- ✓$80.0096\,cm$
- B$80.0272\,cm$
- C$1\,cm$
- D$25.2\,cm$
Hence length of copper wire w.r.t. steel scale or apparent length of copper wire after rise in temperature
${L_{app}} = L{'_{cu}} - \,L{'_{steel}} = [{L_0}(1 + {\alpha _{Cu}}\Delta \theta ) - {L_0}(1 + {\alpha _s}\Delta \theta )$
==> ${L_{app}} = {L_0}({\alpha _{Cu}} - {\alpha _s})\Delta \theta $
$ = 80(17 \times {10^{ - 6}} - 11 \times {10^{ - 6}}) \times 20 = 80.0096 cm$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| List-$I$ | List-$II$ |
| $P$ $\dot{r}(t)=\alpha t \hat{t}+\beta t \hat{j}$ | $1$ $\overrightarrow{ p }$ |
| $Q$ $\dot{r}(t)=\alpha \cos \omega t \hat{i}+\beta \sin \omega t \hat{j}$ | $2$ $\overrightarrow{ L }$ |
| $R$ $\dot{r}(t)=\alpha(\cos \omega t \hat{i}+\sin \omega t \hat{j})$ | $3$ $K$ |
| $S$ $\dot{r}(t)=\alpha t \hat{i}+\frac{\beta}{2} t^2 \hat{j}$ | $4$ $U$ |
| $5$ $E$ |
$(A)$ Final temperature of system will be $0^{\circ} C$.
$(B)$ Final temperature of the system will be greater than $0^{\circ} C$.
$(C)$ The final system will have a mixture of ice and water in the ratio of $5: 1$.
$(D)$ The final system will have a mixture of ice and water in the ratio of $1: 5$.
$(E)$ The final system will have water only.
Choose the correct answer from the options given below: