Solved Questions
In this section provided Solved Questions with Solutions for IIT JEE and Engineering Entrance & Boards Exams. IIT JEE Questions on Units Dimensions and Vectors.
1. If A = 4i – 3j and B = 6i + 8j, find the scalar magnitude and direction of A, B, A + B, A – B and B – A with respect to positive direction of x-axis.
SolutionB = Bx i + By j = 6i + 8j
|A| = \sqrt {{{(4)}^2} + {{(3)}^2}} = \sqrt {25} = 5
|B| = \sqrt {{{(6)}^2} + {{(8)}^2}} = \sqrt {100} = 10
A + B = (Ax + Bx)i + (Ay + By)j = 4i – 3j + 6i + 8j = 10i + 5j
|A + B| = \sqrt {{{(10)}^2} + {{(5)}^2}} = \sqrt {125} = 11.2
A – B = 4i – 3j – 6i – 8j = –2i – 11j
|A – B| = \sqrt {{{( - 2)}^2} + {{( - 11)}^2}} = 11.2
B – A = 6i + 8j – 4i + 3j = 2i + 11j
|B – A| = \sqrt {{{(2)}^2} + {{(11)}^2}} = 11.2
Let θ1, θ2, θ3, θ4 and θ5 be angles which A, B, A + B, A – B and B – A make with the positive direction of x-axis.
tanθ1 = \frac{{{A_y}}}{{{A_x}}} = - \frac{3}{4} = –0.75,
tanθ1 = –tan–1 (0.75) = –37º 
tanθ2 = \frac{{{B_y}}}{{{B_x}}} = \frac{8}{6} = 1.33, θ2 = tan–1 (1.33) = 53.1º
tanθ3 = \frac{{({A_y} + {B_y})}}{{({A_x} + {B_x})}} = \frac{5}{{10}} = 0.5, θ3 = tan–1(1.33) = 26.5º
tanθ4 = \frac{{({A_y} - {B_y})}}{{({A_x} - {B_x})}} = \frac{{ - 11}}{{ - 2}} = 5.5, θ4 = tan–1(5.5) ~ 80º
As (Ay – By) and (Ax – Bx) are both negative, A – B is in third quadrant, the angle which vector A – B makes with the positive direction of x-axis, measured in anticlockwise direction, is equal to 180 + θ4 = 180 + 80 = 260º.
tan θ5 = \frac{{{B_y} - {A_y}}}{{{B_x} - {A_x}}} = \frac{{11}}{2} = 5.5, θ5 = tan–1 (5.5) = 80º.
2. If vectors A = 3i + j – 2k, B = –i + 3j + 4k and C = 4i –2j –6k can be represented by the sides of a triangle, find the two vectors which will give the third vector as a resultant.
Solutionresultant.
A + B = (3i + j – 2k) + (–i + 3j + 4k) = 2i + 4j + 2k ≠ C
A + C = (3i + j – 2k) + (4i –2j –6k) = (7i – j – 8k) ≠ B
B + C = (–i + 3j + 4k) + (4i –2j –6k) = (3i + j –2k) = A
∴ vectors B and C combine to give resultant A
3. Find the scalar product and angle between vectors A = 3i – 5j and B = 6i + 2j
A.B = |A| |B| cosθ
or cosθ = \frac{{{\bf{A}} \cdot {\bf{B}}}}{{|{\bf{A}}|\,|{\bf{B}}|}}
|A| = \sqrt {A_x^2 + A_y^2} = \sqrt {9 + 25} =\sqrt {34}
|B| = \sqrt {B_x^2 + B_y^2} = \sqrt {36 + 4} = \sqrt {40}
∴ cos θ = \frac{8}{{(\sqrt {34} )(\sqrt {40} )}} = \frac{8}{{36.8}} = 0.217
θ = cos–1(0.217) = 77.5º
4. A partical moves from the position 2i + 3j – 5k to position 12i + 15j + 8k in metre units and a uniform force 5i + 2j + 3k , Newton acts on it. Calculate the work done by this force.
SolutionW = F . S
S = r2 – r1 = (12i + 15j + 8k) – (2i + 3j – 5k) = 10i + 12j + 13k
W = (5i + 2j + 3k) . (10i + 12j + 13k) = 50 + 24 + 39 = 113 J
5. If a force F = 4i – 10j Newton is applied to the rim of a disc at a distance or r = –5i – 4j from its axis of rotation, find the torque applied.
