AMU BSc PYQ 2026 | AMU BSc Previous Year Question with explanation - Page 14 of 20

66. A bus covers the first half of the distance between two places at 40 km/h and another half at 60 km/h. The average speed of the bus is :

(a) 50 km/h

(b) 48 km/h

(d) 45 km/h

Answer:

Explanation:

When equal distances are covered at two different speeds, the average speed is found using the harmonic mean formula:

v_{avg} = \frac{2v_1v_2}{v_1+v_2} = \frac{2(40)(60)}{40+60} = \frac{4800}{100} = 48 \text{ km/h}

67. The force required to just move a body up an inclined plane is double the force required to just prevent it from sliding down. If $\phi$ is the angle of friction and $\theta$ is the angle which the plane makes with horizontal, then :

(a) $\tan\phi = 2\tan\theta$

(b) $\tan\theta = 2\tan\phi$

(d) $\tan\phi = 3\tan\theta$

Answer:

(c)

\tan\theta = 3\tan\phi

Explanation:

The upward force needed is

F_1 = mg(\sin\theta + \mu\cos\theta)

. The force to prevent sliding down is

F_2 = mg(\sin\theta – \mu\cos\theta)

. Given

F_1 = 2F_2

, substituting gives

\sin\theta + \mu\cos\theta = 2\sin\theta – 2\mu\cos\theta

, leading to

\sin\theta = 3\mu\cos\theta

. Since

\mu = \tan\phi

, we get

\tan\theta = 3\tan\phi

68. A particle acted upon by constant forces $\vec{F}_1=(-2\hat{i}+3\hat{j}-4\hat{k})$ and $\vec{F}_2=(\hat{i}-2\hat{j}+3\hat{k})$ is displaced from the coordinate point A $(2,1,0)$ to the coordinate point B $(-3,-4,-3)$ . The total work done by the forces in same units is :

(a) 2 units

(b) 2.5 units

(d) 4 units

Answer:

Explanation:

The net force

\vec{F} = \vec{F}_1 + \vec{F}_2 = -\hat{i} + \hat{j} – \hat{k}

. The displacement vector

\Delta\vec{r} = B – A = -5\hat{i} – 5\hat{j} – 3\hat{k}

. Work

W = \vec{F} \cdot \Delta\vec{r} = (-1)(-5) + (1)(-5) + (-1)(-3) = 5 – 5 + 3 = 3

units.

69. A force $\vec{F}=C\hat{i}+3\hat{j}+6\hat{k}$ is acting at a point $\vec{r}=2\hat{i}-6\hat{j}-12\hat{k}$ . The value of C for which angular momentum about origin is conserved is :

(a) m

(b) 2

(d) -2

Answer:

Explanation:

For angular momentum to be conserved, torque must be zero (

\vec{\tau} = \vec{r} \times \vec{F} = 0

). Calculating the cross product yields

\vec{\tau} = 0\hat{i} – (12+12C)\hat{j} + (6+6C)\hat{k} = 0

. Solving

12+12C=0

gives

C = -1

70. A planet has mass $\frac{1}{8}$ of the mass of earth but the average density of the planet is same as that of earth. If the acceleration due to gravity at the surface is $g_p$ and $g_e$ respectively for the planet and earth, then :

(a) $g_p = 2g_e$

(b) $g_p = g_e / 2$

(d) $g_p = g_e / 4$

Answer:

(b)

g_p = g_e / 2

Explanation:

Since density is constant, mass

M \propto R^3

, meaning

R \propto M^{1/3}

. Gravity

g \propto M/R^2 \propto M / M^{2/3} = M^{1/3}

. Given the planet has 1/8 the mass of Earth,

g_p / g_e = (1/8)^{1/3} = 1/2

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

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