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

10 months ago

Class 11th Straight Lines

8. Find the value of x for which the points (x, – 1), (2,1) and (4, 5) are collinear.

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

8. Let P (x, –1), Q (2, 1) and R (4, 5) be the collinear points. Then,

Slope of PQ = Slope of QR

$\Rightarrow \frac{1 - (- 1)}{2 - x} = \frac{5 - 1}{4 - 2}$

$\Rightarrow \frac{1 + 1}{2 - x} = \frac{4}{2}$

$\Rightarrow 2 * 2 = 4 (2 - x)$

$\Rightarrow 4 = 8 - 4 x$

$\Rightarrow x = \frac{4}{4}$

$\Rightarrow x = 1$

0 0

New answer posted

10 months ago

Class 11th Straight Lines

7. Find the slope of the line, which makes an angle of 30°with the positive direction of y-axis measured anticlockwise.

0 Follower 20 Views

alok kumar singh

Contributor-Level 10

Let l be the line making 30° with y-axis as shown in figure. Then,

Angle a = $∠ b$ + 90° (Sum of exterior angle of a triangle)

$\Rightarrow ∠ a = 30 ° + 90 °$ ( $30 ° = ∠ b$ vertically opposite angle)

$\Rightarrow ∠ a = 120 °$

So, slope of line l = tan a

= m = tan 120°

= tan (180° – 60°)

= –tan 60°

$= -\sqrt3$

0 0

New answer posted

10 months ago

Class 11th Straight Lines

6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

. Let the given point be A (4, 4), B (3, 5) and C (–1, –1)

Then, slope of AB, m₁ = $\frac{5 - 4}{3 - 4} = \frac{1}{- 1} = - 1$

Slope of AC, m₂ = $\frac{- 1 - 4}{- 1 - 4} = \frac{- 5}{- 5} = 1$

And slope of BC, m₃ = $= \frac{- 1 - 5}{- 1 - 3} = \frac{- 6}{- 4} = \frac{3}{2}$

As m₁ – m₂ = –1 * 1 = –1

$\Rightarrow m_{1} = \frac{- 1}{m_{2}}$

We conclude that AB and AC are perpendicular to each other.

Hence, ABC is a right-angle triangle right-angled at A

0 0

New answer posted

10 months ago

Class 11th Straight Lines

5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

Let 0 (0, 0) be the origin and A be the mid-point of line joining P (0, –4) and B (8, 0)

Then, co-ordinate of A = $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) = (\frac{0 + 8}{2}, \frac{- 4 + 0}{2}) = (4, - 2)$

$∴$ Slope of OA, m = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{- 2 - 0}{4 - 0} = - \frac{2}{4} = \frac{- 1}{2}$

0 0

New answer posted

10 months ago

Class 11th Straight Lines

4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

0 Follower 9 Views

alok kumar singh

Contributor-Level 10

Let A (x, 0) be the point on x-axis when is equidistant from P (7, 6) and Q (3, 4)

Then, PA = QA

Squaring both sides, we get,

$\Rightarrow x^{2} - 14 x + 49 + 36 = x^{2} - 6 x + 9 + 16$

$\Rightarrow 49 + 36 - 9 - 16 = 14 x - 6 x$

$\Rightarrow 60 = 8 x$

$\Rightarrow x = \frac{60}{8} = \frac{15}{2}$

$∴$ The required point on x-axis is $(\frac{15}{2}, 0) .$

0 0

New answer posted

10 months ago

Class 11th Straight Lines

3. Find the distance between P (x1, y1) and Q (x2, y2) when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

0 Follower 18 Views

alok kumar singh

Contributor-Level 10

0 0

New question posted

10 months ago

Class 11th Straight Lines

3. Find the distance between P (x1, y1) and Q (x2, y2) when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

0 Follower 11 Views

New answer posted

10 months ago

Class 11th Straight Lines

2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

0 Follower 49 Views

alok kumar singh

Contributor-Level 10

2. Let ABC be the equilateral triangle of side 2a and 0 be the origin. Then

AB = BC = AC = 2a

O is the mid-point of AB we have AO = a

BO = a

We know that A and B lies on y-axis so they have co-ordinate of the form (0, y).

Hence, co-ordinate of A is (0, a) and that of B is (0, –a)

Since OC, bisects AB at right angle, by Pythagoras theorem,

AC²= OA² + OC²

$\Rightarrow {(2 a)}^{2} = {(a)}^{2} +OC^{2}$

$\RightarrowOC^{2} = 4 a^{2} - a^{2}$

And as C we on x-axis it has co-ordinate of the form (x, 0)

0 0

New answer posted

10 months ago

Class 11th Straight Lines

Q1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

0 Follower 83 Views

alok kumar singh

Contributor-Level 10

Exercise 9.1

1. Let the given points be A(–4, 5), B(0, 7), C(5, –5) and D(–4, –2).

Then quadrilated ABCD can be plotted on the graph by joining the points A, B, C and D.