Solution= (–5i – 4j) × (4i – 10j)
= (–20) (i × i) + (50) (i × j) + (–16) (j × i) + (40) (j × j)
= 0 + 50 k + (–16) (–k) + 0
= 50k + 16k = 66k
6. A plane is flying with speed 700 km/h in north-eastern direction and wind is blowing at 100 km/h north to south. Find the resultant displacement of plane in 2hours.
Solution v1 = 700 (cos 45)i + 700 (sin 45)j 
= \frac{{700}}{{\sqrt 2 }}i +\frac{{700}}{{\sqrt 2 }} j
v2 = –100j
If v is resultant velocity, v = v1 + v2 = \frac{{700}}{{\sqrt 2 }}i + \frac{{700}}{{\sqrt 2 }}j – 100j
= 496i + 496j – 100j = 496i + 396j km/h
∴ Displacement, S = v × t = (496 × 2)i + (396 × 2)j = 992i + 792j, S = {(992)2 + (792)2}1/2 = 1268.1 km
Objective Questions
In this section provided Objective Questions with Answers for IIT JEE, Engineering Entrance & Boards Exams. IIT JEE Questions on Units Dimensions and Vectors.
- The only pair of physical quantities that does not have identical dimensions are
(a) angular momentum and Planck’s constant
(b) torque and energy
(c) moment of inertia and moment of force
(d) impulse and momentum
Answer- The dimensions of the quantities in one of the following pairs are the same. Identify the pair.
(a) angular momentum and energy
(b) torque and work
(c) energy and Young’s modulus
(d) light year and frequency
Answer- The dimensional formula for gas constant is
(a) [M0 L2 T–2 K–2 mol–1]
(b) [M2 L3 T–1 K–2 mol–1]
(c) [M L0 T–2 K–3 mol–1]
(d) [M L2 T–2 K–1 mol–1] Answer
- The velocity of sound v in gas is v = \sqrt {\frac{{\gamma \,P}}{\rho }} where P is the pressure and r is density of the gas. The dimensional formula for ϒ is
(a) [M0 L0 T0]
(b) [M0 L1 T2]
(c) [M1 L2 T0]
(d) [M1 L1 T0] Answer
- The vectors A and B are such that |A + B| = |A – B|, the angle between these two vectors is
(a) 0º
(b) 30º
(c) 90º
(d) 180º
Answer- If B + C = A and A = B = C, the angle between vectors B and C is
(a) 30º
(b) 90º
(c) 120º
(d) 180º
Answer- If a force F = 2i + 5k Newton acts through a displacement s = 3j + 4k, work done is equal to
(a) 20 J
(b) 26 J
(c) 29 J
(d) 39 J
Answer- If a force F = 8i – 20j Newton acts at the rim of a disc rotating at a distance r = –10i – 8j m from its centre, its torque T = r × F is equal to
(a) –80 N.m
(b) 80 N.m
(c) –264 N.m
(d) 264 N.m
Answer- It is given that P + Q = R, R is perpendicular to P and |P| = |R|, the angle between P and Q is
(a) \frac{\pi }{4}
(b) \frac{\pi }{2}
(c) \frac{{3\pi }}{4}
(d) {\pi } Answer
- A vector is not changed if
(a) it is multiplied to an arbitrary scalar
(b) it is rotated through an arbitrary angle
(c) it is cross multiplied by a unit vector
(d) it is slid parallel to itself.
Answer- If A = 4i – 2j + 6k and B = i – 2j – 3k, the angle which A + B makes with x-axis is
(a) Zero
(b) \frac{\pi }{3}
(c) \frac{\pi }{4}
(d) \frac{\pi }{6}
Answer- The coordinates of a particle moving at any time t is given by x = 2At2 and y = 2Bt2 where A and B are constants. The speed of the particle is
(a) 4t (A + B)
(b) 4t\sqrt {{A^2} + {B^2}}
(c) 4t (A2 + B2)
(d) 2t \sqrt {{A^2} + {B^2}} Answer
- A particle is moving eastwards with a velocity of 10 m/s. Its velocity change to 10 m/s northwards in 20s. The average acceleration in this time is
(a) \frac{1}{{\sqrt 2 }}m/s2 towards north east
(b) \frac{1}{{\sqrt 2 }}m/s2 towards north west
(c) \frac{1}{2}m/s2 towards north west
(d) 1 m/s2 towards north east
Answer- The position vector of a particle changes from 10 m east to 10 m north in 2 seconds. The averge velocity of the particle is
(a) 20 m/s
(b) 5 m/s
(c) 10\sqrt 2 m/s
(d) 5 \sqrt 2 m/s
Answer- \int {\frac{{dx}}{{\sqrt {2\;ax\;-\;{x^2}\;} }}\; = \;{a^n}\;{\text{si}}{{\text{n}}^{-1}}\;\left[ {\frac{x}{a}\;-\;1} \right]} . The value of n is
(a) 0
(b) –1
(c) 1
(d) none of these.
