NEET Practice Questions (MCQs) with Answers & Solutions

(b) Let specific gravities of concrete and saw dust  are ${\mathrm{\rho }}_1$ and ${\mathrm{\rho }}_2$ respectively.
According to principle of floatation weight of whole sphere = upthrust on the sphere

$$\begin{gathered} \frac{4}{3}\mathrm{\pi }({\mathrm{R}}^3-{\mathrm{r}}^3){\mathrm{\rho }}_1\mathrm{g}+\frac{4}{3}{\mathrm{\pi r}}^3{\mathrm{\rho }}_2\mathrm{g}=\frac{4}{3}{\mathrm{\pi R}}^3\times 1\times \mathrm{g} \\ \Rightarrow {\mathrm{R}}^3{\mathrm{\rho }}_1-{\mathrm{r}}^3{\mathrm{\rho }}_1 +{\mathrm{r}}^3{\mathrm{\rho }}_2={\mathrm{R}}^3 \\ \Rightarrow {\mathrm{R}}^3({\mathrm{\rho }}_1-1)={\mathrm{r}}^3({\mathrm{\rho }}_1-{\mathrm{\rho }}_2)\Rightarrow \frac{{\mathrm{R}}^3}{{\mathrm{r}}^3}=\frac{{\mathrm{\rho }}_1-{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1-1} \\ \Rightarrow \frac{{\mathrm{R}}^3-{\mathrm{r}}^3}{{\mathrm{r}}^3}=\frac{{\mathrm{\rho }}_1-{\mathrm{\rho }}_2-{\mathrm{\rho }}_1+1}{{\mathrm{\rho }}_1-1} \\ \Rightarrow \frac{({\mathrm{R}}^3-{\mathrm{r}}^3){\mathrm{\rho }}_1}{{\mathrm{r}}^3{\mathrm{\rho }}_2}=\left(\frac{1-{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1-1}\right)\frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2} \\ \Rightarrow \frac{\mathrm{Mass} \mathrm{of} \mathrm{concrete}}{\mathrm{Mass} \mathrm{of} \mathrm{saw} \mathrm{dust}}=\left(\frac{1-0.3}{2.4-1}\right)\times \frac{2.4}{0.3}=4 \end{gathered}$$

(d) Apparent weight 

$$\begin{gathered} =\mathrm{V}(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g} =\mathrm{l}\times \mathrm{b}\times \mathrm{h}\times (5-1)\times \mathrm{g} \\ =5\times 5\times 5\times 4\times \mathrm{g} \mathrm{Dyne} = 4\times 5\times 5\times 5 \mathrm{gf} \end{gathered}$$

(b) Apparent weight 

$=\mathrm{V}(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g}=\frac{\mathrm{M}}{\mathrm{\rho }}(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g}$

$$\begin{gathered} =\mathrm{M}\left(1-\frac{\mathrm{\sigma }}{\mathrm{\rho }}\right)\mathrm{g}=2.1\left(1-\frac{0.8}{10.5}\right)\mathrm{g}=1.94\mathrm{g} \mathrm{N} \\ =1.94 \mathrm{kg}-\mathrm{wt} \end{gathered}$$

(c) If two different bodies A and B are floating in the same liquid then

$\frac{{\mathrm{\rho }}_{\mathrm{A}}}{{\mathrm{\rho }}_{\mathrm{B}}}=\frac{({\mathrm{f}}_{\mathrm{in}}{)}_{\mathrm{A}}}{({\mathrm{f}}_{\mathrm{in}}{)}_{\mathrm{B}}}=\frac{1/2}{2/3}=\frac{3}{4}$

(c) For the floatation  ${\mathrm{V}}_0{\mathrm{d}}_0\mathrm{g}={\mathrm{V}}_{\mathrm{in}}\mathrm{dg}$

${\mathrm{V}}_{\mathrm{in}}={\mathrm{V}}_0\frac{{\mathrm{d}}_0}{\mathrm{d}}$

$$\begin{gathered} \therefore {\mathrm{V}}_{\mathrm{out}}={\mathrm{V}}_0-{\mathrm{V}}_{\mathrm{in}}={\mathrm{V}}_0-{\mathrm{V}}_0\frac{{\mathrm{d}}_0}{\mathrm{d}}={\mathrm{V}}_0\left[\frac{\mathrm{d}-{\mathrm{d}}_0}{\mathrm{d}}\right] \\ \Rightarrow \frac{{\mathrm{V}}_{\mathrm{out}}}{{\mathrm{V}}_0}=\frac{\mathrm{d}-{\mathrm{d}}_0}{\mathrm{d}} \end{gathered}$$

(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)

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