There is a long cylinder of radius R having a cylindrical cavity of radius R/2 as shown in the figure. A part from the cavity, the entire space inside the cylinder has a uniform magnetic field parallel to the axis of the cylinder. The magnetic field starts changing at a uniform rate of $\frac{\mathrm{dB}}{\mathrm{dt}}=\mathrm{k}\mathrm{T}/\mathrm{s}$. If the induced electric field at a point inside cavity is kR/p. Then P is? 
 

We can consider the entire system of magnetic field to be superposition of fields as shows below. $\frac{{\mathrm{dB}}_1}{\mathrm{dt}}=\frac{{\mathrm{dB}}_2}{\mathrm{dt}}=\mathrm{\alpha }$
 

Say, both fields are increasing

Induced electric field inside a cylindrical region having time changing magnetic field is given by $\mathrm{E}=\frac{\mathrm{r}}{2}\left(\frac{\mathrm{dB}}{\mathrm{dt}}\right)$ where r is distance from the axis of the cylinder. Induced field at point P(inside cavity)
i) Due to changing $\overrightarrow{{\mathrm{B}}_1}$ is $\overrightarrow{{\mathrm{E}}_1}$ (perpendicular to ${\mathrm{O}}_1\mathrm{P}$ )  ${\overrightarrow{\mathrm{E}}}_1=\frac{{\mathrm{r}}_1}{2}\frac{{\mathrm{dB}}_1}{\mathrm{dt}}=\frac{{\mathrm{r}}_1}{2}\mathrm{k}$
ii) Due to changing ${\overrightarrow{\mathrm{B}}}_2\text{ is }{\overrightarrow{\mathrm{E}}}_2$ (perpendicular to $\left.{\mathrm{O}}_1\mathrm{P}\right){\overrightarrow{\mathrm{E}}}_2=\frac{{\mathrm{r}}_2}{2}\frac{{\mathrm{dB}}_2}{\mathrm{dt}}=\frac{{\mathrm{r}}_2}{2}\mathrm{k}$
 

${\mathrm{E}}_1\mathrm{sin\alpha } = \frac{{\mathrm{r}}_1}{2}\mathrm{ksin\alpha }$

${\mathrm{E}}_2\mathrm{sin\beta }=\frac{{\mathrm{r}}_2}{2}\mathrm{ksin\beta }$

${\mathrm{E}}_1\mathrm{cos\alpha } = \frac{{\mathrm{r}}_1}{2}\mathrm{kcos\alpha }$

${\mathrm{E}}_2\mathrm{cos\beta }=\frac{{\mathrm{r}}_2}{2}\mathrm{kcos\beta }$

From geometry, ${\mathrm{r}}_1\mathrm{cos\alpha }+{\mathrm{r}}_2\mathrm{cos\beta }=\frac{\mathrm{R}}{2} \mathrm{and} {\mathrm{r}}_1\mathrm{sin\alpha }={\mathrm{r}}_2\mathrm{sin\beta }$

Thus, ${\mathrm{E}}_1\mathrm{sin\alpha } = {\mathrm{E}}_2\mathrm{sin\beta }$

and ${\mathrm{E}}_{\mathrm{net}} = {\mathrm{E}}_1\mathrm{cos\alpha } +{\mathrm{E}}_2\mathrm{cos\beta } =\frac{\mathrm{k}}{2} \left({\mathrm{r}}_1\mathrm{cos\alpha } +{\mathrm{r}}_2\mathrm{cos\beta }\right)=\frac{\mathrm{kR}}{4}$