Tags Class 11th

Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The line $x + 3 y = 0$ is a diameter of the circle $x^{2} + y^{2} + 6 x + 2 y = 0$ . True or False?

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a true and false Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f t h e c i r c l e i s \\ x^{2} + y^{2} + 6 x + 2 y = 0 \\ C e n t r e i s (- 3, - 1) \\ I f x + 3 y = 0 i s t h e e q u a t i o n o f d i a m e t e r, t h e n t h e c e n t r e (- 3, - 1) w i l l l i e o n x + 3 y = 0 \\ - 3 + 3 (- 1) = 0 \\ \Rightarrow - 6 \neq 0 \\ S o, x + 3 y = 0 i s n o t t h e d i a m e t e r o f t h e c i r c l e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

0 0

New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the hyperbola with:

Vertices $(\pm 5, 0)$ , foci $(\pm 7, 0)$ .

Vertices $(0, \pm 7)$ , eccentricity $e = \sqrt[]{2}$ .

Foci $(0, \pm 10)$ , passing through (2, 3).

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) G i v e n t h a t v e r t i c e s (\pm 5, 0), f o c i (\pm 7, 0) \\ V e r t e x o f h y p e r b o l a = (\pm a, 0) a n d f o c i (\pm a e, 0) \\ ∴ a = 5 a n d a e = 7 \Rightarrow 5 * e = 7 \Rightarrow e = \frac{7}{5} \\ N o w b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = 25 (\frac{49}{25} - 1) \Rightarrow b^{2} = 25 * \frac{24}{25} \Rightarrow b^{2} = 24 \\ T h e e q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{x^{2}}{25} - \frac{y^{2}}{24} = 1 \\ (i i) G i v e n t h a t v e r t i c e s (0, \pm 7), e = \frac{4}{3} \\ C l e a r l y, t h e h y p e r b o l a i s v e r t i c a l . \\ ∴ a = 5 a n d e = \frac{4}{3} \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = 49 (\frac{16}{9} - 1) \Rightarrow b^{2} = 49 * \frac{7}{9} \Rightarrow b^{2} = \frac{343}{9} \\ T h e e q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{y^{2}}{49} - \frac{9 x^{2}}{343} = 1 \Rightarrow 9 x^{2} - 7 y^{2} + 343 = 0 \\ (i i i) G i v e n t h a t : f o c i (0, \pm \sqrt[]{10}) \\ ∴ a e = \sqrt[]{10} \Rightarrow a^{2} e^{2} = 10 \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = a^{2} e^{2} - a^{2} \Rightarrow b^{2} = 10 - a^{2} \\ E q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \Rightarrow \frac{y^{2}}{a^{2}} - \frac{x^{2}}{10 - a^{2}} = 1 \\ I f i t p a s s e s t h r o u g h t h e (2, 3) t h e n \\ \Rightarrow \frac{9}{a^{2}} - \frac{4}{10 - a^{2}} = 1 \Rightarrow \frac{90 - 9 a^{2} - 4 a^{2}}{a^{2} (10 - a^{2})} = 1 \\ \Rightarrow 90 - 13 a^{2} = 10 a^{2} - a^{4} \Rightarrow a^{4} - 23 a^{2} + 90 = 0 \\ \Rightarrow a^{4} - 18 a^{2} - 5 a^{2} + 90 = 0 \\ \Rightarrow a^{2} (a^{2} - 18) - 5 (a^{2} - 18) = 0 \\ \Rightarrow (a^{2} - 18) (a^{2} - 5) = 0 \\ \Rightarrow a^{2} = 18, a^{2} = 5 \\ ∴ b^{2} = 10 - 18 = - 8 a n d b^{2} = 10 - 5 = 5 \\ ∴ b \neq - 8 a n d b^{2} = 5 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s \frac{y^{2}}{5} - \frac{x^{2}}{5} = 1 o r y^{2} - x^{2} = 5 . \end{array}$

0 0

New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Show that the set of all points such that the difference of their distances from (4, 0) and (-4, 0) is always equal to 2 represents a hyperbola.

