भौतिक विज्ञान (Physics) – Physics MCQ
भौतिक विज्ञान (Physics) – इन्जिनियरिङ प्रवेश
Physics · Engineering Entrance – Mathematics, Physics, Chemistry & English · १,३९८ प्रश्न
१,३९८ प्रश्न उपलब्ध। मोड छान्नुहोस्:
भौतिक विज्ञान (Physics): 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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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)
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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)
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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)
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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)
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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)
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Four particles, each of mass 1 kg, are at the corners of a square of side 2 m. Their moment of inertia about an axis through the centre perpendicular to the square is:
सही उत्तर: ख) 8 kg m²
Each particle is at distance a/√2 = √2 m from the centre, so I = 4 × 1 × 2 = 8 kg m².
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A conical pendulum has a string of length 0.8 m making an angle of 60° with the vertical. Its period of revolution is (g = 10 m/s²):
सही उत्तर: क) 0.4π s
T = 2π√(L cos θ/g) = 2π√(0.8 × 0.5/10) = 2π√0.04 = 2π(0.2) = 0.4π s.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A car of mass 1000 kg goes over the top of a hump of radius 40 m at 10 m/s. The normal force on the car at the top is (g = 10 m/s²):
सही उत्तर: ग) 7500 N
mg − N = mv²/r ⇒ N = 10000 − 1000 × 100/40 = 10000 − 2500 = 7500 N.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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For a stone of mass m whirled in a vertical circle on a string, the difference between the tensions at the lowest and highest points is:
सही उत्तर: घ) 6mg
T_low − T_high = m(v_low² − v_high²)/r + 2mg = m(4gr)/r + 2mg = 6mg, using energy conservation v_low² − v_high² = 4gr.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 2 kg block sliding at 10 m/s on a horizontal floor comes to rest because of friction (μ = 0.5). The distance it slides is (g = 10 m/s²):
सही उत्तर: क) 10 m
Work–energy: μmg·s = ½mv², so s = v²/(2μg) = 100/10 = 10 m.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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