DC Pandey Solutions: Units, Dimensions & Vectors - 3 - NEET PDF Download

Section-II
Subjective Questions 

Ques 1: Young’s modulus of steel is 2.0 x 10¹¹ N / m². Express it in dyne/cm².
Ans:



Ques 2: Surface tension of water in the CGS system is 72 dynes/cm . What is its value in SI units?
Ans:  

Ques 3: 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.
Ans: [a] = [y] = [L]
Sol: [wt] = [M⁰L⁰ T⁰] ∴ [ω] = [T^-1]

[θ] = [M⁰L⁰ T⁰]

Ques 4: 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.
Ans:


Ques 5: Write the dimensions of a and b in the relation.

where P is power, x is distance and t is time.
Ans: [b] = [x²] = [L²]



Ques 6: Check the correctness of the relation  where u is initial velocity, a is acceleration and S_t is the displacement of by the body in t^th second.
Ans: 

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

Ques 7: Let x and a stand for distance.  dim ensionally correct?
Ans: LHS is dimensionless. While RHS has the dimensions [L^-1].

Ques 8: In the equation 

Find the value of n.
Ans: LHS is dimensionless. Hence n = 0.

Ques 9: Show dimensionally that the expression,  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.
Ans: Just write the dimension of different physical quantities.

Ques 10: 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.
Ans: E = km^xn^ya^z.

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

Ques 11: 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.

Ans:
(k = a dimensionless constant) 

Solving we get,
x = 1, y = 2 and z = - 1
∴

Ques 12: 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.
Ans: [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

Ques 13: Find the cosine of the angle between the vectors
Ans: 


Ques 14: Obtain the angle between
Ans: 

Angle between


Ques 15: Under what conditions will the vectors  be perpendicular to each other ?
Ans: Their dot product should be zero.

Ques 16: Deduce the condition for the vectors
Ans: Ratio of coefficients of  should be same.

Ques 17: Three vectors which are coplanar with respect to a certain rectangular co-ordinate system are given by

Find

(c) Find the angle between
Ans: No solution is required.

Ques 18: Find the components of a vector  along the directions of 

Ans:

Ques 19: If vectors  be respectively equal to  Find the unit vector parallel to

Ans:

Ques 20: If two vectors are  By calculation, prove that 

 is perpendicular to both
Ans: 


Ques 21: Find the area of the parallelogram whose sides are represented by
Ans: Area of parallelogram

Ques 22: The resultant of two vectors  is at right angles to and its magnitude is half of Find the angle between
Ans: 



Ques 23: The x and y-components of vector  are 4 m and 6 m respectively. The x and y-components of vector  are 10 m and 9 m respectively. Calculate for the vector  the following
(a) its x andy-components 
(b) its length 
(c) the angle it makes with x-axis
Ans: 


Ques 24: Prove by the method of vectors that in a triangle 

Ans: 

Applying sine law, we have

Ques 25: 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.
Ans: 


Ques 26:
R² + S² = 2(P² + Q²)
Ans: R² = P² + Q² + 2PQ cos θ
S² = P² + Q² - 2PQ cos θ
∴ R² + S² = 2 (P² + Q²)