Water flows through a horizontal tube as shown. The difference in heights of water columns is 5 cm. Areas of cross-section at points A and B are 6 cm² and 3 cm² respectively. Find the rate of flow of water. (g = 10 m/s²)

Options:

$( \dfrac{200}{3} )$ cm³/s
$( \dfrac{200}{6} )$ cm³/s
$( \dfrac{200}{3} )$ cm³/s
$( \dfrac{100}{3} )$ cm³/s

Solution:

From continuity equation: $[ 6V_A = 3V_B \Rightarrow V_B = 2V_A ]$ Using Bernoulli’s equation: $[ \rho gh = \frac{1}{2}\rho(V_B^2 - V_A^2) ]$ Solving: $[ V_A = \frac{100}{3}\ \text{cm/s} ]$ Flow rate: $[ Q = A_AV_A = \frac{200}{3}\ \text{cm}^3/s ]$ ✅ Correct Answer: (3)

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IAT 2025 Question Paper

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The magnitudes of power of a biconvex lens (refractive index 1.5) and a plano-concave lens (refractive index 1.7) are equal. If the curvature of the concave surface of the plano-concave lens exactly matches the curvature of the back surface of the biconvex lens, find the ratio of radii of curvature of the front and back surfaces of the biconvex lens

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