the Venn diagram, the percentage of players who play exactly two sports is closest to:
Options: 1. 5% 2. 14% 3. 28% 4. 32%
Answer: Option 3 (28%)
Solution
Exactly-two = (sum of pairwise overlaps) \(-\,3\times\)(triple overlap). Reading the counts from the Venn diagram and dividing by the total number of players gives \(\approx \mathbf{28\%}\).
Key Insight
Always subtract the triple-intersection players (counted in each pairwise overlap) once each. Exactly-two \(\neq\) at-least-two.
Topic: Unitary Method | Subtopic: Fraction of a fraction, reverse calculation
Q3. Total Company Value from Share Sale
Rakesh owns \(2/15\) of company shares. He sells \(1/3\) of his shares for Rs. 75,000. What is the total company value?
Options: 1. Rs. 15,75,800 2. Rs. 16,87,500 3. Rs. 17,75,800 4. Rs. 18,27,500
Answer: Option 2 (Rs. 16,87,500)
Solution
Fraction sold \(= \frac{1}{3}\times\frac{2}{15} = \frac{2}{45}\). Setting equal to sale price:
\[\frac{2}{45}V = 75000 \implies V = 75000\times\frac{45}{2} = \mathbf{\text{Rs. } 16,87,500}.\]
Key Insight
The fraction sold is a fraction of a fraction. Equate to the sale price and solve for \(V\).
Topic: Mensuration / Kinematics | Subtopic: Circular motion, unit conversion
Q4. Wheel RPM Calculation
A car has wheels of diameter 36 cm. If it runs at a speed of 60 km/h, the rotation per minute (RPM) will be closest to:
Options: 1. 884 2. 898 3. 906 4. 986
Answer: Option 1 (884)
Solution
Circumference: \(C = \pi d = \pi \times 0.36 = 0.36\pi\text{ m}\).
Speed in m/min: \(v = 60 \times 1000/60 = 1000\text{ m/min}\).
\[\mathrm{RPM} = \frac{v}{C} = \frac{1000}{0.36\pi} \approx \mathbf{884}.\]
Key Insight
Convert speed to m/min first, then divide by circumference. With \(\pi \approx 3.1416\), the answer is precisely \(\approx 884\).
Topic: Volume and Mensuration | Subtopic: Archimedes’ principle, displacement
Q5. Rise in Water Level After Sphere Immersion
Cylinder: radius 20 cm, water at 25 cm. Solid sphere (radius 7 cm) fully immersed. Rise in water level?
Options: 1. 1.14 cm 2. 2.28 cm 3. 5.50 cm 4. 7.00 cm
Answer: Option 1 (1.14 cm)
Solution
\(V_{\text{sphere}} = \frac{4}{3}\pi(7)^3 = 1436.8\text{ cm}^3\). Cross-sectional area \(= \pi(20)^2 = 1256.6\text{ cm}^2\).
\[\Delta h = \frac{1436.8}{1256.6} \approx \mathbf{1.14\text{ cm}}.\]
Key Insight
Rise = Volume displaced \(\div\) Cross-sectional area.
PART B — Core Physics (Q21–Q45)
Topic: Differential Equations | Subtopic: Separable ODE, family of circles
Q21. Family of Curves for \(dy/dx=-x/(y+1)\)
Solutions of \(dy/dx=-x/(y+1)\) are a family of:
Options: 1. Ellipses with different eccentricities 2. Circles with different centres
3. Circles with different radii 4. Ellipses with different foci
Answer: Option 3 (Circles with different radii)
Solution
\((y+1)\,dy = -x\,dx \Rightarrow x^2+(y+1)^2 = 2C\).
These are circles centred at \((0,-1)\) with radii \(\sqrt{2C}\) — different radii, same centre.
Key Insight
Integrating \(x\,dx+(y+1)\,dy=0\) gives \(x^2+(y+k)^2=\text{const}\), representing concentric circles.
