Class 11-science H C VERMA Solutions Physics Chapter 2 - Physics and Mathematics

The angle between A and B from the x-axis are 20° and 110° respectively.

Their magnitudes are 3 units and 4 units respectively.

Thus the angle between A and B is = 110 - 20 = 90° 

Now, R² = A² + B² + 2ABcosѲ 

 = 3² + 4² + 2.3.4 Cos(90)

   = 5²

 Or, R = 5

Let ϕ is the angle between R and A,

Then tan ϕ = , or ϕ = 53°.

The resultant makes an angle of (53+20)° = 73° with the x axis. 

A and B are inclined at angles of 30 degrees and 60 degrees with respect to the x axis

Angle between them = (60 -30) = 90 degrees

Given that |A| = |B| = 10 units, we get

R² = A² + B² +2ABcosϴ 

 = 10² + 10² + 2.10.10 Cos(30)

 R = 20cos15° 

Or, R = 19.3 units

And tan ϕ = , or ϕ = 15°.

Therefore, this resultant makes an angle of (15+30) = 45 degrees with the x axis

Vectors A, B and C are oriented at 45°, 135° and 315° respectively.

|A|=|B|=|C|=100 units

Let A = A_xi + A_yj +A_zk, B = B_xi + B_yj +B_zk, and C = C_xi + C_yj +C_zk, and we can write that,

A_x= C_x=100cos(45°)=100/√2 , by considering their components

B_x=-100/√2

Now Ay = 100sin(45°)= 100/√2,

By= 100sin(135⁰)= 100/√2

Similarly, Cy= -100/√2

Net x component = 100/√2+100/√2-100/√2=100/√2

Net y component = 100/√2+100/√2-100/√2=100/√2

R²=x²+y²=100²

R=100 and tan ϕ = (100/√2)/( 100/√2)=1, and ϕ = 45° 

a = 4i+3j, b = 3i+4j

|a|=|b|= =5

a+b = 7i+7j and a-b = i-j

|a+b|

=  7√2 and |a-b| =  = √2

x component of OA = 2cos30° = √3

x component of BC = 1.5 cos 120° = -0.75

x component of DE = 1 cos 270° = 0

y component of OA = 2 sin 30° = 1

component of BC = 1.5 sin 120° = 1.3

component of DE = 1 sin 270° = -1

Rx= x component of resultant = 3 - 0.75 + 0 = 0.98 m

Ry = resultant y component = 1 + 1.3 - 1 = 1.3 m

So, R = (R_x²+R_y²)^1/2 = 1.6 m

It makes and angle ϴ with positive x-axis then tan ϴ = R_y/R_x = 1.32 ϴ = tan^-1(1.32)

|a| = 3 and |b| = 4

Let ϴ be the angle between them.

Then, using the relation R² = A² + B² +2ABcosϴ,

a)

We get for R = 1,

1 = 9+16+24Cos ϴ 

Or, ϴ = 180° 

b)

For, R = 5, we have

25 = 9+16+24Cos ϴ 

Or, cos ϴ = 0;

ϴ = 90° 

c)For R =7,

49= 9+16+24Cos ϴ,

Or cos ϴ = 1,

And ϴ = 0° 

AB = 2i + 0.5j + 4i = 6i + 0.5j

As the car went forward, took a left and then a right.

So, AB = (6²+0.5²)^1/2= 6.02km

And ϕ = tan^-1(BE\AE) = tan^-1(0.5/6) = tan^-1(1/12) 

In ⊗ABC, tan⎩ = x/2 and in ⊗DCF, tan⎩= (2 - x)/4, So, (x/2) = (2 - x)/4. Solving,

 4 - 2x = 4x

Or, 6x = 4

 Or, x = 2/3 ft

a) In ⊗ABC, AC = (AB²+ BC²)^1/2 =  ft

b) In ⊗CFD, DF = 1 - (2/3) = 4/3 ft

So, CF = 4 ft. Now, CD = (CF²+ FD²)^1/2 = 4√10/3 ft

c) In ⊗ADE, AE = (AE²+ ED²)^1/2 = 2√2 ft. 

The displacement vector is given by r= 7i+4j +3k

Magnitude of displacement of the mosquitoes= (7²+4²+3²)^1/2 = 74ft

The components of displacement along x axis, y axis and z axis are 7 units, 4 units and 3 units respectively. 

a = 4.5 n, where n is unit vector in north direction

a) 3a = 4.5 X 3n = 13.5 in north direction

b) -4a = 4.5 X -4n = 18 in south direction

We have a = 2m, b = 3m.

ϴ = 60⁰ is the angle between the two vectors

Scalar product between the two vectors = a.b =

2×3×cos(60⁰)=3m²

Vector product between the two vectors = a×b =

2×3×sin(60⁰)=3√3m²

From polygon law of vector addition, the resultant of the six vectors can be affirmed to be zero. Here their magnitudes are the same.

That is, A = B = C = D = E = F.

Rx = A cos→ + A cos ⇒/3 + A cos 2⇒/3 + A cos 3⇒/3 + A cos

4⇒/4 + A cos 5⇒/5 = 0 [As resultant is zero, x component of

resultant is also 0]

Now taking A common and putting Rx=0,

cos ⇒ + cos ⇒/3 + cos 2⇒/3 + cos 3⇒/3 + cos 4⇒/3 + cos 5⇒/3 = 0

a = 2i +3j+4k and b = 3i + 4j +5k

Let angle between them is ϴ 

Then a.b =2.3 +3.4 + 4.5 = 38

|a| =  

|b| =  

Now cos ϴ = a.b/|a||b| = 38/  

(A×B)= ABsinϴ ȗ, where ȗ is a unit vector perpendicular to both A and B.

