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8 months ago

Class 11th Kinetic Theory

13.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.

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Payal Gupta

Contributor-Level 10

13.1 Diameter of the oxygen molecule, d = 3Å

Radius, r = $\frac{d}{2}$ = 1.5 Å = 1.5 $* 10^{- 8}$ cm

Actual volume occupied by 1 mole of oxygen gas at STP = 22.4 lit = 22400 ${c m}^{3}$

Molecular volume of oxygen gas, V = $\frac{4}{3} ? r^{3}$ N,

where N = Avogadro's number = 6.023 $* 10^{23}$ molecules/mole

Therefore, V = $\frac{4}{3}$ $* 3.1416 * ({1.5 * 10^{- 8})}^{3}$ $*$ 6.023 $* 10^{23}$ = 8.51 ${c m}^{3}$

Ratio of the molecular volume to the actual volume of oxygen = $\frac{8.51}{22400}$ = 3.8 $* 10^{- 4}$

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8 months ago

Class 11th Motion in a Plane

4.7 Given a + b + c + d = 0, which of the following statements are correct:

(a) a, b, c, and d must each be a null vector

(b) The magnitude of (a + c) equals the magnitude of (b + d)

(c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d

(d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?

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Vishal Baghel

Contributor-Level 10

4.7

(a) This statement is In order to make a + b+ c + d = 0, it is not necessary to have all four given vectors to be null vectors.

(b) a + b + c +d = 0, ( a + c ) = - (b + d)

Taking the magnitude of both LHS and RHS, | a + c| = | -(b + d)| = | b + d|

So the statement is True.

(c) Given a + b + c + d = 0, a = - (b + c + d)

Taking magnitude of both LHS and RHS, |a| = | -( b + c+ d)| = |b + c +d|

|a| $\leq |b| + |c| + | d |$ ……….(1)

Equation (1) shows that the magnitude of a can never be greater than the sum of magnitudes of b,c and d. So the statement is True.

(d) In the given expression a + b + c + d = 0, the sum of 3 vectors a, (b+c), d ca

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4.7

(a) This statement is In order to make a + b+ c + d = 0, it is not necessary to have all four given vectors to be null vectors.

(b) a + b + c +d = 0, ( a + c ) = - (b + d)

Taking the magnitude of both LHS and RHS, | a + c| = | -(b + d)| = | b + d|

So the statement is True.

(c) Given a + b + c + d = 0, a = - (b + c + d)

Taking magnitude of both LHS and RHS, |a| = | -( b + c+ d)| = |b + c +d|

|a| $\leq |b| + |c| + | d |$ ……….(1)

Equation (1) shows that the magnitude of a can never be greater than the sum of magnitudes of b,c and d. So the statement is True.

(d) In the given expression a + b + c + d = 0, the sum of 3 vectors a, (b+c), d can be zero only if (b+c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle. If a and d are collinear , then it implies that the vector (b+c) is in the line of a and d. This statement holds good only then the sum of all three vectors will be zero. The statement is True.

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8 months ago

Class 11th Motion in a Plane

4.6 Establish the following vector inequalities geometrically or otherwise:

(a) |a+b| < |a| + |b|

(b) |a+b| > a| −|b

(c) |a−−b| < |a| + |b|

(d) |a−−b| > a| − |b

When does the equality sign above apply?

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Vishal Baghel

Contributor-Level 10

(a)

We can write

$\overset{?}{|O M|}$ = $|\overset{?}{a}| \dots \dots \dots . (i)$

$\overset{?}{|M N|}$ = $|\overset{?}{b}|$ = $\overset{?}{|O P|}$ ………(ii)

$\overset{?}{|O N|}$ = $|\overset{?}{a} + \overset{?}{b}| \dots \dots . (i i i)$

In a triangle, each side is smaller than the sum of the other two sides. Therefore in ΔOMN, we have ON < (OM + MN)

$|\overset{?}{a} + \overset{?}{b}|$ < $|\overset{?}{a}|$ + $|\overset{?}{b}|$ ….(iv)

If the two vectors $\overset{?}{a}$ and $\overset{?}{b}$ act along a straight line in the same direction, then we can write $|\overset{?}{a} + \overset{?}{b}|$ = $|\overset{?}{a}|$ + $|\overset{?}{b}| \dots . . (v)$

Combining equations (iv) and (v), we get

$|\overset{?}{a} + \overset{?}{b}|$ $\leq$ $|\overset{?}{a}|$ + $|\overset{?}{b}|$

(b)

Let two vectors $\overset{?}{a}$ and $\overset{?}{b}$ be represented by the adjacent sides of a parallelogram OMNP.

