MCQ
$A\xrightarrow{{}}{(C{H_3})_2}C = CHCOC{H_3}\,\,\,\,A$ is
- ✓Acetone
- BAcetaldehyde
- CPropionaldehyde
- DFormaldehyde
$\begin{array}{*{20}{c}}
{C{H_3}} \\
{C{H_3}}
\end{array} > C = O\, + \,$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{{H_2} - CH - C - C{H_3}}
\end{array}$ $\xrightarrow[{ - \,{H_2}O}]{{HCl}}$ $\mathop {{{(C{H_3})}_2}C = CH.COC{H_3}}\limits_{Mesityl\,\,oxide} $
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$\rho(r)=\left\{\begin{array}{ll}\rho_{0}\left(\frac{3}{4}-\frac{r}{R}\right) & \text { for } r \leq R \\ \text { Zero } & \text { for } r>R\end{array}\right.$
Where, $r ( r < R )$ is the distance from the centre $O$ (as shown in figure). The electric field at point $P$ will be.

