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Test: Dimensional Analysis & Its Applications (NCERT) - NEET MCQ


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10 Questions MCQ Test Physics Class 11 - Test: Dimensional Analysis & Its Applications (NCERT)

Test: Dimensional Analysis & Its Applications (NCERT) for NEET 2024 is part of Physics Class 11 preparation. The Test: Dimensional Analysis & Its Applications (NCERT) questions and answers have been prepared according to the NEET exam syllabus.The Test: Dimensional Analysis & Its Applications (NCERT) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Dimensional Analysis & Its Applications (NCERT) below.
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Test: Dimensional Analysis & Its Applications (NCERT) - Question 1

Checking the correctness of equations using the method of dimensions is based on

Detailed Solution for Test: Dimensional Analysis & Its Applications (NCERT) - Question 1

Only the dimensional accuracy can be checked with the help of principle of homogeneity of dimensions. This is limitation of principle of homogeneity of dimensions.

Test: Dimensional Analysis & Its Applications (NCERT) - Question 2

Using the principle of homogeneity of dimensions, which of the following is correct?

Detailed Solution for Test: Dimensional Analysis & Its Applications (NCERT) - Question 2


Taking dimensions on both sides, we get

∴ LHS = RHS
Now, T2 = 4π2r2
Taking dimensions on both sides
[T]2 = [L]2 
∴ LHS ≠ RHS

Test: Dimensional Analysis & Its Applications (NCERT) - Question 3

Which of the following relations is dimensionally incorrect?

Detailed Solution for Test: Dimensional Analysis & Its Applications (NCERT) - Question 3

As, we know, E = mc2 and energy released can be given as:
E = 931MeV
for, m = 1u
1u = 931.5MeV/c2 - Dimensionally correct [Dimension of mass]
1u = 1.67 × 10−27 kg - Dimensionally correct [Dimension of mass]
but, 931.5 MeV is having the dimension of energy so it can't be equated to dimension of mass or dimension of 1u.
Hence, dimensionally correct data is given in option(A)

Test: Dimensional Analysis & Its Applications (NCERT) - Question 4

Which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct dimensionally?

Detailed Solution for Test: Dimensional Analysis & Its Applications (NCERT) - Question 4

Dimensions on RHS must be displacement [L].
Arguments of sine and cosine are dimensionless. Hence, Option (c) is not correct. 

Test: Dimensional Analysis & Its Applications (NCERT) - Question 5

The displacement of a progressive wave is represented by y = A sin(ωt - kx) where x is distance and t is time. The dimensions of ω/k are same as those of the

Detailed Solution for Test: Dimensional Analysis & Its Applications (NCERT) - Question 5

y = Asin(ωt − kx)
As (ωt − kx) represents an angle which is dimensionless, therefore

Test: Dimensional Analysis & Its Applications (NCERT) - Question 6

If velocity of light c, Planck's constant h and gravitational constant G are taken as fundamental quantities then the dimensions of length will be:

Detailed Solution for Test: Dimensional Analysis & Its Applications (NCERT) - Question 6

Let l ∝ cxhy Gz ;l = kcx hy Gz
where k is a dimensionless constant and x, y and z are the exponents.
Equating dimensions on both sides, we get
[M0 LT0] =[LT−1] x [MLT−1] y [M−1 L3 T−2]z
= [My−z Lx+2y+3z T −x−y−2z]
Applying the principle of homogeneity of dimensions, we get
y − z = 0             ....(i)
x + 2y + 3z = 1     ....(ii)
− x − y − 2z = 0     ...(iii)
On solving Eqs. (i), (ii) and (iii), we get

Test: Dimensional Analysis & Its Applications (NCERT) - Question 7

A new system of units is proposed in which unit of mass is α kg, unit of length is β m and unit of time is γ s. What will be value of 5 J in this new system?

Detailed Solution for Test: Dimensional Analysis & Its Applications (NCERT) - Question 7

Joule is a unit of energy.


Dimensional formula of energy is [ML2 T−2].
Comparing with [Ma Lb Tc], we get a = 1, b = 2,c = -2

Test: Dimensional Analysis & Its Applications (NCERT) - Question 8

If the energy, E = Gp hq cr where G is the universal gravitational constant, h is the Planck's constant and c is the velocity of light, then the values o f p, q and r are, respectively

Detailed Solution for Test: Dimensional Analysis & Its Applications (NCERT) - Question 8

E = Gp hcr…(i)
[M1L2T−2] = [M−1L3T−2]p[ML2T−1]q[LT−1]r
= [M−p+q L3p+2q+r T−2p−q−r]
Applying principle of homogeneity of dimensions, we get
−p + q = 1…(ii)
3p + 2q + r = 2…(iii)
−2p − q − r = −2…(iv)
Adding (iii) and (iv),
we get p + q = 0…(v)
Adding (ii) and (v), we get q = 1/2
From (ii), we get p = q − 1

Test: Dimensional Analysis & Its Applications (NCERT) - Question 9

The equation of state of a gas is given by where p,V,T are pressure, volume and temperature respectively and a,b,c are constants. The dimensions of a and b are respectively

Detailed Solution for Test: Dimensional Analysis & Its Applications (NCERT) - Question 9


Dimension of a/V3 = Dimensions of P
∴ Dimensions of a = dimensions of PV3 

Dimensions of b2  = dimensions of V
∴ [b] = [V]1/2 = [L3]1/2 or [b] = [L3/2]
 

Test: Dimensional Analysis & Its Applications (NCERT) - Question 10

The velocity of a particle (v) at an instant t is given by v = at + bt2. The dimension of b is the

Detailed Solution for Test: Dimensional Analysis & Its Applications (NCERT) - Question 10

v = at + bt2
[v] = [bt2] or [LT−1] = [bT2] or [b] = [LT−3]

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