We can write

$\overset{?}{|O M|}$ = $|\overset{?}{a}| \dots \dots \dots . (i)$

$\overset{?}{|M N|}$ = $|\overset{?}{b}|$ = $\overset{?}{|O P|}$ ………(ii)

$\overset{?}{|O N|}$

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(a)

We can write

$\overset{?}{|O M|}$ = $|\overset{?}{a}| \dots \dots \dots . (i)$

$\overset{?}{|M N|}$ = $|\overset{?}{b}|$ = $\overset{?}{|O P|}$ ………(ii)

$\overset{?}{|O N|}$ = $|\overset{?}{a} + \overset{?}{b}| \dots \dots . (i i i)$

In a triangle, each side is smaller than the sum of the other two sides. Therefore in ΔOMN, we have ON < (OM + MN)

$|\overset{?}{a} + \overset{?}{b}|$ < $|\overset{?}{a}|$ + $|\overset{?}{b}|$ ….(iv)

Combining equations (iv) and (v), we get

$|\overset{?}{a} + \overset{?}{b}|$ $\leq$ $|\overset{?}{a}|$ + $|\overset{?}{b}|$

(b)

Let two vectors $\overset{?}{a}$ and $\overset{?}{b}$ be represented by the adjacent sides of a parallelogram OMNP.

We can write

$\overset{?}{|O M|}$ = $|\overset{?}{a}| \dots \dots \dots . (i)$

$\overset{?}{|M N|}$ = $|\overset{?}{b}|$ = $\overset{?}{|O P|}$ ………(ii)

$\overset{?}{|O N|}$ = $|\overset{?}{a} + \overset{?}{b}| \dots \dots . (i i i)$

In a triangle, each side is smaller than the sum of the other two sides. Therefore in ΔOMN, we have ON + MN > OM and ON + OM > MN

$\overset{?}{|O N|} > |\overset{?}{O M - O P}|$

$|\overset{?}{a} + \overset{?}{b}| >$ $|\overset{?}{a}|$ - $|\overset{?}{b}|$ ……(iv)

Combining (iv) and (v), we get:

$|\overset{?}{a} + \overset{?}{b}| \geq$ $|\overset{?}{a}|$ - $|\overset{?}{b}|$

(c)

Let two vectors $\overset{?}{a}$ and $\overset{?}{b}$ be represented by the adjacent sides of a parallelogram PORS.

We can write

$\overset{?}{|O R|}$ = $|\overset{?}{b}|$ = $\overset{?}{|P S|}$ ………(i)

$\overset{?}{|O P|}$ = $|\overset{?}{a}| \dots \dots \dots . (i i)$

In a triangle, each side is smaller than the sum of the other two sides. Therefore in ΔOPS, we have OS < OP + PS

$|\overset{?}{a} - \overset{?}{b}|$ $<$ $|\overset{?}{a}|$ + $|\overset{?}{- b}|$

$|\overset{?}{a} - \overset{?}{b}|$ $<$ $|\overset{?}{a}|$ + $|\overset{?}{b}|$ …….(iii)

If the two vectors $\overset{?}{a}$ and $\overset{?}{b}$ act along a straight line but in the opposite direction, then we can write $|\overset{?}{a} - \overset{?}{b}|$ = $|\overset{?}{a}|$ + $|\overset{?}{b}| \dots . . (i v)$

Combining (iii) and (iv), we get

$|\overset{?}{a} - \overset{?}{b}|$ $\leq$ $|\overset{?}{a}|$ + $|\overset{?}{b}|$

(d)

We can write

OS + PS > OP …….(i)

OS > PS – OP …….(ii)

$|\overset{?}{a} - \overset{?}{b}|$ $>$ $|\overset{?}{a}|$ + $|\overset{?}{b}|$ ….(iii)

The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus of both sides as:

$||\overset{?}{a} - \overset{?}{b}||$ > $||\overset{?}{a}| - |\overset{?}{b}||$ ….(iv)

Combining (iv) and (v), we get

$|\overset{?}{a} - \overset{?}{b}|$ $\geq$ $|\overset{?}{a}|$ - $|\overset{?}{b}|$

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New answer posted

8 months ago

Class 11th Motion in a Plane

4.5 Read each statement below carefully and state with reasons, if it is true or false:

(a) The magnitude of a vector is always a scalar

(b) Each component of a vector is always a scalar

(c) The total path length is always equal to the magnitude of the displacement vector of a particle

