MCQ
For frictionless surfaces in given arrangement tension $T_2$ is :-


- ✓$mg/3$
- B$2\,mg/3$
- C$3\,mg/2$
- D$5\,mg/3$

$\mathrm{T}_{2}=\mathrm{m} \times \mathrm{a}=\mathrm{m} \times \mathrm{g} / 3$
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${y_1} = 8\,\cos\, \omega t;\,{y_2} = 4\,\cos \,\left( {\omega t + \frac{\pi }{2}} \right)$ ;
${y_3} = 2\cos \,\left( {\omega t + \pi } \right);\,{y_4} = \,\cos \,\left( {\omega t + \frac{{3\pi }}{2}} \right)$ ,
are superposed on each other. The resulting amplitude and phase are respectively;

Reason : $N = {N_0}\,{\left( {\frac{1}{2}} \right)^n}$
where, $n = \frac{{{\rm{time\, elapsed}}}}{{{\rm{half \,life \,period}}}}$