The position vector of a particle is $\vec{r}=4 t ^2 \hat{i}+2 t \hat{j}+3 t \hat{k}$ The acceleration of a particle is having only -
✓
$X – component$
B
$Y – component$
C
$Z – component$
D
$X – Y component$
Answer
Correct option: A.
$X – component$
$X-component$
$4 t ^2 \hat{i}+2 t \hat{j}+3 t \hat{k} \\\vec{v}=\frac{d \vec{r}}{d t}=8 t \hat{i}+2 \hat{j} \\a =\frac{d^2 r}{d t^2}=8 \hat{i} $
a is having only $X {-component. }$
A 100 m long train is traveling from North to South at a speed of 30 ms-1. A bird is flying from South to North at a speed of 10-1. How long will the bird take to, cross the train?
A
3 s
✓
2.5 s
C
10 s
D
5 s
Answer
Correct option: B.
2.5 s
Length of train $=100 m$ Relative velocity $=30+10=40 ms ^{-1}$ Time taken to cross the train $( t )=\frac{\text { distance }}{\text { R.velocity }}=\frac{100}{40}=2.5 s$
A swimmer can swim in still water at of $10 ms-1$ While crossing a river his average speed is $6 ms-1.$ If he crosses the river in the shortest possible time, what is the speed of flow of water?
A
$16 ms ^{-1}$
B
$4 ms ^{-1}$
C
$60 ms ^{-1}$
✓
$8 ms ^{-1}$
Answer
Correct option: D.
$8 ms ^{-1}$
The resultant velocity of swimmer must be perpendicular to speed of water to cross the river in a shortest time
$\therefore v_s^2=v^2+v_w^2$
$v_w^2=v_s^2-v^2$
$=100-36=64$
$\therefore V =8 m / s ^{-1}$
A car covers half of its journey with a speed of 10 ms-1 and the other half by 20 ms-1. The average speed of car during the total journey is –
A
$70 ms ^{-1}$
B
$15 ms ^{-1}$
✓
$13.33 ms ^{-1}$
D
$7.5 ms ^{-1}$
Answer
Correct option: C.
$13.33 ms ^{-1}$
Let $x$ is the total distance Time to cover 1 st half $=\frac{x / 2}{10}$ Time to cover 2 nd half $=\frac{x / 2}{20}$ Average speed $=$$\frac{x}{\frac{x}{20}+\frac{x}{40}}=\frac{1}{\left(\frac{3}{40}\right)}=13.33 ms ^{-1}$
A car starting from rest, accelerates at a constant rate x for sometime after which it decelerates at a constant rate v to come to rest. If the total time elapsed is t, the maximum velocity attained by the car is given by –
A person moving horizontally with velocity $\overrightarrow{ V _m}$. Rain falls vertically with velocity $\overrightarrow{ V _R}$ To save himself from the rain, he should hold an umbrella with vertical at an angle of -
A
$\tan ^{-1}\left(\frac{V_R}{V_m}\right)$
✓
$\tan ^{-1}\left(\frac{V_m}{V_R}\right)$
C
$\tan \theta= V _m+ V _{ R }$
D
$\tan ^{-1}\left( V _{ R }+ V _m / V _{ R }- V _m\right)$
If two objects A and B are moving along a straight line in the opposite direction with the velocities VA and VB respectively, then relative velocity is-
If two objects A and B are moving along a straight line in the same direction with the velocities $v_A$ and $v_B$
respectively, then the relative velocity is -
A particle moves in a circular path of radius 2 cm. If a particle completes 3 rounds, then the distance and displacement of the particle are –
A
0 and 37.7
B
37.7 and 0
✓
0 and 0
D
37.7 and 37.7
Answer
Correct option: C.
0 and 0
Radius $=2 cm$ Circumference of the circle $=2 nr =4 n cm$ Distance covered in 3 rounds $=127 rcm =37.7 cm$ Initial and final positions are same $\therefore$ Displacement $=0$
If $|\overrightarrow{ P }+\overrightarrow{ Q }|=|\overrightarrow{ P }||\overrightarrow{ Q }|$, then angle between $\overrightarrow{ P }$ and $\overrightarrow{ Q }$ will be -
A
$0°$
B
$45°$
✓
$90°$
D
$180°$
Answer
Correct option: C.
$90°$
$|\vec{P}+\vec{Q}|=|\vec{P}||\vec{Q}|$
Square on both side, and the resultants become,
$P ^2+ Q ^2+2 PQ \cos 0$
$= P ^2+ Q ^2-2 PQ \cos \theta 4 PQ \cos \theta=0 $
$\theta=90^{\circ}$
If $|\overrightarrow{ P }+\overrightarrow{ Q }=| \overrightarrow{ P }|-| \overrightarrow{ P } \mid$, then the angle between the vectors $\overrightarrow{ P }$ and $\overrightarrow{ Q }$
A
$0°$
B
$90°$
✓
$180°$
D
$360°$
Answer
Correct option: C.
$180°$
$|\vec{P}+\vec{Q}|=|\vec{P}||\vec{P}|^{\prime}$
Square on both side, and the resultant becomes
$P ^2+ Q ^2+2 PQ \cos \theta= P ^2+ Q ^2-2 PQ $
$\cos \theta=-1$
$\theta=180^{\circ}$
If $|\overrightarrow{ P }+\overrightarrow{ Q }|=|\overrightarrow{ P }|+|\overrightarrow{ Q }|$. The angle between the vectors $\overrightarrow{ P }$ and $\overrightarrow{ Q }$ is -
✓
$0°$
B
$180°$
C
$60°$
D
$90°$
Answer
Correct option: A.
$0°$
$0°$
$|\overrightarrow{ P }+\overrightarrow{ Q }|=|\overrightarrow{ P }|+|\overrightarrow{ Q }|$
Square on both sides and the resultant becomes
$P ^2+ Q ^2+2 PQ \cos \theta= P ^2+ Q ^2+2 PQ \cos \theta=1 $
$\theta=0$
The momentum of a particle is $\overrightarrow{ P }=\cos \theta \hat{i}+\sin \theta \hat{j}$. The angle between momentum and the force acting on a body is -
If two vectors $\vec{A}$ and $\vec{B}$ form adjacent sides of parallelogram, then the magnitude of $|\vec{A} \times \vec{B}|$ will give of parallelogram -