WB JEE 2010 | Differential Equations Question 20 | Mathematics

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WB JEE 2010

MCQ (Single Correct Answer)

-0.25

Solution of the differential equation xdy $$-$$ ydx = 0 represents a

parabola

circle

hyperbola

straight line

WB JEE 2010

MCQ (Single Correct Answer)

-0.25

If the displacement, velocity and acceleration of a particle at time t be x, v and f respectively, then which one is true?

$$f = {v^3}{{{d^2}t} \over {d{x^2}}}$$

$$f = - {v^3}{{{d^2}t} \over {d{x^2}}}$$

$$f = {v^2}{{{d^2}t} \over {d{x^2}}}$$

$$f = - {v^2}{{{d^2}t} \over {d{x^2}}}$$

WB JEE 2010

MCQ (Single Correct Answer)

-0.25

The displacement x of a particle at time t is given by x = At² + Bt + C, where A, B, C are constants and v is velocity of a particle, then the value of 4Ax $$-$$ v² is

4AC + B²

4AC $$-$$ B²

2AC $$-$$ B²

2AC + B²

WB JEE 2010

MCQ (Single Correct Answer)

-0.25

The displacement of a particle at time t is x, where x = t⁴ $$-$$ kt³. If the velocity of the particle at time t = 2 is minimum, then

k = 4

k = $$-$$4

k = 8

k = $$-$$8

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