Vectors important questions of physics || Important question of PHYSICS || Important question of Basic Mathematics || MOTION IN A PLANE || Physics||

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Ques 1: A vector “A” makes an angle of 20° and vector B makes an angle of 110°  with the x-axis. The magnitudes of these two vectors are 3m and 4m respectively. Find the resultant. 

Ans: The angle between two vectors,  A vector and  B vector is 90°.

Magnitude of | A| and | A |, |A|= 3 and |B|=4

The resultant vector, say R= √(A² + B² + 2𝐴𝐵𝐶𝑜𝑠𝜃) = 5 m 

Let angle β be the angle between A and B then, 

β= tan^ -1(4 sin 90° /3+4 cos 90°)= tan^-1(4/3) =53°

The angle between resultant vector and x axis = 53°+ 20° = 73°.

Ques 2: Add vectors A, B and C each having magnitude of 100 unit and inclined to the x-axis at angles 45°, 135° and 315° respectively. 

Ans: Let A , B and C are three vectors of magnitude 100 unit each. 

|A|=|B|=|C|= 100 unit 

x-components of vector A = 100 cos 45° = 100/ √2 unit. 

x-components of vector B = 100 cos 135° = -100/ √2 unit

x-components of vector C = 100 cos 315°= 100/ √2 unit. 

So, resultant x-component = (100/ √2 - 100/ √2 + 100/ √2)unit = 100/ √2 unit. 

y-components of vector A = 100 sin 45° = 100/ √2 unit. 

y -components of vector B = 100 sin 135° = 100/ √2 unit . 

y -components of vector C = 100 sin 315 °= -100/ √2 unit . 

Tan α = 1 , α= 45°

So, resultant y-component = (100/ √2 + 100/ √2 - 100/ √2)unit = 100/ √2 unit

Now,  

The resultant is 100 unit at 45° with x-axis.

Ques 3: Let A and B be the two vectors of magnitude 10 unit each. If they are inclined to the x-axis at angles 30° and 60°. Find the resultant.

Ans: Let A and B be the two vectors of magnitude 10 unit each. 

                |A|=|B|= 10 units

Let θ be the angle between vectors A and B, and β be the angle between vector A and resultant vector.

                    =>  θ = 30°

Then, 

R =√(A² + B² + 2𝐴𝐵𝐶𝑜𝑠𝜃)

R = √ 10²+10²+2×10×10 cos 30° 

R = 19.3

tan β = 10 sin 30°/10+10 cos 30°

β = 15°

Ques 4: A spy report about a suspected car reads as follows. "The car moved 2 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4 km and stopped". Find the displacement of the car. 

Ans: First convert all distance in same unit. 

2 km, 500 m = 0.5 km and 4 km

Vector A D = 2i + 0.5j + 4k

Using Pythagoras Theorem ,

A D = √(AE² + D E²) = 6.02 km

Now, tan θ = (D E/AE) = 1/12 

θ = tan⁻¹(1/12)

The displacement of the car along the distance tan⁻1(1/12) with positive x-axis is 6.02.

Ques 5: A mosquito net over a 7 ft x 4 ft bed is 3 ft high. The net has a hole at one corner of the bed through which a mosquito can enter the net. It flies and sit at the diagonally opposite upper corner of the net. 

(a) Find the magnitude of the displacement of the mosquitos. 

(b) Taking the hole as the origin, the length of the bed as the x-axis, its width as the y- axis, and vertically up as the z-axis, write the components of the displacement vector. 

Ans: Write the displacement vector using given information’s. Let vector r be the displacement vector, 

Vector, r = 7i + 4j + 3k

(a) Magnitude of displacement = √(7² + 4² + 3²) = √(49 + 16 +9) = √(74) ft.

(b) The components of the displacement vector r are 7 ft. , 4 ft. , and 3 ft. 

Ques 6: Two vectors have magnitude 2 m and 3 m. The angle between them is 60°. Find 

a) the scalar product of two vector.

b) magnitude of their vector product.

Ans: Consider two vectors a and b. 

Given : | a | = 2m and | b| = 3m 

Angle between vector a and vector b is 60°, say θ = 60°

(a) Scalar product:

 a.b = |a| |b| cos θ 

= 2 x 3 x cos 60°

= 6 x 1/2

= 3 m²

(b) magnitude of vector product:

|a x b| = |a| |b| sin θ

= 2 x 3 x sin 60°

= 6 x √3/2

= 3√3 m² 

Ques 7: If vectors A, B, C are mutually perpendicular, show that C ×(A×B) = 0. Is the converse true?

