A tube one metre long is filled with liquid of mass 1 kg. The tube is closed at both the ends and is revolved about one end in a horizontal plane at 2 rev/s. The force experienced by the lid at the other end is

4 π² N
8 π² N
16 π² N
9.8 N

Solution

F = mrω² = 1 × ¹⁄₂ × 4π² × 2 × 2 = 8 π² N

The moment of inertia of a disc of mass M and radius R about an axis, which is tangential to the circumference of the disc and parallel to its diameter, is

³⁄₂ MR²
²⁄₃ MR²
⁵⁄₄ MR²
⁴⁄₅ MR²

Solution

Moment of inertia of disc about its diameter is I_d ¹⁄₄ MR²

MI of disc about a tangent passing through rim and in the plane of disc is

I = I_G + MR² = ¹⁄₄ MR² + MR² = ⁵⁄₄ MR²

A weightless ladder 20 ft long rests against a frictionless wallat an angle of 60º from the horizontal. A 150 pound man is 4 ft from the top of the ladder. A horizontal force is needed to keep it from slipping. Choose the correct magnitude from the following.

175 lb
100 lb
120 lb
17.3 lb

Solution

In carbon monoxide molecule,the carbon and the oxygen atoms are separated by a distance 1.12 × 10^–10 m. The distance of the centre of mass, from the carbon atom is

0.64 × 10^–10 m
0.56 × 10^–10 m
0.51 × 10^–10 m
0.48 × 10^–10 m

ABC is a triangular plate of uniform thickness. The sides are in the ratio shown in the figure. I_AB, I_BC and I_CA are the moments of inertia of the plate about AB, BC and CA as axes respectively. Which one of the following relations is correct?

I_AB > I_BC
I_BC > I_AB
I_AB + I_BC = I_CA
I_CA is maximum

Solution

The intersection of medians is the centre of mass of the triangle. Since distances of centre of mass from the sides are related as : x_BC < x_AB < x_AC therefore I_BC > I_AB > I_AC or I_BC > I_AB

A uniform bar of mass M and length L is horizontally suspended from the ceiling by two vertical light cables as shown. Cable A is connected 1/4th distance from the left end of the bar. Cable B is attached at the far right end of the bar. What is the tension in cable A?

1/4 Mg
1/3 Mg
2/3 Mg
3/4 Mg

Solution

This is a torque problem. While the fulcrum can be placed anywhere, placing it at the far right end of the bar eliminated cable B from the calculation. There are now only two forces acting on the bar; the weight that produces a counterclockwise rotation and the tension in cable At hat produces a clockwise rotation.Since the bar is in equilibrium, these two torques must sum to zero.

∑τ = T_A (3 / 4L) - Mg(1 / 2L) = 0

Therefore

T_A = (MgL / 2)/(3L / 4) = (MgL / 2)(4 / 3L) = 2Mg / 3

A cylinder rolls down an inclined plane of inclination 30°,the acceleration of cylinder is

^g⁄₃
g
^g⁄₂
^2g⁄₃

Solution

Two rings of radius R and nR made up of same material have the ratio of moment of inertia about an axis passing through centre as 1 :8. The value of n is

2
2√2
4
¹⁄₂

Solution

A particle is confined to rotate in a circular path decreasing linear speed, then which of the following is correct?

\(\overrightarrow{L}\)(angular momentum)is conserved about the centre
only direction of angular momentum \(\overrightarrow{L}\) is conserved
It spirals towards the centre
its acceleration is towards the centre.

Solution

Since v is changing (decreasing), L is not conserved in magnitude. Since it is given that a particle is confined to rotate in a circular path, it can not have spiral path.Since the particle has two accelerations a_c and a_t therefore the net acceleration is not towards the centre.

The direction of \(\overrightarrow{L}\) remains same even when the speed decreases.

A particle moves in a circle of radius 4 cm clockwise at constant speed 2 cm s–1. If \(\hat{x}\) and \(\hat{y}\) are unit acceleration vectors along X and Y respectively (in cm s^-2), the acceleration of the particle at the instant half way between P and Q is given by

\(-4(\hat{x}+\hat{y})\)
\(4(\hat{x}+\hat{y})\)
\(-(\hat{x}+\hat{y})/\sqrt{2}\)
\((\hat{x}-\hat{y})/4\)

Solution

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