(d) The average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time

(e) Three vectors not lying in a plane can never add up to give a null vector

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Vishal Baghel

Contributor-Level 10

4.5

(a) The magnitude of a vector is a number only, hence always a scalar – TRUE

(b) Each component of a vector is also a vector – FALSE

(c) The total path length is a scalar quantity, whereas displacement is a vector quantity. – FALSE

(d) The total path length is always greater than or equal to the magnitude of displacement of a particle – TRUE

(e) Three vectors which do not lie in a plane can never add up to give a null vector – TRUE

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8 months ago

Class 11th Motion in a Plane

4.4 State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:

(a) Adding any two scalars

(b) Adding a scalar to a vector of the same dimensions

(c) Multiplying any vector by any scalar

(d) Multiplying any two scalars

(e) Adding any two vectors

(f) Adding a component of a vector to the same vector

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Vishal Baghel

Contributor-Level 10

4.4

(a) The addition of two scalar quantities is meaningful – if they both represent the same physical quantity.

(b) The addition of a vector quantity with a scalar quantity is not meaningful.

(c) Multiplying a scalar with a vector is meaningful.

(d) The multiplication of two scalar quantities is meaningful.

(e) The addition of two vectors is meaningful.

(f) Adding a component of a vector to the same vector is meaningful.

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8 months ago

Class 11th Motion in a Plane

4.3 Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy

gravitational potential, coefficient of friction, charge.

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Vishal Baghel

Contributor-Level 10

4.3 Impulse is the only vector quantity from the above list. Impulse is the product of force and time. Force is a vector quantity and time is a scalar quantity. The product of 1 vector and 1 scalar is always a vector quantity.

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Class 11th Motion in a Plane

4.2 Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

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Vishal Baghel

Contributor-Level 10

4.2 Work & Current are scalar quantities. Work is the product of Force and displacement. Since the product of two vectors quantities are always scalar. Current is independent of direction, it is only magnitude, hence scalar.

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Class 11th Motion in a Plane

4.1 State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

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Vishal Baghel

Contributor-Level 10

4.1 A scalar quantity depends only on the magnitude – volume, mass, speeds, density, number of moles, angular frequency are all scalar quantities.

A vector quantity depends on magnitude and direction – velocity, acceleration, displacement and angular velocity are all vector quantities.

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8 months ago

Class 11th Motion in a Straight Line

3.28 The velocity-time graph of a particle in one-dimensional motion is shown in Fig 3.29 :

Which of the following formulae are correct for describing the motion of the particle

over the time-interval t₁ to t₂:

(a) x(t₂ ) = x(t₁) + v (t₁) (t₂ – t₁) +(½) a (t₂ – t₁)²

(b) v(t₂ ) = v(t₁) + a (t₂ – t₁)

(c) v_average = (x(t₂) – x(t₁))/(t₂ – t₁)

(d) a_average = (v(t₂) – v(t₁))/(t₂ – _t1)

(e) x(t₂ ) = x(t₁) + v_average (t₂ – t₁) + (½) a_average (t₂ – t₁)²

(f) x(t₂ ) – x(t₁) = area under the v-t curve bounded by the t-axis and the dotted line

shown.

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Vishal Baghel

Contributor-Level 10

3.28 The graph has a non-uniform slope between the intervals t₁ and t₂ – the graph is not a straight line. The equations (a), (b), and (e) do not describe the motion of the particle. Only the relations (c), (d) and (f) are correct.

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Class 11th Motion in a Straight Line

3.27 The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s (b) t = 2 s to 6 s.

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Vishal Baghel

Contributor-Level 10

3.27

(a) Distance travelled by the particle between t = 0 s and t = 10 s is the area of the triangle = (1/2) x base x height = (1/2) x 10 x 12 = 60 m

The average speed of the particle is 60/10 m/s = 6 m/s

(b) Distance travelled by the particle between t = 2 s and t = 6 s

Let S₁ be the distance travelled by the particle in time 2 to 5 s and S₂ be the distance travelled between 5 to 6s.

For the motion from 0 to 5 sec, u = 0, t = 5, v = 12 m/s

From the equation v = u + at we get a = (v-u)/t = 12/5 = 2.4 m/s²

Distance covered from 2 to 5 sec, S₁ = distance covered in 5 s – distance covered in 2 s

From the equation s = ut + at² we

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