Ans: Given: A , B , C are mutually perpendicular vectors. 

Vector A and B is a vector whose direction is perpendicular to the plane containing A and B. 

Also vector C is perpendicular to vector A and vector B.

Angle between vector C and A x B is 180° or 0° .

C ×(A×B) = 0

But, the converse is not true. Because, If two vector are parallel then,

 C ×(A×B) = 0

 So, they need not be mutually perpendicular. 

Ques 8: The force on a charged particle due to electric and magnetic fields is given by F = q E + q v x B. Suppose E is along the X-axis and B along the Y-axis. In what direction and with what minimum speed v should have a positively charged particle be sent so that the net force on it is zero? 

Ans: Given: F= q E + q(v x B) = 0 

=> E = -(v x B)

Direction of v x B should be opposite to the direction of E. Vector v should be in positive y z plane. 

Now, E = v B sin θ

or v = E / (B sin θ)

For v to be minimum, we have θ = 90° and v= F/B

We can conclude that the particle must be projected at a minimum speed of E/B along +v e z-axis in which θ=90°, so that force is zero.

Ques 9: Given an example for which A. B = C. B but A ≠ C. 

Ans: Vector A perpendicular to the vector B and Vector B perpendicular to the vector C .

[Here Vector A along south side, B along west side, and vector C along north]

So dot product of vector A and B is zero => A.B=0

Therefore, A.B = B.C or B.C = 0 

But B ≠ C. 

Ques 10: A curve is represented by y = sin x. If x is changed from π/3 to π/3 + π/100 , find approximately the change in y. 

Ans: Given equation of curve is y = sin x  

Change in variables, we have

y + Δ y = sin (x + Δ x)

or ∆y = sin (x + Δ x) - sin x

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

           = 0.0157

Ques 11: Find the area enclosed by the curve y = sin x and the X-axis between x = 0 and x = π.

Ans: Equation of the curve, y = sin x.

The required area can found by integrating y w.r.t. x :

Area = ∫y d x = ∫ sin x d x

          = -[cos x]0 ^π =2

Ques 12: Find the area bounded by the curve y = e-x ,  the X-axis and the Y-axis. 

Ans: Equation of curve is  

When x = 0, y = e−0 = 1

x increases, the value of y decrease.

Also, only when x = ∞, y = 0

So, the required area can be found by integrating the function from 0 to ∞.

Area = ∫ dA = ∫ y d x = ∫ e-x d x = e-x + C

Integration between o to ∞

= (0 + 1)

= 1   

Ques 13: A rod of length L is placed along the X-axis between x = 0 and x = L. The linear density (mass/length) ρ of the rod varies with the distance x from the origin as ρ = a + b x. 

(a) Find the SI units of a and b. 

(b) Find the mass of the rod in terms of a, b and L. 

Ans: Given: Density = ρ = mass/length = a + b x 

So, the SI unit of ρ is kg/m.

(a) SI unit of a = kg/m 

SI unit of b = kg/m²

[(From the principle of homogeneity of dimensions]

(b) Let us consider a small element of length d x at a distance x from the origin., we have, 

d m = mass of the element = ρ d x = (a + b x)d x

Therefore, mass of rod = m = ∫d m = ∫(a+b x) = ax + b x²/2 

[Integration between o to L]

= a L + b L² /2

Ques 14: The momentum p of a particle changes with time t according to the relation d p/d t m = (10N) + (2N/s)t. If the momentum is zero at t = 0, what will the  momentum be at t = 10 s? 

Ans: The momentum at time t 

p = ∫ d p = ∫ [(10N) + (2N/s)t] d t

= (10 t + t2 ) + C N/s 

Where C = constant of integration

Given, at t = 0, p = 0.

So, above equation is turn as 0 = C or C = 0 

=> p = (10 x 100 + 10²) N/s = 200 N s or 200 kg m/s.

Ques 15: Write the number of significant digits’ in

(a) 1001

(b) 100.1

(c) 100.10

(d) 0.001001

Ans: (a)1001 

Number of significant digits = 4

(b) 100.1

Number of significant digits = 4

(c) 100.10

Number of significant digits = 5

(d) 0.001001

Number of significant digits = 4

Ques 16: Round the following numbers to 2 significant (a) 3472 (b) 84.16 (c) 2.55 and (d) 28.5 . 

