85588
A wire of length \(20 \mathrm{~cm}\) is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is
1 25 sq. \(\mathrm{cm}\)
2 20 sq. \(\mathrm{cm}\)
3 30 sq. \(\mathrm{cm}\)
4 10 sq. \(\mathrm{cm}\)
Explanation:
(A) : Given, Length \(=20\) Perimeter of sector of a circle- \(2 \mathrm{r}+l\) \(20=2 \mathrm{r}+l \Rightarrow l=20-2 \mathrm{r}\) \(\operatorname{Area}(\mathrm{A})=\frac{\theta}{360} \pi \mathrm{r}^{2}\) \(\mathrm{A}=\frac{\pi \mathrm{r}^{2}}{360} \cdot \frac{l}{\mathrm{r}}=\frac{\pi}{360} l \mathrm{r}=\frac{\pi \mathrm{r}(20-2 \mathrm{r})}{360}\) \(\mathrm{~A}=\frac{\pi}{180}\left(10 \mathrm{r}-\mathrm{r}^{2}\right) \Rightarrow \frac{\mathrm{dA}}{\mathrm{dr}}=\frac{\pi}{180}(10-2 \mathrm{r})\) For maximum \(\frac{\mathrm{dA}}{\mathrm{dr}}=0\) \(\frac{\pi}{180}(10-2 r)=0\) \(r=5\) \(\text { Then } \quad l=20-2 \times 5=10\) \(\text { Area, } \quad=\frac{\pi}{360} \times 5 \times 10=25\)
Karnataka CET-2010
Application of Derivatives
85589
A stone is thrown vertically upwards from the top of a tower 64 metres high according to the law \(s=48 t-16 t^{2}\). The greatest height attained by the stone above the ground is
1 100 metres
2 64 metres
3 36 metres
4 32 metres
Explanation:
(A) : Given, \(\mathrm{s}=48 \mathrm{t}-16 \mathrm{t}^{2}\) On differentiating both sides w. r. t. t, we get- \(\frac{\mathrm{ds}}{\mathrm{dt}}=48-32 \mathrm{t}\) For maximum, \(\frac{\mathrm{ds}}{\mathrm{dt}}=0\) \(48-32 \mathrm{t}=0\) \(\mathrm{t}=\frac{3}{2}\) Hence, \(\mathrm{s}=48 \times \frac{3}{2}-16 \times \frac{9}{4}=72-36=36\) \(\therefore\) The total height \(=64+36=100\)
Karnataka CET-2009
Application of Derivatives
85590
A stone is thrown vertically upwards and the height \(x\) ft. reached by the stone in \(t\) seconds is given by \(x=80 t-16 t^{2}\). The stone reaches the maximum height in
1 3 seconds
2 1.5 seconds
3 2 seconds
4 2.5 seconds
Explanation:
(D) : Given, \(\mathrm{x}=80 \mathrm{t}-16 \mathrm{t}^{2}\) On differentiating both sides w.r.t \(t\), we get- \(\frac{\mathrm{dx}}{\mathrm{dt}}=80-32 \mathrm{t}\) For maximum, \(\frac{\mathrm{dx}}{\mathrm{dt}}=0\) \(80-32 \mathrm{t}=0\) \(\mathrm{t}=\frac{80}{32}=\frac{5}{2}\) \(\mathrm{t}=2.5 \text { seconds }\) Hence, the stone reached the maximum height in 2.5 seconds.
