Physics MCQs for NEET — Practice Questions with Answers

(d)

According to Pascal's principle, the pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid with the walls of the containing vessel too. A common application of this principle is in hydraulic lift, to lift the car. Here, a small amount of force applied to the small-area piston, which will be transformed into a large force at a large-area piston.

(b)

$Upthrust reduces from Vdg to Vd\left(g-a\right) which is responsible for the extension in the spring.$

(a) When body (sphere) is half immersed, then
upthrust = weight of sphere 

$\Rightarrow \frac{\mathrm{V}}{2}\times {\mathrm{\rho }}_{\mathrm{liq}}\times \mathrm{g}=\mathrm{V}\times \mathrm{\rho }\times \mathrm{g} \therefore \mathrm{\rho }=\frac{{\mathrm{\rho }}_{\mathrm{liq}}}{2}$

When body (sphere) is fully immersed then,

Upthrust = wt. of sphere + wt. of water poured in sphere

$$\begin{gathered} \Rightarrow \mathrm{V}\times {\mathrm{\rho }}_{\mathrm{liq}}\times \mathrm{g}=\mathrm{V}\times \mathrm{\rho }\times \mathrm{g} + \mathrm{V}\text{'}\times {\mathrm{\rho }}_{\mathrm{liq}}\times \mathrm{g} \\ \Rightarrow \mathrm{V}\times {\mathrm{\rho }}_{\mathrm{liq}}=\frac{\mathrm{V}\times {\mathrm{\rho }}_{\mathrm{liq}}}{2}+\mathrm{V}\text{'}\times {\mathrm{\rho }}_{\mathrm{liq}}\Rightarrow \mathrm{V}\text{'}=\frac{\mathrm{V}}{2} \end{gathered}$$

(b) The velocity of ball before entering the water surface

$\mathrm{v}=\sqrt{2\mathrm{gh}}=\sqrt{2\mathrm{g}\times 9}$

When ball enters into water, due to upthrust of water the velocity of ball decreases (or retarded)

The retardation, $\mathrm{a}=\frac{\mathrm{apparent} \mathrm{weight}}{\mathrm{mass} \mathrm{of} \mathrm{ball}}$

$=\frac{\mathrm{V}(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g}}{\mathrm{V\rho }}=\left(\frac{\mathrm{\rho }-\mathrm{\sigma }}{\mathrm{\rho }}\right)\mathrm{g}=\left(\frac{0.4-1}{0.4}\right)\times \mathrm{g}=-\frac{3}{2}\mathrm{g}$

If h be the depth upto which ball sink, then,

$$\begin{gathered} 0-{\mathrm{v}}^2=2\times \left(-\frac{3}{2}\mathrm{g}\right)\times \mathrm{h}\Rightarrow 2\mathrm{g}\times 9=3\mathrm{gh} \\ \therefore \mathrm{h}=6 \mathrm{cm} \end{gathered}$$

(c)

$$\begin{gathered} {\rho }_{ball} \times V_{ball} \times g = {\rho }_{water} \times V_{displaced} \times g \\ As {\rho }_{ball} > {\rho }_{water} \\ {V}_{displaced} > V_{ball} \\ So water level falls \end{gathered}$$

(c) Since, upthrust (F) $=\mathrm{V\sigma g}$   i.e. F∝V

Submarine has a large volume ,so buoyant force  is more than the weight. Hence it floats on water surface. When it has to go inside water, it is filled with water so that weight is more than the buoyant force acting on submarine. So it sinks.

(b) For streamline flow, Reynold's number ${\mathrm{N}}_{\mathrm{R}}\propto \frac{\mathrm{r\rho }}{\mathrm{\eta }}$ should be less. For less value of ${\mathrm{N}}_{\mathrm{R}}$, radius and density should be small and viscosity should be high.

(a) ${\mathrm{d}}_{\mathrm{A}}=2\mathrm{cm}$and ${\mathrm{d}}_B=4 \mathrm{cm}$               $\therefore {\mathrm{r}}_{\mathrm{A}}=1\mathrm{cm}$ and ${\mathrm{r}}_{\mathrm{B}}=2 \mathrm{cm}$

From equation of continuity , av= constant

$\therefore \frac{{\mathrm{v}}_{\mathrm{A}}}{{\mathrm{v}}_{\mathrm{B}}}=\frac{{\mathrm{a}}_{\mathrm{B}}}{{\mathrm{a}}_{\mathrm{A}}}=\frac{\mathrm{\pi }({\mathrm{r}}_{\mathrm{B}}{)}^2}{\mathrm{\pi }({\mathrm{r}}_{\mathrm{A}}{)}^2}={\left(\frac{2}{1}\right)}^2\Rightarrow {\mathrm{v}}_{\mathrm{A}}=4{\mathrm{v}}_{\mathrm{B}}$

(a) Velocity head $\mathrm{h}=\frac{{\mathrm{v}}^2}{2\mathrm{g}}=\frac{(5{)}^2}{2\times 10}=1.25\mathrm{m}$