Answer- A physical quantity is measured and the result is expressed as nu where u is the unit used and n is the numerical value. If the result is expressed in various units then
(a) n \propto size of u
(b) n \propto u2
(c) n \propto \sqrt u
(d) n \propto \frac{1}{u} Answer
- A vector \overrightarrow a is turned through θ about its initial point. \left| {\Delta \overrightarrow a } \right| of the vector is
(a) 0
(b) 2|\vec a|\theta
(c) 2|\vec a|\sin \theta /2
(d) 2|\vec a|\cos \theta /2 Answer
- The component of the vector \vec A\; = \;2\;\hat i\; + \;3\;\hat j along the vector \hat i\; + \;\hat j is
(a) 5
(b) \frac{5}{{\sqrt 2 }}
(c) 10\sqrt 2
(d) 2\sqrt 5
Answer- Suppose a quantity x can be dimensionally represented in terms of M, L and T, that is, [x] = Ma LbTc. The quantity mass
(a) can always be dimensionally represented in terms of L, T and x,
(b) can never be dimensionally represented in terms of L, T and x,
(c) may be represented in terms of L, T and x if a = 0,
(d) may be represented in terms of L, T and x if a ≠ 0.
Answer- A unitless quantity
(a) never has a nonzero dimension,
(b) always has a nonzero dimension,
(c) may have a nonzero dimension
(d) does not exist
Answer- The magnitude of i + j + k is
(a) 6
(b) 3
(c) 1
(d)\sqrt 3 Answer
- Two forces F1 and F2 are acting at right angles to each other. Then their resultant is
(a) F1 + F2
(b) \sqrt {{F_1}^2\, + {F_2}^2}
(c) \sqrt {{F_1}^2\, - {F_2}^2}
(d) \frac{{{F_1}\, + {F_2}}}{2}
Answer- The x and y components of a force are 2N and –3N. The force is
(a) 2i – 3j
(b) 2i + 3j
(c) –2i – 3j
(d) 3i + 2j
Answer- Which of the following quantities are dependent on the choice of coordinate axes?
(a) a + b
(b) a – b
(c) a × b
(d) ax + by
Answer- The angle between i + j + k and 2i + 2j + 2k is
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Answer- Identify the vector quantity
(a) work
(b) time
(c) heat
(d) angular momentum
Answer
Miscellaneous Questions
In this section provided Miscellaneous Questions with Answers for IIT JEE, Engineering Entrance & Boards Exams. IIT JEE Questions on Units Dimensions and Vectors.
Comprehension
In the study of physics, we often have to measure the physical quantities. The numerical value of a measured quantity can only be approximate as it depends upon the least count of the measuring instrument used. The number of significant figure in any measurement indicates the degree of precision of that measurement. The importance of significant figure lies in calculation. A mathematical calculation cannot increase the precision of a physical measurement. Therefore, the number of significant figure in the sum or product of a group of numbers cannot be greater than the number that has the least number of significant figures. A chain cannot be stronger than its weakest link.
- A bee of mass 0.000087 kg sits on a flower of mass 0.0123 kg. What is the total mass supported by the stem of the flower upto appropriate significant figures?
(a) 0.012387 kg
(b) 0.01239 kg
(c) 0.0124 kg
(d) 0.012 kg
Answer- The radius of a uniform wire is r = 0.021 cm. The value π is given to be 3.142. What is the area of cross-section of the wire upto appropriate significant figures?
(a) 0.0014 cm2
(b) 0.00139 cm2
(c) 0.001386 cm2
(d) 0.0013856 cm2
Answer- A man runs 100.5 m in 10.3 s. Find his average speed upto appropriate significant figures.
(a) 9.71 ms–1
(b) 9.708 ms–1
(c) 9.7087 ms–1
(d) 9.70874 ms–1
AnswerAssertion : Reason
Each of the questions given below consists of two statements, an assertion (A) and reason (R). Select the number corresponding to the appropriate alternative as follows
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true
- A: Light year is a unit of time.
