CTEVT डिप्लोमा इन्जिनियरिङ डिप्लोमा इन सिभिल/कम्प्युटर/इलेक्ट्रिकल प्रवेश तयारी
CTEVT डिप्लोमा इन्जिनियरिङ – डिप्लोमा इन सिभिल/कम्प्युटर/इलेक्ट्रिकल प्रवेश
CTEVT Diploma Entrance · CTEVT Diploma in Engineering
१,५७१ प्रश्न उपलब्ध। मोड छान्नुहोस्:
CTEVT डिप्लोमा इन्जिनियरिङ प्रवेश: प्रश्न र उत्तर
सही लागेको विकल्पमा थिच्नुहोस्। सही भए हरियो, गलत भए रातो देखिन्छ; ३ सेकेन्डपछि सही उत्तर र व्याख्या देखिन्छ।
सही: ० / ०
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If the velocity of a moving body is doubled (mass unchanged), its kinetic energy becomes:
सही उत्तर: ख) Four times
KE = ½mv², so KE ∝ v²; doubling v gives 2² = 4 times. Momentum, not KE, would only double.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The kinetic energy of a 2 kg body moving at 10 m/s is:
सही उत्तर: ख) 100 J
KE = ½mv² = ½ × 2 × 10² = 100 J. 200 J results from forgetting the ½.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A boy lifts a 50 kg load to a height of 4 m in 10 s (g = 10 m/s²). His power is:
सही उत्तर: घ) 200 W
Work = mgh = 50 × 10 × 4 = 2000 J; Power = 2000/10 = 200 W. 2000 is the work in joules, not the power.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A stone weighs 10 N in air and 6 N when fully immersed in water. Its relative density is:
सही उत्तर: क) 2.5
Relative density = weight in air / loss of weight in water = 10/4 = 2.5. 1.67 comes from wrongly dividing 10 by 6.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A stone weighs 10 N in air and 6 N when fully immersed in water. The upthrust acting on it is:
सही उत्तर: घ) 4 N
Upthrust = loss in weight = 10 − 6 = 4 N. By Archimedes' principle this equals the weight of water displaced.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In a hydraulic press, a force of 50 N is applied on a small piston of area 0.01 m². The large piston has an area of 0.5 m². The force on the large piston is:
सही उत्तर: घ) 2500 N
By Pascal's law F₂ = F₁ × A₂/A₁ = 50 × 0.5/0.01 = 2500 N. The pressure is same on both pistons, so the larger area gets the larger force.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The pressure due to water alone at a depth of 10 m (density 1000 kg/m³, g = 10 m/s²) is:
सही उत्तर: ख) 1 × 10⁵ Pa
P = hρg = 10 × 1000 × 10 = 1 × 10⁵ Pa. This is roughly equal to one atmospheric pressure.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A force of 200 N acts uniformly on an area of 0.5 m². The pressure produced is:
सही उत्तर: क) 400 Pa
P = F/A = 200/0.5 = 400 Pa. Multiplying (200 × 0.5 = 100) is a common mistake.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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If the masses of both bodies are doubled while the distance between them is kept the same, the gravitational force becomes:
सही उत्तर: ख) Four times
F ∝ m₁m₂, so (2m₁)(2m₂) = 4m₁m₂ and the force becomes 4 times. Doubling only one mass would double the force.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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If the distance between two bodies is doubled, the gravitational force between them becomes:
सही उत्तर: ख) One-fourth
F ∝ 1/d², so doubling d makes F = F/2² = F/4. 'Half' would be true only if F ∝ 1/d.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
यी अभ्यास प्रश्न हुन्, आधिकारिक प्रश्न होइनन्। पाठ्यक्रम, विज्ञापन र नतिजाका लागि सम्बन्धित निकायको आधिकारिक वेबसाइट हेर्नुहोस्। उत्तरमा गल्ती भेटे प्रश्नमुनिको "टिप्पणी / गल्ती रिपोर्ट" मा लेख्नुहोस्।
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