85136
A particle moves in a straight line with a velocity given by \(\frac{d x}{d t}=x+1\), where \(x\) is the distance travelled in time \(t\). The time taken by the particle to travel a distance of 999 metres is
1 \(\log _{1000} \mathrm{e}\)
2 \(\log _{\mathrm{e}} 100\)
3 \(3 \log _{\mathrm{e}} 10\)
4 \(4 \log _{\mathrm{e}} 10\)
Explanation:
(C) : \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{x}+1\) \(\frac{d x}{x+1}=d t\) \(\int \frac{1}{x+1} d x=\int d t\) \(\log _{e}(x+1)=t+c\) Initially when \(t=0\) then \(x=0\) \(0=0+c\) \(c=0\) \(\therefore \quad \log _{\mathrm{e}}(\mathrm{x}+1)=\mathrm{t}\) When \(\mathrm{x}=999\) Then, \(\mathrm{t}=\log _{\mathrm{e}}(\mathrm{x}+1), \mathrm{t}=\log _{\mathrm{e}}(1000)\) \(\mathrm{t}=\log _{\mathrm{e}} 10^{3,} \mathrm{t}=3 \log _{\mathrm{e}} 10 end\)
AMU-2012
Application of Derivatives
85137
If the radius of a sphere is mentioned as \(7 \mathrm{~m}\) with an error of \(0.02 \mathrm{~m}\), then the approximate error in calculating its volumes is
85139
There is an error of \(\pm 0.04 \mathrm{~cm}\) in the measurement of the diameter of a sphere. When the radius is \(10 \mathrm{~cm}\), the percentage error in the volume of the sphere is-
85141
The volume of a spherical balloon is increasing at the rate of 0.05 cubic meters per seconds. How fast is the surface area increasing when the radius is 2 meters?
1 0.05 square meters per second
2 0.25 square meters per second
3 1.05 square meters per second
4 1.25 square meters per second
Explanation:
(A) : Given, \(\frac{\mathrm{dV}}{\mathrm{dt}}=0.05 \mathrm{~m}^{3} / \mathrm{sec}, \mathrm{r}=2 \mathrm{~m}\) Let \(\mathrm{r}\) radius, \(\mathrm{V}\) volume and \(\mathrm{S}\) is the surface area of spherical balloon. \(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\) And, \(\quad S=4 \pi r^{2}\) Then rate of change of volume with respect to \(t\) Then, \(\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi 3 \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}=0.05\) \(4 \pi r^{2} \frac{d r}{d t}=0.05\) \(\frac{d r}{d t}=\frac{5}{400 \pi r^{2}}\) Now rate of change of surface area, \(\frac{\mathrm{dS}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(4 \pi \mathrm{r}^{2}\right)\) \(=8 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=8 \pi \mathrm{r} \times \frac{5}{400 \pi \mathrm{r}^{2}}\) \(=\frac{1}{\mathrm{r} \times 10}=\frac{1}{2 \times 10}=\frac{1}{20}\) \(\frac{\mathrm{dS}}{\mathrm{dt}}=0.05 \text { Square meter } / \mathrm{sec}\)
85136
A particle moves in a straight line with a velocity given by \(\frac{d x}{d t}=x+1\), where \(x\) is the distance travelled in time \(t\). The time taken by the particle to travel a distance of 999 metres is
1 \(\log _{1000} \mathrm{e}\)
2 \(\log _{\mathrm{e}} 100\)
3 \(3 \log _{\mathrm{e}} 10\)
4 \(4 \log _{\mathrm{e}} 10\)
Explanation:
(C) : \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{x}+1\) \(\frac{d x}{x+1}=d t\) \(\int \frac{1}{x+1} d x=\int d t\) \(\log _{e}(x+1)=t+c\) Initially when \(t=0\) then \(x=0\) \(0=0+c\) \(c=0\) \(\therefore \quad \log _{\mathrm{e}}(\mathrm{x}+1)=\mathrm{t}\) When \(\mathrm{x}=999\) Then, \(\mathrm{t}=\log _{\mathrm{e}}(\mathrm{x}+1), \mathrm{t}=\log _{\mathrm{e}}(1000)\) \(\mathrm{t}=\log _{\mathrm{e}} 10^{3,} \mathrm{t}=3 \log _{\mathrm{e}} 10 end\)
AMU-2012
Application of Derivatives
85137
If the radius of a sphere is mentioned as \(7 \mathrm{~m}\) with an error of \(0.02 \mathrm{~m}\), then the approximate error in calculating its volumes is
85139
There is an error of \(\pm 0.04 \mathrm{~cm}\) in the measurement of the diameter of a sphere. When the radius is \(10 \mathrm{~cm}\), the percentage error in the volume of the sphere is-
85141
The volume of a spherical balloon is increasing at the rate of 0.05 cubic meters per seconds. How fast is the surface area increasing when the radius is 2 meters?