We connect diagonal AC such that

area (ABCD) = (ΔABC) + (ΔADC)

Now,

$(ΔABC) = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | - 4 [7 - (- 5)] + 0 [- 5 - 5] + 5 [5 - 7] | unit^{2}$

$= \frac{1}{2} | - 4 (7 + 5) + 0 + 5 (5 - 7) | unit^{2}$

$= \frac{1}{2} | (- 4) * 12 + 5 * (- 2) |unit^{2}$

$= \frac{1}{2} | - 48 - 10 |unit^{2}$

$= \frac{1}{2} | - 58 | unit^{2}$

$= \frac{58}{2} unit^{2} = 29 unit^{2}$

Similarly, $(ΔACD) = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | - 4 [- 5 - (2)] + 5 [- 2 - 5] + (- 4) [5 - (- 5)] | unit^{2}$

$= \frac{1}{2} | - 4 (- 5 + 2) + 5 (- 2 - 5) - 4 (5 + 5) |unit^{2}$

$= \frac{1}{2} | (- 4) * (- 3) + 5 * (- 7) - 4 * (10) |unit^{2}$

$= \frac{1}{2} | 12 - 35 - 40 |unit^{2}$

$= \frac{1}{2} | - 63 | unit^{2}$

$= \frac{63}{2} unit^{2}$

Hence, area (ABCD) = $29 + \frac{63}{2} unit^{2}$

$= \frac{58 + 63}{2} unit^{2} = \frac{121}{2} unit^{2}$

0 0

New answer posted

11 months ago

Class 11th Mechanical Properties of Fluids

10.31 (a) It is known that density $ρ$ of air decreases with height y as

where $ρ = ρ_{o} e^{\frac{- y}{y_{o}}}$ = 1.25 kg m^–3 is the density at sea level, and $ρ_{o}$ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.

(b) A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?

[Take $y_{o}$ = 8000 m and $y_{o}$ = 0.18 kg m–3].

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

Volume of the balloon, V = 1425 $ρ_{H e}$

Mass of the payload, m = 400 kg

Acceleration due to gravity, g = 9.8 m/ $m^{3}$

$s^{2}$ = 8000 m

$y_{o}$ = 0.18 kg m^–3

$ρ_{H e}$ = 1.25 kg m^–3

Density of the balloon = $ρ_{o}$

Height to which the balloon will rise = y

Density of air decreases with height and the relationship is given by:

$ρ$ = $ρ = ρ_{o} e^{\frac{- y}{y_{o}}}$ ……(i)

Differentiating equation (i), we get

$\frac{ρ}{ρ_{o}}$ $e^{\frac{- y}{y_{o}}}$

$- \frac{d ρ}{d y}$ , where k is the constant of proportionality

$α ρ$ , height changes from 0 to y, while density changes from $\frac{d ρ}{d y} = - k ρ$ to $\frac{d ρ}{ρ} = - k d y$ . Integrating both sides between the limits, we get:

$ρ_{o}$

$ρ$ = -ky

$\int_{ρ_{o}}^{ρ} \frac{d ρ}{ρ} = - \int_{0}^{y} k d y$ = ${[\log_{e} ? ρ]}_{ρ_{o}}^{ρ}$ ….(ii)

From equation (i) and (ii), we get

$\frac{ρ}{ρ_{o}}$ =&nbs

...more

Volume of the balloon, V = 1425 $ρ_{H e}$

Mass of the payload, m = 400 kg

Acceleration due to gravity, g = 9.8 m/ $m^{3}$

$s^{2}$ = 8000 m

$y_{o}$ = 0.18 kg m^–3

$ρ_{H e}$ = 1.25 kg m^–3

Density of the balloon = $ρ_{o}$

Height to which the balloon will rise = y

Density of air decreases with height and the relationship is given by:

$ρ$ = $ρ = ρ_{o} e^{\frac{- y}{y_{o}}}$ ……(i)

Differentiating equation (i), we get

$\frac{ρ}{ρ_{o}}$ $e^{\frac{- y}{y_{o}}}$

$- \frac{d ρ}{d y}$ , where k is the constant of proportionality

$α ρ$ , height changes from 0 to y, while density changes from $\frac{d ρ}{d y} = - k ρ$ to $\frac{d ρ}{ρ} = - k d y$ . Integrating both sides between the limits, we get:

$ρ_{o}$

$ρ$ = -ky

$\int_{ρ_{o}}^{ρ} \frac{d ρ}{ρ} = - \int_{0}^{y} k d y$ = ${[\log_{e} ? ρ]}_{ρ_{o}}^{ρ}$ ….(ii)

From equation (i) and (ii), we get

$\frac{ρ}{ρ_{o}}$ = $e^{- k y}$ , k = $y_{0}$

Substituting the value of k in equation (ii), we get

$\frac{1}{k}$

Density $\frac{1}{y_{o}}$ = $ρ = ρ_{o} e^{\frac{- y}{y_{o}}} \dots \dots (i i i)$

= $ρ = \frac{M a s s}{V o l u m e} = \frac{M a s s o f t h e p a y l o a d + M a s s o f h e l i u m}{V o l u m e}$ = 0.461 kg/ $\frac{m + V ρ_{H e}}{V}$

From equation (iii), taking log on both sides

$\frac{400 + 1425 * 0.18}{1425}$ = - $m^{3}$

y = -8000 $\log_{e} ? \frac{ρ}{ρ_{o}}$ = 8000 m = 8 km

The balloon will rise to 8 km

less

0 0

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

On Shiksha, get access to

66k Colleges
1.2k Exams
688k 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

Exercise 9.1

Share Your College Life Experience