Question
Two simple pendulum whose lengths are $1\,m$ and $121\, cm$ are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum will the two be in phase again
$\Rightarrow \mathrm{N} \sqrt{121}=(\mathrm{N}+1) \sqrt{100}$
$\Rightarrow \mathrm{N}(11)+(\mathrm{N}+1) 10$
$\Rightarrow 11 \mathrm{N}=10 \mathrm{N}+10$
$\Rightarrow \mathrm{N}=10$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

