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Units & Measurements

JEE Physics · 184 questions · Page 17 of 19 · Click an option or "Show Solution" to reveal answer

Q161

Match with . table, th, th, (Number) (Significant figure)

List - IList - II
(A) 1001 (I) 3
(B) 010.1 (II) 4
(C) 100.100 (III) 5
(D) 0.0010010 (IV) 6
A (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
B (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
C (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
D (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Correct Answer
Option A
Solution

Significant figures in a number represent the digits that carry meaning contributing to its precision.

This includes all digits except: All leading zeros.

Trailing zeros only when they are merely placeholders to indicate the scale of the number (they must not be significant if there is no decimal point).

Here is the explanation for the significant figures of each number in List I: (A) 1001 - All the digits in this number are significant because there are no leading or trailing zeros acting as placeholders.

Hence, this number has 4 significant figures.

(B) 010.1 - The leading zero is not significant, but the zero between 1 and the decimal point, the 1 itself, and the 1 after the decimal point are all significant.

So, this number has 3 significant figures.

(C) 100.100 - Here, the number has zeros which are significant since they are between significant digits or after the decimal point.

This makes all the zeros and the 1s significant.

Therefore, the number has 6 significant figures.

(D) 0.0010010 - The leading zeros are not significant, but the three zeros within and at the end of the number are significant because they are between significant digits or at the end of the number after the decimal.

Therefore, this number has 5 significant figures.

Given List I and List II, the correct matching based on the explanation of significant figures is as follows: Option A (A) - II (1001 has 4 significant figures) (B) - I (010.1 has 3 significant figures) (C) - IV (100.100 has 6 significant figures) (D) - III (0.0010010 has 5 significant figures) Thus, the correct answer is Option A.

Q162
The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :
A 5.755 mm
B 5.950 mm
C 5.725 mm
D 5.740 mm
Correct Answer
Option C
Solution

We know, Least count (LC) =

Pitchno. of divisions{{Pitch} \over {no.\,of\,divisions}}

∴\therefore LC =

0.5100{{0.5} \over {100}}

= 0.5 ×\times 10−-2 mm Reading = MSR + CSR −- positive error Given, Main scale reading (MSR) = 5.5 mm Circular scale reading (CSR) = 48 ×\times 0.5 ×\times 10−-2 mm = 0.24 As zero of its circular scale lines 3 division below the mean line, it means error is position error. ∴\therefore positive error = 3 ×\times 0.5 ×\times 10−-2 mm = 0.015 mm ∴\therefore Reading = 5.5 + 0.24 −- 0.015 = 5.725 mm

Q163
A student determined Young's Modulus of elasticity using the formula Y=MgL34bd3δY = {{Mg{L^3}} \over {4b{d^3}\delta }}. The value of g is taken to be 9.8 m/s2, without any significant error, his observation are as following. .tg .tg Physical Quantity Least count of the Equipment used for measurement Observed value Mass (M) 1 g 2 kg Length of bar (L) 1 mm 1 m Breadth of bar (b) 0.1 mm 4 cm Thickness of bar (d) 0.01 mm 0.4 cm Depression (δ\delta ) 0.01 mm 5 mm Then the fractional error in the measurement of Y is :
A 0.0083
B 0.0155
C 0.155
D 0.083
Correct Answer
Option B
Solution
y=MgL34bd3δy = {{Mg{L^3}} \over {4b{d^3}\delta }}
Δyy=ΔMM+3ΔLL+Δbb+3Δdd+Δδδ{{\Delta y} \over y} = {{\Delta M} \over M} + {{3\Delta L} \over L} + {{\Delta b} \over b} + {{3\Delta d} \over d} + {{\Delta \delta } \over \delta }
Δyy=10−32+3×10−31+10−24+3×10−24+10−25{{\Delta y} \over y} = {{{{10}^{ - 3}}} \over 2} + {{3 \times {{10}^{ - 3}}} \over 1} + {{{{10}^{ - 2}}} \over 4} + {{3 \times {{10}^{ - 2}}} \over 4} + {{{{10}^{ - 2}}} \over 5}
=10−3[0.5+3+2.5+7.5+2]=0.0155= {10^{ - 3}}[0.5 + 3 + 2.5 + 7.5 + 2] = 0.0155

