MCQ
The length of a seconds pendulum is .... $cm$
- A$99.8$
- ✓$99$
- C$100$
- DNone of these
$\Rightarrow l = \frac{{g{T^2}}}{{4{\pi ^2}}} = \frac{{9.8 \times 4}}{{4 \times {\pi ^2}}} = 99\;cm$
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$(I)$ $P$ runs with speed $1 \,ms ^{-1}$ towards $Q$, while $Q$ remains stationary.
$(II)$ $Q$ runs with speed $1 \,ms ^{-1}$ towards $P$, while $P$ remains stationary.
Note That irrespective of speed of $P$, ball always leaves $P$ s hand with speed $2 \,ms ^{-1}$ with respect to the ground. Ignore gravity. Balls will be received by $Q$.