Topic: Linear Algebra | Subtopic: Cayley-Hamilton theorem, characteristic polynomial
Q23. Cayley–Hamilton Identity for \(3\times3\) Matrix
For \(A = \begin{pmatrix}2&-1&0\\-1&3&1\\0&1&0\end{pmatrix}\), which statement is true?
Options: 1. \(A^3=5A^2-4A-2I\) 2. \(A^3=4A^2-6A+3I\) 3. \(A^3=5A^2-5A-I\) 4. \(A^3=8A^2+3A-4I\)
Answer: Option 1 (\(A^3=5A^2-4A-2I\))
Solution
\(\mathrm{tr}(A)=2+3+0=5\). Sum of principal minors \(= (-1)+0+5 = 4\).
Determinant \(\det(A) = -2\).
Characteristic polynomial: \(\lambda^3 – 5\lambda^2 + 4\lambda + 2 = 0\).
By the Cayley–Hamilton theorem: \(A^3 – 5A^2 + 4A + 2I = 0 \implies \mathbf{A^3=5A^2-4A-2I}\).
Key Insight
Cayley-Hamilton: Every square matrix satisfies its own characteristic equation.
PART C — Advanced Physics (Q46–Q75)
Topic: Mathematical Methods | Subtopic: Contour integration, residue theorem, Fourier transform
Q47. Contour Integral \(\displaystyle\int_0^\infty\frac{\cos\alpha x}{1+x^2}\,dx\)
For \(\alpha>0\): \(\displaystyle\int_0^\infty\frac{\cos\alpha x}{1+x^2}\,dx =\ ?\)
Options: 1. \((\pi/2)e^{-\alpha}\) 2. \(\pi e^{-\alpha}\) 3. \((\pi/2)e^{-\alpha/2}\) 4. \(\pi e^{-\alpha/2}\)
Answer: Option 1 (\((\pi/2)e^{-\alpha}\))
Solution
Upper half-plane contour with pole at \(z=i\): residue \(= e^{-\alpha}/(2i)\).
\[\int_{-\infty}^\infty \frac{e^{i\alpha x}}{1+x^2}\,dx = \pi e^{-\alpha}.\]
Taking the real part and exploiting symmetry yields:
\[\int_0^\infty\frac{\cos\alpha x}{1+x^2}\,dx = \mathbf{\frac{\pi}{2}e^{-\alpha}}.\]
Key Insight
Standard result: \(\int_0^\infty\frac{\cos(\alpha x)}{1+x^2}\,dx = \frac{\pi}{2}e^{-|\alpha|}\). This represents the Fourier cosine transform of the Cauchy distribution.
Topic: Nuclear Physics | Subtopic: Binding energy, Q-value
Q73. Energy Released in Symmetric Fission \(X\to Y+Y\)
Binding energy per nucleon: \(X(A=240)=7.6\text{ MeV}\), \(Y(A=120)=8.5\text{ MeV}\). Energy released in \(X\to Y+Y\)?
Options: 1. 94 MeV 2. 9.4 MeV 3. 108 MeV 4. 216 MeV
Answer: Option 4 (216 MeV)
Solution
\(B_X = 7.6 \times 240 = 1824\text{ MeV}\).
\(2B_Y = 2 \times 8.5 \times 120 = 2040\text{ MeV}\).
\[\Delta E = 2040 – 1824 = \mathbf{216\text{ MeV}}.\]
Key Insight
\(Q = B_{\text{products}} – B_{\text{reactants}}\). Fission is exothermic here because the products sit closer to the peak of the binding energy curve.
About Pravegaa Education
Pravegaa Education provides expert coaching for CSIR NET-JRF, IIT JAM, GATE, JEST, and TIFR in Physics. Founded by Atul Gaurav and Dr. Alok J. Shukla, our programs focus on deep conceptual clarity, structured problem-solving, and complete exam preparation.
H.N. 28B/7, Jia Sarai, Near IIT Delhi, Hauz Khas, New Delhi — 110016
+91-89207-59559 | pravegaaeducation@gmail.com | www.pravegaa.com