Option (b)

Q164
Match List I with List II .tg .tg LIST I LIST II A. Torque I. ML−2T−2\mathrm{ML^{-2}T^{-2}} B. Stress II. ML2T−2\mathrm{ML^2T^{-2}} C. Pressure gradient III. ML−1T−1\mathrm{ML^{-1}T^{-1}} D. Coefficient of viscosity IV. ML−1T−2\mathrm{ML^{-1}T^{-2}} Choose the correct answer from the options given below:
A A-II, B-I, C-IV, D-III
B A-II, B-IV, C-I, D-III
C A-IV, B-II, C-III, D-I
D A-III, B-IV, C-I, D-II
Correct Answer
Option B
Solution

Let's analyze the SI units of each quantity from List I: Torque: Torque (τ) is given by the cross product of the radius (r) and the force (F).

Therefore, its SI units are Newton meter (Nm), which translates to

ML2T−2\mathrm{ML^2T^{-2}}

in fundamental units.

Stress: Stress is force per unit area.

The SI unit for force is the Newton (N) and for area is meter squared (m²).

Therefore, the SI unit for stress is Pascal (Pa), which translates to

ML−1T−2\mathrm{ML^{-1}T^{-2}}

in fundamental units.

Pressure gradient: The pressure gradient is the rate of increase or decrease in pressure.

It has units of pressure per distance.

In SI units, that's Pascal per meter (Pa/m), which translates to

ML−2T−2\mathrm{ML^{-2}T^{-2}}

in fundamental units.

Coefficient of viscosity: This is a measure of a fluid's resistance to shear or flow, and its SI units are the Pascal second (Pa.s), which translates to

ML−1T−1\mathrm{ML^{-1}T^{-1}}

in fundamental units. Therefore, the correct matches are: A - II B - IV C - I D - III

Q165
Match List I with List II .tg .tg LIST I LIST II A. Angular momentum I. [ML2 T−2]\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right] B. Torque II. [ML−2 T−2]\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right] C. Stress III [ML2 T−1]\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right] D. Pressure gradient IV. [ML−1 T−2]\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] Choose the correct answer from the options given below:
A A - I, B - IV, C - III, D - II
B A - III, B - I, C - IV, D - II
C A - IV, B - II, C - I, D - III
D A - II, B - III, C - IV, D - I
Correct Answer
Option B
Solution

L⃗=r⃗×p⃗⇒[L]=[M0 L1 T0][M1 L1 T−1]\vec{L}=\vec{r} \times \vec{p} \Rightarrow[\mathrm{L}]=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\right]\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\right]

=[M1 L2 T−1]=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]
τ⃗=r⃗×F⃗⇒[τ]=[L1][MLT−2]=[ML2 T−2]\begin{aligned} \vec{\tau}=\vec{r} \times \vec{F} \Rightarrow[\tau] & =\left[\mathrm{L}^{1}\right]\left[\mathrm{MLT}^{-2}\right] \\\\ & =\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right] \end{aligned}

Stress ≡\equiv Pressure =FA⇒[=\dfrac{F}{A} \Rightarrow[ Stress ]=[ML−1 T−2]]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] Pressure Gradient =dPdx⇒[=\dfrac{d P}{d x} \Rightarrow[ Pressure Gradient ]]

=[ML−2 T−2]=\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]
Q166
Match List I with List II : .tg .tg List I (Physical Quantity) List II (Dimensional Formula) A. Pressure gradient I. [M∘L2 T−2]\left[\mathrm{M}^{\circ} \mathrm{L}^{2} \mathrm{~T}^{-2}\right] B. Energy density II. [M1L−1 T−2]\left[\mathrm{M}^{1} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right] C. Electric Field III. [M1L−2 T−2]\left[\mathrm{M}^{1} \mathrm{L}^{-2} \mathrm{~T}^{-2}\right] D. Latent heat IV. [M1 L1 T−3 A−1]\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right] Choose the correct answer from the options given below:
A A-III, B-II, C-IV, D-I
B A-III, B-II, C-I, D-IV
C A-II, B-III, C-IV, D-I
D A-II, B-III, C-I, D-IV
Correct Answer
Option A
Solution

Pressure gradient =dpdx=[ML−1 T−2][L]=\dfrac{d p}{d x}=\dfrac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}{[\mathrm{L}]}