1 0.05 square meters per second
2 0.25 square meters per second
3 1.05 square meters per second
4 1.25 square meters per second
Explanation:
(A) : Given, \(\frac{\mathrm{dV}}{\mathrm{dt}}=0.05 \mathrm{~m}^{3} / \mathrm{sec}, \mathrm{r}=2 \mathrm{~m}\) Let \(\mathrm{r}\) radius, \(\mathrm{V}\) volume and \(\mathrm{S}\) is the surface area of spherical balloon. \(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\) And, \(\quad S=4 \pi r^{2}\) Then rate of change of volume with respect to \(t\) Then, \(\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi 3 \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}=0.05\) \(4 \pi r^{2} \frac{d r}{d t}=0.05\) \(\frac{d r}{d t}=\frac{5}{400 \pi r^{2}}\) Now rate of change of surface area, \(\frac{\mathrm{dS}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(4 \pi \mathrm{r}^{2}\right)\) \(=8 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=8 \pi \mathrm{r} \times \frac{5}{400 \pi \mathrm{r}^{2}}\) \(=\frac{1}{\mathrm{r} \times 10}=\frac{1}{2 \times 10}=\frac{1}{20}\) \(\frac{\mathrm{dS}}{\mathrm{dt}}=0.05 \text { Square meter } / \mathrm{sec}\)
85136
A particle moves in a straight line with a velocity given by \(\frac{d x}{d t}=x+1\), where \(x\) is the distance travelled in time \(t\). The time taken by the particle to travel a distance of 999 metres is
1 \(\log _{1000} \mathrm{e}\)
2 \(\log _{\mathrm{e}} 100\)
3 \(3 \log _{\mathrm{e}} 10\)
4 \(4 \log _{\mathrm{e}} 10\)
Explanation:
(C) : \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{x}+1\) \(\frac{d x}{x+1}=d t\) \(\int \frac{1}{x+1} d x=\int d t\) \(\log _{e}(x+1)=t+c\) Initially when \(t=0\) then \(x=0\) \(0=0+c\) \(c=0\) \(\therefore \quad \log _{\mathrm{e}}(\mathrm{x}+1)=\mathrm{t}\) When \(\mathrm{x}=999\) Then, \(\mathrm{t}=\log _{\mathrm{e}}(\mathrm{x}+1), \mathrm{t}=\log _{\mathrm{e}}(1000)\) \(\mathrm{t}=\log _{\mathrm{e}} 10^{3,} \mathrm{t}=3 \log _{\mathrm{e}} 10 end\)
AMU-2012
Application of Derivatives
85137
If the radius of a sphere is mentioned as \(7 \mathrm{~m}\) with an error of \(0.02 \mathrm{~m}\), then the approximate error in calculating its volumes is
85139
There is an error of \(\pm 0.04 \mathrm{~cm}\) in the measurement of the diameter of a sphere. When the radius is \(10 \mathrm{~cm}\), the percentage error in the volume of the sphere is-
85141
The volume of a spherical balloon is increasing at the rate of 0.05 cubic meters per seconds. How fast is the surface area increasing when the radius is 2 meters?
