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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.

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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)

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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.

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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) = $29 + \frac{63}{2} unit^{2}$

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

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6 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].

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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

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6 months ago

Class 11th Mechanical Properties of Fluids

10.30 Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 * 10^–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 * 103 kg m^–3 (g = 9.8 m s^–2)

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Vishal Baghel

Contributor-Level 10

Diameter of the 1st bore, $d_{1}$ = 3 mm = 3 $* 10^{- 3}$ m

Radius of the first bore, $r_{1}$ = 1.5 mm = 1.5 $* 10^{- 3}$ m

Diameter of the 2nd bore, $d_{2}$ = 6 mm = 6 $* 10^{- 3}$ m

Radius of the 2nd bore, $r_{2}$ = 3.0 mm = 3 $* 10^{- 3}$ m

Surface tension of water, s= 7.3 $* 10^{- 2}$ N/m

Angle of contact between the bore surface and water, $θ = 0$

Density of water, $ρ = 1.0 * 10^{3}$ kg/ $m^{3}$

Acceleration due to gravity. g = 9.8 m/ $s^{2}$

Let $h_{1}$ and $h_{2}$ be the heights to which water rises in 1st and 2nd tubes respectively. These heights are given by the relations:

$h_{1} = {(2 s c o s ? θ) / ρ g r}_{1}$ …(i)

$h_{2} = {(2 s c o s ? θ) / ρ g r}_{2}$ …(ii)

The difference in level of water in the 2 li

...more

Diameter of the 1st bore, $d_{1}$ = 3 mm = 3 $* 10^{- 3}$ m

Radius of the first bore, $r_{1}$ = 1.5 mm = 1.5 $* 10^{- 3}$ m

Diameter of the 2nd bore, $d_{2}$ = 6 mm = 6 $* 10^{- 3}$ m

Radius of the 2nd bore, $r_{2}$ = 3.0 mm = 3 $* 10^{- 3}$ m

Surface tension of water, s= 7.3 $* 10^{- 2}$ N/m

Angle of contact between the bore surface and water, $θ = 0$

Density of water, $ρ = 1.0 * 10^{3}$ kg/ $m^{3}$

Acceleration due to gravity. g = 9.8 m/ $s^{2}$

Let $h_{1}$ and $h_{2}$ be the heights to which water rises in 1st and 2nd tubes respectively. These heights are given by the relations:

$h_{1} = {(2 s c o s ? θ) / ρ g r}_{1}$ …(i)

$h_{2} = {(2 s c o s ? θ) / ρ g r}_{2}$ …(ii)

The difference in level of water in the 2 limbs

$h_{1} - h_{2}$ = $\frac{2 s c o s ? θ}{ρ g}$ ( $\frac{1}{r_{1}}$ - $\frac{1}{r_{2}})$

= $\frac{2 * 7.3 * 10^{- 2} * c o s ? 0}{1.0 * 10^{3} * 9.8}$ ( $\frac{1}{1.5 * 10^{- 3}}$ - $\frac{1}{3 * 10^{- 3}})$ = 4.965 $* 10^{- 3}$ m = 4.965 mm

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Class 11th Mechanical Properties of Fluids

10.29 Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 N m^–1. Density of mercury = 13.6 * 10³kg m^–3.

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Vishal Baghel

Contributor-Level 10

Radius of the narrow tube, r = 1 mm= 1 $* 10^{- 3}$ m

Surface tension of mercury at the given temperature, s= 0.465 N/m

Density of mercury $ρ = 13.6 * 10^{3}$ kg/ $m^{3}$

Dipping depth = h

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

Surface tension related with angle of contact and dipping depth is given by:

s = $\frac{h ρ g r}{2 \cos ? θ}$ or h = $\frac{2 scos ? θ}{ρ g r}$ = $\frac{2 * 0.465 * \cos ? 140 °}{13.6 * 10^{3} * 9.8 * 1 * 10^{- 3}}$ = -5.345 $* 10^{- 3}$ m= -5.345 mm

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6 months ago

Class 11th Mechanical Properties of Fluids

10.28 In Millikan's oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 * 10^–5 m and density 1.2 * 10³ kg m^–3. Take the viscosity of air at the temperature of the experiment to be 1.8 * 10^–5 Pa s. How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop due to air.

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Vishal Baghel

Contributor-Level 10

Radius of the uncharged drop, r = 2.0 $* 10^{- 5}$ m

Density of the uncharged drop, $ρ$ = 1.2 $* 10^{3}$ kg/ $m^{3}$

Viscosity of air, $η$ = 1.8 $* 10^{- 5}$ Pas

Density of air $ρ_{o}$ , can be taken as zero in order to neglect the buoyancy of air

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

Terminal velocity, v can be written as

v = $\frac{2 * r^{2} * (ρ - ρ_{o}) g}{9 * η}$ = $\frac{2 * ({2.0 * 10^{- 5})}^{2} * (1.2 * 10^{3} - 0) * 9.8}{9 * 1.8 * 10^{- 5}}$ = 0.05807 m/s = 5.8 cm/s

The viscous force on the drop is given by

F = 6 $π η r v$ = 6 $* 3.1416 *$ 1.8 $* 10^{- 5} *$ 2.0 $* 10^{- 5} *$ 0.05807

= 3.9 $* 10^{- 10}$ N

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Class 11th Mechanical Properties of Fluids

10.27 A plane is in level flight at constant speed and each of its two wings has an area of 25 m². If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane's mass. (Take air density to be 1 kg m^–3).

