Class 11 Physics | System of Particles and Rotational Motion MCQs with Answers – Part 3

Explanation: Torque measures the turning effect of a force about a chosen point or axis. It depends on the force and also on where the force is applied relative to that point or axis. This is why the relation is written as \(\vec{\tau}=\vec{r}\times\vec{F}\). The cross product includes the perpendicular part of the force through the factor \(rF\sin\theta\). A force applied at the same point but in a different direction can produce a different turning effect.

Explanation: Torque is defined by the cross product \(\vec{\tau}=\vec{r}\times\vec{F}\). The magnitude of any cross product is the product of the two magnitudes times the sine of the angle between them. Therefore, \(\tau=rF\sin\theta\). The factor \(\sin\theta\) selects the part of the force perpendicular to \(\vec{r}\). This is why a tangential force gives maximum torque, while a force along the line of \(\vec{r}\) gives zero torque.

Explanation: \( \textbf{Given data:} \) \(r=0.50\,\text{m}\), \(F=20\,\text{N}\), and \(\theta=90^\circ\). \( \textbf{Required quantity:} \) Torque magnitude \(\tau\). \( \textbf{Torque relation:} \) \[ \tau=rF\sin\theta \] \( \textbf{Reason for using it:} \) Torque is the magnitude of \(\vec{r}\times\vec{F}\). \( \textbf{Substitution:} \) \[ \tau=(0.50)(20)\sin90^\circ \] \( \textbf{Use the perpendicular-angle sine value:} \) \[ \tau=(0.50)(20) \] \( \textbf{Calculation:} \) \[ \tau=10\,\text{N m} \] \( \textbf{Final answer:} \) The torque magnitude is \(10\,\text{N m}\).

Explanation: Torque about a pivot is \(\vec{\tau}=\vec{r}\times\vec{F}\). Its magnitude is \(\tau=rF\sin\theta\), where \(\theta\) is the angle between the position vector and the force. If the force acts along the same line as \(\vec{r}\), the angle is \(0^\circ\) or \(180^\circ\). In both cases, the sine factor is zero, so the torque is zero. Such a force may push or pull along the line, but it has no perpendicular lever arm to produce rotation about the pivot.

Explanation: Torque is defined as \(\vec{\tau}=\vec{r}\times\vec{F}\). Its magnitude has the form \(\tau=rF\sin\theta\), so its unit is the product of the unit of distance and the unit of force. Distance is measured in \(\text{m}\), and force is measured in \(\text{N}\). Therefore, torque is measured in \(\text{N m}\). Although \(\text{N m}\) is dimensionally the same as joule, torque is not normally written in joule because torque represents turning effect, not energy transfer.

Explanation: The turning effect of a force depends not only on the magnitude of the force but also on the perpendicular distance of its line of action from the axis. This perpendicular distance is called the moment arm or lever arm. For the same applied force, increasing the moment arm increases the torque. A hand placed farther from the nut usually gives a larger perpendicular distance from the turning axis. If the line of action passed through the axis, the torque would become zero, not larger.

Explanation: \( \textbf{Given data:} \) \(r=0.40\,\text{m}\), \(F=50\,\text{N}\), and \(\theta=30^\circ\). \( \textbf{Required quantity:} \) Torque magnitude \(\tau\). \( \textbf{Relevant formula:} \) \[ \tau=rF\sin\theta \] \( \textbf{Why this formula applies:} \) The torque is the magnitude of the cross product \(\vec{r}\times\vec{F}\). \( \textbf{Substitution:} \) \[ \tau=(0.40)(50)\sin30^\circ \] \( \textbf{Use the sine value:} \) \[ \tau=(0.40)(50)\left(\frac{1}{2}\right) \] \( \textbf{Intermediate product:} \) \[ (0.40)(50)=20 \] \( \textbf{Calculation:} \) \[ \tau=20\left(\frac{1}{2}\right)=10\,\text{N m} \] \( \textbf{Physical check:} \) The torque is half of the maximum value \(rF=20\,\text{N m}\) because the force is at \(30^\circ\), not \(90^\circ\). \( \textbf{Final answer:} \) The torque magnitude is \(10\,\text{N m}\).

Explanation: The torque direction is the direction of the cross product \(\vec{r}\times\vec{F}\). Here, \(\vec{r}\) is along \(+\hat{i}\), and \(\vec{F}\) is along \(+\hat{j}\). The unit-vector product is \(\hat{i}\times\hat{j}=\hat{k}\). Since \(+\hat{k}\) is upward in the chosen convention, the torque points upward. Reversing the order to \(\vec{F}\times\vec{r}\) would give the opposite direction, so the order in torque is essential.

Explanation: Torque about a pivot depends on the perpendicular distance from the pivot to the line of action of the force. If the line of action passes through the pivot, this perpendicular distance is zero. Equivalently, the angle between \(\vec{r}\) and \(\vec{F}\) is \(0^\circ\) or \(180^\circ\), so \(\tau=rF\sin\theta=0\). The force may still produce translation of the body if not balanced by other forces. It produces no turning effect about that particular pivot because the moment arm is absent.

Explanation: Torque magnitude is \(\tau=rF\sin\theta\). At \(\theta=0^\circ\), the sine factor is zero, so row P is not suitable. At \(\theta=90^\circ\), \(\sin90^\circ=1\), giving the maximum value \(rF\), so row Q is suitable. At \(\theta=180^\circ\), the torque is again zero. At \(\theta=30^\circ\), the torque is \(\frac{rF}{2}\), not zero.

