==>$\frac{{\Delta T}}{T} = \frac{1}{2}\frac{{\Delta l}}{l} = \frac{1}{2}\alpha \Delta \theta $
Also according to thermal expansion $l' = (1 + \alpha \Delta \theta )$
$\frac{{\Delta l}}{l} = \alpha + \theta $. Hence $\frac{{\Delta T}}{T} = \frac{1}{2}\frac{{\Delta l}}{l} = \frac{1}{2}\alpha \Delta \theta $
$ = \frac{1}{2} \times 12 \times {10^{ - 6}} \times (40 - 20) = 12 \times {10^{ - 5}}$
$ \Rightarrow \Delta T = 12 \times {10^{ - 5}} \times 86400\, seconds / day$
$\Delta T \approx 10.3\, seconds/day$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Column - $\mathrm{I}$ | Column - $\mathrm{II}$ |
| $(a)$Rain drops moves downwards with constant velocity. | $(i)$ Viscous liquids |
| $(b)$ Floating clouds at a height in air. | $(ii)$ Viscosity |
| $(iiii)$ Less density |
$(A)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01 \ mm$.
$(B)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005 \ mm$.
$(C)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01 \ mm$.
$(D)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005 \ mm$.
| $(i)$ Suspension fibre of galvanometer | $(a)$ Linear |
| $(ii)$ Bending of beam | $(b)$ Shear |
| $(iii)$ cutting piece of paper | $(c)$ Bulk |
| $(iv)$ mechanical waves in fluid | $(d)$ Shear |
$y = A\,\cos \,\omega t\,\cos \,2\omega t + A\,\sin \,\omega t\,\sin \,2\omega t$.
Than the nature of the function is