NEET Practice Questions (MCQs) with Answers & Solutions

$$\begin{gathered} \left(\mathrm{b}\right) \mathrm{Given} : \mathrm{kinetic} \mathrm{energy} \mathrm{K} = 360 \mathrm{J} \\ \mathrm{Angular} \mathrm{speed} \mathrm{\omega } = 20 \mathrm{rad}/\mathrm{s} \\ \therefore \mathrm{K} = \frac{1}{2}{\mathrm{I\omega }}^2 \\ \mathrm{Where} \mathrm{I} = \mathrm{moment} \mathrm{of} \mathrm{inertia} \\ \Rightarrow \mathrm{I} = \frac{2\mathrm{K}}{{\mathrm{\omega }}^2}=\frac{2\times 360}{20\times 20}=1.8 \mathrm{kg} {\mathrm{m}}^{-2}. \end{gathered}$$

$L=I\omega =I\left(2\pi f\right)$

Frequency f is doubled. Hence angular momentum will become 2L.

$\begin{array}{rl}\left(\mathrm{KE}\right)& =(\mathrm{KE}{)}_{\text{vrensialonal}}+(\mathrm{KE}{)}_{\text{retational}}\\ & =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\\ & =\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{1}{2}m{r}^2{\omega }^2\\ & =\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2=\frac{3}{4}m{v}^2\\ & =\frac{3}{4}\times 20\times 4^2=240\mathrm{J}\end{array}$

Acceleration of a rolling body down an inclined plane is $\mathrm{a}=\frac{\mathrm{gsin} \mathrm{\theta }}{1+{\mathrm{k}}^2/{\mathrm{r}}^2}=\frac{\mathrm{gsin} \mathrm{\theta }}{1+\frac{2}{5}}(1=\frac{2}{5}{\mathrm{mr}}^2={\mathrm{mk}}^2$

$\frac{k^2}{r^2}=\frac{2}{5}=\frac{5}{7}g\sin \theta$ Choice A is correct.

$\begin{array}{rl}& {\mathrm{u}}_{\mathrm{CM}}=\frac{{\mathrm{m}}_1{\mathrm{u}}_1+{\mathrm{m}}_2{\mathrm{u}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}=0\\ & \text{as }{\mathrm{m}}_1{\mathrm{u}}_1+{\mathrm{m}}_2{\mathrm{u}}_2=0 \text{is given.}\end{array}$

Hence velocity of the composite system will be zero.

$$\begin{gathered} \overrightarrow{r_{cm}}=\frac{m_1{\overrightarrow{r}}_1+m_4{\overrightarrow{r}}_2}{m_1+m_2}=\frac{60\times \left(0i+0j+0k\right)+40{\overrightarrow{r}}_2}{100} \\ {\overrightarrow{r}}_2=\frac{5}{2}\left(3i+3j+3k\right)=7.5i+7.5j+7.5k \end{gathered}$$

$\left(a\right) \overrightarrow{L} = \overrightarrow{r}\times \overrightarrow{p}= \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -1\\ 3& 4& -2\end{array}\right|=-\hat{j}-2\hat{k}$

and the X-axis is given by $\hat{i} + 0\hat{j} + 0\hat{k}$ 

The dot product of these two vectors is zero i.e. angular momentum is perpendicular to X-axis.

3.

In the absence of air resistance, the only external force acting on the body is the force of gravity. As for the breaking of the body Into two unequal parts occurs, It is purely due to Internal forces that are exerted mutually by parts of the system and cancel. Hence the center of mass continues to move In the same direction and there is no cause of shift of the centre of mass of the falling body (parts of the body after breaking) Hence choice 3 is correct.

1.

$\mathrm{Because} \mathrm{its} \mathrm{M}.\mathrm{I}. \left(\mathrm{or} \mathrm{value} \mathrm{of} \frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}\right) \mathrm{is} \mathrm{minimum} \mathrm{for} \mathrm{sphere}$

$$\begin{gathered} KE=\frac{1}{2}I{\omega }^2; L=I\omega \\ KE=\frac{I^2{\omega }^2}{2I}=\frac{L^2}{2I} \\ \left(KE\right)=\frac{L^2}{2I}; \left(KE\right)\text{'}=\frac{{\left({\displaystyle \frac{110L}{100}}\right)}^2}{2I} \\ \frac{KE\text{'}-KE}{KE}=\frac{{\left({\displaystyle \frac{110}{100}}\right)}^2{\displaystyle \frac{L^2}{2I}}-{\displaystyle \frac{L^2}{2I}}}{{\displaystyle \frac{L^2}{2I}}} \\ =\frac{121}{100}-1=\frac{21}{100} \\ \frac{21}{100}\times 100=21\% \\ (Choice 3) \end{gathered}$$