मेडिकल प्रवेश (CEE) – Physics, Chemistry, Zoology, Botany, MAT
मेडिकल प्रवेश (CEE) – Physics, Chemistry, Zoology, Botany, MAT
Medical Entrance (CEE) – Physics, Chemistry, Biology & MAT · २,०३९ प्रश्न
२,०३९ प्रश्न उपलब्ध। मोड छान्नुहोस्:
CEE 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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Which of the following represents a characteristic feature of cnidarians (e.g., jellyfish)? (a) A closed circulatory system with fully formed blood vessels. (b) Radial symmetry and specialized stinging cells called nematocysts. (c) Segmented body plan with paired appendages for locomotion. (d) Bilateral symmetry with a defined head region and complex nervous system.
सही उत्तर: ख) (b) Radial symmetry and specialized stinging cells called nematocysts.
Cnidarians are characterized by radial symmetry and the presence of nematocysts (stinging cells), found in their tentacles. A closed circulatory system is not typical (a); segmentation (c) and bilateral symmetry (d) are incorrect.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A beam of light passes from air into glass at an angle of incidence of 30 degrees. The refractive index of glass is 1.5. What is the angle of refraction?
सही उत्तर: ग) (c) 60 degrees
Snell's law states: n₁sinθ₁ = n₂sinθ₂, where n₁ and n₂ are the refractive indices of air and glass, respectively, and θ₁ and θ₂ are the angles of incidence and refraction. In this case, n₁ = 1, n₂ = 1.5, and θ₁ = 30°. Therefore, sinθ₂ = (n₂/n₁)sinθ₁ = (1.5/1)sin30° = 1.5 * 0.5 = 0.75. Thus, θ₂ = arcsin(0.75) ≈ 45 degrees. Option (a) is incorrect because it’s half the correct angle.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The magnetic field at a point P due to an infinite current-carrying wire is perpendicular to the direction of the wire. If the distance between the wire and the point P is doubled, what happens to the magnetic field strength at point P?
सही उत्तर: घ) (d) Decreases by a factor of 2
The magnetic field strength due to an infinite wire is given by B = μ₀I/r, where r is the distance from the wire. If the distance ‘r’ doubles (becomes 2r), then the new magnetic field strength becomes B' = μ₀I/(2r) = (1/2) * (μ₀I/r) = (1/2)B. Therefore, the magnetic field strength decreases by a factor of 2. Option (a) is incorrect as it doesn't account for distance.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A capacitor consists of two parallel plates separated by an air gap. If the dielectric material between the plates is replaced with a rubber sheet having a dielectric constant of k = 3, what will be the capacitance?
सही उत्तर: घ) (d) Depends on the area and distance only
When a dielectric material with a dielectric constant ‘k’ is inserted between the plates of a capacitor, the capacitance increases. The new capacitance (C’) becomes C’ = k * C, where C is the original capacitance. Thus, the correct option is (c). Using the formula C=ε₀A/d, we replace ε₀ with ε₀k to find the increased capacitance.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The period of oscillation of a simple pendulum is given by T = 2π√(L/g), where L is the length and g is the acceleration due to gravity. If the length of the pendulum is increased by 20% and the gravitational acceleration remains constant, what is the percentage change in its period?
सही उत्तर: ख) (b) 20% decrease
The new period will be T' = 2π√(L’/g), where L’ = 1.2L. Therefore, T’/T = √(1.2/1) = √1.2 ≈ 1.095. This represents a 9.5% increase over the original period, closest to 10%. Hence the correct answer is (a). Option (b) is incorrect as it would result in a decrease.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A particle undergoing simple harmonic motion has a maximum velocity at a point where its displacement from equilibrium is zero. This is because…
सही उत्तर: ख) (b) The force acting on the particle is always perpendicular to its velocity vector
The correct answer is (b). In Simple Harmonic Motion (SHM), the restoring force is proportional to the displacement and acts along the direction of velocity. Therefore, the force is always perpendicular to the velocity vector, leading to maximum velocity at the equilibrium point. Option (a) is incorrect because velocity is inversely proportional to displacement in SHM. Options (c) and (d) describe energy states but don't explain the specific condition for maximum velocity.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Find the next digit in the series: 2, 6, 18, 54, 162, ?
सही उत्तर: क) 384
The pattern is multiplication by 3. 2 * 3 = 6; 6 * 3 = 18; 18 * 3 = 54; 54 * 3 = 162; therefore, the next number is 162 * 3 = 486.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A current-carrying wire is placed in a uniform magnetic field. What is the direction of the force experienced by the wire?
सही उत्तर: ख) Always perpendicular to both the magnetic field and the current
The force on a current-carrying wire in a magnetic field is given by F = I L x B, where I is the current, L is the length of the wire, and B is the magnetic field. The cross product indicates that the force is always perpendicular to both the current and the magnetic field. This means it’s at 90 degrees to both.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A uniform magnetic field of 0.4 Tesla is applied perpendicular to the plane of a circular coil with a radius of 10 cm and 100 turns. If the frequency of the AC source driving the current in the coil is 50 Hz, what is the induced emf in one turn of the coil?
सही उत्तर: क) 20 cos(100πt) V
The induced emf in a coil is given by Faraday’s Law: ε = -N dΦ/dt, where N is the number of turns, Φ is the magnetic flux, and dt is the time interval. The magnetic flux is Φ = B A cos(ωt), so ε = -BAncos(ωt). For a frequency of 50 Hz, ω = 2π * 50 = 100π rad/s. With N=100 turns, r=10cm (0.1m) and B=0.4T, ε = -0.4 * 100 * cos(100πt) V = -40cos(100πt)V which is equal to 20cos(100πt) V in the options. The most tempting option is 40V because it gives a larger magnitude. However, this value represents the instantaneous EMF and not the RMS value.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In plants, what is the primary significance of CAM (Crassulacean Acid Metabolism) photosynthesis? (a) Rapid carbon dioxide uptake during daylight hours (b) Water conservation in arid environments (c) Increased sugar production for rapid growth (d) Enhanced chlorophyll synthesis under intense sunlight
सही उत्तर: ख) (b) Water conservation in arid environments
CAM plants initially fix CO2 at night to minimize water loss, then use that stored CO2 for photosynthesis during the day. This adaptation is crucial in drought-prone conditions. (a), (c), and (d) describe processes not directly related to the CAM strategy.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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