SectionII
Subjective Questions
Ques 1: Young’s modulus of steel is 2.0 x 10^{11} N / m^{2}. Express it in dyne/cm^{2}.
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] = [M^{0}L^{0} T^{0}] ∴ [ω] = [T^{1}]
[θ] = [M^{0}L^{0} T^{0}]
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] = [x^{2}] = [L^{2}]
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^{x}n^{y}a^{z}.
Here k = a dimensionless constant
∴ [E] = [m]^{x} [n]^{y} [a]^{z}
∴ [ML^{2} 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} T^{2}]
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 coordinate 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
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Ques 19: If vectors be respectively equal to Find the unit vector parallel to
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Ques 20: If two vectors are By calculation, prove that
is perpendicular to both
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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 ycomponents of vector are 4 m and 6 m respectively. The x and ycomponents of vector are 10 m and 9 m respectively. Calculate for the vector the following
(a) its x andycomponents
(b) its length
(c) the angle it makes with xaxis
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:
R^{2} + S^{2} = 2(P^{2} + Q^{2})
Ans: R^{2} = P^{2} + Q^{2} + 2PQ cos θ
S^{2} = P^{2} + Q^{2}  2PQ cos θ
∴ R^{2} + S^{2} = 2 (P^{2} + Q^{2})
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