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t (x, y) b e a n ypoint \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ \sqrt[]{{(x - 4)}^{2} + {(y - 0)}^{2}} - \sqrt[]{{(x + 4)}^{2} + {(y - 0)}^{2}} = 2 \\ \Rightarrow \sqrt[]{x^{2} + 16 - 8 x + y^{2}} - \sqrt[]{x^{2} + 16 + 8 x + y^{2}} = 2 \\ P u t t i n g x^{2} + y^{2} + 16 = z \dots (i) \\ \Rightarrow \sqrt[]{z - 8 x} - \sqrt[]{z + 8 x} = 2 \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ \Rightarrow z - 8 x + z + 8 x - 2 \sqrt[]{(z - 8 x) (z + 8 x)} = 4 \\ \Rightarrow 2 z - 2 \sqrt[]{z^{2} - 64 x^{2}} = 4 \\ \Rightarrow z - \sqrt[]{z^{2} - 64 x^{2}} = 2 \\ \Rightarrow (z - 2) = \sqrt[]{z^{2} - 64 x^{2}} \\ A g a i n s q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow z^{2} + 4 - 4 z = z^{2} - 64 x^{2} \\ \Rightarrow 4 - 4 z + 64 x^{2} = 0 \\ P u t t i n g t h e v a l u e o f z, w e h a v e \\ \Rightarrow 4 - 4 (x^{2} + y^{2} + 16) + 64 x^{2} = 0 \\ \Rightarrow 4 - 4 x^{2} - 4 y^{2} - 64 + 64 x^{2} = 0 \\ \Rightarrow 60 x^{2} - 4 y^{2} - 60 = 0 \\ \Rightarrow 60 x^{2} - 4 y^{2} = 60 \\ \Rightarrow \frac{60 x^{2}}{60} - \frac{4 y^{2}}{60} = 1 \\ \Rightarrow \frac{x^{2}}{1} - \frac{y^{2}}{15} = 1 \\ W h i c h r e p r e s e n t a h p e r b o l a . H e n c e p r o v e d . \end{array}$

0 0

New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the set of all points whose distance from (0, 4) equals their distance from the line $y = 9$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t (x, y) b e a n y \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ \sqrt[]{{(x - 0)}^{2} + {(y - 4)}^{2}} = \frac{2}{3} | \frac{y - 9}{1} | \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ \Rightarrow x^{2} + {(y - 4)}^{2} = \frac{4}{9} (y^{2} + 81 - 18 y) \\ \Rightarrow 9 x^{2} + 9 {(y - 4)}^{2} = 4 y^{2} + 324 - 72 y \\ \Rightarrow 9 x^{2} + 9 y^{2} + 144 - 72 y = 4 y^{2} + 324 - 72 y \\ \Rightarrow 9 x^{2} + 5 y^{2} + 144 - 324 = 0 \\ \Rightarrow 9 x^{2} + 5 y^{2} - 180 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 9 x^{2} + 5 y^{2} - 180 = 0 . \end{array}$