1 0.05 square meters per second
2 0.25 square meters per second
3 1.05 square meters per second
4 1.25 square meters per second
Explanation:
(A) : Given, \(\frac{\mathrm{dV}}{\mathrm{dt}}=0.05 \mathrm{~m}^{3} / \mathrm{sec}, \mathrm{r}=2 \mathrm{~m}\) Let \(\mathrm{r}\) radius, \(\mathrm{V}\) volume and \(\mathrm{S}\) is the surface area of spherical balloon. \(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\) And, \(\quad S=4 \pi r^{2}\) Then rate of change of volume with respect to \(t\) Then, \(\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi 3 \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}=0.05\) \(4 \pi r^{2} \frac{d r}{d t}=0.05\) \(\frac{d r}{d t}=\frac{5}{400 \pi r^{2}}\) Now rate of change of surface area, \(\frac{\mathrm{dS}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(4 \pi \mathrm{r}^{2}\right)\) \(=8 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=8 \pi \mathrm{r} \times \frac{5}{400 \pi \mathrm{r}^{2}}\) \(=\frac{1}{\mathrm{r} \times 10}=\frac{1}{2 \times 10}=\frac{1}{20}\) \(\frac{\mathrm{dS}}{\mathrm{dt}}=0.05 \text { Square meter } / \mathrm{sec}\)
85136
A particle moves in a straight line with a velocity given by \(\frac{d x}{d t}=x+1\), where \(x\) is the distance travelled in time \(t\). The time taken by the particle to travel a distance of 999 metres is
1 \(\log _{1000} \mathrm{e}\)
2 \(\log _{\mathrm{e}} 100\)
3 \(3 \log _{\mathrm{e}} 10\)
4 \(4 \log _{\mathrm{e}} 10\)
Explanation:
(C) : \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{x}+1\) \(\frac{d x}{x+1}=d t\) \(\int \frac{1}{x+1} d x=\int d t\) \(\log _{e}(x+1)=t+c\) Initially when \(t=0\) then \(x=0\) \(0=0+c\) \(c=0\) \(\therefore \quad \log _{\mathrm{e}}(\mathrm{x}+1)=\mathrm{t}\) When \(\mathrm{x}=999\) Then, \(\mathrm{t}=\log _{\mathrm{e}}(\mathrm{x}+1), \mathrm{t}=\log _{\mathrm{e}}(1000)\) \(\mathrm{t}=\log _{\mathrm{e}} 10^{3,} \mathrm{t}=3 \log _{\mathrm{e}} 10 end\)
AMU-2012
Application of Derivatives
85137
If the radius of a sphere is mentioned as \(7 \mathrm{~m}\) with an error of \(0.02 \mathrm{~m}\), then the approximate error in calculating its volumes is
85139
There is an error of \(\pm 0.04 \mathrm{~cm}\) in the measurement of the diameter of a sphere. When the radius is \(10 \mathrm{~cm}\), the percentage error in the volume of the sphere is-
85141
The volume of a spherical balloon is increasing at the rate of 0.05 cubic meters per seconds. How fast is the surface area increasing when the radius is 2 meters?
