WBCHSE Class 11 Physics Statics Long Answer Questions

Statics Long Answer Type Questions

Question 1. Describe the moment of a force about a point as a vector quantity.
Answer:

O is the point of rotation, P is the point of application of a force \(\vec{F}\), \(\vec{r}=\overrightarrow{O P}\), position vector of the point \(\vec{P}\), with respect to the point O.

Then, the moment of the force \(\vec{F}\) about the point O, is \(\vec{G}\) = \(\vec{r}\) x \(\vec{F}\)

According to the properties of the cross product, the direction of the vector \(\vec{G}\) is along the direction of advance of a righthanded screw, when it is rotated from \(\vec{r}\) towards \(\vec{F}\). The direction of \(\vec{G}\), as shown in the figure, is perpendicular to the plane of rotation, i.e., always along the axis of rotation.

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Question 2. What is the significance of a moment o fa force about a point?
Answer:

The moment of a force on a body about a point is an external influence that can rotate the body about that point. Higher torque would produce more rotation if the body remains the same.

Question 3. It is easier to open or close a door by pushing it at the edges than by pushing it closer to the hinges. Explain.
Answer:

A door is fitted to the door frame using hinges. Opening or closing of a door means a rotation about these hinges. The application of a higher moment of force will facilitate this rotation. If the force applied remains the same, the moment of force is higher when the arm is longer. So the door is usually pushed farthest from the hinges, i.e., at the edges.

WBBSE Class 11 Statics Long Answer Questions

Question 4. State whether three forces of magnitude 1 dyn, 2 dyn and 3 dyn acting simultaneously on a body can keep it In equilibrium.
Answer:

If 1 dyn and 2 dyn forces act on a body in parallel, then the resultant force on the body becomes (1+2)dyn or 3 dyn. Now if the given 3 dyn force acts on the body along the same line as the resultant force but in the opposite direction to, then the resultant force on the body becomes zero. Hence the body will be in equilibrium.

Question 5. A bomb Is thrown vertically upwards. At its topmost position, it explodes into a number of fragments. What would be the locus of the centre of gravity of the bomb?
Answer:

Forces developed due to an explosion are internal forces and therefore the motion of the centre of gravity does not get affected. Hence, the centre of gravity would come down vertically like a freely falling body.

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Question 6. A ping pong ball is floating at the vertex of a vertical fountain. State whether the equilibrium of the ball is vertically stable, unstable or neutral.
Answer:

The equilibrium of the ball, in the vertical direction, is stable. At the top of the fountainhead, the thrust of water acts vertically upwards and the weight of the ball acts downwards. The two forces acting on the ball, being equal and opposite, keep the ball in equilibrium at A.

• If the ball is lifted upward a little to B, the thrust is less than the weight and so the ball returns to the equilibrium position A.
• On the other hand, if the ball is taken down to position C, the thrust being greater than the weight of the ball, pushes it up to the equilibrium position A . Hence, the pingpong ball is in a stable equilibrium.

Step-by-Step Solutions to Statics Problems

Question 7. A rectangular parallelepiped of mass m has three sides of length l, 2l and 3l respectively. State with reason, in which position the block should be kept at rest on a horizontal floor so as to attain the highest stability of equilibrium.
Answer:

The longer sides 2l and 3l should form the horizontal base of the parallelepiped. In this position, maximum stability of equilibrium can be achieved. Centre of gravity is at its lowest possible position, i.e., at a height of \(\frac{l}{2}\) from the base and the base area is maximum.

Question 8. Two particles, initially at rest, are approaching each other due to their mutual force of attraction. What is the velocity of the centre of mass of the system when the particles have a relative velocity v?
Answer:

For a two-particle system, the mutual attraction is an internal force and has no effect on the motion of the centre of mass. Hence, the velocity of the centre of mass is zero.

Question 9. A piece of stone thrown vertically upwards comes to rest momentarily at its maximum height. Is the stone in equilibrium at this position?
Answer:

The stone, though momentarily at rest at the maximum height, is not in equilibrium there. It has a downward acceleration, due to the earth’s gravitational force.

Question 10. Should there be any change in the position of the centre of gravity of a hollow sphere when it is

1. Half filled,
2. Completely filled with water?

Answer:

The centre of gravity of an empty hollow sphere is at its centre.

1. When it is half filled with water, the centre of gravity is lowered.
2. When it is completely filled with water the centre of gravity again shifts to the centre.

Question 11. It is easier to stand on two legs than to stand on a single leg. Why?
Answer:

A body remains in stable equilibrium when the vertical line through its centre of gravity remains well within its base. Equilibrium is disturbed when this vertical line falls outside the base. It is difficult to maintain the vertical line passing through one foot only.

This reduces the stability while standing on one leg. Standing on two legs increases the base area through which the vertical line is more likely to pass and hence, a higher stability of equilibrium is achieved.