0 0

New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t (x, y) b e a n y \\ G i v e n a r e (3, 0) a n d (9, 0) \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ \sqrt[]{{(x - 3)}^{2} + {(y - 0)}^{2}} + \sqrt[]{{(x - 9)}^{2} + {(y - 0)}^{2}} = 12 \\ \Rightarrow \sqrt[]{x^{2} + 9 - 6 x + y^{2}} + \sqrt[]{x^{2} + 81 - 18 x + y^{2}} = 12 \\ P u t t i n g x^{2} + 9 - 6 x + y^{2} = k \\ \Rightarrow \sqrt[]{k} + \sqrt[]{72 - 12 x + k} = 12 \\ \Rightarrow \sqrt[]{72 - 12 x + k} = 12 - \sqrt[]{k} \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ \Rightarrow 72 - 12 x + k = 144 + k - 24 \sqrt[]{k} \\ \Rightarrow 24 \sqrt[]{k} = 144 - 72 + 12 x \\ \Rightarrow 24 \sqrt[]{k} = 72 + 12 x \\ \Rightarrow 2 \sqrt[]{k} = 6 + x \\ A g a i n s q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow 4 k = 36 + x^{2} + 12 x \\ P u t t i n g t h e v a l u e o f k, w e h a v e \\ 4 (x^{2} + 9 - 6 x + y^{2}) = 36 + x^{2} + 12 x \\ \Rightarrow 4 x^{2} + 36 - 24 x + 4 y^{2} = 36 + x^{2} + 12 x \\ \Rightarrow 3 x^{2} + 4 y^{2} - 36 x = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 3 x^{2} + 4 y^{2} - 36 x = 0 . \end{array}$

0 0

New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of each of the following parabolas:

Directrix $x = 0$ , focus at (6, 0).

Vertex at (0, 4), focus at (0, 2).

Focus at (-1,-2), directrix $x - 2 y + 3 = 0$ .

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) G i v e n t h a t d i r e c t r i x = 0 a n d f o c u s (6, 0) \\ ∴ T h e e q u a t i o n o f t h e p a r a b o l a i s \\ {(x - 6)}^{2} + y^{2} = x^{2} \\ \Rightarrow x^{2} + 36 - 12 x + y^{2} = x^{2} \\ \Rightarrow y^{2} - 12 x + 36 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s y^{2} - 12 x + 36 = 0 \\ (i i) G i v e n t h a t v e r t e x a t (0, 4) a n d f o c u s a t (0, 2) . \\ S o, e q u a t i o n o f d i r e c t r i x i s y - 6 = 0 \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a \\ P F = P M \\ \sqrt[]{{(x - 0)}^{2} + {(y - 2)}^{2}} = | y - 6 | \\ \Rightarrow \sqrt[]{x^{2} + y^{2} + 4 - 4 y} = | y - 6 | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ x^{2} + y^{2} + 4 - 4 y = y^{2} + 36 - 12 y \\ \Rightarrow x^{2} + 4 - 4 y = 36 - 12 y \\ \Rightarrow x^{2} + 8 y - 32 = 0 \\ \Rightarrow x^{2} = 32 - 8 y \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} = 32 - 8 y \\ (i i i) G i v e n t h a t f o c u s a t (- 1, - 2) a n d d i r e c t r i x x - 2 y + 3 = 0 \\ L e t (x, y) b e a n y o n t h e p a r a b o l a . \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a \\ P F = P M \\ \sqrt[]{{(x + 1)}^{2} + {(y + 2)}^{2}} = | \frac{x - 2 y + 3}{\sqrt[]{{(1)}^{2} + {(- 2)}^{2}}} | \\ \Rightarrow \sqrt[]{x^{2} + 1 + 2 x + y^{2} + 4 + 4 y} = | \frac{x - 2 y + 3}{\sqrt[]{5}} | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ x^{2} + 1 + 2 x + y^{2} + 4 + 4 y = \frac{x^{2} + 4 y^{2} + 9 - 4 x y - 12 y + 6 x}{5} \\ \Rightarrow 5 x^{2} + 5 + 10 x + 5 y^{2} + 20 + 20 y = x^{2} + 4 y^{2} + 9 - 4 x y - 12 y + 6 x \\ \Rightarrow 4 x^{2} + y^{2} + 4 x y + 4 x + 32 y + 16 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 4 x^{2} + y^{2} + 4 x y + 4 x + 32 y + 16 = 0 . \end{array}$

0 0

New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of a circle passing through the point (7, 3), having a radius of 3 units, and whose center lies on the line $y = x - 1$ .