1 0.05 square meters per second
2 0.25 square meters per second
3 1.05 square meters per second
4 1.25 square meters per second
Explanation:
(A) : Given, \(\frac{\mathrm{dV}}{\mathrm{dt}}=0.05 \mathrm{~m}^{3} / \mathrm{sec}, \mathrm{r}=2 \mathrm{~m}\) Let \(\mathrm{r}\) radius, \(\mathrm{V}\) volume and \(\mathrm{S}\) is the surface area of spherical balloon. \(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\) And, \(\quad S=4 \pi r^{2}\) Then rate of change of volume with respect to \(t\) Then, \(\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi 3 \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}=0.05\) \(4 \pi r^{2} \frac{d r}{d t}=0.05\) \(\frac{d r}{d t}=\frac{5}{400 \pi r^{2}}\) Now rate of change of surface area, \(\frac{\mathrm{dS}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(4 \pi \mathrm{r}^{2}\right)\) \(=8 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=8 \pi \mathrm{r} \times \frac{5}{400 \pi \mathrm{r}^{2}}\) \(=\frac{1}{\mathrm{r} \times 10}=\frac{1}{2 \times 10}=\frac{1}{20}\) \(\frac{\mathrm{dS}}{\mathrm{dt}}=0.05 \text { Square meter } / \mathrm{sec}\)
85136
A particle moves in a straight line with a velocity given by \(\frac{d x}{d t}=x+1\), where \(x\) is the distance travelled in time \(t\). The time taken by the particle to travel a distance of 999 metres is
1 \(\log _{1000} \mathrm{e}\)
2 \(\log _{\mathrm{e}} 100\)
3 \(3 \log _{\mathrm{e}} 10\)
4 \(4 \log _{\mathrm{e}} 10\)
Explanation:
(C) : \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{x}+1\) \(\frac{d x}{x+1}=d t\) \(\int \frac{1}{x+1} d x=\int d t\) \(\log _{e}(x+1)=t+c\) Initially when \(t=0\) then \(x=0\) \(0=0+c\) \(c=0\) \(\therefore \quad \log _{\mathrm{e}}(\mathrm{x}+1)=\mathrm{t}\) When \(\mathrm{x}=999\) Then, \(\mathrm{t}=\log _{\mathrm{e}}(\mathrm{x}+1), \mathrm{t}=\log _{\mathrm{e}}(1000)\) \(\mathrm{t}=\log _{\mathrm{e}} 10^{3,} \mathrm{t}=3 \log _{\mathrm{e}} 10 end\)
AMU-2012
Application of Derivatives
85137
If the radius of a sphere is mentioned as \(7 \mathrm{~m}\) with an error of \(0.02 \mathrm{~m}\), then the approximate error in calculating its volumes is
85139
There is an error of \(\pm 0.04 \mathrm{~cm}\) in the measurement of the diameter of a sphere. When the radius is \(10 \mathrm{~cm}\), the percentage error in the volume of the sphere is-
85141
The volume of a spherical balloon is increasing at the rate of 0.05 cubic meters per seconds. How fast is the surface area increasing when the radius is 2 meters?
1 0.05 square meters per second
2 0.25 square meters per second
3 1.05 square meters per second
4 1.25 square meters per second
Explanation:
(A) : Given, \(\frac{\mathrm{dV}}{\mathrm{dt}}=0.05 \mathrm{~m}^{3} / \mathrm{sec}, \mathrm{r}=2 \mathrm{~m}\) Let \(\mathrm{r}\) radius, \(\mathrm{V}\) volume and \(\mathrm{S}\) is the surface area of spherical balloon. \(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\) And, \(\quad S=4 \pi r^{2}\) Then rate of change of volume with respect to \(t\) Then, \(\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi 3 \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}=0.05\) \(4 \pi r^{2} \frac{d r}{d t}=0.05\) \(\frac{d r}{d t}=\frac{5}{400 \pi r^{2}}\) Now rate of change of surface area, \(\frac{\mathrm{dS}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(4 \pi \mathrm{r}^{2}\right)\) \(=8 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=8 \pi \mathrm{r} \times \frac{5}{400 \pi \mathrm{r}^{2}}\) \(=\frac{1}{\mathrm{r} \times 10}=\frac{1}{2 \times 10}=\frac{1}{20}\) \(\frac{\mathrm{dS}}{\mathrm{dt}}=0.05 \text { Square meter } / \mathrm{sec}\)
