Units Measurements and Dimensions Test

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Units Measurements and Dimensions Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

Convert 1 kWh into joules.

A
33 10⁵ Joule

B
36 10⁵ Joule

C
36 10⁴ Joule

D
None of these

Solution

We know,
Watt = joule/second
1 kW = 1 kjoule/second
And 1 kWh = 1 kjoule × hour/second

1 kW hour = 1000 × 60 × 60 = 3600 ×1000 Joules

1 kW hour = 36 ×10⁵ Joule
Question 2
1 / -0

If the atmospheric pressure is 10⁶ dyne/cm², then what is its value in SI units?

A
10⁵ N/m²

B
10⁶ N/m²

C
10⁷N/m²

D
10⁸ N/m²

Solution

Pressure = 10⁶ dyne/cm²1 Newton = 10⁵ dyne, 1 m² = 10⁴cm²Hence Pressure = 10⁶dyne/cm²=10⁶× (10^-5N)/(10^-⁴ m²) Pressure = 10⁽^{6 - 5 + 4)} N/m² = 10⁵ N/m²
Question 3
1 / -0

What is the dimensional formula of thermal conductivity?

A
[MLT^-1θ^-1]

B
[MLT^-3θ^-1]

C
[M²LT^-3θ^-2]

D
[ML²T^-2θ]

Solution

According to law of thermal conductivity,
[K] =
K = [MLT^-3θ^-1]

Question 4

1 / -0

Match the following:

List - I	List - II
(1) Planck's constant	(A) [ML^-1T^-2]
(2) Gravitational constant	(B) [ML^-1T^-1]
(3) Bulk modulus	(C) [ML²T^-1]
(4) Coefficient of viscosity	(D) [M^-1L³T^-2]

(1) - (C), (2) - (D), (3) - (A), (4) - (B)

(1) - (B), (2) - (A), (3) - (C), (4) - (D)

(1) - (C), (2) - (B), (3) - (A), (4) - (D)

(1) - (C), (2) - (D), (3) - (B), (4) - (A)

Solution

Dimensions of Planck's constant = [ML²T^-1]
Dimensions of gravitational constant = [M^-1L³T^-2]
Dimensions of bulk modulus = [ML^-1T^-2]
Dimensions of coefficients of viscosity = [ML^-1T^-1]
Hence, option (1) is correct.

Question 5
1 / -0

If force (F), work (W) and velocity (v) are taken as fundamental quantities, then what is the dimensional formula of time (T)?

A
[WFv]

B
[WFv^-1]

C
[W^-1F^-1v]

D
[WF^-1v^-1]

Solution

Work done = Force × distance

As distance = Velocity × time

∵ V =
W = F.V.t

∴ T =
∴ [T] = [WF^-1V^-1]
Question 6
1 / -0

Which of the following sets of quantities has the same dimensional formula?

A
Frequency, angular frequency and angular momentum

B
Surface tension, stress and spring constant

C
Acceleration, momentum and retardation

D
Work, energy and torque

Solution

As the work is defined as the dot product of force and displacement.
Torque is represented as the cross product of radius and the force.
Energy is defined as the capacity of doing work so work, energy and the torque has same dimensions.
Question 7
1 / -0

The dimensions of ε₀E²( where ε₀permittivity of free space, E : electric field) are

A
[MLT^-1]

B
[ML²T^-2]

C
[ML^-1T^-²]

D
[ML²T^-1]

Solution

Dimensions = [M^-1L^-3A²T ⁴][MLA^-1T^-3]² = [ML^-1T^-2]
Question 8
1 / -0

The displacement of a particle moving along x-axis is given as x = at + bt²- ct³, where t is time. The respective dimensions of a, b and c are

A
L², T, LT ²

B
LT², LT, LT^-2

C
LT^-1, LT^-2, LT^-3

D
L, LT, T ²

Solution

x = at + bt² - ct³[L] = a[T] + b[T²] - c[T³]
[a] = LT^-1[b] = LT^-2[c] = LT^-3
Question 9
1 / -0

The density of material of a cube is measured by measuring its mass and length of its side. If the maximum errors in the measurement of mass and length are 3% and 2%, respectively, then the maximum error in the measurement of density is

A
1%

B
5%

C
7%

D
9%

Solution

, where m is mass of the block and L is length of the cube.

Percentage change in value of the density,

%

Or, maximum percentage error in the measurement of the density is 9%.
Question 10
1 / -0

The length, breadth and thickness of a block are given as l = 12 cm, b = 6 cm and t = 2.45 cm, respectively. The volume of the block, according to the idea of significant figures, should be

A
1 × 10² cm³

B
2 × 10² cm³

C
1.763 × 10² cm³

D
1.83 × 10²cm³

Solution

Volume, V = l × b × t = 12 × 6 × 2.45
= 176.4 cm³
or V = 1.764 × 10² cm³Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure.
Hence, V = 2 × 10² cm³
Question 11
1 / -0

Time period 'T' of a simple pendulum may depend on 'm', the mass of the bob, '', the length of the string, and 'g', the acceleration due to gravity. This means that 'T' ∝ m^a g^c. What are the respective values of 'a', 'b' and 'c'?

