Q. Three vectors $\overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}}$ and $\overrightarrow{\mathrm{R}}$ are shown in the figure. Let S be any point on the vector $\overrightarrow{\mathrm{R}}$. The distance between the points P and S is b $|\overrightarrow{\mathrm{R}}|$. The general relation among vectors $\overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}}$ and $\overrightarrow{\mathrm{S}}$ is :
$(\mathrm{A}) \overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b}^{2} \overrightarrow{\mathrm{Q}}$
(B) $\overrightarrow{\mathrm{S}}=(b-1) \overrightarrow{\mathrm{P}}+b \overrightarrow{\mathrm{Q}}$
(C) $\overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}}$
$(\mathrm{D}) \overrightarrow{\mathrm{S}}=\left(1-\mathrm{b}^{2}\right) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}}$
[JEE Advanced – 2017]
$(\mathrm{A}) \overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b}^{2} \overrightarrow{\mathrm{Q}}$
(B) $\overrightarrow{\mathrm{S}}=(b-1) \overrightarrow{\mathrm{P}}+b \overrightarrow{\mathrm{Q}}$
(C) $\overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}}$
$(\mathrm{D}) \overrightarrow{\mathrm{S}}=\left(1-\mathrm{b}^{2}\right) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}}$
[JEE Advanced – 2017]
Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...
Sol. (C) Let vector from point P to point S be $\overrightarrow{\mathrm{c}}$ $\Rightarrow \overrightarrow{\mathrm{c}}=\mathrm{b}|\overrightarrow{\mathrm{R}}| \hat{\mathrm{R}}=\mathrm{b}|\overrightarrow{\mathrm{R}}|\left(\frac{\overrightarrow{\mathrm{R}}}{|\overrightarrow{\mathrm{R}}|}\right)=\mathrm{b} \overrightarrow{\mathrm{R}}=\mathrm{b}(\overrightarrow{\mathrm{Q}}-\overrightarrow{\mathrm{P}})$ from triangle rule of vector addition $\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{S}}$ $\overrightarrow{\mathrm{P}}+\mathrm{b}(\overrightarrow{\mathrm{Q}}-\overrightarrow{\mathrm{P}})=\overrightarrow{\mathrm{S}}$ $\Rightarrow \overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}}$
Q. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density remains uniform throughout the volume. The rate of fractional change in density $\left(\frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}}\right)$ is constant. The velocity v of any point on the surface of the expanding sphere is proportional to :
(A) $\mathrm{R}^{3}$
(B) $\frac{1}{\mathrm{R}}$
(C) R
(D) $\mathrm{R}^{2 / 3}$
[JEE Advanced – 2017]
Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...
Sol. (C) Density of sphere is $\rho=\frac{\mathrm{m}}{\mathrm{v}}=\frac{3 \mathrm{m}}{4 \pi \mathrm{R}^{3}}$ $\Rightarrow \frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}}=-\frac{3}{\mathrm{R}} \frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}}$ since $\Rightarrow \frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}}$ is constant $\therefore \frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}} \propto \mathrm{R}$ Velocity of any point on the circumfrence V is equal to $\frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}}$ (rate of change of radius of outer layer)