Explanation: Statement I is true because the position vector \(\vec{r}\) is measured from the chosen point or axis. Changing that point can change the torque of the same force. Statement II is true because torque is defined as the cross product of position vector and force. Statement III is false because a force whose line of action passes through the pivot has zero moment arm. Maximum torque occurs when the force is perpendicular to \(\vec{r}\), not when it passes through the pivot.

Explanation: A door rotates about an axis through its hinges. For a force applied at the handle, the torque is largest when the force is perpendicular to the position vector from the hinge to the handle. Pushing perpendicular to the door near the handle gives a large moment arm and a large perpendicular force component. Pulling or pushing along the line toward the hinge gives little or no turning effect. The door example shows why both distance from axis and force direction matter.

Explanation: \( \textbf{Given data:} \) Force \(F=12\,\text{N}\), and perpendicular distance \(d_{\perp}=0.25\,\text{m}\). \( \textbf{Required quantity:} \) Torque magnitude about the chosen axis. \( \textbf{Moment-arm relation:} \) \[ \tau=Fd_{\perp} \] \( \textbf{Why this relation applies:} \) The perpendicular distance is already given, so the sine factor has been included through the moment arm. \( \textbf{Substitution:} \) \[ \tau=(12)(0.25)\,\text{N m} \] \( \textbf{Calculation:} \) \[ \tau=3.0\,\text{N m} \] \( \textbf{Unit check:} \) Force multiplied by distance gives \(\text{N m}\). \( \textbf{Interpretation:} \) A smaller perpendicular distance would reduce the turning effect even if the force magnitude stayed the same. \( \textbf{Final answer:} \) The torque magnitude is \(3.0\,\text{N m}\).

Explanation: The symbol \(d_{\perp}\) represents the moment arm of the force. It is the perpendicular distance from the chosen axis or pivot to the line along which the force acts. This is not necessarily the length of the body or the distance from the force point to the centre of mass. The same force applied at the same point can have different torque if its direction changes, because its line of action changes. The moment arm is a geometrical distance tied to the chosen axis and the force direction.

Explanation: \( \textbf{Relation used:} \) \[ \tau=Fd_{\perp} \] \( \textbf{Graph axes:} \) The vertical axis is \(\tau\), and the horizontal axis is \(d_{\perp}\). \( \textbf{Straight-line form:} \) \[ \tau=(F)d_{\perp} \] \( \textbf{Comparison with straight-line graph form:} \) The slope is the constant multiplying \(d_{\perp}\). \( \textbf{Slope value:} \) \[ \text{slope}=F \] \( \textbf{Unit check:} \) The slope has unit \(\frac{\text{N m}}{\text{m}}=\text{N}\), matching force. \( \textbf{Final answer:} \) The slope represents the force magnitude \(F\).

Explanation: \( \textbf{Given data:} \) \(r=0.60\,\text{m}\), \(F=15\,\text{N}\), and \(\theta=60^\circ\). \( \textbf{Required quantity:} \) Torque magnitude about the pivot. \( \textbf{Torque formula:} \) \[ \tau=rF\sin\theta \] \( \textbf{Substitution:} \) \[ \tau=(0.60)(15)\sin60^\circ \] \( \textbf{Use the sixty-degree sine value:} \) \[ \tau=(9.0)(0.866)\,\text{N m} \] \( \textbf{Calculation:} \) \[ \tau\approx7.794\,\text{N m} \] \( \textbf{Rounded value:} \) \[ \tau\approx7.8\,\text{N m} \] \( \textbf{Check against maximum:} \) The maximum possible torque would be \(rF=9.0\,\text{N m}\), so \(7.8\,\text{N m}\) is reasonable for \(60^\circ\). \( \textbf{Final answer:} \) The torque magnitude is approximately \(7.8\,\text{N m}\).

Explanation: Torque about a fixed axis can be assigned a sign according to a chosen convention. If anticlockwise torque is taken as positive, then clockwise torque is taken as negative. The sign does not mean the force magnitude is negative; it identifies the sense of rotation produced about the axis. A different convention would reverse the signs, but it must be used consistently throughout the problem. Sign convention is a bookkeeping tool for rotational direction.

Explanation: \( \textbf{Sign convention:} \) Anticlockwise torque is positive, and clockwise torque is negative. \( \textbf{Torque values:} \) The anticlockwise torque is \(+8\,\text{N m}\). \( \textbf{Clockwise torque:} \) The clockwise torque is \(-5\,\text{N m}\). \( \textbf{Net torque relation:} \) \[ \tau_{\text{net}}=\tau_1+\tau_2 \] \( \textbf{Substitution:} \) \[ \tau_{\text{net}}=8+(-5) \] \( \textbf{Calculation:} \) \[ \tau_{\text{net}}=3\,\text{N m} \] \( \textbf{Direction meaning:} \) The positive sign means the net torque is anticlockwise. \( \textbf{Final answer:} \) The net torque is \(3\,\text{N m}\) anticlockwise.

Explanation: Torque is defined as \(\vec{\tau}=\vec{r}\times\vec{F}\), where \(\vec{r}\) is measured from the chosen point to the point of application of the force. If the reference point is changed, \(\vec{r}\) can change even though \(\vec{F}\) is the same. Since torque depends on \(\vec{r}\), the calculated torque may also change. This is why torque must always be stated about a specified point or axis. A force does not have a single universal torque independent of reference point.

Explanation: The position vector from the pivot to the point of application points to the right. The force acts downward at the right end. A downward push on the right side of a horizontal rod tends to lower the right end and rotate the rod clockwise about the left pivot. In vector form, with right as \(+\hat{i}\) and upward as \(+\hat{j}\), the force is along \(-\hat{j}\), so \(\vec{r}\times\vec{F}\) points along \(-\hat{k}\). That direction corresponds to clockwise rotation when viewed from the usual front side of the page.