NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85132
For the curve \(y=5 x-2 x^{3}\), if ' \(x^{\prime}\) increases at the rate of 2 units/sec. the rate of change in the slope of the curve at \(x=3\) is /sec
85133
The radius of a sphere initially at zero increases at the rate of \(5 \mathrm{~cm} / \mathrm{sec}\). Then its volume after \(1 \mathrm{sec}\) is increasing at the rate 0
85134
For a particle moving on a straight line it is observed that the distance ' \(S\) ' at a time ' \(t\) ' is given by \(S=6 t-\frac{t^{3}}{2}\). The maximum velocity during the motion is
1 3
2 6
3 9
4 12
Explanation:
(B) : Given, \(S=6 t-\frac{t^{3}}{2}\) Velocity \((\mathrm{v})=\frac{\mathrm{dS}}{\mathrm{dt}}=6-\frac{3}{2} \mathrm{t}^{2} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=0-3 \mathrm{t}\) For maximum velocity. \(\frac{\mathrm{dv}}{\mathrm{dt}}=0 \Rightarrow-3 \mathrm{t}=0\) Velocity will be maximum at \(\mathrm{t}=0\) \(\mathrm{v}=6-\frac{3}{2} \mathrm{t}^{2}, \quad \mathrm{v}=6-0, \quad \mathrm{v}=6\)
AP EAMCET-2022-06.07.2022
Application of Derivatives
85135
The radius of a sphere increases at the rate of \(0.04 \mathrm{~cm} / \mathrm{sec}\). The rate of increase in the volume of that sphere with respect to its surface area, when its radius is \(10 \mathrm{~cm}\) is
85132
For the curve \(y=5 x-2 x^{3}\), if ' \(x^{\prime}\) increases at the rate of 2 units/sec. the rate of change in the slope of the curve at \(x=3\) is /sec
85133
The radius of a sphere initially at zero increases at the rate of \(5 \mathrm{~cm} / \mathrm{sec}\). Then its volume after \(1 \mathrm{sec}\) is increasing at the rate 0
85134
For a particle moving on a straight line it is observed that the distance ' \(S\) ' at a time ' \(t\) ' is given by \(S=6 t-\frac{t^{3}}{2}\). The maximum velocity during the motion is
1 3
2 6
3 9
4 12
Explanation:
(B) : Given, \(S=6 t-\frac{t^{3}}{2}\) Velocity \((\mathrm{v})=\frac{\mathrm{dS}}{\mathrm{dt}}=6-\frac{3}{2} \mathrm{t}^{2} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=0-3 \mathrm{t}\) For maximum velocity. \(\frac{\mathrm{dv}}{\mathrm{dt}}=0 \Rightarrow-3 \mathrm{t}=0\) Velocity will be maximum at \(\mathrm{t}=0\) \(\mathrm{v}=6-\frac{3}{2} \mathrm{t}^{2}, \quad \mathrm{v}=6-0, \quad \mathrm{v}=6\)
AP EAMCET-2022-06.07.2022
Application of Derivatives
85135
The radius of a sphere increases at the rate of \(0.04 \mathrm{~cm} / \mathrm{sec}\). The rate of increase in the volume of that sphere with respect to its surface area, when its radius is \(10 \mathrm{~cm}\) is
85132
For the curve \(y=5 x-2 x^{3}\), if ' \(x^{\prime}\) increases at the rate of 2 units/sec. the rate of change in the slope of the curve at \(x=3\) is /sec
85133
The radius of a sphere initially at zero increases at the rate of \(5 \mathrm{~cm} / \mathrm{sec}\). Then its volume after \(1 \mathrm{sec}\) is increasing at the rate 0
85134
For a particle moving on a straight line it is observed that the distance ' \(S\) ' at a time ' \(t\) ' is given by \(S=6 t-\frac{t^{3}}{2}\). The maximum velocity during the motion is
1 3
2 6
3 9
4 12
Explanation:
(B) : Given, \(S=6 t-\frac{t^{3}}{2}\) Velocity \((\mathrm{v})=\frac{\mathrm{dS}}{\mathrm{dt}}=6-\frac{3}{2} \mathrm{t}^{2} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=0-3 \mathrm{t}\) For maximum velocity. \(\frac{\mathrm{dv}}{\mathrm{dt}}=0 \Rightarrow-3 \mathrm{t}=0\) Velocity will be maximum at \(\mathrm{t}=0\) \(\mathrm{v}=6-\frac{3}{2} \mathrm{t}^{2}, \quad \mathrm{v}=6-0, \quad \mathrm{v}=6\)
AP EAMCET-2022-06.07.2022
Application of Derivatives
85135
The radius of a sphere increases at the rate of \(0.04 \mathrm{~cm} / \mathrm{sec}\). The rate of increase in the volume of that sphere with respect to its surface area, when its radius is \(10 \mathrm{~cm}\) is
85132
For the curve \(y=5 x-2 x^{3}\), if ' \(x^{\prime}\) increases at the rate of 2 units/sec. the rate of change in the slope of the curve at \(x=3\) is /sec
85133
The radius of a sphere initially at zero increases at the rate of \(5 \mathrm{~cm} / \mathrm{sec}\). Then its volume after \(1 \mathrm{sec}\) is increasing at the rate 0
85134
For a particle moving on a straight line it is observed that the distance ' \(S\) ' at a time ' \(t\) ' is given by \(S=6 t-\frac{t^{3}}{2}\). The maximum velocity during the motion is
1 3
2 6
3 9
4 12
Explanation:
(B) : Given, \(S=6 t-\frac{t^{3}}{2}\) Velocity \((\mathrm{v})=\frac{\mathrm{dS}}{\mathrm{dt}}=6-\frac{3}{2} \mathrm{t}^{2} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=0-3 \mathrm{t}\) For maximum velocity. \(\frac{\mathrm{dv}}{\mathrm{dt}}=0 \Rightarrow-3 \mathrm{t}=0\) Velocity will be maximum at \(\mathrm{t}=0\) \(\mathrm{v}=6-\frac{3}{2} \mathrm{t}^{2}, \quad \mathrm{v}=6-0, \quad \mathrm{v}=6\)
AP EAMCET-2022-06.07.2022
Application of Derivatives
85135
The radius of a sphere increases at the rate of \(0.04 \mathrm{~cm} / \mathrm{sec}\). The rate of increase in the volume of that sphere with respect to its surface area, when its radius is \(10 \mathrm{~cm}\) is
