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DC Pandey Solutions Units, Dimensions & Vectors - 2 - DC Pandey Solutions

Q31. Which one is a vector quantity?
(a) Time
(b) Temperature
(c) Flux density
(d) Magnetic field intensity

Magnetic field intensity.
Option (d) is correct.Vector Quantity has both Magnitude and DirectionVector Quantity has both Magnitude and Direction


Q32. Given that P = 12,Q = 5 and R =13 also DC Pandey Solutions: Units, Dimensions & Vectors - 2 then the angle between DC Pandey Solutions: Units, Dimensions & Vectors - 2 will be
(a) π 
(b) π/2
(c) zero
(d) π/4

DC Pandey Solutions: Units, Dimensions & Vectors - 2
∴ Angle between DC Pandey Solutions: Units, Dimensions & Vectors - 2
Option (b) is correct.


Q33. The forces, which meet at one point but their lines of action do not lie in one plane, are called 
(a) non-coplanar non-concurrent forces 
(b) non-coplanar concurrent forces 
(c) coplanar concurrent forces 
(d) coplanar non-concurrent forces

non-coplanar concurrent forces


Q34. Given that DC Pandey Solutions: Units, Dimensions & Vectors - 2 Two out of the three vectors are equal in magnitude. The magnitude of the third vector is √2 times that of the other two. Which of the following can be the angles between these vectors? 
(a) 90°, 135°, 135° 
(b) 45°, 45°, 90° 
(c) 30°, 60°, 90° 
(d) 45°, 90°, 135°

DC Pandey Solutions: Units, Dimensions & Vectors - 2
or DC Pandey Solutions: Units, Dimensions & Vectors - 2
or DC Pandey Solutions: Units, Dimensions & Vectors - 2
or DC Pandey Solutions: Units, Dimensions & Vectors - 2 ...(i) 
Let Q2 = P2 and R = P√2
Thus, Eq. (i) takes the form
P2 + P2 + 2PQcos θ = 2P2 
or 2PQcos θ = 0
or cosθ = 0
or  θ = 90°
∴ Angle between DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
or DC Pandey Solutions: Units, Dimensions & Vectors - 2
or P2 + R2 + 2PR cos φ = Q2
or 2PR cos φ = Q2 - P2 - R2
or   2PR cos φ = - R2
or   2P cos φ = - R
or    2P cos φ = - P√2
or DC Pandey Solutions: Units, Dimensions & Vectors - 2
∴ φ = 135°

∴  Angle between DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
Option (a) is correct.


Q35. The angle between DC Pandey Solutions: Units, Dimensions & Vectors - 2
(a) 90° 

(b) between 0° and 180°
(c) 180° only
(d) None of these

Angle (φ) between DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
Angle φ between DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
This implies that angle between DC Pandey Solutions: Units, Dimensions & Vectors - 2 and DC Pandey Solutions: Units, Dimensions & Vectors - 2 will vary from 0 to π.
Option (b) is correct.


Q36. Two vectors of equal magnitude have a resultant equal to either of them, then the angle between them will be 
(a) 30° 
(b) 120° 
(c) 60° 
(d) 45°

R2 = P2 + Q2 + 2PQcosθ
for R = P = Q
P2 = P2 + P2 + 2PPcos θ
or DC Pandey Solutions: Units, Dimensions & Vectors - 2
or    θ = 120°
Option (b) is correct.


Q37. A forceDC Pandey Solutions: Units, Dimensions & Vectors - 2 newton acts on a body and displaces it by DC Pandey Solutions: Units, Dimensions & Vectors - 2metre. The work done by the force is 
(a) 5 J 
(b) 25 J 
(c) 10 J 
(d) 30 J

DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
= 25 J
Option (b) is correct.


Q38. If the vectors DC Pandey Solutions: Units, Dimensions & Vectors - 2 are perpendicular to each other then the positive value of a is 
(a) zero 
(b) 1 
(c) 2 
(d) 3

DC Pandey Solutions: Units, Dimensions & Vectors - 2
Other value is - ive.
Option (d) is correct.