=[M1 L−2 T−2]=\left[\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{-2}\right]

Energy density = energy  volume =[ML2 T−2][L3]=\dfrac{\text{ energy }}{\text{ volume }}=\dfrac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{3}\right]}

=[M1 L−1 T−2]=\left[\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]

Electric field = Force  ch arge =[ MLT −2][ A.T ]=\dfrac{\text{ Force }}{\text{ ch arge }}=\dfrac{\left[\text{ MLT }^{-2}\right]}{[\text{ A.T }]} =[M1 L1 T−3 A−1]=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right] Latent heat = heat  mass =[ML2 T−2][M]=\dfrac{\text{ heat }}{\text{ mass }}=\dfrac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{M}]} =[M0 L2 T−2]=\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]

Q167
Match List I with List II .tg .tg List I List II A. Young's Modulus (Y) I. [ML−1T−1]\mathrm{[ML^{-1}T^{-1}]} B. Co-efficient of Viscosity (η\eta) II. [ML2T−1]\mathrm{[ML^2T^{-1}]} C. Planck's Constant (h) III. [ML−1T−2]\mathrm{[ML^{-1}T^{-2}]} D. Work function (φ\varphi ) IV. [ML2T−2]\mathrm{[ML^2T^{-2}]} Choose the correct answer from the options given below :
A A-II, B-III, C-IV, D-I
B A-I, B-II, C-III, D-IV
C A-I, B-III, C-IV, D-II
D A-III, B-I, C-II, D-IV
Correct Answer
Option D
Solution
Y= Stress  Strain =F/AΔℓ/ℓ=[MLT−2][L2]=[ML−1 T−2]F=6πηrv⇒η=F6πrv[η]=[MLT−2][L][LT−1]=[ML−1 T−1]E=hv⇒h=Ev=[ML2 T−2][T−1]=[ML2 T−1]\begin{aligned} & \mathrm{Y}=\frac{\text{ Stress }}{\text{ Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \\\\ & \mathrm{F}=6 \pi \eta \mathrm{rv} \Rightarrow \eta=\frac{\mathrm{F}}{6 \pi \mathrm{rv}} \\\\ & {[\eta]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]\left[\mathrm{LT}^{-1}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]} \\\\ & \mathrm{E}=\mathrm{h} v \Rightarrow \mathrm{h}=\frac{\mathrm{E}}{v}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] \end{aligned}

Work function has same dimension as that of energy, so [ϕ]=[ML2 T−2][\phi]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]

Q168
Match List I with List II .tg .tg List-I List-II A. Planck's constant (h) I. [M1 L2 T−2]\mathrm{[{M^1}\,{L^2}\,{T^{ - 2}}]} B. Stopping potential (Vs) II. [M1 L1 T−1]\mathrm{[{M^1}\,{L^1}\,{T^{ - 1}}]} C. Work function (ϕ\phi) III. [M1 L2 T−1]\mathrm{[{M^1}\,{L^2}\,{T^{ - 1}}]} D. Momentum (p) IV. [M1 L2 T−3 A−1]\mathrm{[{M^1}\,{L^2}\,{T^{ - 3}}\,{A^{ - 1}}]} Choose the correct answer from the options given below :
A A-I, B-III, C-IV, D-II
B A-III, B-IV, C-I, D-II
C A-II, B-IV, C-III, D-I
D A-III, B-I, C-II, D-IV
Correct Answer
Option B
Solution

(A) Planck's constant

hv=Eh=Ev=M1 L2 T−2 T−1=M1 L2 T−1...(III)\begin{aligned} & \mathrm{h} v=\mathrm{E} \\\\ & \mathrm{h}=\frac{\mathrm{E}}{v}=\frac{\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}}=\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1} \quad...(\text{III}) \end{aligned}

(B) E=qV\mathrm{E}=\mathrm{qV}

V=Eq=M1L2T−2A1T1=M1L2T−3A−1 (IV) V=\frac{E}{q}=\frac{M^1 L^2 T^{-2}}{A^1 T^1}=M^1 L^2 T^{-3} A^{-1} \text{ (IV) }

(C) ϕ(\phi( work function )=)= energy

=M1 L2 T−2....(I)=\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2} \quad....(\text{I})

