JEE Questions for Maths Applications Of Derivatives Quiz 9 - MCQExams.com

A stone is thrown vertically upwards from the top of a tower 64 metres high according to the law s = 48t – 16t2. The greatest height attained by the stone above the ground is
  • 100 metre
  • 64 metre
  • 36 metre
  • 32 metre

Maths-Applications of Derivatives-9881.png

  • Maths-Applications of Derivatives-9882.png
  • 2)
    Maths-Applications of Derivatives-9883.png

  • Maths-Applications of Derivatives-9884.png

  • Maths-Applications of Derivatives-9885.png
A particle is moving in a straight line according to the formula s = t2 + 8t + 12. If s be measured in metre and t be measured in second, then the average velocity of the particle in third second is
  • 14 m/sec
  • 13 m/sec
  • 15 m/sec
  • None of these
A 10 cm long rod AB moves with its ends on two mutually perpendicular straight lines OX and OY. If the end A be moving at the race of 2 cm/sec, then when the distance of A from O is 8 cm, the rate at which the end B is moving, is

  • Maths-Applications of Derivatives-9887.png
  • 2)
    Maths-Applications of Derivatives-9888.png

  • Maths-Applications of Derivatives-9889.png
  • None of these
A triangular park is enclosed on two side on two sides by a fence and on the third side by straight river bank. The two sides having fence are of same length x. The maximum are enclosed by the park is

  • Maths-Applications of Derivatives-9891.png
  • 2)
    Maths-Applications of Derivatives-9892.png

  • Maths-Applications of Derivatives-9893.png

  • Maths-Applications of Derivatives-9894.png
If y = x3 + 5 and x changes from 3 to 2.99, then the approximate change in y is
  • 2.7
  • –0.27
  • 27
  • None of these
A particle is moving on a straight line, where its position’s (in metre) is a function of time t (in seconds) given by = at2 + bt + 6 ≥ 0. If it is known that the particle comes to rest after 4 seconds at a distance of 16 metre from the starting position (t = 0), then the retardation in its motion is
  • –1 m/s2
  • 2)
    Maths-Applications of Derivatives-9897.png

  • Maths-Applications of Derivatives-9898.png

  • Maths-Applications of Derivatives-9899.png
The equation of particle is x = t3 – 9t2 + 3t + 1 and v = –24, then a =
  • 3
  • 2
  • 1
  • 0
A ball thrown vertically upwards falls back on the ground after 6 second, Assuming that the equation of motion is of the form s = ut – 4.9t2, where is in metre and t is in second, find the velocity at t = 0
  • 0 m/s
  • 1 m/s
  • 29.4 m/s
  • None of these
Radius of a circle is increasing uniformly at the rate of 3 cm/sec. The rate of increasing of area when radius is 10 cm, will be
  • π cm2/s
  • 2π cm2/s
  • 10π cm2/s
  • None of these
Two automobiles start from a point A at the same time. One travels towards west at the speed of 60 miles per hour while the other travels towards North at 35 mph. Their distance apart after 3 hours later is increasing at the rate (in mph)

  • Maths-Applications of Derivatives-9904.png
  • 2)
    Maths-Applications of Derivatives-9905.png

  • Maths-Applications of Derivatives-9906.png
  • None of these
A particle is moving in a straight line. Its displacement at time t is given by s = – 4t2 + 2t, then its velocity and acceleration at time t = 1/2 second are
  • –2, –5
  • 2, 6
  • –2, 8
  • 2, 8
If the distance travelled by a point in time t is s = 180t – 16 t2, then the rate of change in velocity is
  • – 16 t unit
  • 48 unit
  • – 32 unit
  • None of these
Water is flowing into a vertical cylindrical tank of radius 2ft. at the rate of 8 cubic ft. per minute. The water level is rising at the speed of (ft.per minute)
  • None of these

  • Maths-Applications of Derivatives-9910.png
  • 2)
    Maths-Applications of Derivatives-9911.png
  • 2
A rectangle of perimeter 24 inches is rotated about one of its sides so as to form a cylinder. To attain the maximum volume of the cylinder what would be the dimensions of the rectangle (in inches)
  • 8, 4
  • 4, 4
  • 8, 8
  • None of these
A ladder is resting with the wall at an angle of 30o. A man is ascending the ladder at the rate of 3 ft/sec. His rate of approaching the wall is
  • 3 ft/sec
  • 2)
    Maths-Applications of Derivatives-9914.png