Application of Derivatives
85591
The perimeter of a sector is \(P\). The area of the sector is maximum when its radius is :
1 \(1 / \sqrt{\mathrm{P}}\)
2 \(\mathrm{P} / 2\)
3 \(\mathrm{P} / 4\)
4 \(\sqrt{\mathrm{P}}\)
Explanation:
(C) : Now perimeter of sector \(\mathrm{P}\) \(\mathrm{P}=2 \mathrm{r}+l\) \(\mathrm{P}=2 \mathrm{r}+\mathrm{r} \theta\) \(\theta=\frac{\mathrm{P}-2 \mathrm{r}}{\mathrm{r}}\) \(\operatorname{Area}(\mathrm{A}) =\frac{\pi \mathrm{r}^{2}}{360} \theta\) \(\mathrm{A} =\frac{\pi \mathrm{r}^{2}}{2 \pi}\left(\frac{\mathrm{P}-2 \mathrm{r}}{\mathrm{r}}\right)=\frac{\mathrm{r}}{2}(\mathrm{P}-2 \mathrm{r})\) On differentiating both sides w.r.t \(r\), we get- \(\frac{\mathrm{dA}}{\mathrm{dr}}=\frac{\mathrm{P}-4 \mathrm{r}}{2}\) For maximum area, \(\frac{P-4 r}{2}=0 \Rightarrow r=\frac{P}{4}\)
85588
A wire of length \(20 \mathrm{~cm}\) is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is
1 25 sq. \(\mathrm{cm}\)
2 20 sq. \(\mathrm{cm}\)
3 30 sq. \(\mathrm{cm}\)
4 10 sq. \(\mathrm{cm}\)
Explanation:
(A) : Given, Length \(=20\) Perimeter of sector of a circle- \(2 \mathrm{r}+l\) \(20=2 \mathrm{r}+l \Rightarrow l=20-2 \mathrm{r}\) \(\operatorname{Area}(\mathrm{A})=\frac{\theta}{360} \pi \mathrm{r}^{2}\) \(\mathrm{A}=\frac{\pi \mathrm{r}^{2}}{360} \cdot \frac{l}{\mathrm{r}}=\frac{\pi}{360} l \mathrm{r}=\frac{\pi \mathrm{r}(20-2 \mathrm{r})}{360}\) \(\mathrm{~A}=\frac{\pi}{180}\left(10 \mathrm{r}-\mathrm{r}^{2}\right) \Rightarrow \frac{\mathrm{dA}}{\mathrm{dr}}=\frac{\pi}{180}(10-2 \mathrm{r})\) For maximum \(\frac{\mathrm{dA}}{\mathrm{dr}}=0\) \(\frac{\pi}{180}(10-2 r)=0\) \(r=5\) \(\text { Then } \quad l=20-2 \times 5=10\) \(\text { Area, } \quad=\frac{\pi}{360} \times 5 \times 10=25\)
Karnataka CET-2010
Application of Derivatives
85589
A stone is thrown vertically upwards from the top of a tower 64 metres high according to the law \(s=48 t-16 t^{2}\). The greatest height attained by the stone above the ground is
1 100 metres
2 64 metres
3 36 metres
4 32 metres
Explanation:
(A) : Given, \(\mathrm{s}=48 \mathrm{t}-16 \mathrm{t}^{2}\) On differentiating both sides w. r. t. t, we get- \(\frac{\mathrm{ds}}{\mathrm{dt}}=48-32 \mathrm{t}\) For maximum, \(\frac{\mathrm{ds}}{\mathrm{dt}}=0\) \(48-32 \mathrm{t}=0\) \(\mathrm{t}=\frac{3}{2}\) Hence, \(\mathrm{s}=48 \times \frac{3}{2}-16 \times \frac{9}{4}=72-36=36\) \(\therefore\) The total height \(=64+36=100\)
Karnataka CET-2009
Application of Derivatives
85590
A stone is thrown vertically upwards and the height \(x\) ft. reached by the stone in \(t\) seconds is given by \(x=80 t-16 t^{2}\). The stone reaches the maximum height in
1 3 seconds
2 1.5 seconds
3 2 seconds
4 2.5 seconds
Explanation:
(D) : Given, \(\mathrm{x}=80 \mathrm{t}-16 \mathrm{t}^{2}\) On differentiating both sides w.r.t \(t\), we get- \(\frac{\mathrm{dx}}{\mathrm{dt}}=80-32 \mathrm{t}\) For maximum, \(\frac{\mathrm{dx}}{\mathrm{dt}}=0\) \(80-32 \mathrm{t}=0\) \(\mathrm{t}=\frac{80}{32}=\frac{5}{2}\) \(\mathrm{t}=2.5 \text { seconds }\) Hence, the stone reached the maximum height in 2.5 seconds.
Application of Derivatives
85591
The perimeter of a sector is \(P\). The area of the sector is maximum when its radius is :
1 \(1 / \sqrt{\mathrm{P}}\)
2 \(\mathrm{P} / 2\)
3 \(\mathrm{P} / 4\)
4 \(\sqrt{\mathrm{P}}\)
Explanation:
(C) : Now perimeter of sector \(\mathrm{P}\) \(\mathrm{P}=2 \mathrm{r}+l\) \(\mathrm{P}=2 \mathrm{r}+\mathrm{r} \theta\) \(\theta=\frac{\mathrm{P}-2 \mathrm{r}}{\mathrm{r}}\) \(\operatorname{Area}(\mathrm{A}) =\frac{\pi \mathrm{r}^{2}}{360} \theta\) \(\mathrm{A} =\frac{\pi \mathrm{r}^{2}}{2 \pi}\left(\frac{\mathrm{P}-2 \mathrm{r}}{\mathrm{r}}\right)=\frac{\mathrm{r}}{2}(\mathrm{P}-2 \mathrm{r})\) On differentiating both sides w.r.t \(r\), we get- \(\frac{\mathrm{dA}}{\mathrm{dr}}=\frac{\mathrm{P}-4 \mathrm{r}}{2}\) For maximum area, \(\frac{P-4 r}{2}=0 \Rightarrow r=\frac{P}{4}\)
85588
A wire of length \(20 \mathrm{~cm}\) is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is
1 25 sq. \(\mathrm{cm}\)
2 20 sq. \(\mathrm{cm}\)
3 30 sq. \(\mathrm{cm}\)
4 10 sq. \(\mathrm{cm}\)
Explanation:
(A) : Given, Length \(=20\) Perimeter of sector of a circle- \(2 \mathrm{r}+l\) \(20=2 \mathrm{r}+l \Rightarrow l=20-2 \mathrm{r}\) \(\operatorname{Area}(\mathrm{A})=\frac{\theta}{360} \pi \mathrm{r}^{2}\) \(\mathrm{A}=\frac{\pi \mathrm{r}^{2}}{360} \cdot \frac{l}{\mathrm{r}}=\frac{\pi}{360} l \mathrm{r}=\frac{\pi \mathrm{r}(20-2 \mathrm{r})}{360}\) \(\mathrm{~A}=\frac{\pi}{180}\left(10 \mathrm{r}-\mathrm{r}^{2}\right) \Rightarrow \frac{\mathrm{dA}}{\mathrm{dr}}=\frac{\pi}{180}(10-2 \mathrm{r})\) For maximum \(\frac{\mathrm{dA}}{\mathrm{dr}}=0\) \(\frac{\pi}{180}(10-2 r)=0\) \(r=5\) \(\text { Then } \quad l=20-2 \times 5=10\) \(\text { Area, } \quad=\frac{\pi}{360} \times 5 \times 10=25\)
Karnataka CET-2010
Application of Derivatives
85589
A stone is thrown vertically upwards from the top of a tower 64 metres high according to the law \(s=48 t-16 t^{2}\). The greatest height attained by the stone above the ground is
1 100 metres
2 64 metres
3 36 metres
4 32 metres
Explanation:
(A) : Given, \(\mathrm{s}=48 \mathrm{t}-16 \mathrm{t}^{2}\) On differentiating both sides w. r. t. t, we get- \(\frac{\mathrm{ds}}{\mathrm{dt}}=48-32 \mathrm{t}\) For maximum, \(\frac{\mathrm{ds}}{\mathrm{dt}}=0\) \(48-32 \mathrm{t}=0\) \(\mathrm{t}=\frac{3}{2}\) Hence, \(\mathrm{s}=48 \times \frac{3}{2}-16 \times \frac{9}{4}=72-36=36\) \(\therefore\) The total height \(=64+36=100\)
Karnataka CET-2009
Application of Derivatives
85590
A stone is thrown vertically upwards and the height \(x\) ft. reached by the stone in \(t\) seconds is given by \(x=80 t-16 t^{2}\). The stone reaches the maximum height in
1 3 seconds
2 1.5 seconds
3 2 seconds
4 2.5 seconds
Explanation:
(D) : Given, \(\mathrm{x}=80 \mathrm{t}-16 \mathrm{t}^{2}\) On differentiating both sides w.r.t \(t\), we get- \(\frac{\mathrm{dx}}{\mathrm{dt}}=80-32 \mathrm{t}\) For maximum, \(\frac{\mathrm{dx}}{\mathrm{dt}}=0\) \(80-32 \mathrm{t}=0\) \(\mathrm{t}=\frac{80}{32}=\frac{5}{2}\) \(\mathrm{t}=2.5 \text { seconds }\) Hence, the stone reached the maximum height in 2.5 seconds.
Application of Derivatives
85591
The perimeter of a sector is \(P\). The area of the sector is maximum when its radius is :
1 \(1 / \sqrt{\mathrm{P}}\)
2 \(\mathrm{P} / 2\)
3 \(\mathrm{P} / 4\)
4 \(\sqrt{\mathrm{P}}\)
Explanation:
(C) : Now perimeter of sector \(\mathrm{P}\) \(\mathrm{P}=2 \mathrm{r}+l\) \(\mathrm{P}=2 \mathrm{r}+\mathrm{r} \theta\) \(\theta=\frac{\mathrm{P}-2 \mathrm{r}}{\mathrm{r}}\) \(\operatorname{Area}(\mathrm{A}) =\frac{\pi \mathrm{r}^{2}}{360} \theta\) \(\mathrm{A} =\frac{\pi \mathrm{r}^{2}}{2 \pi}\left(\frac{\mathrm{P}-2 \mathrm{r}}{\mathrm{r}}\right)=\frac{\mathrm{r}}{2}(\mathrm{P}-2 \mathrm{r})\) On differentiating both sides w.r.t \(r\), we get- \(\frac{\mathrm{dA}}{\mathrm{dr}}=\frac{\mathrm{P}-4 \mathrm{r}}{2}\) For maximum area, \(\frac{P-4 r}{2}=0 \Rightarrow r=\frac{P}{4}\)
85588
A wire of length \(20 \mathrm{~cm}\) is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is
1 25 sq. \(\mathrm{cm}\)
2 20 sq. \(\mathrm{cm}\)
3 30 sq. \(\mathrm{cm}\)
4 10 sq. \(\mathrm{cm}\)
Explanation:
(A) : Given, Length \(=20\) Perimeter of sector of a circle- \(2 \mathrm{r}+l\) \(20=2 \mathrm{r}+l \Rightarrow l=20-2 \mathrm{r}\) \(\operatorname{Area}(\mathrm{A})=\frac{\theta}{360} \pi \mathrm{r}^{2}\) \(\mathrm{A}=\frac{\pi \mathrm{r}^{2}}{360} \cdot \frac{l}{\mathrm{r}}=\frac{\pi}{360} l \mathrm{r}=\frac{\pi \mathrm{r}(20-2 \mathrm{r})}{360}\) \(\mathrm{~A}=\frac{\pi}{180}\left(10 \mathrm{r}-\mathrm{r}^{2}\right) \Rightarrow \frac{\mathrm{dA}}{\mathrm{dr}}=\frac{\pi}{180}(10-2 \mathrm{r})\) For maximum \(\frac{\mathrm{dA}}{\mathrm{dr}}=0\) \(\frac{\pi}{180}(10-2 r)=0\) \(r=5\) \(\text { Then } \quad l=20-2 \times 5=10\) \(\text { Area, } \quad=\frac{\pi}{360} \times 5 \times 10=25\)
Karnataka CET-2010
Application of Derivatives
85589
A stone is thrown vertically upwards from the top of a tower 64 metres high according to the law \(s=48 t-16 t^{2}\). The greatest height attained by the stone above the ground is
1 100 metres
2 64 metres
3 36 metres
4 32 metres
Explanation:
(A) : Given, \(\mathrm{s}=48 \mathrm{t}-16 \mathrm{t}^{2}\) On differentiating both sides w. r. t. t, we get- \(\frac{\mathrm{ds}}{\mathrm{dt}}=48-32 \mathrm{t}\) For maximum, \(\frac{\mathrm{ds}}{\mathrm{dt}}=0\) \(48-32 \mathrm{t}=0\) \(\mathrm{t}=\frac{3}{2}\) Hence, \(\mathrm{s}=48 \times \frac{3}{2}-16 \times \frac{9}{4}=72-36=36\) \(\therefore\) The total height \(=64+36=100\)
Karnataka CET-2009
Application of Derivatives
85590
A stone is thrown vertically upwards and the height \(x\) ft. reached by the stone in \(t\) seconds is given by \(x=80 t-16 t^{2}\). The stone reaches the maximum height in
1 3 seconds
2 1.5 seconds
3 2 seconds
4 2.5 seconds
Explanation:
(D) : Given, \(\mathrm{x}=80 \mathrm{t}-16 \mathrm{t}^{2}\) On differentiating both sides w.r.t \(t\), we get- \(\frac{\mathrm{dx}}{\mathrm{dt}}=80-32 \mathrm{t}\) For maximum, \(\frac{\mathrm{dx}}{\mathrm{dt}}=0\) \(80-32 \mathrm{t}=0\) \(\mathrm{t}=\frac{80}{32}=\frac{5}{2}\) \(\mathrm{t}=2.5 \text { seconds }\) Hence, the stone reached the maximum height in 2.5 seconds.
Application of Derivatives
85591
The perimeter of a sector is \(P\). The area of the sector is maximum when its radius is :
1 \(1 / \sqrt{\mathrm{P}}\)
2 \(\mathrm{P} / 2\)
3 \(\mathrm{P} / 4\)
4 \(\sqrt{\mathrm{P}}\)
Explanation:
(C) : Now perimeter of sector \(\mathrm{P}\) \(\mathrm{P}=2 \mathrm{r}+l\) \(\mathrm{P}=2 \mathrm{r}+\mathrm{r} \theta\) \(\theta=\frac{\mathrm{P}-2 \mathrm{r}}{\mathrm{r}}\) \(\operatorname{Area}(\mathrm{A}) =\frac{\pi \mathrm{r}^{2}}{360} \theta\) \(\mathrm{A} =\frac{\pi \mathrm{r}^{2}}{2 \pi}\left(\frac{\mathrm{P}-2 \mathrm{r}}{\mathrm{r}}\right)=\frac{\mathrm{r}}{2}(\mathrm{P}-2 \mathrm{r})\) On differentiating both sides w.r.t \(r\), we get- \(\frac{\mathrm{dA}}{\mathrm{dr}}=\frac{\mathrm{P}-4 \mathrm{r}}{2}\) For maximum area, \(\frac{P-4 r}{2}=0 \Rightarrow r=\frac{P}{4}\)
