Ncert Exercises & Solutions

The figure shows a lamina in $\text{XY}$-plane. Two axes $\text{z}$ and $z'$ pass perpendicular to its plane. A force $\vec{F}$ acts in the plane of the lamina at point P as shown. (The point $\text{p}$ is closer to the $z'$-axis than the $\text{z}$-axis.)

(a)	torque $\vec{\tau}$ caused by $\vec{F}$ about $\text{z}$-axis is along - $\hat{k}$
(b)	torque $\vec{\tau}'$ caused by $\vec{F}$ about $z'$-axis is along - $\hat{k}$
(c)	torque caused by $\vec{F}$ about the $\text{z}$-axis is greater in magnitude than that about the z'-axis
(d)	total torque is given by $\vec{\tau}_{net}=\vec{\tau}+\vec{\tau}'$

Choose the correct option:
1. (c, d)
2. (a, c)
3. (b, c)
4. (a, b)

Hint: $\tau=\vec{r}\times \vec{F}$.

Step 1: Find the direction of the torque aabout the two axes.
Consider the adjacent diagram, where $r>r'$.
Torque $\vec{\tau}$ about $\text{z}$-axis = $\vec{r}\times \vec{F}$ which is along $\hat{k}$
Torque $\vec{\tau}$ about $\text{z'}$-axis = $\vec{r'}\times \vec{F}$ which is along $-\hat{k}$

Step 2: Find the magnitude of the torque about the two axes.

(c) $| \vec{τ} |_{z} = {Fr}_{⊥} =$ the magnitude of the torque about the z-axis where $r_{⊥}$ is the perpendicular distance between F and z-axis.
$\begin{array}{l} Similarly, | \vec{τ} |_{z^{'}} = {Fr}_{⊥}^{'} \\ Clearly, r_{⊥} > r_{⊥}^{'} \Rightarrow | \vec{τ} |_{z} > | \vec{τ} |_{z^{'}} \end{array}$

(d) We are always calculating resultant torque about a common axis.

Hence, total torque ${\vec{τ}}_{net} \neq \vec{τ} + {\vec{τ}}^{'}$ , because $\vec{τ}$ and $\vec{τ}$ ' are not about a common axis.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions

(a)	torque \(\vec{\tau}\) caused by \(\vec{F}\) about \(\text{z}\)-axis is along - \(\hat{k}\)
(b)	torque \(\vec{\tau}'\) caused by \(\vec{F}\) about \(z'\)-axis is along - \(\hat{k}\)
(c)	torque caused by \(\vec{F}\) about the \(\text{z}\)-axis is greater in magnitude than that about the z'-axis
(d)	total torque is given by \(\vec{\tau}_{net}=\vec{\tau}+\vec{\tau}'\)