Q39. The angles which the vector DC Pandey Solutions: Units, Dimensions & Vectors - 2 makes with the co-ordinate axes are
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
(d) none of the above

If a vector makes angles α, β and γ with the co-ordinate axes, then
cos2 α + cos2 β + cos2 γ = 1
DC Pandey Solutions: Units, Dimensions & Vectors - 2
∴ Option (a) is correct.


Q40. Unit vector parallel to the resultant of vectors DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
(d) None of these

DC Pandey Solutions: Units, Dimensions & Vectors - 2 and DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
Option (b) is correct.


Q41. The value of n so that vectors DC Pandey Solutions: Units, Dimensions & Vectors - 2 may be coplanar, will be 
(a) 18 
(b) 28 
(c) 9 
(d) 36

DC Pandey Solutions: Units, Dimensions & Vectors - 2
∴ Vectors DC Pandey Solutions: Units, Dimensions & Vectors - 2 will be coplanar if their scalar triple product is zero i.e.,

DC Pandey Solutions: Units, Dimensions & Vectors - 2
or 65 - 4n + 7 = 0
or  n = 18
Option (a) is correct.


Q42. Which one of the following statement is false? 
(a) A vector has only magnitude, whereas a scalar has both magnitude and direction 
(b) Distance is a scalar quantity but displacement is a vector quantity 
(c) Momentum, force, torque are vector quantities 
(d) Mass, speed and energy are scalar quantities

A vector has only magnitude, whereas a scalar has both magnitude and direction


Q43.DC Pandey Solutions: Units, Dimensions & Vectors - 2 are two vectors then the value of DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2

DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
Option (a) is correct.


Q44. The angle between the two vectors DC Pandey Solutions: Units, Dimensions & Vectors - 2
(a) 60° 
(b) 0° 
(c) 90° 
(d) None of these

DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
= 0
⇒ θ = 90°
Option (c) is correct.


Q45. Maximum and minimum values of the resultant of two forces acting at a point are 7 N and 3 N respectively. The smaller force will be equal to 
(a) 5 N 
(b) 4 N 
(c) 2 N 
(d) 1 N

A + B = 7
A - B = 3
∴ B = 2 N
Option (c) is correct.


Q46. The component of vector DC Pandey Solutions: Units, Dimensions & Vectors - 2 along the vectorDC Pandey Solutions: Units, Dimensions & Vectors - 2
(a) 5/√2
(b) 10/√2
(c) 5 √2
(d) 5

Angle between DC Pandey Solutions: Units, Dimensions & Vectors - 2
and  DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
Component of  DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
Option (a) is correct.


Q47. The resultant of two forces 3P and 2P is R. If the first force is doubled then the resultant is also doubled. The angle between the two forces is 
(a) 60° 
(b) 120° 
(c) 70° 
(d) 180°

DC Pandey Solutions: Units, Dimensions & Vectors - 2
Option (b) is correct.


Q48. The resultant of two forces, one double the other in magnitude, is perpendicular to the smaller of the two forces. The angle between the two forces is 
(a) 120° 
(b) 60° 
(c) 90° 
(d) 150°

DC Pandey Solutions: Units, Dimensions & Vectors - 2
As   θ = 90°, tan α = ∞
∴ P + Q cos α  = 0
i.e., DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
∴ α = 120°
Option (a) is correct.


Q49. Three vectors satisfy the relation DC Pandey Solutions: Units, Dimensions & Vectors - 2 is parallel to
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2

DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2 ...(i)
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2 ...(ii)
From Eq. (i) and Eq. (ii), we conclude that DC Pandey Solutions: Units, Dimensions & Vectors - 2 is perpendicular to the plane containing 

DC Pandey Solutions: Units, Dimensions & Vectors - 2
This implies that DC Pandey Solutions: Units, Dimensions & Vectors - 2 is perpendicular to DC Pandey Solutions: Units, Dimensions & Vectors - 2
Option (c) is correct.


