Physics MCQs for NEET — Practice Questions with Answers

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Equations for mechanical equilibrium, $\sum \vec{F_i} = 0$ and $\sum \vec{\tau_i} = 0$, are vector equations. How many scalar equations do they represent in total for a general rigid body?

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Explanation

The NCERT text clarifies: 'Eq. (6.30a) and Eq. (6.30b), both, are vector equations. They are equivalent to three scalar equations each. Eq. (6.30a) corresponds to $\sum F_{ix} = 0$, $\sum F_{iy} = 0$ and $\sum F_{iz} = 0$. Similarly, Eq. (6.30b) is equivalent to three scalar equations $\sum \tau_{ix} = 0$, $\sum \tau_{iy} = 0$ and $\sum \tau_{iz} = 0$.'

What happens to the rotational state of motion of a rigid body if the total torque on the body does not vanish?

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Explanation

The text states: 'The total torque on the body may not vanish. Such a torque changes the rotational state of motion of the rigid body, i.e. it changes the total angular momentum of the body in accordance with Eq. (6.28 b).'

When considering the equilibrium of a rigid body, the term 'force' refers to:

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Explanation

The NCERT text explicitly states: 'Henceforth we shall omit the adjective 'external' because unless stated otherwise, we shall deal with only external forces and torques.'

Which of the following is true for a rigid body rotating about a fixed axis?

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Explanation

The NCERT summary point 3 states: 'In rotation about a fixed axis, every particle of the rigid body moves in a circle which lies in a plane perpendicular to the axis and has its centre on the axis. Every Point in the rotating rigid body has the same angular velocity at any instant of time.'

A rigid body is subjected to a system of forces. It is observed to be in rotational equilibrium, but not in translational equilibrium. Which of the following conditions correctly describes this situation?

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Explanation

For rotational equilibrium, the total torque must be zero ($\sum \vec{\tau_i} = 0$). For 'not in translational equilibrium', the total force must be non-zero ($\sum \vec{F_i} \neq 0$). This combination corresponds to the description of partial equilibrium provided in the text, specifically the example with Fig. 6.20(a) where the rod was in rotational equilibrium ($net moment = 0$) but not translational equilibrium ($\sum F \neq 0$).

Which of the following statements about unit vectors is INCORRECT?

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Explanation

According to the NCERT text, 'A unit vector is a vector of unit magnitude and points in a particular direction. It has no dimension and unit. It is used to specify a direction only.' Therefore, the statement that unit vectors have specific dimensions and units is incorrect.

A vector $\vec{A}$ lies in the x-y plane. If its x-component is $A_x$ and its y-component is $A_y$, how can $\vec{A}$ be expressed in terms of unit vectors $\hat{i}$ and $\hat{j}$?

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Explanation

The NCERT text states, 'Thus, $\vec{A} = A_x \hat{i} + A_y \hat{j}$.' This equation correctly represents a vector in terms of its components along the x and y axes using unit vectors.

If a vector $\vec{A}$ has a magnitude $|A|$ and a unit vector along its direction is $\hat{n}$, which of the following expressions is correct?

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Explanation

According to equation 3.10 in the NCERT text, 'In general, a vector A can be written as $\vec{A} = |A| \hat{n}$ where $\hat{n}$ is a unit vector along A.'

Which of the following is NOT a characteristic of a unit vector?

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Explanation

A unit vector is used to specify direction. Multiplying a unit vector by a scalar changes its magnitude, not its direction (unless the scalar is negative, which reverses the direction). The NCERT states: 'If we multiply a unit vector, say $\hat{n}$ by a scalar $\lambda$, the result is a vector $\lambda \vec{n}$.'

A vector $\vec{A}$ is resolved into two component vectors, $\vec{A_1}$ and $\vec{A_2}$. If $\vec{A_1}$ is parallel to $\hat{i}$ and $\vec{A_2}$ is parallel to $\hat{j}$, which condition must be met for this resolution?

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Explanation

The NCERT text explicitly states: 'We draw lines from the head of A perpendicular to the coordinate axes as in Fig. 3.9(b), and get vectors $\vec{A_1}$ and $\vec{A_2}$ such that $\vec{A_1} + \vec{A_2} = \vec{A}$'.

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