Rotational Motion MCQs for NEET — Physics Questions with Answers

The NCERT text states, 'For rotation about a fixed axis, the angular velocity vector lies along the axis of rotation, and points out in the direction in which a right handed screw would advance, if the head of the screw is rotated with the body.'

The linear velocity is given by $\mathbf{v} = \mathbf{\omega} \times \mathbf{r}$. $\mathbf{v} = (2\hat{\mathbf{k}}) \times (3\hat{\mathbf{i}} + 4\hat{\mathbf{j}})$ $\mathbf{v} = (2)(3)(\hat{\mathbf{k}} \times \hat{\mathbf{i}}) + (2)(4)(\hat{\mathbf{k}} \times \hat{\mathbf{j}})$ $\mathbf{v} = 6\hat{\mathbf{j}} + 8(-\hat{\mathbf{i}})$ $\mathbf{v} = -8\hat{\mathbf{i}} + 6\hat{\mathbf{j}}$ m/s. Wait, let me double check the cross product: $\hat{\mathbf{k}} \times \hat{\mathbf{i}} = \hat{\mathbf{j}}$, and $\hat{\mathbf{k}} \times \hat{\mathbf{j}} = -\hat{\mathbf{i}}$. So, $\mathbf{v} = 6\hat{\mathbf{j}} - 8\hat{\mathbf{i}}$. This is option 1, if written as $8\hat{\mathbf{i}} - 6\hat{\mathbf{j}}$ after reordering. However, my calculation is $-8\hat{\mathbf{i}} + 6\hat{\mathbf{j}}$. Let me retry. Oh, there seems to be a mistake in the options provided based on my correct calculation. If option 'o1' was meant to be $8\hat{\mathbf{j}} - 6\hat{\mathbf{i}}$, then that'd be correct. Let's assume the order changed or there's a typo in the options. Assuming $-8\hat{\mathbf{i}} + 6\hat{\mathbf{j}}$ is the calculated answer, I'll select the closest possible option or point out the error if it's a generated question for a quiz. Reconsidering the provided options, if it needs to match one exactly: $v = (2\hat{k}) \times (3\hat{i} + 4\hat{j}) = 6(\hat{k} \times \hat{i}) + 8(\hat{k} \times \hat{j}) = 6\hat{j} - 8\hat{i}$. So the correct vector is $-8\hat{i} + 6\hat{j}$. None of the options correctly represent this. Let me re-evaluate, Perhaps, the question meant a different vector for r. Or I should pick the option which shares the same numerical values but may have a sign error if I am to pick one. Let's re-verify the cross product. $\hat{k} \times \hat{i} = \hat{j}$ and $\hat{k} \times \hat{j} = -\hat{i}$. So $2\hat{k} \times (3\hat{i} + 4\hat{j}) = 6\hat{j} - 8\hat{i}$. This is $-8\hat{i} + 6\hat{j}$. If I must pick from the given options, and since this is a practice question, let's look for common mistakes. A reversal might lead to $8\hat{\mathbf{i}} - 6\hat{\mathbf{j}}$. This implies $-( -8\hat{\mathbf{i}} + 6\hat{\mathbf{j}})$ or $(2\hat{k}) \times (-(3\hat{i} + 4\hat{j}))$. Let's assume it was intended as positive $8\hat{i} - 6\hat{j}$ which is not what I got. I'll maintain my calculated answer $ -8\hat{i} + 6\hat{j} $. If the question implies a magnitude based check for instance, all options have magnitudes of $\sqrt{8^2+6^2} = 10$. Since I have to provide one of the options as correct, and acknowledging my calculation: $6\hat{j} - 8\hat{i}$, which is $-8\hat{i} + 6\hat{j}$. Option o1 is $8\hat{i} - 6\hat{j}$. Option o2 is $-8\hat{i} + 6\hat{j}$. My calculation yields o2. So, 'o2' is the correct answer.

The NCERT text states, 'We know from our study of circular motion that the magnitude of linear velocity v of a particle moving in a circle is related to the angular velocity of the particle $\omega$ by the simple relation $v = \omega r$, where r is the radius of the circle'.

The NCERT text explicitly makes the analogy: 'We have already noted the analogy between angular velocity $\omega$ (in respect of rotational motion about a fixed axis) and linear velocity v (in respect of linear motion)'. Therefore, angular velocity is the rotational analogue of linear velocity.

According to the NCERT text, 'Two important quantities in the study of rotational motion, namely, moment of a force and angular momentum, are defined as vector products.'

The NCERT text states: '(i) magnitude of $c = c = ab \sin\theta$ where a and b are magnitudes of $\vec{a}$ and $\vec{b}$ and $\theta$ is the angle between the two vectors.'

The NCERT text specifies: '(ii) $\vec{c}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$'.

The NCERT text describes: '(iii) if we take a right handed screw... and if we turn the head in the direction from $\vec{a}$ to $\vec{b}$, then the tip of the screw advances in the direction of $\vec{c}$. This right handed screw rule is illustrated in Fig. 6.15a.' It also mentions the right-hand rule with curling fingers.

The NCERT text states: 'While applying either of the above rules, the rotation should be taken through the smaller angle ($<180^\circ$) between $\vec{a}$ and $\vec{b}$'.

The NCERT text explicitly states: 'Because of the cross ($\times$) used to denote the vector product, it is also referred to as cross product.'