Q50. The sum of two forces at a point is 16 N. If their resultant is normal to the smaller force and has a magnitude of 8 N. Then two forces are 
(a) 6N, 10N 
(b) 8 N, 8 N 
(c) 4 N, 12N 
(d) 2 N, 14N

P2 + Q2 + 2PQ cos α = R2 
or P2 + Q2 + 2PQ cos α = 82
or P2 + Q2 + 2PQ + 2PQ cos α - 2PQ = 64
or (P + Q)2 + 2 PQ (cos α - 1) = 64
or   (16)2 + 2 PQ (cos α - 1) = 64
or   2 PQ (cos α - 1) = - 192
or   PQ cos α - PQ = - 96 ...(i)

DC Pandey Solutions: Units, Dimensions & Vectors - 2
(as θ = 90°)
∴ P + Q cos α = 0
Qcos α = -P ...(ii)
Using Eq. (ii) and Eq. (i),
P (- P) - PQ = - 96

or - P (P + Q) = - 96
or DC Pandey Solutions: Units, Dimensions & Vectors - 2
P = + 6 N
∴ Q = 10 N
Option (a) is correct.


Q51.DC Pandey Solutions: Units, Dimensions & Vectors - 2 then the value of DC Pandey Solutions: Units, Dimensions & Vectors - 2
(a) (A2 + B2 + AB)1/2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
(c) (A + B)
DC Pandey Solutions: Units, Dimensions & Vectors - 2

DC Pandey Solutions: Units, Dimensions & Vectors - 2
⇒ tan θ = √3
⇒ θ = 60°
DC Pandey Solutions: Units, Dimensions & Vectors - 2
= A2 + B2 + 2AB cos 60°
= A2 + B2 + AB
DC Pandey Solutions: Units, Dimensions & Vectors - 2


Q52. If the angle between the vectorsDC Pandey Solutions: Units, Dimensions & Vectors - 2 the value of the product DC Pandey Solutions: Units, Dimensions & Vectors - 2is equal to
(a) BA2 cos θ
(b) BA2 sin θ
(c) BA2 sin θ cos θ
(d) zero

DC Pandey Solutions: Units, Dimensions & Vectors - 2 is perpendicular to both DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
or DC Pandey Solutions: Units, Dimensions & Vectors - 2
Option (d) is correct.


Q53. If a vector DC Pandey Solutions: Units, Dimensions & Vectors - 2 is perpendicular to the vector DC Pandey Solutions: Units, Dimensions & Vectors - 2 then the value of α is
(a) -1 
(b) 1/2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
(d) 1

DC Pandey Solutions: Units, Dimensions & Vectors - 2
⇒ - 8 + 12 + 8a = 0
DC Pandey Solutions: Units, Dimensions & Vectors - 2
Option (c) is correct.


Q54. Minimum number of vectors of unequal magnitudes which can give zero resultant are 
(a) two 
(b) three 
(c) four 
(d) more than four

DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2


Q55. The (x, y, z) coordinates of two points A and B are given respectively as (0, 3, - 1) and (- 2, 6, 4). The displacement vector from A to B is given by
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2

DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
DC Pandey Solutions: Units, Dimensions & Vectors - 2
Option (c) is correct.


Q56. The sum of two vectors DC Pandey Solutions: Units, Dimensions & Vectors - 2 is at right angles to their difference. Then 
(a) A = B 
(b) A = 2B
(c) B = 2A
DC Pandey Solutions: Units, Dimensions & Vectors - 2 have the same direction

Using answer to questions no. 35, as angle between DC Pandey Solutions: Units, Dimensions & Vectors - 2
A2 + B2 cos 2θ = 0
or  A2 = - B2 cos 2θ
or DC Pandey Solutions: Units, Dimensions & Vectors - 2
or  A2 = - B2 cos π
or   A2 = B2
⇒ A = B
Option (a) is correct.