Now, A.(A×B) is basically a dot product between two vectors which are perpendicular to each other. Then cos90° = 0, and thus

A.(A×B) = 0

A = 2i + 3j + 4k, B = 4i + 3j + 2k

A×B =  = -6i + 12j-6k 

A, B and C are mutually perpendicular vectors. Now, if we take cross product between any two vectors, the resultant vector will be in parallel to the third vector, as there are only three axis perpendicular to each other.

So if we consider (A×B), then it is parallel to C, and so angle between the resultant vector and C is 0°, and sin(0°)=0. So, C×(A×B) = 0

The particle moves from PP' in a straight line with a constant speed v.

From the figure, we see that OP×v = (OP)v sinθ ȗ, where ȗ is a unit vector perpendicular to v and OP. Now,

We know, OQ= OP sinθ = OP'sinθ'

So, the position of the particle may vary, but the magnitude and direction of OP×v will be constant.

OP×v is independent of P.

F =q(E+v×B)

Now, for net force to be 0, we must have E = -(v×B)

So, the direction of E must be opposite to that of (v×B), so v must be in Z axis and its magnitude is E/(B sinθ). For v to be minimum. θ= 90°, so v = E/B. 

Let us assume that B is along Y axis, and A is along positive x axis and C is along negative X axis. Now, A.B = B.C = 0. But A ⇏C

To find the slope at any point, we draw a tangent and we extend it to meet the X axis. Then we can find Ѳ as shown in the figure.

We can use another way, which is by differentiation. We write,

y = sin(x)

Let y1 = sin(π/3) and y2 = sin(π/3 + π/100)

Change in y = y2 -y1 = sin(π/3 + π/100)- sin(π/3)

= sin(π/3 +( π/3 + π/100 - π/3)) -sin(π/3)

= 0.0157

We have, i= i₀ e^-t/RC

Rate of change of current  

a) When t = 0, di/dt = -i₀/RC

b) When t = RC, di/dt = -i₀/RCe

c) When t = 10RC, di/dt = -  

i= i₀ e^{- t/RC}

Here, i₀ = 2A, R = 6×10⁵Ω, C=5×10^-6 F= 5×10^-7F

a) i|_t=0.3= 2×e(-0.3/0.3)=2/e Amperes.

b) di/dt| _t=0.3 = -2e(-0.3/0.3)/0.3 = -20/3e

c) i|_t=0.31 = 5.8/3e Amperes

y= 3x² +6x+7

Area bounded under the curve within x = 5 and x = 10 is calculated by the method of integration.

Area = ¹⁰₅= 1135 sq. units 

y = sin(x)

Area under the curve from x =0 t x= π is calculated by the method of integration.

Area =  

The equation of the curve is y = e^{- x}. And when x =0, then y = e^-0 = 1. Now, y = 0 when x = ∞. Using these boundaries, we get, Area = . 

It is given that linear density (mass per unit length) 𝘱 = a + bx, where x is distance from origin.

a) S.I. unit of a = kg/m

S.I. unit of b = kg/m²

b) To find the mass of the rod, we have to take a small segment dx placed at a distance x from the origin.

or, dm = 𝘱dx,

or,= [ax+bx²/2]^L₀ = aL + 0.5bL² 

It is given that  

Now, p=0 at t = 0. So,

= 10(10-0)+ (100-0)=200N/s = 200 kg m/s 

It is given that  

We can write dy = x²dx

or, ∫ 

The number of significant digits are as follows:

a) 4

b) 4

c) 5

d) 4 

It is stated that the meter scale is graduated at every millimeter. Now, we know 1m = 1000mm. So minimum measurement can be of 1 digit value(say, 3mm, 5mm), then there can be up to two digits and three digits(say, 76mm, 305 mm) and the maximum value can be 1000mm. So, there can be 1, 2, 3 or 4 significant digits. 

(a) Here after 4, the number is 7, which is greater than 5. So, the next two digits are neglected and the 4 becomes 5. Hence, rounded value = 3500

(b) Here after 4, the number is 1, which is less than 5. So, the next two digits are neglected and the 4 doesn't change. Hence, rounded value = 84

(c) Here after 5, the number is 5. So, the next two digits are neglected and the 5 is increased by 1. Hence, rounded value 2.6

(d) Here, after 8 the number is 5, and hence 8 is increased by 1. So the rounded value is 29

Length L = 4.54cm, radius r = 1.75cm

Volume = πr²L = π × (4.54) × (1.75)² = 43.6577cm³.

Now, it has to be rounded off to 3 significant digits, so we see that after 6, the value is 5. So 6 becomes 7, and thus the rounded off value is given by V= 43.7 cm³. 

Average thickness

Now we have to round it off to three significant figures. So, after 7, the digit is 3. Hence, Average thickness = 2.17mm.

Effective Length = (90.0+2.13) cm

In the measurement of the 90.0cm, there are only two significant digits. Keeping this in mind, we have to consider only two significant digits during addition. So,

Effective Length = (90.0+2.1)cm= 92.1cm