Three non zero vectors  satisfy the relation . Then $\stackrel{\to }{\mathrm{A}}$ can be parallel to:

(A)  $\stackrel{\to }{\mathrm{B}}$

(B)  $\stackrel{\to }{\mathrm{C}}$

(C)  $\stackrel{\to }{\mathrm{B}}·\stackrel{\to }{\mathrm{C}}$

(D)  $\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}$

Concept Questions :-

Scalar Product
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The scalar product of two vectors is 8 and the magnitude of vector product is $8\sqrt{3}$. The angle between them is:

(A)  $30°$

(B)  $60°$

(C)  $120°$

(D)  $150°$

Concept Questions :-

Vector Product
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Given: $\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}=0$. Out of the three vectors  and $\stackrel{\to }{\mathrm{c}}$ two are equal in magnitude. The magnitude of the third vector is $\sqrt{2}$ times that of either of the two having equal magnitude. The angles between the vectors are:

(A)  90$°$, 135$°$, 135$°$

(B)  30$°$, 60$°$, 90$°$

(C)  45$°$, 45$°$, 90$°$

(D)  45$°$, 60$°$, 90$°$

Concept Questions :-

Resultant of Vectors
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Vector $\stackrel{\to }{\mathrm{A}}$ is of length 2 cm and is 60$°$ above the x-axis in the first quadrant. Vector $\stackrel{\to }{\mathrm{B}}$ is of length 2 cm and 60$°$ below the x-axis in the fourth quadrant. The sum $\stackrel{\to }{\mathrm{A}}$+$\stackrel{\to }{\mathrm{B}}$ is a vector of magnitude -

(A)  2 along + y-axis

(B)  2 along + x-axis

(C)  1 along - x-axis

(D)  2 along - x-axis

Concept Questions :-

Resultant of Vectors
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Six forces, 9.81 N each, acting at a point are coplanar. If the angle between neighboring forces are equal, then the resultant is

(A)  0 N

(B)  9.81 N

(C)  2$×$9.81 N

(D)  3$×$9.81 N

Concept Questions :-

Resultant of Vectors
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Temperature of a body varies with time as , where ${\mathrm{T}}_{0}$ is the temperature in Kelvin at , then the rate of change of temperature $\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)$ at  is-

1.

2.

3.

4.

Concept Questions :-

Differentiation
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If the distance 's' travelled by a body in time 't' is given by $\mathrm{s}=\frac{\mathrm{a}}{\mathrm{t}}+{\mathrm{bt}}^{2}$ then the acceleration equals

(A)  $\frac{2\mathrm{a}}{{\mathrm{t}}^{3}}+2\mathrm{b}$

(B)  $\frac{2\mathrm{s}}{{\mathrm{t}}^{3}}$

(C)  $2\mathrm{b}-\frac{2\mathrm{a}}{{\mathrm{t}}^{3}}$

(D)  $\frac{\mathrm{s}}{{\mathrm{t}}^{2}}$

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The velocity of a particle moving on the x-axis is given by $\mathrm{v}={\mathrm{x}}^{2}+\mathrm{x}$ where v is in m/s and x is in m. Find its acceleration in $\mathrm{m}/{\mathrm{s}}^{2}$ when passing through the point x=2m.

1.  0

2.  5

3.  11

4.  30

Concept Questions :-

Differentiation
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A particle moves in the XY plane and at time t is at the point whose coordinates are . Then at what instant of time, will its velocity and acceleration vectors be perpendicular to each other?

(A)  1/3 sec

(B)  2/3 sec

(C)  3/2 sec

(D)  never

Concept Questions :-

Differentiation
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A particle moves in the x-y plane with velocity ${\mathrm{v}}_{\mathrm{x}}=8\mathrm{t}-2$ and ${\mathrm{v}}_{\mathrm{y}}=2$. If it passes through the point x=14 and y=4 at t=2 sec. The equation of the path is

(A)  $\mathrm{x}={\mathrm{y}}^{2}-\mathrm{y}+2$

(B)  $\mathrm{x}=\mathrm{y}+2$

(C)  $\mathrm{x}={\mathrm{y}}^{2}+2$

(D)  $\mathrm{x}={\mathrm{y}}^{2}+\mathrm{y}+2$

Concept Questions :-

Integration
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