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
-
A particle's displacement is x = 3 sin(2t) + 4 cos(2t) (in cm). The amplitude of its motion is:
सही उत्तर: क) 5 cm
Two perpendicular-phase components combine: A = √(3² + 4²) = 5 cm.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A satellite of kinetic energy K moves in a circular orbit. The minimum additional energy needed for it to escape from the Earth's gravity is:
सही उत्तर: ख) K
In a circular orbit the total energy is −K, so raising it to zero (escape) needs an additional energy K.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A uniform rod of mass 2 kg and length 1 m rotates about a perpendicular axis through one end at 6 rad/s. Its rotational kinetic energy is:
सही उत्तर: ग) 12 J
I = ML²/3 = 2/3 kg m², so K = ½Iω² = ½ × (2/3) × 36 = 12 J.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
The minimum speed with which a body must be projected vertically upward from the Earth's surface to reach a height equal to the Earth's radius R is (g = surface gravity):
सही उत्तर: घ) √(gR)
½mv² = GMm/R − GMm/(2R) = GMm/(2R) = mgR/2, so v = √(gR).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A planet has 1/8 of the Earth's mass and 1/2 of its radius. If the escape velocity from the Earth is 11.2 km/s, that from the planet is:
सही उत्तर: ख) 5.6 km/s
vₑ ∝ √(M/R) = √((1/8)/(1/2)) = 1/2, so vₑ = 5.6 km/s.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A satellite close to the Earth's surface has a period of about 84 minutes. A satellite in a circular orbit of radius 4R (R = Earth's radius) has a period of about:
सही उत्तर: घ) 672 minutes
By Kepler's third law T ∝ r^(3/2): T = 84 × 4^(3/2) = 84 × 8 = 672 minutes.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
At what depth below the Earth's surface is the acceleration due to gravity 25% less than at the surface? (R = 6400 km)
सही उत्तर: क) 1600 km
g′ = g(1 − d/R) = 0.75g gives d = R/4 = 1600 km.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A wheel rotating at 600 rpm is brought to rest uniformly after making 50 revolutions. Its angular retardation is:
सही उत्तर: ग) 2π rad/s²
ω₀ = 20π rad/s and θ = 100π rad; α = ω₀²/(2θ) = 400π²/(200π) = 2π rad/s².
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A torque brings a flywheel of moment of inertia 0.5 kg m² from 20 rad/s to rest. The work done by the torque has magnitude:
सही उत्तर: घ) 100 J
W = ΔK = ½Iω² = ½ × 0.5 × 400 = 100 J.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A turntable of moment of inertia 2 kg m² spins freely at 10 rad/s. A 1 kg lump of clay is dropped onto it at 1 m from the axis and sticks. The new angular speed is:
सही उत्तर: ग) 20/3 rad/s
Angular momentum is conserved: 2 × 10 = (2 + 1 × 1²)ω, so ω = 20/3 ≈ 6.67 rad/s.
💬 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।
नमस्ते! पाठ्यक्रम, सामान्य ज्ञान, भाषा परीक्षा वा कुनै प्रश्नको उत्तर बुझ्न सोध्नुहोस्।
AI ले गल्ती गर्न सक्छ। महत्वपूर्ण कुरा आधिकारिक स्रोतमा जाँच्नुहोस्।