Hello students! If you are aiming for top marks in the upcoming UP Board Exam 2026, then this post is exactly what you need. While many students search for Class 10 Science Important Questions 2026 UP Board, they struggle to find complete, accurate solutions for specific board sets, especially in English.
That is why we have brought you the complete UP Board Class 10 Science Solved Paper 2026 Set 824 EJ. Whether you study in Hindi medium or English medium (you can use the Translate button on our site), these solutions are crucial. In this post, we provide the detailed Science paper set 824 (EJ) solution, covering everything from Physics numericals and ray diagrams to Chemistry equations and Biology MCQs.
Based on previous year papers and the latest syllabus, this fully solved UP Board 10th Science Paper Set 824 (EJ) will give your last-minute preparation a massive boost. Let's start solving!
PART - A: Multiple Choice Questions (MCQs)
Sub-Section - (1) | Physics
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Ans: (D) Between the pole of mirror and principal focus.Explanation: A concave mirror forms a virtual, erect, and magnified image only in one specific case: when the object is placed very close to the mirror, strictly between its Pole (P) and Principal Focus (F). This property is utilized in shaving mirrors and dentist mirrors.
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Ans: (C) A convex lens of focal length 5 cm.Explanation: To read small letters, a magnifying glass is required, which is a convex lens. The magnification produced by a simple convex lens is higher when its focal length is shorter. Therefore, a convex lens of \(5 \text{ cm}\) focal length will provide better magnification than one with \(50 \text{ cm}\).
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Ans: (A) both concave.Explanation: According to the New Cartesian Sign Convention, the focal length of a concave mirror and a concave lens is always considered negative. Since both values are given as \(-15 \text{ cm}\), both must be concave.
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Ans: (D) Retina.Explanation: The eye lens focuses the incoming light rays to form a real and inverted image on the light-sensitive screen at the back of the eye, which is called the retina.
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Ans: (C) 1 : 4.Explanation: Let resistance of one wire = \(R\).
Equivalent resistance in series (\(R_s\)) = \(R + R = 2R\)
Equivalent resistance in parallel (\(R_p\)) = \(\frac{R \cdot R}{R + R} = \frac{R}{2}\)
Heat generated (\(H\)) for same voltage (\(V\)) and time (\(t\)) is given by \(H = \frac{V^2 t}{R}\)
Ratio = \(\frac{H_s}{H_p} = \frac{R_p}{R_s} = \frac{R/2}{2R} = \frac{1}{4}\).
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Ans: (C) increases very large.Explanation: During a short circuit, the live wire and neutral wire come into direct contact. This drops the resistance of the circuit to almost zero, causing a sudden and massive increase in electric current, which can cause fires.
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Ans: (A) Generator.Explanation: An electric generator works on the principle of electromagnetic induction. It converts mechanical energy into electrical energy, thereby producing an electric current.
Sub-Section - (2) | Chemistry
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Ans: (C) metalloid.Explanation: Antimony (Sb) shows properties of both metals and non-metals, therefore it is classified as a metalloid.
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Ans: (B) \(\text{Pb(NO}_3\text{)}_2\).Explanation: Lead usually exhibits a valency of +2 (\(\text{Pb}^{2+}\)), and the nitrate ion has a valency of -1 (\(\text{NO}_3^-\)). Criss-crossing the valencies gives the formula \(\text{Pb(NO}_3\text{)}_2\).
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Ans: (B) \(\text{C}_2\text{H}_4\).Explanation: Unsaturated hydrocarbons contain double or triple bonds between carbon atoms (alkenes or alkynes). \(\text{C}_2\text{H}_4\) is ethene (an alkene with a double bond). The others are alkanes (saturated).
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Ans: (B) ethanoic acid.Explanation: Acetic acid has the formula \(\text{CH}_3\text{COOH}\). It contains two carbon atoms, so the root word is "eth-". The functional group is a carboxylic acid (-oic acid). Hence, Ethanoic acid.
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Ans: (B) \(\text{MO}\).Explanation: The formula \(\text{MCl}_2\) implies that the valency of element M is +2 (since Chlorine has a valency of -1). Oxygen has a valency of -2. Therefore, combining \(\text{M}^{2+}\) and \(\text{O}^{2-}\) gives the simplest ratio \(\text{MO}\).
Sub-Section - (3) | Biology
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Ans: (C) Excretion.Explanation: Kidneys are the primary organs of the human excretory system. They filter blood to remove nitrogenous waste like urea and excrete it in the form of urine.
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Ans: (C) Bryophyllum.Explanation: Bryophyllum reproduces vegetatively through its leaves. Adventitious buds develop on the margins of the leaf, which fall to the ground and grow into new plants.
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Ans: (C) Gregor Johann Mendel.Explanation: Gregor Mendel formulated the fundamental laws of inheritance based on his extensive experiments on pea plants, earning him the title "Father of Genetics".
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Ans: (D) Hydrogen bonds.Explanation: The two polynucleotide strands of a DNA double helix are held together by weak hydrogen bonds forming between the complementary nitrogenous bases (A with T, and C with G).
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Ans: (A) Hydra.Explanation: Hydra reproduces asexually by budding. A small outgrowth (bud) develops on the parent body due to repeated cell division, grows into a tiny individual, and detaches when mature.
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Ans: (B) Garden pea.Explanation: Mendel conducted his hybridization experiments on the Garden pea plant (Pisum sativum) because it had distinct contrasting traits and a short life span.
