Physics 2010 Jamb Examination
Physics 2010 Jamb Examination
Question 41
A hydrometer is an instrument used in measuring
A hydrometer is a floating instrument used to measure the relative density (specific gravity) of liquids. It works on Archimedes’ principle, which states that the weight of the liquid displaced by the immersed part of the hydrometer equals the weight of the instrument. The level to which it sinks indicates the relative density. Liquids with higher densities cause it to float higher, while lighter ones allow it to sink deeper.
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Question 42
One special advantage of alcohol over mercury as a thermometric liquid is its
Alcohol remains liquid at very low temperatures, giving it a low freezing point (around −115°C). This makes it suitable for thermometers used in extremely cold regions, unlike mercury, which freezes at −39°C. Although alcohol expands more uniformly and can measure lower temperatures, it is not ideal for very high temperatures because it boils at around 78°C. Hence, its low freezing point is its greatest advantage over mercury.
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Question 43
The density of a certain oil on frying becomes
0.4 kgm-3 with the volume of 20m3. what will be its initial volume when its initial density is 0.8 kgm-3 assuming no loss of oil due to spillage?
0.4 kgm-3 with the volume of 20m3. what will be its initial volume when its initial density is 0.8 kgm-3 assuming no loss of oil due to spillage?
Mass of the oil remains constant (no spillage). The relationship between density and volume for a fixed mass is:
ρ₁V₁ = ρ₂V₂
Let the initial density be ρ₁ = 0.8 kgm⁻³, final density ρ₂ = 0.4 kgm⁻³ and final volume V₂ = 20 m³. Solving for the initial volume V₁:
V₁ = (ρ₂V₂) / ρ₁ = (0.4 × 20) / 0.8 = 10 m³
Therefore, the initial volume was 10 m³.
ρ₁V₁ = ρ₂V₂
Let the initial density be ρ₁ = 0.8 kgm⁻³, final density ρ₂ = 0.4 kgm⁻³ and final volume V₂ = 20 m³. Solving for the initial volume V₁:
V₁ = (ρ₂V₂) / ρ₁ = (0.4 × 20) / 0.8 = 10 m³
Therefore, the initial volume was 10 m³.
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Question 44
Heat is radiated by hot objects in the form
Hot objects emit electromagnetic radiation across a spectrum determined by their temperature. For ordinary temperatures, the dominant emission is infrared radiation, which carries thermal energy away from the object. Hence, heat is radiated in the form of infrared rays.
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Question 45
A car moves with an initial velocity of 25 ms-1 and reaches a velocity of 45 ms-1 in 10s. what is the acceleration of the car?
Using a = (v - u) / t
a = (45 - 25) / 10 = 2 ms⁻²
Therefore, the car’s acceleration is 2 ms⁻².
a = (45 - 25) / 10 = 2 ms⁻²
Therefore, the car’s acceleration is 2 ms⁻².
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Question 46
The surfaces of conveyor belts are made rough so as to
A rough surface increases friction between the belt and the load, preventing slipping while the belt moves. Thus, the surfaces are made rough to prevent loads from slipping.
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Question 47
A machine of velocity ratio 6 requires an effort of 400 N to raise a load of 800 N through 1 m. Find the efficiency of the machine.
Mechanical advantage (MA) = Load / Effort = 800 / 400 = 2
Efficiency = (MA / VR) × 100 = (2 / 6) × 100 = 33.3%
Hence, the machine’s efficiency is 33.3%.
Efficiency = (MA / VR) × 100 = (2 / 6) × 100 = 33.3%
Hence, the machine’s efficiency is 33.3%.
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Question 48
If a wire 30 cm long is extended to 30.5 cm by a force of 300 N, find the strain energy of the wire.
Extension x = 30.5 - 30 = 0.5 cm = 0.005 m
Strain energy = ½ × F × x = ½ × 300 × 0.005 = 0.75 J
Therefore, the strain energy is 0.75 J.
Strain energy = ½ × F × x = ½ × 300 × 0.005 = 0.75 J
Therefore, the strain energy is 0.75 J.
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Question 49
In a hydraulic press, the pump piston exerts a pressure of 100 Pa on the liquid. What force is exerted in the second piston of cross-sectional area 3 m²?
Force = Pressure × Area = 100 × 3 = 300 N
Hence, the force on the second piston is 300 N.
Hence, the force on the second piston is 300 N.
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Question 50
Two cars moving in the same direction have speeds of 100 kmh-1 and 130 kmh-1. What is the velocity of the faster car as measured by an observer in the slower car?
Relative velocity = 130 - 100 = 30 kmh⁻¹
Thus, the faster car moves at 30 kmh⁻¹ relative to the slower one.
Thus, the faster car moves at 30 kmh⁻¹ relative to the slower one.
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| Entrance fees | N1200 |
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