The volume of sphere is increasing at the rate of 1200 cu cm/s. T

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381.

The volume of sphere is increasing at the rate of 1200 cu cm/s. The rate of increase in its surface area when the radius is 10 cm is

  • 120 sqcm/s

  • 240 sqcm/s

  • 200 sqcm/s

  • 100 sqcm/s


B.

240 sqcm/s

Let V be the volume, S be the surface area and r be the radius of the sphereIt is given that, dVdt = 1200cucm/sand r = 10cmNow, V = 43πr3 and S = 4πr2 dVdt = 43π3r2drdtand dSdt = 4π . 2rdSdt 1200 = 4π . 102drdtand dSdt = 4π . 210dSdt drdt = 3π and dSdt = 4π 2 × 103π dSdt = 240 sqcm/s


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382.

The slope of the tangent to the curve

y = 0x11 + t3dt at the point, where x = 1 is 

  • 14

  • 13

  • 12

  • 1


383.

The solution of dydx + 1x = eyx2 is

  • 2x = 1 +Cx2 ey

  • x = 1 +Cx2 ey

  • 2x2 = 1 +Cx2 e - y

  • x2 = 1 +Cx2 e - y


384.

The differential equation dydx = 1ax +by +c ,where a, b, c are all non-zero real numbers, is

  • linear in y

  • linear in x

  • linear in both x and y

  • homogeneous equation


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385.

Let D be the domain of a twice differentiable function f.For all x  D, f"(x) + f(x) = 0 and f(x)= g(x)dx + constant.If h(x) = f(x)2 + g(x)2  and h(0) = 5, then h(2015) - h(2014) is equal to

  • 5

  • 3

  • 0

  • 1


386.

The solution of the differential equation 1 + y2 + x - etan-1ydydx = 0, is

  • xetan-1y = tan-1y + C

  • xe2tan-1y = tan-1y + C

  • 2xetan-1y = e2tan-1y + C

  • x2etan-1y = 4e2tan-1y + C


387.

The solution of the differential equation 2x - 4y + 3dydx + x - 2y + 1 = 0 is

  • log2x - 4y + 3 = x - 2y + C

  • log22x - 4y + 3 = 2x - 2y+ C

  • log2x - 2y + 5 = 2x + y + C

  • log4x - 2y + 5 = 4x + 2y + C


388.

For x2 - 4  0, the value of ddxlogexx - 2x + 234 at x = 3 is

  • 85

  • 2

  • 1

  • 8e35


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389.

If  y = sinh-1x1 + x2, then 1 + x2y2 + 3xy1 + y = ?

  • 2

  • 1

  •  - 1 

  • 0


390.

The solution of the equation

x - 4y3dydx - y = 0, y > 0 is

  • x = y3 + cy

  • x + 2y3 = cy

  • y = x3 + cx

  • y + 2x3 = cx


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