(D) Momentum (p) = F.t

=M1 L1 T−2 T1=M1 L1 T−1....(II)\begin{aligned} & =\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2} \mathrm{~T}^1 \\\\ & =\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-1} \quad....(\text{II}) \end{aligned}
Q169
Match List I with List II .tg .tg LIST I LIST II A. Spring constant I. [T−1]\mathrm{[T^{-1}]} B. Angular speed II. [MT−2]\mathrm{[MT^{-2}]} C. Angular momentum III. [ML2]\mathrm{[ML^2]} D. Moment of inertia IV. [ML2T−1]\mathrm{[ML^2T^{-1}]} Choose the correct answer from the options given below:
A A-IV, B-I, C-III, D-II
B A-II, B-III, C-I, D-IV
C A-II, B-I, C-IV, D-III
D A-I, B-III, C-II, D-IV
Correct Answer
Option C
Solution

Let's analyze each item in List I and find the corresponding dimensional formula in List II.

A.

Spring constant (k) The spring constant is related to Hooke's Law, which states that the force exerted by a spring is proportional to its displacement:

F=kxF = kx

. The dimensional formula for force is

[MLT−2]\mathrm{[MLT^{-2}]}

, and for displacement is

[L]\mathrm{[L]}

. Dimensional formula of spring constant, k:

[k]=[MLT−2][L]=[MT−2]\mathrm{[k]} = \frac{\mathrm{[MLT^{-2}]}}{\mathrm{[L]}} = \mathrm{[MT^{-2}]}

So, A matches with II.

B.

Angular speed (ω) Angular speed is the rate of change of angular displacement with respect to time.

The dimensional formula for angular displacement is the same as the plane angle, which is dimensionless.

Therefore, the dimensional formula for angular speed is the reciprocal of the dimensional formula for time.

Dimensional formula of angular speed, ω:

[ω]=[T−1]\mathrm{[\omega]} = \mathrm{[T^{-1}]}

So, B matches with I.

C.

Angular momentum (L) Angular momentum is the product of the moment of inertia (I) and the angular velocity (ω).

The dimensional formula for the moment of inertia is

[ML2]\mathrm{[ML^2]}

, and the dimensional formula for angular velocity is

[T−1]\mathrm{[T^{-1}]}

. Dimensional formula of angular momentum, L:

[L]=[ML2]⋅[T−1]=[ML2T−1]\mathrm{[L]} = \mathrm{[ML^2]}\cdot\mathrm{[T^{-1}]} = \mathrm{[ML^2T^{-1}]}

So, C matches with IV.

D.

Moment of inertia (I) The moment of inertia is a measure of an object's resistance to rotational motion.

It depends on the mass of the object and its distribution around the axis of rotation.

Dimensional formula of moment of inertia, I:

[I]=[ML2]\mathrm{[I]} = \mathrm{[ML^2]}

So, D matches with III. The correct matching is A-II, B-I, C-IV, D-III

Q170

Match with . \mathrm{L}^2 \mathrm{~T}^{-2}\right]$$

List - IList - II
(B) Surface tension (II) [ML2 T−1]\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right]
(C) Angular momentum (III) [ML−1 T−1]\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-1}\right]
(D) Rotational kinetic energy (IV) [ML0 T−2]\left[\mathrm{M} \mathrm{L}^0 \mathrm{~T}^{-2}\right]
A (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
B (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
C (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
D (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Correct Answer
Option D
Solution
F=ηAdvdy[MLT−2]=η[L2][T−1]η=[ML−1T−1]S.T=Fℓ=[MLT−2][L]=[ML0T−2]L=mvr=[ML2T−1]K.E=12Iω2=[ML2T−2]\begin{aligned} & F=\eta A \frac{d v}{d y} \\ & {\left[M L T^{-2}\right]=\eta\left[L^2\right]\left[T^{-1}\right]} \\ & \eta=\left[M L^{-1} T^{-1}\right] \\ & S . T=\frac{F}{\ell}=\frac{\left[M L T^{-2}\right]}{[L]}=\left[M L^0 T^{-2}\right] \\ & L=m v r=\left[M L^2 T^{-1}\right] \\ & K . E=\frac{1}{2} I \omega^2=\left[M L^2 T^{-2}\right] \end{aligned}
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