NEET Practice Questions (MCQs) with Answers & Solutions

(d) Let ${\mathrm{M}}_0$=mass of the body in the vacuum.
Apparent weight of the body in air = Apparent weight of standard weights in air
$\Rightarrow$Actual weight – upthrust due to displaced air = Actual weight – upthrust due to displaced air 

$$\begin{gathered} \Rightarrow {\mathrm{M}}_0\mathrm{g}-\left({\mathrm{V}}_0\right)\mathrm{dg}=\mathrm{Mg}-\left(\mathrm{V}\right)\mathrm{dg} \\ \Rightarrow {\mathrm{M}}_0\mathrm{g}-\left(\frac{{\mathrm{M}}_0}{{\mathrm{d}}_1}\right)\mathrm{dg}=\mathrm{Mg}-\left(\frac{\mathrm{M}}{{\mathrm{d}}_2}\right)\mathrm{dg} \\ \Rightarrow {\mathrm{M}}_0=\frac{\mathrm{M}\left[1-{\displaystyle \frac{\mathrm{d}}{{\mathrm{d}}^2}}\right]}{\left[1-{\displaystyle \frac{\mathrm{d}}{{\mathrm{d}}_1}}\right]} \end{gathered}$$

(a) $\mathrm{h}=\frac{\mathrm{P}}{\mathrm{\rho g}} \therefore \mathrm{h}\propto \frac{1}{\mathrm{g}} (\mathrm{P} \mathrm{and} \mathrm{\rho } \mathrm{are} \mathrm{constant})$

If value of g decreased by 2% then h will increase by 2%.

(d) $\mathrm{h}=\frac{\mathrm{P}}{\mathrm{\rho g}} \therefore \mathrm{h}\propto \frac{1}{\mathrm{g}}$. If lift moves upward with some acceleration then effective g increases. So the value of h decreases i.e. reading will be less than 76 cm.

(b) Thrust on lamina = pressure at centroid × Area

                               $=\frac{\mathrm{h\rho g}}{3}\times \mathrm{A}=\frac{1}{3}\mathrm{A\rho gh}$

(c)  $\mathrm{\rho }=\frac{\mathrm{Total} \mathrm{mass}}{\mathrm{Total} \mathrm{volume}}=\frac{2\mathrm{m}}{{\mathrm{V}}_1+{\mathrm{V}}_2}=\frac{2\mathrm{m}}{\mathrm{m}\left({\displaystyle \frac{1}{{\mathrm{\rho }}_1}}+{\displaystyle \frac{1}{{\mathrm{\rho }}_2}}\right)}$

$\therefore \mathrm{\rho }=\frac{2{\mathrm{\rho }}_1{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1+{\mathrm{\rho }}_2}$

(a) $\mathrm{\rho }=\frac{\mathrm{Total} \mathrm{mass}}{\mathrm{Total} \mathrm{volume}}=\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{2\mathrm{V}}=\frac{\mathrm{V}({\mathrm{\rho }}_1+{\mathrm{\rho }}_2)}{2\mathrm{V}}=\frac{{\mathrm{\rho }}_1+{\mathrm{\rho }}_2}{2}$

(b) Since, with increase in temperature, volume of given body increases, while mass remains constant so that density will decrease. 

i.e. $$\begin{gathered} \frac{\mathrm{\rho }}{{\mathrm{\rho }}_0}=\frac{\mathrm{m}/\mathrm{V}}{\mathrm{m}/{\mathrm{V}}_0}=\frac{{\mathrm{V}}_0}{\mathrm{V}}=\frac{\mathrm{V}}{{\mathrm{V}}_0(1+\mathrm{\gamma }∆\mathrm{\theta })}=(1-\mathrm{\gamma }∆\mathrm{\theta }) \\ \therefore \mathrm{\rho }={\mathrm{\rho }}_0(1-\mathrm{\gamma } ∆\mathrm{\theta }) \end{gathered}$$

(c) Let the total volume of ice-berg is V and its density is $\mathrm{\rho }$. If this ice-berg floats in water with volume ${\mathrm{V}}_{in}$ 

inside it then ${\mathrm{V}}_{\mathrm{in}}\mathrm{\sigma g}=\mathrm{V\rho g}\Rightarrow {\mathrm{V}}_{\mathrm{in}}=\left(\frac{\mathrm{\rho }}{\mathrm{\sigma }}\right)\mathrm{V}$

or $$\begin{gathered} {\mathrm{V}}_{\mathrm{out}}=\mathrm{V}-{\mathrm{V}}_{\mathrm{in}}=\left(\frac{\mathrm{\sigma }-\mathrm{\rho }}{\mathrm{\sigma }}\right)\mathrm{V} \\ \Rightarrow \frac{{\mathrm{V}}_{\mathrm{out}}}{\mathrm{V}}=\left(\frac{\mathrm{\sigma }-\mathrm{\rho }}{\mathrm{\sigma }}\right)=\frac{1000-900}{1000}=\frac{1}{10} \\ \therefore {\mathrm{V}}_{\mathrm{out}}= 10\% \mathrm{of} \mathrm{V} \end{gathered}$$

(c) Weight of the bowl = mg

=$\mathrm{V\rho g}=\frac{2}{3}\mathrm{\pi }\left[{\left(\frac{\mathrm{D}}{2}\right)}^3-{\left(\frac{\mathrm{d}}{2}\right)}^3\right]\mathrm{\rho g}$

where D = Outer diameter ,

d = Inner diameter

$\mathrm{\rho }$ = Density of bowl                               

Weight of the liquid displaced by the bowl

$=\mathrm{V\sigma g}=\frac{2}{3}\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^3\mathrm{\sigma g}$

where $\mathrm{\sigma }$ is the density of the liquid

For the floatation    $$\begin{gathered} \frac{2}{3}\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^2\mathrm{\sigma g}=\frac{2}{3}\mathrm{\pi }\left[{\left(\frac{\mathrm{D}}{2}\right)}^3-{\left(\frac{\mathrm{d}}{2}\right)}^3\right]\mathrm{\rho g} \\ \Rightarrow {\left(\frac{1}{2}\right)}^3\times 1.2\times {10}^3=\left[{\left(\frac{1}{2}\right)}^3-{\left(\frac{\mathrm{d}}{2}\right)}^3\right]2\times {10}^4 \end{gathered}$$

By solving we get d = 0.98 m.