  • Maths-Applications of Derivatives-9915.png

  • Maths-Applications of Derivatives-9916.png
If the distance ‘s’ travelled by a particle in time t is s = a sin t + b cos 2t, then the acceleration at t = 0 is
  • a
  • – a
  • 4b
  • – 4b
A particle moves so that S = 6 + 48t – t3. The direction of motion reverses after moving a distance of
  • 63
  • 104
  • 134
  • 288
If the volume of a spherical balloon is increasing at the rate 900 cm3/sec, then the rate of charge of radius of balloon instant when radius is 15 cm [in cm/sec]

  • Maths-Applications of Derivatives-9920.png
  • 22

  • Maths-Applications of Derivatives-9921.png
  • None of these
If the path of a moving point is the curve x = at, y = b sin at, then its acceleration at any instant
  • Is constant
  • Varies as the distance from the axis of y
  • Varies as the distance from the axis of y
  • Varies as the distance of the point from the origin
If the rate of increase of area of a circle is not constant but the rate of increase of perimeter is constant, then the rate of increase of area varies
  • As the square of the perimeter
  • Inversely as the perimeter
  • As the radius
  • Inversely as the radius
A stone thrown vertically upwards rises ‘s’ metre in t seconds, where s = 80t – 16t2, then the velocity after 2 second is
  • 8 m/sec
  • 16 m/sec
  • 32 m/sec
  • 64 m/sec

Maths-Applications of Derivatives-9926.png
  • 2
  • 2)
    Maths-Applications of Derivatives-9927.png

  • Maths-Applications of Derivatives-9928.png
  • –3
The speed v of a particle moving along a straight line is given by a + bv2 = x2 (where x is its distance from the origin). The acceleration of the particle is

  • Maths-Applications of Derivatives-9930.png
  • 2)
    Maths-Applications of Derivatives-9931.png

  • Maths-Applications of Derivatives-9932.png

  • Maths-Applications of Derivatives-9933.png
A particle is moving along the curve x = at2 + bt + c. If ac = b2, then the particle would be moving with uniform
  • Rotation
  • Velocity
  • Acceleration
  • retardation
The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when the side is 10 cm is

  • Maths-Applications of Derivatives-9936.png
  • 10 sq. unit/sec

  • Maths-Applications of Derivatives-9937.png

  • Maths-Applications of Derivatives-9938.png
The rate of the change of the surface area of a sphere of radius r when the radius is increasing at the rate of 2cm/sec is proportional to

  • Maths-Applications of Derivatives-9940.png
  • 2)
    Maths-Applications of Derivatives-9941.png
  • r
  • r2
Moving along the x - axis are two points with x = 10 + 6t; x = 3 + t2. The speed with which they are reaching from each other at the time of encounter is (x is in cm and it is in seconds)
  • 16 cm/sec
  • 20 cm/sec
  • 8 cm/sec
  • 12 cm/sec
The position of a point in time ‘t’ is given by x = a + bt – ct2, y = at + bt2. Its acceleration at time ‘t’ is
  • b – c
  • b + c
  • 2b – 2c

  • Maths-Applications of Derivatives-9944.png
Gas is being pumped into a spherical balloon at the rate of 30 ft3/min. Then the rate at which the radius increases when it reaches the value 15 ft. is

  • Maths-Applications of Derivatives-9946.png
  • 2)
    Maths-Applications of Derivatives-9947.png

  • Maths-Applications of Derivatives-9948.png

  • Maths-Applications of Derivatives-9949.png
If the distance ‘s’ metre traversed by a particle in t seconds is given by s = t3 – 3t2, then the velocity of the particle when the acceleration is zero, in metre/sec is
  • 3
  • –2
  • –3
  • 2
A particle moves in a straight line so that s = √t, then its acceleration is proportional to
  • velocity
  • (velocity)3/2
  • (velocity)3
  • (velocity)2
A point on the parabola y2 = 18x at which the ordinate increases at twice the rate of the abscissa is

  • Maths-Applications of Derivatives-9953.png
  • 2)
    Maths-Applications of Derivatives-9954.png

