Is stress a vector quantity?
$$\text { Stress }=\frac{\text { Magnitude of internal reaction force }}{\text { Area of cross }- \text { section }}$$
Therefore, stress is a scalar quantity not a vector quantity.
Identical springs of steel and copper are equally stretched. On which, more work will have to be done?
Work done in stretching a wire is given by $W=\frac{1}{2} F \times \Delta l$
[where $F$ is applicd force and $\Delta l$ is extension in the wire]
As springs of steel and copper are equally stretched. Therefore, for same force $(F)$,
$$W \propto \Delta l\quad \text{... (i)}$$
$$\text { Young's modulus }(Y)=\frac{F}{A} \times \frac{l}{\Delta l} \Rightarrow \Delta l=\frac{F}{A} \times \frac{l}{Y}$$
As both springs are identical,
$$\Delta l \propto \frac{1}{Y}\quad \text{... (ii)}$$
From Eqs. (i) and (ii), we get
$$W \propto \frac{1}{y}$$
$$\begin{aligned} &\begin{array}{lll} \therefore & \frac{W_{\text {steel }}}{W_{\text {copper }}}=\frac{Y_{\text {copper }}}{Y_{\text {steel }}}< 1 & \left(\text { As, } Y_{\text {steel }}>Y_{\text {copper }}\right) \\ \Rightarrow & W_{\text {steel }}< W_{\text {copper }} \end{array}\\ &\text { Therefore, more work will be done for stretching copper spring. } \end{aligned}$$
What is the Young's modulus for a perfect rigid body?
Young's modulus $(Y)=\frac{F}{A} \times \frac{l}{\Delta l}$
For a perfectly rigid body, change in length $\Delta l=0$
$$Y=\frac{F}{A} \times \frac{l}{0}=\infty$$
Therefore, Young's modulus for a perfectly rigid body is infinite( $(\infty)$.
What is the Bulk modulus for a perfect rigid body?
Bulk modulus $(K)=\frac{p}{\Delta V / V}=\frac{p V}{\Delta V}$
For perfectly rigid body, change in volume $\Delta V=0$
$$\therefore\quad K=\frac{p V}{0}=\infty$$
Therefore, bulk modulus for a perfectly rigid body is infinity ( $\infty$ ).
A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$, its length increases by $l$. Another wire of the same material of length $2 L$ and radius $2 r$, is pulled by a force $2 f$. Find the increase in length of this wire.
The situation is shown in the diagram.
Now, $$\quad$$ Young's modulus $(Y)=\frac{f}{A} \times \frac{L}{l}$
For first wire, $$Y=\frac{f}{\pi r^2} \times \frac{L}{l} \quad \text{... (i)}$$
For second wire,
$$\begin{aligned} Y & =\frac{2 f}{\pi(2 r)^2} \times \frac{2 L}{l^{\prime}} \\ & =\frac{f}{\pi r^2} \times \frac{L}{l^{\prime}}\quad \text{... (ii)} \end{aligned}$$
From Eqs. (i) and (ii),
$\quad \frac{f}{\pi r^2} \times \frac{L}{l}=\frac{\mathrm{f}}{\pi r^2} \times \frac{L}{l^{\prime}}$
$\therefore$ $l=l^{\prime}$
$$\text { ( } \because \text { Both wires are of same material, hence, Young's modulus will be same.) }$$