Ans: (a) In Value 3472, after digit 4, 7 is there which is greater than 5. So, neglect next two digits and the values of 4 is increased by 1. Value is 3500 

  (b) 84.16  

Required number is 84

(c) 2.55

Required number is 2.6

(d) 28.5

Required number is 28.

Ques 17: A meter scale is graduated at every millimeter. How many significant digits will be there in a length measurement with this scale? 

Ans: We know, 1 m = 1000 mm

The minimum number of significant digits can be 1 (for example, for measurements like 2 mm and 6 mm) and the maximum number of significant digits can be 4 (for example, for measurements like 1000 mm).  

The number of significant digits may be 1, 2, 3 or 4.

Ques 18: The length and the radius of a cylinder measured with a slide calipers are found to be 4.54 cm and 1.75 cm respectively. Calculate the volume of the cylinder. 

Ans: From given information’s, 

Length of the cylinder, l = 4.54 cm

Radius of the cylinder, r = 1.75 cm

Now, Volume, V = π r²

l = π (1.75)² (4.54)  

= 3.14 x 1.75 x 1.75 x 4.54

= 43.6577

The minimum number of significant digits in a particular term is three. Therefore,

rounded off the result into three significant digits, we have:

V = 43.7 cm³.

Ques 19: The thickness of a glass plate is measured 2.17 mm, 2.17 mm and 2.18 mm at three different places. Find the average thickness of the plate from this data. 

Ans: The thickness of a glass plate is measured 2.17 mm, 2.17 mm and 2.18 mm at  three different places.  (Given) 

Average thickness = (2.17+2.17+2.18)/3 mm = 2.1733 mm

Round off above result to three significant digits, the average thickness becomes 2.17 mm.

Ques 20: The length of the string of a simple pendulum is measured with a meter scale to be 90.0 cm. The radius of the bob plus the length of the hook is calculated to be 2.13 cm using measurements with a slide calipers. What is the effective length of the pendulum?

Ans: The length of the simple pendulum = 90 cm and 

The radius of bob and hook = 2.13 cm

Actual effective length of the pendulum = (90.0 + 2.13) cm

However, in the measurement 90.0 cm, the number of significant digits is only 2.

So, the effective length should contain only two significant digits.

Therefore, the effective length of the pendulum = 90.0 + 2.13 = 92.1 cm.

However, in the measurement 90.0 cm, the number of significant digits is only 2.

So, the effective length should contain only two significant digits.

Therefore, the effective length of the pendulum = 90.0 + 2.13 = 92.1 cm.

Ques 21: A vector has magnitude and direction. (a) Does it have a location in the space? (b) Can it vary with time? (c)Will two equal vectors a and b at different locations in space necessarily have identical physical effects? Explain in details. 

Ans: (a)A vector in general has no definite location in space because a vector remains unaffected whenever is displaced anywhere in space provided its magnitude and direction do not change. However a positive vector has a definite location in space.

(b)  vector can vary with time e.g. the velocity vector of an accelerated particle varies with  time. 

(c) Two equal vectors at different locations in space do not necessarily have same physical effects For example, two equal forces acting at two different points on a body which can cause the rotation of a body about an axis will not produce equal tuning effect.

Ques 22: A vector has both magnitude and direction. Does that mean anything that has magnitude and direction is necessarily a vector ? The rotation of a body can be specified by the direction of the axis of rotation and the angle of rotation about the axis. Does that make any rotation a vector ?

Ans: No. There are certain physical quantities which have both magnitude and direction, but they are not vector as they do not follow the laws of vectors addition, which is essential for vectors. The finite rotation of body about an axis is not a vector because the finite rotations do not obey the laws of vectors addition However, the small rotation of a body (i.e., small angle of rotation) is a vector quantity as it obeys the law of vectors addition. 

Ques 23: Can you associate vectors with (a) the length of a wire bent into a loop (b) a plane area (e) a sphere? Explain.

Ans: (a) We cannot associate a vector with the length of a wire bent into a loop. 

(b) We can associate a vector with a plane area. Such a vector is called area vector and its directions represented by outward drawn normal to the area. 

(c) We can not associate a vector with volume of sphere however a vector can be associated with the of sphere.