Embibe Experts Solutions for Chapter: Mathematical Tools in Physics, Exercise 3: Exercise-3

$3$ boys are pulling a heavy trolley by means of $3$ ropes. The boy in the middle is exerting a pull of $100 \mathrm{N}$. The other $2$ boys, whose ropes both make an angle of $30°$ with the centre rope, are pulling with forces of $50\sqrt{3} \mathrm{N}$ and $100\sqrt{3} \mathrm{N}$. Consider all the forces in the same plane. The magnitude of the resultant pull on the trolley is:

A radius vector of point A relative to the origin varies with time $t$ as $\overrightarrow{r}=\mathrm{at} \widehat{\mathrm{i}}-\mathrm{b}{\mathrm{t}}^2\widehat{\mathrm{j}}$ where $\mathrm{a}$ and $\mathrm{b}$ are constants. If the equation of point's t as trajectory is $\mathrm{y}=\frac{-\mathrm{b}}{{\mathrm{a}}^{\mathrm{q}}}{\mathrm{x}}^{\mathrm{p}}$, then value of $\mathrm{p}+\mathrm{q}$ is:

A room has dimension $3 \mathrm{m}\times 4 \mathrm{m}\times 5 \mathrm{m}$. A fly starting at one corner ends up at diametrically opposite corner. The magnitude of its displacement is:

A room has dimensions $3 \mathrm{m}\times 4 \mathrm{m}\times 5 \mathrm{m}$. A fly starting at one corner ends up ta diametrically opposite corner. If the fly does not fly but walks, then the length of the shortest path it can take is:

Vector $\overrightarrow{\mathrm{a}}$ has components ${\mathrm{a}}_{\mathrm{x}}=3, {\mathrm{a}}_{\mathrm{y}}=4$. If the components of a vector $\overrightarrow{\mathrm{c}}$ which is perpendicular to $\overrightarrow{\mathrm{a}}$ and has a magnitude of $5$ units, are ${\mathrm{c}}_{\mathrm{x}}$ and ${\mathrm{c}}_{\mathrm{y}}$, then $\left|{\mathrm{c}}_{\mathrm{x}}\right|+\left|{\mathrm{c}}_{\mathrm{y}}\right|$ is:

A car travels due east on a level road for $30 \mathrm{km}$. It then turns due north at an intersection and travels $40 \mathrm{km}$ before stopping. The resultant displacement of the car is:

The kinetic energy of a particle of mass $200 \mathrm{g}$ moving the velocity $3\widehat{\mathrm{i}}-5\widehat{\mathrm{j}}+4\widehat{\mathrm{k}} {\mathrm{ms}}^{-1}$ is:

Acceleration of particle moving in straight line can be written as $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dx}}$. From the given graph the acceleration at $\mathrm{x}=20 \mathrm{m}$ is: