If the volume of spherical ball is increasing at the rate of 4&pi

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 Multiple Choice QuestionsMultiple Choice Questions

611.

If the line lx + my + n = 0 is tangent to the parabola y2 = 4ax, then

  • mn = al2

  • lm = an2

  • ln = am2

  • None of the above


612.

All the points on the curve y2 = 4ax + asinxa, where the tangent is parallel to the axis of x are lies on

  • circle

  • parabola

  • straight line

  • None of these


613.

The length of normal at any point to the curve y = c coshxc is

  • fixed

  • y2c2

  • y2c

  • yc2


614.

The height of right circular cylinder of maximum volume inscribed in a sphere of diameter 2a is

  • 23a

  • 3a

  • 2a3

  • a3


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615.

The approximate value of f(x) = x3 + 5x2 - 7x + 9 at x = 1.1 is

  • 8.6

  • 8.5

  • 8.4

  • 8.3


616.

The point on the curve 6y = x3 + 2 at which y-coordinate is changing 8 times as fast as x-coordinate is

  • (4, 11)

  • (4, - 11)

  • (- 4, 11)

  • (- 4, - 11)


617.

The maximum value of fx = logxxx  0, x  1 is

  • e

  • 1e

  • e2

  • 1e2


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618.

If the volume of spherical ball is increasing at the rate of 4π cm3/s, then the rate of change of its surface area when the volume is 288 π cm3, is

  • 43π cm2/s

  • 23π cm2/s

  • 4π cm2/s

  • 2π cm2/s


A.

43π cm2/s

Let V and r be the volume and radius of spherical ball, respectively.

Volume of spherical ball = 43πr3

       V = 43πr3     ...i 288π = 43πr3   given, V = 288 cm3    288 = 43r3       r3 = 72 × 3 = 8 × 27       r3 = 23 × 33             taking cube roots both sides        r = 6

On differentiating Eq. (i) w.r.t. 't', we get

     dVdt = 4πr2drdt  4π = 4πr2drdt       given dVdt = 4π cubic cm/s    1 = 62drdt      r = 6 drdt = 136Now, surface area of spherical ball, (s) = 4πr2    s = 4πr2

On differentiating both sides, w.r.t. 't', we get

    dsdt = 4 × 2πrdrdt           = 8 × π × 6 × 136           r = 6 and drdt = 136 dsdt = 43π cm2/s


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619.

The equation of displacement of a particle is x(t) = 5t2 - 7t + 3. The acceleration at the moment when its velocity becomes 5 m/sec is

  • 3 m/sec2

  • 7 m/sec2

  • 10 m/sec2

  • 8 m/sec2


620.

The mean value of the function fx = 2ex + 1 on the interval [0, 2] is

  • 2 - loge2e2 + 1

  • 2 + loge2e2 + 1

  • 2 + loge2e2 - 1

  • - 2 + loge2e2 - 1


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