Physics MCQs for NEET — Practice Questions with Answers

The NCERT text states, 'every particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis.' Since the origin is chosen on the axis, the center of the circle must also be on the axis.

The NCERT text states, 'The linear velocity of the particle at P is $\mathbf{v} = \mathbf{\omega} \times \mathbf{r}$. It is perpendicular to both $\mathbf{\omega}$ and $\mathbf{r}$ and is directed along the tangent to the circle described by the particle.' This clarifies that $\mathbf{v}$ is perpendicular to both.

The NCERT text explicitly states, 'Every Point in the rotating rigid body has the same angular velocity at any instant of time.' and also 'we may characterise pure rotation by all parts of the body having the same angular velocity at any instant of time'.

The NCERT text explains, 'For particles on the axis, $r = 0$, and hence $v = \omega r = 0$. Thus, particles on the axis are stationary.'

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 provided text, 'In fact, the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current.' This highlights the fundamental definition and significance of RMS current in terms of power dissipation.

The text states: '$I = \frac{I_m}{\sqrt{2}} = 0.707 I_m$'. This is the standard definition of the RMS value for a sinusoidal alternating current.