R: Light year is the distance travelled by light in vacuum in one year.
Answer- A: nm is not same as mN
R: 1 nm = 10–9 m and 1 mN = 10–3 N
Answer- A: Planck’s constant has the dimensions of angular momentum.
R: Both represent product of energy and time
Answer- A: Parallax method is used for measuring distances of nearby stars only.
R: With increase of distance of star, parallactic angle becomes too small to be measured accurately.
Answer- A: Pressure has the dimensions of energy density.
R: Energy density = \frac{{energy}}{{volume}}\; = \;\frac{{M{L^2}{T^{ - 2}}}}{{{L^3}}} = [ML–1T–2] = pressure
Answer- A: The dimensions of rate of flow are [M0L3T–1]
R: Rate of flow is velocity/sec.
Answer- A: Specific gravity of liquid is a dimensionless quantity.
R: It is the ratio of density of fluid to the density of water.
Answer
Basic Questions
In this section provided Basic level Questions with Answers for IIT JEE and Engineering Entrance & Boards Exams. IIT JEE Questions on Units Dimensions and Vectors.
- The value of acceleration due to gravity is 980 cm/s2. Find its value if the unit of length is kilometre and that of time is minute using method of dimensional equation.
- It is given that the time period of oscillation, T, of a gas bubble from an explosion under water depends upon pressure, p, density of water, d, and total energy of explosion, E, as T \propto padb Ec . Derive a relation for T in terms of various dimensions (powers) of p, d and E.
- Assuming that the mass M of the largest stone that can be moved by a flowing river depends upon the velocity v of water, acceleration due to gravity g and density d of water. Find the dependence of M on velocity v using dimensional equation.
- It is experimentally observed that the frequency f of a tuning fork depends upon the length l of the prongs of the tuning fork and density d and Young’s modulus Y of its material. Derive a relation for the frequency of tuning fork in terms of l, d and Y using principle of homogeneity of dimensions.
- A planet moves around the sun in a circular orbit. The time period of revolution t of the planet depends upon radius of orbit R, mass of sun M and Gravitational constant G. Derive a relation for t in terms of R, M and G using dimensional equation.
- Find the scalar product and angle between the vectors A = i + j –2k and B = –i + 2j – k
- A particle suffers three displacements by 4m northwards, 2m in south-east direction and 1m in southwest direction. Find the displacement of the particle and distance covered by it.
- A force F = 5i – 3j + 2k Newton is applied to a particle so that it moves from a distance r1= 2i + 7j + 4k m to r2 = –5i + 2j + 3k m. Find the work done.
- If vectors A = 3i –2j + k, B = i – 3j + 5k and c = 2i + j – 4k form a triangle. Find
(a) Which two vectors combine to form the third vector as their resultant ?
(b) Which vectors are perpendicular to each other.
Answer- Find the unit vector which is perpendicular to each of the vectors A = 3i + j + 2k and B = 2i –2j + 4k . Also find the sine of angle between vectors A and B.
- Find the angle between vectors A = i –5j and B = 2i – 10j
- Find the area of a parallelogram formed from the vectors A = (i – 2j + 3k)m and B = (3i –2j + k)m which are the adjacent sides of the parallelogram.
- Find the derivative of the following functions with respect to x:
(a) \sqrt {ax + b}
(b) sin(x2)
(c) sin2(x2)
(d) log (ax + b)
(e) \frac{{\sin x}}{x}
(f) ex sinx
Answer(a) \frac{a}{{2\sqrt {ax + b} }}
(b) 2x cos(x2)
(c) 4x sin x2 cos x2
(d) \frac{a}{{ax + b}}
(e) \frac{{x\cos x - \sin x}}{{{x^2}}}
(f) ex (sinx + cosx)
- Integrate the following functions w.r.t. x :
(a) \int\limits_0^1 {\frac{{dx}}{{3x + 4}}}
(b) \int\limits_0^\pi {\cos (2\theta )d\theta }
(c) \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}\,\,\,\theta \,\,d\theta }
(d) \int {\frac{{{x^2}}}{{1 + {x^2}}}dx}
(e) \int {\sqrt {1 + \cos x\,dx} } Answer
(a) \frac{1}{3}\ln \left( {\frac{7}{4}} \right)
(b) zero
(c) \frac{\pi }{4}
(d) x – tan–1 x + C
(e) 2\sqrt 2 \sin \left( {\frac{x}{2}} \right) + C
- Find derivative of the function y = (ax + b)n.
