The process of combining two light nuclei to form a heavier nucleus is MCQ with Answer and Explanation
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Question #1
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When a small spherical metal body falls through a highly viscous liquid, it attains a terminal velocity. This terminal velocity is directly proportional to:
Correct Answer: Option D
Explanation:
According to Stokes' Law, the terminal velocity (Vt) of a spherical body falling through a viscous medium is given by Vt = 2r^2(rho - sigma)g / 9(eta). Here, r is the radius. The formula clearly shows that the terminal velocity is directly proportional to the square of the radius (r^2).
According to Stokes' Law, the terminal velocity (Vt) of a spherical body falling through a viscous medium is given by Vt = 2r^2(rho - sigma)g / 9(eta). Here, r is the radius. The formula clearly shows that the terminal velocity is directly proportional to the square of the radius (r^2).
Question #2
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The velocity-time graph of a particle moving in a straight line is shown. The displacement of the particle from t=0 to t=4s is: [Graph: triangle from (0,0) to (2,10) to (4,0)]
Correct Answer: Option A
Explanation:
Displacement = area under velocity-time graph. The graph forms a triangle with base 4 s and height 10 m/s. Area = ½ × base × height = ½ × 4 × 10 = 20 m. Since velocity is always positive, displacement equals distance travelled. Graphical analysis is powerful: area gives displacement, slope gives acceleration. This triangle represents motion with uniform acceleration followed by uniform retardation. Exam tip: For piecewise linear v-t graphs, calculate area of each geometric segment separately. Such graph-based questions assess conceptual clarity in motion analysis.
Displacement = area under velocity-time graph. The graph forms a triangle with base 4 s and height 10 m/s. Area = ½ × base × height = ½ × 4 × 10 = 20 m. Since velocity is always positive, displacement equals distance travelled. Graphical analysis is powerful: area gives displacement, slope gives acceleration. This triangle represents motion with uniform acceleration followed by uniform retardation. Exam tip: For piecewise linear v-t graphs, calculate area of each geometric segment separately. Such graph-based questions assess conceptual clarity in motion analysis.
Question #3
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One Parsec is a unit of astronomical distance. It is defined as the distance at which a radius of one astronomical unit (AU) subtends an angle of one second of arc. One parsec is approximately equal to:
Correct Answer: Option A
Explanation:
The parsec (parallax second) is the largest standard unit of distance used in astronomy. 1 Astronomical Unit (AU) is approx 1.5 x 10^11 m. 1 Light Year (ly) is approx 9.46 x 10^15 m. 1 Parsec is approximately 3.08 x 10^16 m. Converting parsecs to light years: (3.08 x 10^16) / (9.46 x 10^15) ≈ 3.26 light years.
The parsec (parallax second) is the largest standard unit of distance used in astronomy. 1 Astronomical Unit (AU) is approx 1.5 x 10^11 m. 1 Light Year (ly) is approx 9.46 x 10^15 m. 1 Parsec is approximately 3.08 x 10^16 m. Converting parsecs to light years: (3.08 x 10^16) / (9.46 x 10^15) ≈ 3.26 light years.
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