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

2 along + y-axis

2 along + x-axis

1 along - x-axis

2 along - x-axis

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Explanation

Vector A has a magnitude of 2 cm at an angle of 60Â° above the x-axis. Vector B has the same magnitude of 2 cm but at 60Â° below the x-axis. Their sum A + B will be a vector along the positive x-axis with a magnitude of 2âˆš3 â‰ˆ 3.46 cm.

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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

0 N

9.81 N

2 $\times$ 9.81 N

3 $\times$ 9.81 N

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Explanation

When six equal forces of 9.81 N each are acting at a point in the same plane, and the angles between neighboring forces are equal, the forces will form a closed polygon. The resultant of such a system of coplanar forces is always zero, as the vector sum of all forces is zero.

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Temperature of a body varies with time as $T = (T_{0} + {at}^{2} + b sint) K$ , where $T_{0}$ is the temperature in Kelvin at $t = 0 \sec & a = 2 / π K / s^{2} & b = - 4 K$ , then the rate of change of temperature $(\frac{dT}{dt})$ at $t = π \sec$ is-

$8 K$

$8^{0} K$

$8 K / \sec$

$8^{0} K / \sec$

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Explanation

$T = (T_{0} + {at}^{2} + b \sin t) K \frac{dT}{dt} = 2 at + b \cos t At t = π \sec \frac{dT}{dt} = 2 (\frac{2}{π}) π + (- 4) cosπ = 8 K / \sec$

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

$\frac{2 a}{t^{3}} + 2 b$

$\frac{2 s}{t^{3}}$

$2 b - \frac{2 a}{t^{3}}$

$\frac{s}{t^{2}}$

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Explanation

(A)

$s = \frac{a}{t} + {bt}^{2} v = \frac{ds}{dt} = \frac{- a}{t^{2}} + 2 bt a = \frac{dv}{dt} = \frac{2 a}{t^{3}} + 2 b$

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

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Explanation

$v = x^{2} + x Differentiating both sides w . r . t t \frac{dv}{dt} = \frac{d (x^{2})}{dt} + \frac{dx}{dt} a = 2 x (\frac{dx}{dt}) + \frac{dx}{dt} a = (2 x + 1) v [∵ v = \frac{dx}{dt}] = (2 x + 1) (x^{2} + x) At x = 2 m a = [2 (2) + 1] [2^{2} + 2] = 30 {ms}^{- 2}$

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A particle moves in the XY plane and at time t is at the point whose coordinates are $(t^{2}, t^{3} - 2 t)$ . Then at what instant of time, will its velocity and acceleration vectors be perpendicular to each other?

1/3 sec

2/3 sec

3/2 sec

never

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Explanation

(B)

$\vec{r} = t^{2} \hat{i} + (t^{3} - 2 t) \hat{j} \vec{v} = \frac{d \vec{r}}{dt} = 2 t \hat{i} + (3 t^{2} - 2) \hat{j} \vec{a} = 2 \hat{i} + (6 t) \hat{j} When \vec{v} ⊥ \vec{a}, \vec{v} . \vec{a} = 0 [2 t \hat{i} + (3 t^{2} - 2) \hat{j}] . [2 \hat{i} + 6 t \hat{j}] = 0 4 t + 6 t (3 t^{2} - 2) = 0 2 t [2 + 3 (3 t^{2} - 2)] = 0 t = 0 or 2 + 9 t^{2} - 6 = 0 t^{2} = \frac{4}{9} or t = \frac{2}{3} \sec$

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

$x = y^{2} - y + 2$

$x = y + 2$

$x = y^{2} + 2$

$x = y^{2} + y + 2$

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Explanation

$v_{x} = 8 t - 2 v_{y} = 2 \frac{dx}{dt} = 8 t - 2 \frac{dy}{dt} = 2 \int dx = \int (8 t - 2) dt \int dy = \int 2 dt x = 4 t^{2} - 2 t + c y = 2 t + c_{1} At t = 2 s, x = 14 At t = 2 s, y = 4 14 = 4 {(2)}^{2} - 2 (2) + c 4 = 2 (2) + c_{1} c = 2 c_{1} = 0 x = 4 t^{2} - 2 t + 2 y = 2 t Putting t = \frac{y}{2} in x x = 4 {(\frac{y}{2})}^{2} - 2 (\frac{y}{2}) + 2 x = y^{2} - y + 2$

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A motor boat of mass m moving along a lake with velocity $V_{0}$ . At t=0, the engine of the boat is shut down. Magnitude of resistance force offered to the boat is equal to rV. (V is instantaneous speed). What is the total distance covered till it stops completely? $[Hint : F (x) = mV \frac{dV}{dx} = - rV]$

${mV}_{0} / r$

$3 {mV}_{0} / 2 r$

${mV}_{0} / 2 r$

$2 {mV}_{0} / r$

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Explanation

$mV \frac{dV}{dx} = - rV \frac{dV}{dx} = - \frac{r}{m} \int_{v_{o}}^{0} d v = \frac{- r}{m} \int_{0}^{s} d x s = \frac{m v_{o}}{r}$

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A particle is moving along positive x-axis. Its position varies as $x = t^{3} - 3 t^{2} + 12 t + 20$ , where x is in meters and t is in seconds.

Initial velocity of the particle is

1 m/s

3 m/s

12 m/s

20 m/s

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Explanation

$v = \frac{d x}{d t} = 3 t^{2} - 6 t + 12 A t t = 0 v = 12 m / s$

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A particle is moving along positive x-axis. Its position varies as $x = t^{3} - 3 t^{2} + 12 t + 20$ , where x is in meters and t is in seconds.

Initial acceleration of the particle is

Zero

$1 m / s^{2}$

$- 3 m / s^{2}$

$- 6 m / s^{2}$

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Explanation

$x = t^{3} - 3 t^{2} + 12 t + 20 v = \frac{d x}{d t} = 3 t^{2} - 6 t + 12 a = \frac{d v}{d t} = 6 t - 6 a t t = 0, a = - 6 m / s^{2}$

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