AMU BSc PYQ 2026 | AMU BSc Previous Year Question with explanation - Page 15 of 20

71. The time period of revolution of a satellite around a planet in a circular orbit of radius R is T. If the radius of the satellite is made 4R then its time period would become:

(a) T

(b) $T^2$

(d) 8T

Answer:

Explanation:

According to Kepler’s Third Law,

T^2 \propto R^3

. When the radius is increased to

4R

, the new period squared

(T’)^2 \propto (4R)^3 = 64R^3

. Thus,

T’ = \sqrt{64}T = 8T

72. The speeds of air flow on the upper and lower surfaces of a wing of an aeroplane are $V_1$ and $V_2$ , respectively. If A is the cross-sectional area of the wing and $\rho$ is the density of air, then the upward lift is :

(a) $\frac{1}{2}\rho A(V_1+V_2)(V_1+V_2)$

(b) $\frac{1}{2}\rho A(V_1+V_2)(V_1-V_2)$

(d) $\frac{1}{2}\rho A(V_1^2+V_2^2)$

Answer:

(b)

\frac{1}{2}\rho A(V_1+V_2)(V_1-V_2)

Explanation:

Using Bernoulli’s principle, the pressure difference creates lift:

\Delta P = \frac{1}{2}\rho (V_1^2 – V_2^2)

. Upward force

F = A \times \Delta P = \frac{1}{2}\rho A (V_1^2 – V_2^2)

, which factors out algebraically into option (b).

73. A liquid drop of radius R breaks into 64 tiny drops each of radius r. If the surface tension of the liquid is T, then the gain in energy is :

(a) $48 \pi R^2T$

(b) $48 \pi r^2T$

(d) $96 \pi r^2T$

Answer:

(c)

192 \pi r^2T

Explanation:

By conservation of volume,

R^3 = 64r^3

, so

R = 4r

. Initial surface area is

4\pi R^2 = 64\pi r^2

. Final surface area of 64 drops is

64 \times 4\pi r^2 = 256\pi r^2

. The increase in area is

192\pi r^2

. Gain in surface energy =

192\pi r^2 T

74. The P-V diagram of a system undergoing thermodynamic transformation is shown in figure. The work done by the system in going from $A \rightarrow B \rightarrow C$ is 30 J, and 40 J heat is given to the system. The change in internal energy between A and C is :
https://canva.link/bik148drmqm0jtn

(a) 10 J

(b) 70 J

(d) Zero

Answer:

Explanation:

According to the First Law of Thermodynamics,

\Delta U = Q – W

. The heat added to the system

Q = 40

J and work done by the system

W = 30

J. Thus,

\Delta U = 40 – 30 = 10

75. If a cylinder containing a gas at high pressure explodes, the gas undergoes:

(a) irreversible adiabatic change and fall in temperature

(b) reversible adiabatic change and rise in temperature

(d) reversible adiabatic change and fall in temperature

Answer:

Explanation:

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AMU BSc PYQ 2026 | AMU BSc Previous Year Question with explanation

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