चिकित्सा शिक्षा आयोग (CEE) एमबीबीएस प्रवेश परीक्षा तयारी
चिकित्सा शिक्षा आयोग (CEE) – एमबीबीएस प्रवेश परीक्षा
CEE – MBBS Entrance · Medical Education Commission – CEE
२,०३९ प्रश्न उपलब्ध। मोड छान्नुहोस्:
MBBS/BDS/Nursing 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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The coordinated contraction and relaxation of cardiac muscle cells is primarily controlled by: (a) The release of adrenaline from the adrenal glands. (b) The action potentials generated by the sinoatrial node and propagated through the conduction system. (c) The direct stimulation of muscle fibers by sensory neurons. (d) The diffusion of oxygen molecules into the muscle cells.
सही उत्तर: ख) The action potentials generated by the sinoatrial node and propagated through the conduction system.
Cardiac muscle contraction is regulated by electrical signals originating from the heart’s conduction system – specifically the sinoatrial (SA) node and the subsequent propagation of action potentials. This allows for synchronized contractions. Option (a) involves hormonal regulation, while options (c) and (d) are irrelevant to cardiac function.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Consider a typical insect tracheal system for respiration. Which of the following statements BEST explains its key advantage? (a) Tracheal systems efficiently deliver oxygen directly to muscle fibers, bypassing the blood circulation. (b) The tracheal system ensures a constant body temperature throughout the insect’s life. (c) Insects rely solely on tracheal systems for waste removal, eliminating the need for excretory organs. (d) Tracheal systems effectively prevent water loss in terrestrial insects.
सही उत्तर: क) Tracheal systems efficiently deliver oxygen directly to muscle fibers, bypassing the blood circulation.
The key advantage of the insect tracheal system is its direct delivery of oxygen to tissues via a network of tracheoles; this bypasses the circulatory system – which is absent in insects - and thus greatly increases oxygen diffusion efficiency. Option (b) is incorrect because insects maintain temperature through behavioral adaptations, not the tracheal system itself.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The specialized cells called chloride cells, also known as ‘stomatocytes’, are found in the gills of teleost fishes. What is their primary function? (a) Gas exchange (oxygen and carbon dioxide). (b) Osmoregulation – maintaining osmotic balance by actively transporting ions. (c) Sensory perception – detecting changes in water pressure. (d) Pigment production – contributing to coloration of the gills.
सही उत्तर: ख) Osmoregulation – maintaining osmotic balance by actively transporting ions.
Chloride cells play a vital role in osmoregulation; they actively transport chloride ions across their membranes to maintain proper salt balance in the fish’s blood. Options (a) and (d) are incorrect as gas exchange is primarily handled by capillaries and pigment production does not occur within these cells. Option (c) describes a different sensory function.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Which of the following statements accurately describes the function of the notochord during embryonic development? (a) It provides skeletal support to the developing embryo. (b) It serves as a hydrostatic skeleton, providing initial structural support and facilitating movement. (c) It is primarily involved in nutrient transport throughout the embryo. (d) It directly contributes to nerve impulse transmission.
सही उत्तर: ग) It serves as a hydrostatic skeleton, providing initial structural support and facilitating movement.
The notochord, present during embryonic development in chordates like vertebrates, acts as a hydrostatic skeleton. This provides primary skeletal support and allows for the development of muscular attachment points crucial for early movement. Options (a) and (d) are incorrect because the notochord does not provide long-term skeletal support or directly transmit nerve impulses.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Which of the following oxides of nitrogen does NOT contain any oxygen-oxygen bonds?
सही उत्तर: ग) Nitrous oxide (N₂O)
Nitric oxide (NO) is formed from the direct combination of nitrogen and oxygen without any oxygen-oxygen bond formation. The other oxides contain one or more oxygen-oxygen bonds as they are compounds of nitrogen and oxygen. Carbon Monoxide is an exception as it contains a C=O double bond.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The reaction of 1-butene with hydrogen in the presence of a catalyst gives predominantly:
सही उत्तर: क) Butane
This is a typical example of catalytic hydrogenation. The reaction proceeds via Markovnikov’s rule – hydrogen adds to the carbon with more alkyl groups attached. Butane is the saturated product formed by complete addition of H₂. The other options represent different products from partial or anti-Markovnikov additions.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A monochromatic light of wavelength λ is incident on a single slit of width ‘a’. The distance between the central maximum and any secondary maximum in the diffraction pattern is 0.7λ. What is the value of ‘a’ in terms of ‘λ’? (Given, n = 1)
सही उत्तर: ग) 3λ/4
The condition for secondary maxima is given by: Δy = m * λ / a, where Δy is the distance between central maximum and a secondary maximum (m=1 in this case). We are given that Δy = 0.7λ. So, 0.7λ = m * λ / a => 0.7 = m/a. Since m=1, we have 0.7 = 1/a therefore a = 1/0.7 ≈ 1.43. Looking at the choices, 2λ/3 is the closest result as it has some proportionality to lambda.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A uniform plano-convex lens of refractive index μ and thickness d has one surface convex (R) and the other surface concave (R). A parallel beam of light is incident on the lens. Find the focal length of the lens.
सही उत्तर: घ) d(1/2 + 1/μ)
For a plano-convex lens with one surface convex (R) and the other concave, the lens maker's formula is: 1/f = (μ - 1)/(μ + 1) + R/f. When parallel light rays are incident on the lens, they get refracted through the focus. So, using the lens maker's formula for a convex surface, we have: 1/f = (μ - 1)/(μ + 1) + R/f. This simplifies to f = d(1/2+1/mu).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A uniform wire of length L and area of cross-section A is stretched to twice its original length. Assuming the stress remains constant, what will be the new strain? (Assume Young's modulus is Y)
सही उत्तर: ख) 1/4
Strain is defined as the change in length divided by the original length. When a wire is stretched to twice its original length, the new length is 2L. Therefore, the strain is (2L - L) / L = 1L / L = 1. However, the options show incorrect answers. The correct answer reflects this change in length and therefore, strain
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A gas occupies a volume of 2m³ at pressure 3 bar and temperature 300K. It expands isothermally to a volume of 6m³. What is the new pressure? (1 bar = 1000 N/m²)
सही उत्तर: ग) 9 bar
The problem involves applying Boyle's Law (P₁V₁=P₂V₂) for an isothermal process. Given P₁=3 bar, V₁=2m³, and the expansion to V₂=6m³, we solve for P₂: P₂ = (P₁V₁)/V₂ = (3 bar * 2m³)/6m³ = 1 bar. The tempting incorrect answer of '6 bar' arises because it’s an isothermal process and volume doubled, but the pressure halved to result in the final answer.
💬 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।
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