0 Follower 10 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f t h e c i r c l e b e \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ I f i t p a s s e s t h r o u g h (7, 3) t h e n \\ {(7 - h)}^{2} + {(3 - k)}^{2} = {(3)}^{2} [? r = 3] \\ \Rightarrow 49 + h^{2} - 14 h + 9 + k^{2} - 6 k = 9 \\ \Rightarrow h^{2} + k^{2} - 14 h - 6 k + 49 = 0 \dots (i) \\ I f c e n t r e (h, k) l i e s o n t h e l i n e y = x - 1 t h e n \\ k = h - 1 \dots (i i) \\ P u t t i n g t h e v a l u e o f k i n e q n . (i) w e g e t \\ h^{2} + {(h - 1)}^{2} - 14 h - 6 (h - 1) + 49 = 0 \\ \Rightarrow h^{2} + h^{2} + 1 - 2 h - 14 h - 6 h + 6 + 49 = 0 \\ \Rightarrow 2 h^{2} - 22 h + 56 = 0 \\ \Rightarrow 2 (h^{2} - 11 h + 28) = 0 \\ \Rightarrow h^{2} - 11 h + 28 = 0 \\ \Rightarrow h^{2} - 7 h - 4 h + 28 = 0 \\ \Rightarrow h (h - 7) - 4 (h - 7) = 0 \\ \Rightarrow (h - 7) (h - 4) = 0 \Rightarrow h = 7, 4 \\ F r o m e q n . (i i) w e g e t k = 4 - 1 = 3 a n d k = 7 - 1 = 6 . \\ S o, t h e c e n t r e s a r e (4, 3) a n d (7, 6) . \\ E q u a t i o n o f t h e c i r c l e i s \\ T a k i n g c e n t r e (4, 3) \\ {(x - 4)}^{2} + {(y - 3)}^{2} = 9 \\ \Rightarrow x^{2} + 16 - 8 x + y^{2} + 9 - 6 y = 9 \\ \Rightarrow x^{2} + y^{2} - 8 x - 6 y + 16 = 0 \\ T a k i n g c e n t r e (7, 6) \\ {(x - 7)}^{2} + {(y - 6)}^{2} = 9 \\ \Rightarrow x^{2} + 49 - 14 x + y^{2} + 36 - 12 y = 9 \\ \Rightarrow x^{2} + y^{2} - 14 x - 12 y + 76 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n s a r e \\ x^{2} + y^{2} - 8 x - 6 y + 16 = 0 \\ a n d x^{2} + y^{2} - 14 x - 12 y + 76 = 0 \end{array}$

0 0

New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of a circle of radius 5 which is touching another circle $x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ at (5, 5).

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n c i r c l e i s = (3, - 1) \\ x^{2} + y^{2} - 2 x - 4 y - 20 = 0 \\ 2 g = - 2 \Rightarrow g = - 1 \\ 2 f = - 4 \Rightarrow f = - 2 \\ ∴ C e n t r e C_{1} = (1, 2) \\ a n d r a d i u s r = \sqrt[]{g^{2} + f^{2} - c} \\ = \sqrt[]{1 + 4 + 20} = 5 \\ L e t t h e c e n t r e o f t h e r e q u i r e d c i r c l e b e (h, k) . \\ C l e a r l y, P i s t h e m i d o f C_{1} C_{2} \\ ∴ 5 = \frac{1 + h}{2} \Rightarrow h = 9 \\ a n d 5 = \frac{2 + k}{2} \Rightarrow k = 8 \\ R a d i u s o f t h e r e q u i r e d c i r c l e = 5 \\ ∴ E q u a t i o n o f t h e c i r c l e i s \\ {(x - 9)}^{2} + {(y - 8)}^{2} = {(5)}^{2} \\ \Rightarrow x^{2} + 81 - 18 x + y^{2} + 64 - 16 y = 25 \\ \Rightarrow x^{2} + y^{2} - 18 x - 16 y + 145 - 25 = 0 \\ \Rightarrow x^{2} + y^{2} - 18 x - 16 y + 120 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 18 x - 16 y + 120 = 0 . \end{array}$

0 0

New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of a circle whose center is (3, -1) and which cuts off a chord of length 6 units on the line $2 x - 5 y + 18 = 0$ .