Advanced Questions
In this section provided Advance Level Questions with Answers for IIT JEE and Engineering Entrance & Boards Exams. IIT JEE Questions on Units Dimensions and Vectors.
- Find the dimension of
(a) electric field E
(b) magnetic field B
(c) magnetic permeability μ0.
The relevant equations are F = q E, F = qv B, and B = \frac{{{\mu _0}\;I}}{{2\;\pi \;\alpha }}, where F is force, q is charge, v is speed, I is current, and α is distance.
Answer(a) MLT–3 A–1
(b) MT–2 A–1
(c) MLT–2 A–2
- Find the dimension of
(a) electric dipole moment p
(b) magnetic dipole moment M. The defining equations are p = q . d. and M = I A , where d is distance, A is area, q is charge and I is current.
Answer(a) LTA
(b) L2A
- Find the dimensions of
(a) the specific heat capacity c
(b) the coefficient of linear expansion α and
(c) the gas constant R.
Some of the equations involving these quantities are Q = mc (T2 – T1), lt = l0 [1 + α (T2 – T1)] and PV = nRT.
Answer(a) L2 T–2 K–1
(b) K–1
(c) ML2 T–2 K–1 (mol)–1
- Taking force, length and time to be the fundamental quantities find the dimensions of
(a) density
(b) pressure
(c) momentum
(d) energy
Answer(a) FL–4 T2
(b) FL–2
(c) FT
(d) FL
- Let x and a stand for distance. Is \int {\;\frac{{d\;x}}{{\sqrt {{a^2}\;--\,{x^2}} }}} = \frac{1}{a}\;{\text{si}}{{\text{n}}^{{\text{--1}}}}\;\frac{a}{x} dimensionally correct ?
- A carrom board (4 ft × 4 ft square) has the queen at the centre. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen
(a) from the centre to the front edge,
(b) from the front edge to the hole
(c) from the centre to the hole.
Answer (a) \frac{2}{3}\,\sqrt {10} \,\,ft
(b) \frac{4}{3}\,\sqrt {10} \,\,ft
(c) 2\sqrt 2 \,\,ft
- Refer to the figure. Find
(a) the magnitude, 
(b) x and y components and
(c) the angle with the X-axis of the resultant of \overrightarrow {OA} ,\,{\text{ }}\overrightarrow {BC} and \overrightarrow {DE} .
Answer(a) 1.6m
(b) 0.98 m and 1.3 m
(c) tan–1 (1.32)
- If a and b are two vectors, then what will be the value of (a + b) × (a – b).
- A particle moves on a given straight line with a constant speed v. At a certain time it is at a point P on its straight line path. O is a fixed point. Show that \overrightarrow {OP} \; \times \;\vec v is independent of the position P.
- Find the unit vector along the vector A × B. Where A = 2i + 3j and B = i + j.
- The electric current in a charging R– C circuit is given by i = i0 e–t/RC where i0, R and C are constant parameters of the circuit and t is time. Find the rate of change of current at (a) t = 0, (b) t = R C, (c) t = 10 R C.
(a) \frac{{{i_0}}}{{RC}}
(b) \frac{{{i_0}}}{{RCe}}
(c) \frac{{{i_0}}}{{RC{e^{10}}}}
- A rod of length L is placed along the X-axis between x = 0 and x = L. The linear density (mass/ length) r 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.
(a) kg/m, kg/m2
(b) aL + bL2/2
- The momentum p of a particle changes with time t according to the relation \frac{{d\;p}}{{d\;t}} = (10 N) + (2 N/s) t. If the momentum is zero at t = 0 what will the momentum be at t = 10 s ?
- Do you think that commutative and associtave laws are valid for vector subtraction also?, if yes write down the correct mathematical form and prove them, if no explain why not?
- It it possible to have the modulus of subtraction of two vectors greater than the modulus of addition of the same two vectors.
- Will it be possible to have the modulus of subtraction of two vectors less than the subtraction of the modulus of the same two vectors.
- A particle is moving eastward with a velocity 10 m/s. In 10 s the velocity changes to 10 m/s northwards. Find the average acceleration during this time.
- A body goes 20 km north and then 10 km due east. Find the displacement of body from its starting point.
- Position of a particle is given as x = 3t + 5 where x in meter and t in seconds. Find velocity (dx/dt) of particle at time t = 5 second.