इन्जिनियरिङ प्रवेश – Mathematics, Physics, Chemistry, English
इन्जिनियरिङ प्रवेश – Mathematics, Physics, Chemistry, English
Engineering Entrance – Mathematics, Physics, Chemistry & English · ३,८४१ प्रश्न
३,८४१ प्रश्न उपलब्ध। मोड छान्नुहोस्:
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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For a perfectly inelastic collision, the coefficient of restitution is:
सही उत्तर: घ) 0
In a perfectly inelastic collision the bodies move together after impact, so the relative speed of separation and hence e is zero.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Two bodies of masses 1 kg and 4 kg have equal kinetic energies. The ratio of their linear momenta is:
सही उत्तर: ग) 1 : 2
p = √(2mK); with equal K, p₁/p₂ = √(1/4) = 1/2.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The work done by a conservative force on a particle moving around any closed path is:
सही उत्तर: ख) zero
For a conservative force the work depends only on the end points, so for a closed path (same start and end) it is zero.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A car engine exerts a forward force of 2000 N while the car moves at a steady 20 m/s. The power delivered is:
सही उत्तर: क) 40 kW
P = F v = 2000 × 20 = 40000 W = 40 kW.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 5 kg block on a horizontal floor (μ_k = 0.2) is pulled by a horizontal force of 20 N. Its acceleration is (g = 10 m/s²):
सही उत्तर: घ) 2 m/s²
Kinetic friction = 0.2 × 5 × 10 = 10 N; net force = 20 − 10 = 10 N, so a = 10/5 = 2 m/s².
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A man of mass m stands on a weighing machine in a lift moving upward with constant velocity. The machine reads:
सही उत्तर: ग) mg
With constant velocity there is no acceleration, so the normal reaction equals mg.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A constant force of 10 N acts on a 2 kg body initially at rest for 3 s. Its final speed is:
सही उत्तर: ख) 15 m/s
Impulse = change in momentum: 10 × 3 = 2v, so v = 15 m/s.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A bob of a conical pendulum hangs by a string of length L making an angle θ with the vertical. Its time period is:
सही उत्तर: क) 2π√(L cos θ/g)
T cos θ = mg and T sin θ = mω²L sin θ give ω² = g/(L cos θ), so the period is 2π√(L cos θ/g).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The propulsion of a rocket is based on the principle of conservation of:
सही उत्तर: घ) linear momentum
The momentum gained by the rocket forwards equals that carried backward by the exhaust gases, so total linear momentum is conserved (Newton's third law).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The area under a force–time graph represents:
सही उत्तर: ग) impulse
Area under F–t graph = ∫F dt = impulse, which equals the change in momentum.
💬 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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