[Hint: To determine the radius of the circle, find the perpendicular distance from the center to the given line.]

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ C e n t r e o f t h e c i r c l e = (3, - 1) \\ L e n g t h o f c h o r d A B = 6 u n i t s \\ C P = | \frac{2 * 3 - 5 * 1 + 18}{\sqrt[]{{(2)}^{2} + {(- 5)}^{2}}} | \\ = | \frac{6 + 5 + 18}{\sqrt[]{29}} | = \sqrt[]{29} \\ N o w A B = 6 u n i t s . \\ ∴ A P = \frac{1}{2} A B = \frac{1}{2} * 6 = 3 u n i t s \\ I n Δ C P A, A C^{2} = C P^{2} + A P^{2} \\ = {(\sqrt[]{29})}^{2} + {(3)}^{2} = 29 + 9 = 38 \\ ∴ A C = \sqrt[]{38} \\ S o, t h e r a d i u s o f t h e c i r c l e, r = \sqrt[]{38} \\ ∴ E q u a t i o n o f t h e c i r c l e i s \\ {(x - 3)}^{2} + {(y + 1)}^{2} = {(\sqrt[]{38})}^{2} \\ \Rightarrow {(x - 3)}^{2} + {(y + 1)}^{2} = 38 \\ \Rightarrow x^{2} + 9 - 6 x + y^{2} + 1 + 2 y = 38 \\ \Rightarrow x^{2} + y^{2} - 6 x + 2 y = 28 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 6 x + 2 y = 28 . \end{array}$

0 0

New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

Find the equation of the circle which passes through the points (2, 3) and (4, 5), and its center lies on the straight line $y - 4 x + 3 = 0$ .

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \dots (i) \\ I f t h e c i r c l e passes t h r o u g h (2, 3) a n d (4, 5) t h e n \\ {(2 - h)}^{2} + {(3 - k)}^{2} = r^{2} \dots (i i) \\ a n d {(4 - h)}^{2} + {(5 - k)}^{2} = r^{2} \dots (i i i) \\ S u b t r a c t i n g e q n . (i i i) f r o m e q n . (i i) w e h a v e \\ {(2 - h)}^{2} - {(4 - h)}^{2} + {(3 - k)}^{2} - {(5 - k)}^{2} = 0 \\ \Rightarrow 4 + h^{2} - 4 h - 16 - h^{2} + 8 h + 9 + k^{2} - 6 k - 25 - k^{2} + 10 k = 0 \\ \Rightarrow 4 h + 4 k - 28 = 0 \\ \Rightarrow h + k = 7 \dots (i v) \\ , t h e c e n t r e (h, k) l i e s o n t h e l i n e y - 4 x + 3 = 0 \\ t h e n k - 4 h - 3 = 0 \Rightarrow k = 4 h - 3 \\ P u t t i n g t h e v a l u e o f k i n e q n . (i v) w e g e t \\ h + 4 h - 3 = 7 \Rightarrow 5 h = 10 \Rightarrow h = 2 \\ F r o m (i v) w e g e t k = 5 \\ P u t t i n g t h e v a l u e o f h a n d k i n e q n . (i i) w e g e t \\ {(2 - 2)}^{2} + {(3 - 5)}^{2} = r^{2} \Rightarrow r^{2} = 4 \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - 2)}^{2} + {(y - 5)}^{2} = 4 \\ \Rightarrow x^{2} + 4 - 4 x + y^{2} + 25 - 10 y = 4 \\ \Rightarrow x^{2} + y^{2} - 4 x - 10 y + 25 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 4 x - 10 y + 25 = 0 . \end{array}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

67k Colleges
1.2k Exams
717k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 11th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience