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Sabse pehle paper ka title poochha jaata hai — jaise “Class 10 Science — Half Yearly”. Phir class, ek ya kai subjects, board, medium, school/institute ka naam, exam type, time aur total marks.

  • Paper ki bhasha — English, हिंदी, ya dono saath
  • Roll No. box chahiye ya nahi — aap chunein
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SectionTypeQMarks
AMCQ + Assertion–Reason2020 × 1 = 20
BVery Short Answer55 × 2 = 10
CShort Answer66 × 3 = 18
DLong Answer44 × 5 = 20
ECase Study33 × 4 = 12
Total3880
  • Apna format — section jodein / hataayein / upar-neeche karein
  • Ek section me kai types — Q.1 MCQ, Q.2 Fill in the Blanks, Q.3 True/False…
  • Choice — “5 me se koi 4” jaisa internal choice; marks apne aap usi hisaab se
  • Marks limit — format ke marks paper se zyada hote hi laal alert, aur download tab tak band
  • Numbering — teen tareeke (neeche “Kya khaas hai” me dekhein)
MCQ (Multiple Choice) Assertion – Reason Fill in the Blanks True / False Match the Following One Word / One Sentence Very Short Answer (VSA) Short Answer (SA) Long Answer (LA) Case Study / Passage Based Numerical Diagram / Map Based
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General Instructions :
(i) This question paper contains 38 questions. All questions are compulsory.
(ii) The question paper is divided into 5 sections — A, B, C, D and E.
(iii) Section A (questions 1 to 20) has 18 Multiple Choice Questions (MCQs) and 2 Assertion-Reason based questions of 1 mark each.
(iv) Section B (questions 21 to 25) has 5 Very Short Answer (VSA) questions of 2 marks each.
… baaki sections bhi isi tarah

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Auto-fill Poora 38-question ka paper ek click me. Chapters chun lein to sirf unhi me se.
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Teen bhasha English, हिंदी, ya dono saath-saath — headings aur instructions sameth.
Teen numbering style Q.1, Q.2… (CBSE) · har section me Q.1 se · Q.1 (i) (ii) (iii) (MP Board).
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Roll No.
EduGrown Public School
Half Yearly Examination 2026-27
Class: 10  |  Subject: Science
Time: 3 Hours Max. Marks: 40
General Instructions :
(i) This question paper contains 9 questions. All questions are compulsory.
(ii) The question paper is divided into 4 sections — A, B, C and D.
(iii) Section A (questions 1 to 3) has 2 Multiple Choice Questions (MCQs) and 1 Assertion-Reason based question of 1 mark each.
(iv) Section B (questions 4 to 5) has 2 Very Short Answer (VSA) questions of 2 marks each.
(v) Section C (questions 6 to 7) has 2 Short Answer (SA) questions of 3 marks each.
(vi) Section D (questions 8 to 9) has 2 Long Answer (LA) questions of 5 marks each.
(vii) Marks for each question are shown against it.
(viii) Write the question number clearly before writing its answer.
SECTION A
(3 × 1 = 3 marks)

Choose the correct option.

Q.1[1]
An electric dipole of moment $p$ is placed in a uniform field $\vec{E}$. The torque on it is maximum when the angle between $\vec{p}$ and $\vec{E}$ is:
(a) $0^\circ$(b) $45^\circ$(c) $90^\circ$(d) $180^\circ$
Answer: (c) $90^\circ$ — torque $\tau = pE\sin\theta$ is maximum at $\theta = 90^\circ$.
Q.2[1]
Four chambered heart is a characteristic feature of which of the following animals?
(a) Crocodile(b) Frog(c) Lizard(d) Snake
Answer: (a) Crocodile

In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Q.3[1]
Assertion (A): The electric field inside a charged hollow conductor is zero.
Reason (R): The charge on a conductor resides entirely on its outer surface.
(a) Both A and R true, R explains A(b) Both A and R true, R does not explain A
(c) A is true but R is false(d) A is false but R is true
Answer: (a) Both A and R are true, and R is the correct explanation of A.
SECTION B
(2 × 2 = 4 marks)

Answer the following questions in brief.

Q.4[2]
State Gauss’s Law and use it to find the electric field near a uniformly charged infinite plane sheet.
Answer: Total flux through any closed surface $= q_{enc}/\varepsilon_0$. Taking a cylindrical Gaussian surface through the sheet, $2EA = \sigma A/\varepsilon_0$, so $E = \frac{\sigma}{2\varepsilon_0}$, directed away from the sheet.
Q.5[2]
Two long straight wires carry linear charge densities $\lambda_1 = 10\ \mu C/m$ and $\lambda_2 = -20\ \mu C/m$. Write the field of wire 1 at a point $10$ cm away.
Answer: $E_1 = \frac{\lambda_1}{2\pi\varepsilon_0(0.10)} = \frac{2k\lambda_1}{0.10}$, directed radially outward.
SECTION C
(2 × 3 = 6 marks)
Q.6[3]
Derive an expression for the potential energy of a dipole $\vec{p}$ in a uniform field $\vec{E}$.
Answer:

Work done in rotating the dipole against the torque: $W = \int_{\theta_1}^{\theta_2} pE\sin\theta\, d\theta = pE(\cos\theta_1 - \cos\theta_2)$.

Taking $\theta_1 = 90^\circ$ as reference, $U(\theta) = -pE\cos\theta = -\vec{p}\cdot\vec{E}$.

Q.7[3]
From the table below, calculate the percentage change in each item:
Particulars2025 (₹)2024 (₹)
Share Capital24,00,00020,00,000
Reserves and Surplus6,00,0005,00,000
Long term borrowings12,00,00010,00,000
Answer: Har item 20% badha hai — yaani poori sheet ek hi anupaat me badhi hai.
SECTION D
(2 × 5 = 10 marks)

Answer the following questions in detail.

Q.8[5]
An electric dipole has charges $-1.0$ pC at $(0,0)$ and $+1.0$ pC at $(3\,mm, 4\,mm)$. A field $\vec{E} = \left(\frac{1000\ V}{m}\right)\hat{i}$ is switched on. Find the torque $\vec{\tau}$ acting on the dipole.
Answer:

$\vec{d} = (3\,mm, 4\,mm)$, so $d = 5\ mm = 5 \times 10^{-3}\ m$.

$\vec{\tau} = \vec{p} \times \vec{E}$, magnitude $\tau = pE\sin\theta = 4 \times 10^{-12}\ N\,m$.

OR
Solve for $x$:   $2x^2 - 7x + 3 = 0$, and verify your answer.
Answer:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{7 \pm \sqrt{49-24}}{4} = \frac{7 \pm 5}{4}$.

So $x = 3$ or $x = \frac{1}{2}$.

Q.9[5]
Explain the working of a moving coil galvanometer with a labelled diagram. Why is a radial magnetic field used in it?
Answer: Poora solution diagram ke saath.
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