Match the Columns

Q1. Column-I shows some vector equations. Match column I with the value of the angle between A and B given in column II.

Column IColumn II
Match the Columns(p) zero
Match the Columns(q) π/2
Match the Columns(r) π/4
Match the Columns(s) 3π/4

Match the Columns
or Match the Columns
or Match the Columns
Match the Columns
Thus, (a) → (r) (s).

Match the Columns (given)
or Match the Columns
or Match the Columns
or sin θ = - sin θ
or   2 sin θ = 0
⇒ θ = 0 rad
Match the Columns
or Match the Columns
Match the Columns
Match the Columns
or  Match the Columns
Match the Columns
Match the Columns
or Match the Columns
or Match the Columns
or Match the Columns
or Match the Columns
Match the Columns
Match the Columns

Section-II
Subjective Questions 

Q1. Young's modulus of steel is 2.0 x 1011 N / m2. Express it in dyne/cm2.

Match the Columns
Match the Columns


Q2. Surface tension of water in the CGS system is 72 dynes/cm . What is its value in SI units

Match the Columns


Q3. In the expression y = a sin (ωt + θ), y is the displacement and t is the time. Write the dim ensions of a, ω and θ.

[a] = [y] = [L]
Sol: [wt] = [M0L0 T0] ∴ [ω] = [T-1]

[θ] = [M0L0 T0]


Q4. The relation between the energy E and the frequency v of a photon is expressed by the equation E = hv, where h is Planck's constant. Write down the SI units of h and its dimensions.

Match the Columns


Q5. Write the dimensions of a and b in the relation.
Match the Columns
where P is power, x is distance and t is time.

[b] = [x2] = [L2]
Match the Columns
Match the Columns


Q6. Check the correctness of the relation Match the Columns where u is initial velocity, a is acceleration and St is the displacement of by the body in tth second.

Match the Columns
Here t in second. Hence the given equation seems to be dimensionally incorrect. But it is correct because 1 is hidden.


Q7. Let x and a stand for distance. Match the Columns dim ensionally correct?

LHS is dimensionless. While RHS has the dimensions [L-1].


Q8. In the equation 
Match the Columns
Find the value of n.

LHS is dimensionless. Hence n = 0.


Q9. Show dimensionally that the expression, Match the Columns is  dimensionally correct, where  Y is Young's modulus of the material of wire, L is length of wire, Mg is the weight applied on the wire and l is the increase in the length of the wire.

Just write the dimension of different physical quantities.


Q10. The energy E of an oscillating body in simple harmonic motion depends on its mass m, frequency n and amplitude a. Using the method of dimensional analysis find the relation between E, m, n and a.

E = kmxnyaz.

Here k = a dimensionless constant
∴ [E] = [m]x [n]y [a]z
∴ [ML2 T-2] = [M]x[T-1]y[L]z
∴ x = 1, y = 2 and z = 2


Q11. The centripetal force F acting on a particle moving uniformly in a circle may depend upon mass (m), velocity (v) and radius r of the circle. Derive the formula for F using the method of dimensions.

Match the Columns
(k = a dimensionless constant) 
Match the Columns
Solving we get,
x = 1, y = 2 and z = - 1
Match the Columns


Q12. Taking force F, length L and time T to be the fundamental quantities, find the dimensions of (a) density, (b) pressure, (c) momentum and (d) energy.

[d] = [F]x [L]y [T]z
∴ [ML-3] = [MLT-2]x[L]y[T]z 
Equating the powers we get,
x = 1, y = - 4, z = 2
∴ [ d] = [FL-4 T2]
Similarly other parts can be solved.

Vectors

Q13. Find the cosine of the angle between the vectors Match the Columns

Match the Columns


Q14. Obtain the angle between Match the Columns

Match the Columns
Angle between Match the Columns
Match the Columns


Q15. Under what conditions will the vectors Match the Columns be perpendicular to each other ?

Their dot product should be zero.


Q16. Deduce the condition for the vectors Match the Columns

Ratio of coefficients of Match the Columns should be same.