  • Maths-Applications of Derivatives-9955.png

  • Maths-Applications of Derivatives-9956.png
A spherical iron ball 10cm in radius is coated with a layer of ice of uniform thickness that metals at a rate of 50 cm3/min. when the thickness of ice is 5 cm, then the rate of which the thickness of ice decreases, is

  • Maths-Applications of Derivatives-9958.png
  • 2)
    Maths-Applications of Derivatives-9959.png

  • Maths-Applications of Derivatives-9960.png

  • Maths-Applications of Derivatives-9961.png

Maths-Applications of Derivatives-9963.png
  • 1100
  • 1250
  • 1050
  • 5250
A ladder 10 m long against a vertical wall with the lower and on the horizontal ground. The lower end of the ladder is pulled along the ground away from the wall at the rate of 3 cm/sec. The height of the upper end while it is descending at the rate of 4 cm/sec is

  • Maths-Applications of Derivatives-9965.png
  • 2)
    Maths-Applications of Derivatives-9966.png

  • Maths-Applications of Derivatives-9967.png

  • Maths-Applications of Derivatives-9968.png

  • Maths-Applications of Derivatives-9969.png
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 4Ax – v2 is
  • 4 AC + B2
  • 4 AC – B2
  • 2 AC – B2
  • 2 AC + B2
A spherical balloon is filled with 450π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72π cubic meters per minute, the the (rate in meter per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is

  • Maths-Applications of Derivatives-9972.png
  • 2)
    Maths-Applications of Derivatives-9973.png

  • Maths-Applications of Derivatives-9974.png

  • Maths-Applications of Derivatives-9975.png

Maths-Applications of Derivatives-9977.png
  • 2 ln 18
  • ln 19
  • 1/2 ln 18
  • ln 18
The point on the curve y = 12xx3 at which the gradient is zero are
  • (0, 2), (2,
  • (0, –2), (2, –
  • (2, –16), (–2,
  • (2, 16), (–2, –
The slope of tangent to the curve x = t2 + 3t – 8 y = 2t2 – 2t – 5 at the point (2, –is

  • Maths-Applications of Derivatives-9980.png
  • 2)
    Maths-Applications of Derivatives-9981.png
  • –6
  • None of these
The line x + y = 2 is tangent to the curve x2 = 3 – 2y at its point
  • (1, 1)
  • (–1, 1)

  • Maths-Applications of Derivatives-9983.png
  • (3, –3)
The equation of the normal to the curve y2 = ax3 at (a, a) is
  • x + 2y = 3a
  • 3x – 4y + a = 0
  • 4x + 3y = 7a
  • 4x – 3y = 0
The equation of tangent to the curve y = 2 cos x at x = π/4 is

  • Maths-Applications of Derivatives-9987.png
  • 2)
    Maths-Applications of Derivatives-9988.png

  • Maths-Applications of Derivatives-9989.png

  • Maths-Applications of Derivatives-9990.png

Maths-Applications of Derivatives-9992.png
  • (0,2)
  • 2)
    Maths-Applications of Derivatives-9993.png
  • (2,5)
  • (7,6)
  • (6,7)
For the curve by2 = (x + a)3 the square of sub tangent is proportional to
  • (subnormal)1/2
  • Subnormal
  • (subnormal)3/2
  • None of these

Maths-Applications of Derivatives-9996.png
  • a
  • 2a

  • Maths-Applications of Derivatives-9997.png
  • None of these
Co-ordinates of a point on the curve y = x log x at which the normal is parallel to the line 2x – 2y = 3 are
  • (0, 0)
  • (e, e)
  • (e2, 2e2)
  • (e–2, 2e–2)
If normal to the curve y = f(x) is parallel to x - axis, then correct statement is

  • Maths-Applications of Derivatives-10000.png
  • 2)
    Maths-Applications of Derivatives-10001.png

  • Maths-Applications of Derivatives-10002.png
  • None of these
The length of normal to the curve x = a (θ + sin θ), y = a(1 – cos θ) at the point θ = π/2 is

  • Maths-Applications of Derivatives-10004.png
  • 2)
    Maths-Applications of Derivatives-10005.png

  • Maths-Applications of Derivatives-10006.png

  • Maths-Applications of Derivatives-10007.png
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