Haryana Board (HBSE) Class 11 Physics SAT-2 Question Paper 2025 Answer Key
Instructions :
• All questions are compulsory.
• Questions (1-8) carry 1 mark each.
• Questions (9-10) carry 2 marks each.
• Question (11) carry 3 marks.
• Question (12) carry 5 marks.
1. SI unit of acceleration due to gravity is :
(a) m/s
(b) m/s2
(c) Nm
(d) Kgm/s
Answer – (b) m/s2
2. Name of the place on Earth where g is maximum :
(a) pole
(b) focus
(c) equator
(d) all of the above
Answer – (a) pole
3. SI unit of stress is :
(a) Newton
(b) Joule
(c) Pascal
(d) Watt
Answer – (c) Pascal
4. What is the dimension of thrust?
(a) MLT
(b) MLT–2
(c) MLT2
(d) MLT3
Answer – (b) MLT–2
5. Principal of transmission of pressure is based on :
(a) Stoke’s law
(b) Bernouli’s law
(c) Newton’s law
(d) Pascal’s law
Answer – (d) Pascal’s law
6. Kepler’s law deals with …………. motion.
Answer – planetary
7. Hooke’s law state that within elastic limit stress is proportional to :
Answer – Strain
8. Value of Poisson’s ratio lies between :
Answer : –1 to 0.5
9. Explain why the blood pressure in humans is greater at the feet than at the brain?
Answer – The height of the blood column is quite larger at feet than at the brain. Consequently the blood pressure in human is greater at the feet than at the brain.
OR
Explain why the angle of contact of Mercury with glass is obtuse while that of water with glass is acute?
Answer – Mercury does not wet glass, so cohesive forces are greater than adhesive forces, hence angle of contact is obtuse. Water wets glass, so adhesive forces are greater, hence angle of contact is acute.
10. The climate of a harbour town is more temperate than that of a town in a desert at the same latitude?
Answer – The relative humidity of a harbour town is more than that of a desert town. Due to high specific heat of water the variation in the temperature of humid air is less. Hence the climate of a harbour town is without the extremes of hot or cold.
OR
Molar volume is the volume occupied by one mole of any ideal gas at standard temperature and pressure. Show that it is 22.4 litre.
Answer – At STP,
Temperature, T = 273 k
Pressure, P = 1 atm
Using ideal gas equation,
PV = nRT
For one mole of gas (n = 1),
V = RT/P = (8.314 × 273) / (1.013 × 105) = 22.4 L
Molar volume at STP = 22.4 litre
11. Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
Answer – Kepler’s third law, T2 ∝ R3
If the planet goes around the Sun twice as fast as Earth, Tp = Te/2
(Tp/Te)2 = (Rp/Re)3
(1/2)2 = (Rp/Re)3
Rp/Re = (1/2)2/3 = 0.63
Rp = 0.63 Re
OR
Compute the fractional change in volume of a glass slab when subjected to a hydraulic pressure of 10 ATM.
Answer – Fractional change in volume is given by ∆V/V = P/K
Where,
P is applied pressure
R is bulk modulus of elasticity of glass
For glass, K = 5 × 1010 Nm–2
Pressure applied,
P = 10 atm = 10 × 1.013 × 105 = 1.013 × 106 Nm–2
∆V/V = (1.013 × 106) / (5 × 1010) = 2.026 × 10–5
Hence, the fractional change in volume of the glass slab is 2.026 × 10–5.
12. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance when displaced and released, oscillates with a period of 0.6 seconds. What is the weight of the body?
Answer – Maximum reading of spring balance = 50 kg
Length of scale = 20 cm = 0.20 m
Force corresponding to 50 kg,
F = mg = 50 × 9.8 = 490 N
So, spring constant:
k = F/x = 490/0.20 = 2450 Nm–1
Time period of oscillation, T = 0.6 s
We know, T = 2π√m/k
m = T2k/4π2
m = (0.6 × 0.6 × 2450) / (4 × 3.14 × 3.14) = 22.36 kg
Weight of the body,
W = mg = 22.36 × 9.8 = 219.13 N
OR
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25 m³ at a temperature of 27° celsius and one atm pressure.
Answer – Air is treated as an ideal gas. Using the relation
PV = NkT
Given,
V = 25 m3
T = 27° C = 300 K
P = 1.013 × 105 Nm–2
k = 1.38 × 10–23 JK–1
N = PV/kT = (1.013 × 105 × 25) / (1.38 × 10–23 × 300) = 6.117 × 1026
Thus, the total number of air molecules in the room is 6.117 × 1026.