Q17. Three vectors which are coplanar with respect to a certain rectangular co-ordinate system are given by
Match the Columns
Find
Match the Columns
(c) Find the angle between Match the Columns

No solution is required.


Q18. Find the components of a vector Match the Columns along the directions of 

Match the Columns
Match the Columns


Q19. If vectors Match the Columns be respectively equal to Match the Columns Find the unit vector parallel to Match the Columns

Match the Columns


Q20. If two vectors are Match the Columns By calculation, prove that 

Match the Columns is perpendicular to both Match the Columns

Match the Columns


Q21. Find the area of the parallelogram whose sides are represented by 

Area of parallelogram Match the Columns


Q22. The resultant of two vectors Match the Columns is at right angles toMatch the Columns and its magnitude is half of Match the ColumnsFind the angle between Match the Columns

Match the Columns
Match the Columns


Q23. The x and y-components of vector Match the Columns are 4 m and 6 m respectively. The x and y-components of vector Match the Columns are 10 m and 9 m respectively. Calculate for the vector Match the Columns the following
(a) its x andy-components 
(b) its length 
(c) the angle it makes with x-axis

Match the Columns


Q24. Prove by the method of vectors that in a triangle 
Match the Columns

Match the Columns
Applying sine law, we have

Match the Columns


Q25. Four forces of magnitude P, 2P, 3P and AP act along the four sides of a square ABCDm cyclic order. Use the vector method to find the resultant force.

Match the Columns


Q26. Match the Columns
R2 + S2 = 2(P2 + Q2)

R2 = P2 + Q2 + 2PQ cos θ
S2 = P2 + Q2 - 2PQ cos θ
∴ R2 + S2 = 2 (P2 + Q2)

The document DC Pandey Solutions: Units, Dimensions & Vectors - 2 is a part of the JEE Course DC Pandey Solutions for JEE Physics.
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FAQs on DC Pandey Solutions: Units, Dimensions & Vectors - 2

1. How do I convert units correctly in physics problems for JEE exams?
Ans. Unit conversion requires identifying the conversion factor between two measurement systems and multiplying systematically. For JEE physics, master SI base units (metre, kilogram, second) and use dimensional analysis to verify conversions. Practise problems involving energy, force, and pressure conversions-these appear frequently in competitive exams and demand accuracy with prefixes like kilo, milli, and micro.
2. What's the difference between dimensions and units, and why does it matter for JEE?
Ans. Units are standardised quantities used to measure physical properties (metre for length); dimensions represent the nature of that property using brackets like [L], [M], [T]. Understanding this distinction helps verify equation correctness through dimensional analysis-a critical JEE problem-solving strategy. An equation can have correct units but wrong dimensions if derived incorrectly.
3. How can I use dimensional analysis to check if my physics equations are correct?
Ans. Dimensional analysis involves comparing dimensions on both sides of an equation to confirm validity. Write each term using fundamental dimensions [M], [L], [T], [A], and verify they match. This technique catches algebraic errors instantly and works for any physical quantity-displacement, acceleration, or momentum-making it invaluable for JEE problem-solving without calculating numerical answers.
4. Why do some vector addition problems give different answers when I use different methods?
Ans. Vector addition results depend on method accuracy, not method choice. Graphical, component, and parallelogram methods yield identical answers if executed correctly. Common mistakes include incorrect angle measurement, forgetting vector direction, or misapplying the Pythagorean theorem for non-perpendicular vectors. Always resolve vectors into horizontal and vertical components for most reliable JEE exam results.
5. What are scalar and vector quantities, and how do I identify them in physics problems?
Ans. Scalars possess magnitude only (temperature, speed, mass); vectors require both magnitude and direction (velocity, force, displacement). Identify vectors in JEE questions by checking if direction matters for the physical phenomenon-motion problems typically involve vector quantities. Refer to flashcards and mind maps on EduRev to distinguish between common scalar-vector pairs and strengthen conceptual clarity for exams.
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