Question 12. Passengers on a boat should not be allowed to stand up on the boat while crossing a river. Why?
Answer:

The centre of gravity of a boat with its passengers should be as low as possible, so that it may always be in a stable equilibrium. But if the passengers stand up on the boat, the centre of gravity of the system also moves up. The system then attains an unstable condition. Then, even for a very small tilting, the boat may capsize. So, the passengers on a boat are not allowed to stand up while crossing a river.

Question 13. Two arms of a common balance are unequal. How can the balance be used to find the correct mass of an object?
Answer:

Suppose the correct mass of the object is m and the lengths of the left and right arms of the balance are x and y respectively. At first, the object is kept on the left pan and the counterpoising weight m₁ on the right pan.

Hence, \(m x=m_1 y \text { or, } \frac{m}{m_1}=\frac{y}{x}\)…(1)

Next, the object is kept on the right pan and the counterpoising weight m₂ on the left pan.

Hence, \(m_2 x=m y \text { or, } \frac{m}{m_2}=\frac{y}{x}\)…(2)

From equations (1) and (2) we get, \(\frac{m}{m_1}=\frac{m_2}{m} \text { or, } m^2=m_1 m_2 \text { or, } m=\sqrt{m_1 m_2}\)

Hence, the true mass m of the object can be calculated.

Real-Life Examples of Static Equilibrium

Question 14. A businessman uses a faulty balance of unequal arms of lengths x and y respectively. He sells W kg of tea to each customer. While selling to the 1st and the 2nd customers, he keeps the counterpoising weight on the left pan and on the right pan respectively. Does he gain or lose?
Answer:

Suppose the lengths of the left and the right arms of the balance are x and y respectively. For the first customer, he placed the counterpoising weight in the left pan. In this case, the tea is placed in the other pan. Suppose W₁ kg of tea balances W kg weight.

Hence, W x x = W₁ x y

or, \(W_1=\frac{W x}{y}\)

For the second customer, he placed the W kg counterpoising weight in the right pan. Suppose, in this case, W₂ kg tea is placed in the other pan.

∴ W₂ x x = W x y

or, \(W_2=\frac{W y}{x}\)

The customers pay for 2 Wkg but get (W₁ + W₂) kg of tea.

Now \(W_1+W_2-2 W=\frac{W x}{y}+\frac{W y}{x}-2 W\)

= \(W\left(\frac{x}{y}+\frac{y}{x}-2\right)=\frac{W}{x y}(x-y)^2\) which is a positive quantity.

∴ (W₁ + W₂) > 2 W. Hence, there is a loss for the businessman.

Question 15. A businessman uses a faulty balance of unequal arms. He buys some old papers from a person and for this, he uses a \(\frac{1}{2}\) kg counterpoising weight. He then readily agrees to weigh the papers alternately by changing the pans of the balance during successive weighings. Show that he gains in every 1 kg of purchase. [Upthrust due to air is neglected.)
Answer:

Let the length of the left and the right arms of the faulty balance be x and y (x>y). Suppose Wy kg of paper on the right pan balances the \(\frac{1}{2}\) kg counterpoising weight on the left pan.

∴ \(\frac{1}{2} \cdot x=W_1 \cdot y \text { or, } W_1=\frac{1}{2} \cdot \frac{x}{y}\)

Suppose W₂ kg of paper is required on the left pan to balance \(\frac{1}{2}\) kg counterpoising weight on the right pan.

∴ \(W_2 x=\frac{1}{2} y \text { or, } W_2=\frac{1}{2} \cdot \frac{y}{x}\)

Paper received by the businessman, \(W_1+W_2=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{1}{2} \cdot\left(\frac{x^2+y^2}{x y}\right)\)

= \(\frac{1}{2} \cdot\left[2+\frac{(x-y)^2}{x y}\right]=1+\frac{1}{2} \frac{(x-y)^2}{x y}\)

Hence the businessman gains in every kilogram of purchase.

Mathematical Problems in Statics

Question 16. A person may stand on the lower steps or at the top of a ladder. In which of these cases does the possibility of sliding of the ladder become maximum?
Answer:

The possibility of sliding becomes maximum when the person stands at the top of the ladder because the combined centre of gravity of the system is very high and the system becomes unstable.

Question 17. Why is it easier to walk with two equal loads in both hands than with the entire load in one hand?
Answer:

While carrying a load in one hand the centre of gravity comes down to some extent, but it also shifts towards the side of the load. As a result, the line of action of the weight tends to shift away from the base. In this state, to maintain equilibrium, the body has to be tilted towards the other side.

It is quite difficult to walk in such a manner. While carrying two equal loads in both hands, the centre of gravity of the system is not displaced sideways, but it only comes’ clown. Hence, the body with two loads becomes more stable and? walking becomes easier.

Question 18. Babies fall down frequently while trying to walk. Why?
Answer:

The centre of gravity of a baby, while crawling, remains quite low and the line of action of its body weight always passes through the base. As the baby tries to stand, the centre of gravity of the body goes up and so the body becomes unstable.

To keep the body stable, the baby should keep its body erect, so that the line of action of the body weight passes through the base. Hence, due to a lack of sufficient practice, babies fall down frequently.