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Vishal Baghel

Contributor-Level 10

The area of the wing of the plane, A = 2 $* 25 m^{2}$ = 50 $m^{2}$

Speed of air over the lower wing, $V_{1}$ = 180 km/h = 50 m/s

Speed of air over the upper wing, $V_{2}$ = 234 km/h = 65 m/s

Density of air, $ρ$ = 1 kg/ $m^{3}$

Let us assume, pressure over the lower wing = $P_{1}$ and pressure over the upper wing = $P_{2}$

The upward force on the plane can be obtained using Bernoulli's equation:

On lower wing= $P_{1}$ + $\frac{1}{2} ρ V_{1}^{2}$ , On upper wing = $P_{2}$ + $\frac{1}{2} ρ V_{2}^{2}$

From equilibrium of momentum

$P_{1}$ + $\frac{1}{2} ρ V_{1}^{2}$ = $P_{2}$ + $\frac{1}{2} ρ V_{2}^{2}$

( $P_{1}$ - $P_{2}) = \frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}$ )

The net upward force = ( $P_{1}$ - $P_{2}) A$ = $\frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}$ ) $* A$

=&nbs

...more

The area of the wing of the plane, A = 2 $* 25 m^{2}$ = 50 $m^{2}$

Speed of air over the lower wing, $V_{1}$ = 180 km/h = 50 m/s

Speed of air over the upper wing, $V_{2}$ = 234 km/h = 65 m/s

Density of air, $ρ$ = 1 kg/ $m^{3}$

Let us assume, pressure over the lower wing = $P_{1}$ and pressure over the upper wing = $P_{2}$

The upward force on the plane can be obtained using Bernoulli's equation:

On lower wing= $P_{1}$ + $\frac{1}{2} ρ V_{1}^{2}$ , On upper wing = $P_{2}$ + $\frac{1}{2} ρ V_{2}^{2}$

From equilibrium of momentum

$P_{1}$ + $\frac{1}{2} ρ V_{1}^{2}$ = $P_{2}$ + $\frac{1}{2} ρ V_{2}^{2}$

( $P_{1}$ - $P_{2}) = \frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}$ )

The net upward force = ( $P_{1}$ - $P_{2}) A$ = $\frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}$ ) $* A$

= $\frac{1}{2} 1 (65^{2} - 50^{2}$ ) $* 50$ = 43125 N

Using the relation F = mg, we get m = F/g

= (43125/9.8) Kg = 4400.51 kg

Hence the mass of the plane is about 4400 kg.

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Class 11th Mechanical Properties of Fluids

10.26 (a) What is the largest average velocity of blood flow in an artery of radius 2*10^–3m if the flow must remain laminar? (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 * 10^–3 Pa s).

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Vishal Baghel

Contributor-Level 10

Radius of the artery, r = 2 $* 10^{- 3}$ m

Diameter of the artery, d = 4 $* 10^{- 3}$ m

Viscosity of blood, $η$ = 2.084 $* 10^{- 3}$ Pa-s

Density of the blood, $ρ$ = 1.06 $* 10^{3}$ kg/ $m^{3}$

Reynolds's number for laminar flow, $R_{e}$ = 2000

The largest velocity is given by the relation: $V_{a v g}$ = $\frac{R_{e} η}{ρ d}$

= $\frac{2000 * 2.084 * 10^{- 3}}{1.06 * 10^{3} * 4 * 10^{- 3}}$ = 0.983 m/s

Flow rate is given by the relation:

R = $π r^{2} V_{a v g}$ = 3.1416 $* ({2 * 10^{- 3})}^{2} * 0.983$ = 1.235 $* 10^{- 5}$ $m^{3}$ /s

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Class 11th Mechanical Properties of Fluids

10.25 In deriving Bernoulli's equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (a) What is the largest average velocity of blood flow in an artery of diameter 2 * 10^–3 m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.

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Vishal Baghel

Contributor-Level 10

Diameter of the artery, d = 2 $* 10^{- 3}$ m

Viscosity of the blood, $η$ = 2.084 $* 10^{- 3}$ Pa-s

Density of the blood, $ρ$ = 1.06 $* 10^{3}$ kg/ $m^{3}$

Reynolds's number for laminar flow, $R_{e}$ = 2000

$V_{a v g}$ = $\frac{R_{e} η}{ρ d}$ = $\frac{2000 * 2.084 * 10^{- 3}}{1.06 * 10^{3} * 2 * 10^{- 3}}$ = 1.966 m/s

As the fluid velocity increases, the dissipative forces become more important. This is because of the rise of turbulence. Turbulent flow causes dissipative loss in a fluid.

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Class 11th Mechanical Properties of Fluids

10.24 During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein? [Use the density of whole blood from Table 10.1].

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Vishal Baghel

Contributor-Level 10

Gauge pressure, P = 2000 Pa

Density of whole blood, $ρ$ = 1.06 $* 10^{3}$ kg/ $m^{3}$

Acceleration due to gravity, g = 9.8 m/s

Let the height of the blood container = h

Pressure on the blood container P = $ρ g h$

By equating, we get h = (2000)/ ( 1.06 $* 10^{3} * 9.8)$ = 0.1925m

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