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Ans: (C) Vas deferens.Explanation: The vas deferens is a duct that transports sperm from the testis; hence, it is a part of the male reproductive system, not the female.
PART - B: Descriptive Questions (Solutions)
Sub-Section - (1) | Physics Long Answers
(a) Between centre of curvature and focus
(b) Between pole and focus.
Solution:
(a) Object placed between C (Centre of curvature) and F (Focus):
- Position of Image: Beyond the centre of curvature (Beyond C).
- Nature of Image: Real, Inverted, and Magnified (larger than the object).
(b) Object placed between P (Pole) and F (Focus):
- Position of Image: Behind the mirror.
- Nature of Image: Virtual, Erect, and Magnified.
Solution:
Since radius of curvature is positive, it is a Convex Mirror.
Height of object (\(h_o\)) = +5.0 cm
Radius of curvature (\(R\)) = +30 cm → Focal length (\(f\)) = R/2 = +15 cm
Object distance (\(u\)) = -20 cm
Using Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{15} + \frac{1}{20} = \frac{4 + 3}{60} = \frac{7}{60}\)
\(v = \frac{60}{7} \approx +8.57\) cm
Position: The image is formed \(8.57\) cm behind the mirror.
Nature: Since \(v\) is positive, the image is Virtual and Erect.
For Size, use Magnification formula: \(m = \frac{h_i}{h_o} = \frac{-v}{u}\)
\(h_i = \frac{15}{7} \approx +2.14\) cm
Size: The height of the image is \(2.14\) cm (diminished).
Solution:
Heat (\(H\)) = 100 Joules
Time (\(t\)) = 1 second
Resistance (\(R\)) = \(4\ \Omega\)
We know the Joule's law of heating formula: \(H = \frac{V^2 \cdot t}{R}\)
\(V^2 = 100 \cdot 4 = 400\)
\(V = \sqrt{400} = 20\text{ Volts}\)
Answer: The potential difference is 20 Volts.
Solution:
Principle: An electric motor works on the principle that when a current-carrying rectangular coil is placed in a magnetic field, it experiences a force (according to Fleming's Left-Hand Rule) which rotates it continuously. It converts electrical energy into mechanical energy.
Working: When current flows from the battery through the brushes into the coil (ABCD), the arm AB experiences a downward force, and the arm CD experiences an upward force due to the magnetic field. This creates a torque, causing the coil to rotate.
Importance of Split Rings (Commutator): The split rings reverse the direction of current in the coil after every half rotation. This ensures that the force acting on the arms remains in the same direction, allowing the coil to rotate continuously in a single direction.
Sub-Section - (2) | Chemistry Long Answers
(a) Methanal: H-CHO
(b) Propyne: \(\text{CH}_3-\text{C} \equiv \text{CH}\)
(c) Pentanone-3: \(\text{CH}_3-\text{CH}_2-\text{CO}-\text{CH}_2-\text{CH}_3\)
(d) Butanone-2: \(\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_3\)
(a) \(\text{Na}_2\text{SO}_4 + \text{BaCl}_2 \rightarrow \text{BaSO}_4 \downarrow + \mathbf{2\text{NaCl}}\) (Double displacement reaction)
(b) \(2\text{HgO} \xrightarrow{\text{Heat}} 2\text{Hg} + \mathbf{\text{O}_2 \uparrow}\) (Thermal decomposition)
(c) \(\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \mathbf{\text{H}_2 \uparrow}\) (Displacement reaction)
(d) \(\mathbf{\text{Fe}} + \text{CuSO}_4 \rightarrow \mathbf{\text{FeSO}_4} + \text{Cu} \downarrow\)
(e) \(\text{CaCO}_3 \xrightarrow{\text{Heat}} \text{CaO} + \mathbf{\text{CO}_2 \uparrow}\)
(f) \(\text{Zn} + 2\text{AgNO}_3 \rightarrow \text{Zn(NO}_3\text{)}_2 + \mathbf{2\text{Ag} \downarrow}\)
Sub-Section - (3) | Biology Long Answers
| Aerobic Respiration | Anaerobic Respiration |
|---|---|
| Occurs in the presence of oxygen. | Occurs in the absence of oxygen. |
| Produces a large amount of energy (38 ATP). | Produces very little energy (2 ATP). |
| Blood | Lymph |
|---|---|
| Red in color due to Hemoglobin (RBCs present). | Colorless or pale fluid (RBCs absent). |
| Flows rapidly through blood vessels. | Flows slowly through lymphatic vessels. |
It is a type of asexual reproduction in plants where new plants grow from vegetative parts like roots, stems, or leaves without the involvement of seeds.
(d) What is the function of heart?The human heart is a muscular pumping organ that pumps oxygenated blood to all parts of the body and receives deoxygenated blood to send it to the lungs for purification.
Conclusion
Dear students, mastering these questions from the UP Board Class 10 Science Solved Paper 2026 Set 824 EJ will definitely elevate your confidence for the board exam. Remember to practice the numericals and diagrams thoroughly. Best of luck!
Frequently Asked Questions (FAQs)
Q1. Will these exact questions appear in the UP Board Exam 2026?
Ans: These questions follow the strict UP Board pattern. While exact duplication isn't guaranteed, practicing them ensures you understand the core concepts likely to be tested.
Q2. Is English medium marking different in UP Board?
Ans: No, the marking scheme is identical for both Hindi and English mediums. Clarity and correct scientific terms fetch full marks.