A
1/2, 0 and 1/2

B
1/2, 0 and -1/2

C
0, 1/2 and -1/2

D
0, 1/2 and 1/2

Solution

T
[T] = [M]^a[L]^b[LT^-2]^c[T] = [M]^a[L]^b+c[T]^-2cHere,
a = 0
b + c = 0
- 2c = 1
c =
b =
a = 0, b = , c =
Question 12
1 / -0

The volume of a cube in m³ is equal to the surface area of the cube in m². The volume of the cube is

A
64 m³

B
216 m³

C
512 m³

D
196 m³

Solution

Let the side of the cube be 'a' metres.
Surface area of the cube = Volume of the cube
⇒ 6 × a × a = a × a × a
⇒ 6a² = a³
⇒ a = 6
⇒ Volume of the cube = a³ = 216 m³
Question 13
1 / -0

If 'C' and 'R' respectively denote capacitance and resistance, then what will be the dimensions of 'CR'?

A
[M⁰L⁰TA⁰]

B
[ML⁰TA^-2]

C
[ML⁰TA²]

D
[MLTA^-2]

Solution

Dimensions of capacitance C and R are
C = [M^-1L^-2T⁴A²], R = [ML²T^-3A^-2]
∴ Dimensions of CR
= [M^-1L^-2T⁴A²] [ML²T^-3A^-2]
= [M⁰L⁰TA⁰]
Question 14
1 / -0

The velocity 'v' of a particle at time 't' is given by 'v' = at + , where 'a', 'b' and 'c' are constants. The dimensions of 'a', 'b' and 'c' are

A
[LT^-2], [L] and [T]

B
[L²], [T] and [LT²]

C
[LT²], [LT] and [L]

D
[L], [LT] and [T²]

Solution

'at' must have the dimensions of velocity.
So, let the dimensions of 'a' be [M^xL^yT^z].
[M^xL^yT^z][T] = [M⁰L¹T^-1]
[M^xL^yT^z] = [M⁰L¹T^-2]
Hence, the dimensions of 'a' = [L¹T^-2]
c should have dimensions of time as c and t are added. (Two physical quantities which are added or subtracted should have same dimensions)
Only option (1) satisfies the given result.
Question 15
1 / -0

If the velocity of light 'c', gravitational constant 'G' and Planck's constant 'h' are chosen as fundamental units, then the dimensions of length 'L' in the new system will be

A
[h¹c¹G^-1]

B
[h^1/2c^1/2G^-1/2]

C
[h¹c^-3G¹]

D
[h^1/2c^-3/2G^1/2]

Solution

Let, L = [h^ac^bG^d]
[L]¹ = [M¹L²T ^-1]^a[LT ^-1]^b[M^-1L³T ^-2]^d[L]¹ = [M^a-dL^2a+b+3dT ^-a-b-2d]Comparing the dimensions of M, L and T on both sides, we have
a - d = 0 ..............1
2a + b + 3d = 1 ..........2
-a - b - 2d = 0 ............3
From equation 1, a = d
and from equation 1 and 3, b = -3a
Then, substituting the value of b and d in 2,

Hence,
Question 16
1 / -0

Which of the following units denotes the dimensions [ML²/Q²], where 'Q' represents the electric charge?

A
Wb/m²

B
Henry (H)

C
H/m²

D
Weber (Wb)

Solution

Magnetic energy = 1/2 LI² = Lq²/2t² [as I = q/t]
where L = inductance, I = Current
Energy has the dimensions = [ML² T^-2]
Equate the dimensions, we have
[ML²T^-2] = [henry] ×
[henry] =
Question 17
1 / -0

What are the dimensions of permittivity of free space ε₀?

A
M^-1L^-3T⁴A²

B
ML¹T³A¹

C
ML^-1T^-2A^-2

D
M^-1L^-2T²A^-1

Solution

= = M^-1L^-3T⁴A²
Question 18
1 / -0

The dimensional formula of Stefan's constant is

A
MT^–3K^–4

B
ML²T²K^–4

C
ML²T^–2

D
MT^–2L⁰

Solution
Question 19
1 / -0

In an experiment, the values of two resistances were measured as R₁ = 5 ± 0.2 ohm and R₂ = 10 ± 0.1 ohm. What is their combined resistance in series?

A
[15 ± 2%]

B
[10 ± 6%]

C
[5 ± 7%]

D
[4 ± 8%]

Solution

In series, R .

In terms of percentage, we have

%

Value of the combination of resistance in series is 15 ± 2%