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

(1)  0 N

(2)  9.81 N

(3)  2×9.81 N

(4)  3×9.81 N

Subtopic:  Resultant of Vectors |
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Level 1: 80%+
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The temperature of a body varies with time as \(T= (T_0 +at^2 +b\sin t)K,\) where \(T_0\) is the temperature in Kelvin at \(t =0 ~\text{s}~\text{&}~a = \frac{2}{\pi}~\text{K/s}^2~\text{&}~b = -4~\text{K}\), then the rate of change of temperature \(\frac{dT}{dt}\) at \(t = \pi ~\text{s}\) is:
1. \(8~\text{K}\)
2. \(80~\text{K}\)
3. \(8~\text{K/sec}\)
4. \(80~\text{K/sec}\)

Subtopic:  Differentiation |
 58%
Level 3: 35%-60%
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If the distance 's' travelled by a body in time 't' is given by s=at+bt2 then the acceleration equals

(1)  2at3+2b

(2)  2st3

(3)  2b-2at3

(4)  st2

Subtopic:  Differentiation |
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Level 2: 60%+
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The velocity of a particle moving on the x-axis is given by v=x2+x where v is in m/s and x is in m. Find its acceleration in m/s2 when passing through the point x=2m.

1.  0

2.  5

3.  11

4.  30

Subtopic:  Differentiation |
Level 3: 35%-60%
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A particle moves in the XY plane and at time t is at the point whose coordinates are t2,t3-2t. Then at what instant of time, will its velocity and acceleration vectors be perpendicular to each other?

(1)  1/3 sec

(2)  2/3 sec

(3)  3/2 sec

(4)  never

Subtopic:  Scalar Product |
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A motor boat of mass m moving along a lake with velocity V0. 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:Fx=mVdVdx=-rV

(1)  mV0/r

(2)  3mV0/2r

(3)  mV0/2r

(4)  2mV0/r

Subtopic:  Integration |
Level 3: 35%-60%
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A particle is moving along positive \(x\text-\)axis. Its position varies as \(x = t^3-3t^2+12t+20,\) where \(x\) is in meters and \(t\) is in seconds. The velocity of the particle when its acceleration zero is:
1. \(1~\text{m/s}\)
2. \(3~\text{m/s}\)
3. \(6~\text{m/s}\)
4. \(9~\text{m/s}\)

Subtopic:  Differentiation |
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Level 2: 60%+
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Two forces F1=2i^+2j^N and F2=3j^+4k^N are acting on a particle.

The resultant force acting on particle is:

(A)  2i^+5j^+4k^

(B)  2i^-5j^-4k^

(C)  i^-3j^-2k^

(D)  i^-j^-k^

Subtopic:  Resultant of Vectors |
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Level 1: 80%+
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A=4i+4j-4k and B=3i+j+4k, then angle between vectors A and B is:

(1)  180°

(2)  90°

(3)  45°

(4)  0°

Subtopic:  Resultant of Vectors |
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If vectors \(\overrightarrow{{A}}=\cos \omega t \hat{{i}}+\sin \omega t \hat{j}\) and \(\overrightarrow{{B}}=\cos \left(\frac{\omega t}{2}\right)\hat{{i}}+\sin \left(\frac{\omega t}{2}\right) \hat{j}\) are functions of time. Then, at what value of \(t\) are they orthogonal to one another?
1. \(t = \frac{\pi}{4\omega}\)
2. \(t = \frac{\pi}{2\omega}\)
3. \(t = \frac{\pi}{\omega}\)
4. \(t = 0\)

Subtopic:  Scalar Product |
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Level 2: 60%+
NEET - 2015
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