Units and Measurements NEET MCQs

Correct Answer: (3)

Solution:

There are seven base quantities,
(i) Mass (ii) Length
(iii) Time (iv) Current
(v) Amount of substance (vi) Luminous intensity
(vii) Temperature

Correct Answer: (4)

Solution:

Light year is the unit of distance
1 light year = 9.46 × 1015 m

Correct Answer: (2)

Solution:

One astronomical unit is the average distance between earth and sun
1 astronomical unit (AU) = 1.496 × 1011 m

Correct Answer: (2)

Solution:

q
Current I = ⇒ q = It
t
q = Ampere second
So, ampere second is the unit of charge.

Correct Answer: (2)

Solution:

Nm → Unit of torque
mN → Milli newton ⇒ 10−3 N
nm → Nano metre
Ns → Unit of momentum

Correct Answer: (2)

Solution:

180º = π radian
π
rad
1º =
π
× 60 rad
60º =
π
60º = rad

Correct Answer: (2)

Solution:

The total plane angle is 360º or 2π rad.

Correct Answer: (3)

Solution:

1 amu = 1.66 × 10 −27 kg
(Measurement of Length, Mass and Time, Accuracy, Precision)

Correct Answer: (3)

Solution:

θ = Arc length θ
Radius D D
d
θ =
D
⇒ d = Dθ d

Correct Answer: (1)

Solution:

Time span of human life = 109 s
Age of universe = 1017 s
So, Age of universe = 10 = 10 8
Time of human 10 9
If, Age of universe = 10 8
⇒ Age of universe = 10 10 s

Correct Answer: (2)

Solution:

Speed of light ⇒ 3 × 108 m s−1
Order of magnitude = 8

Correct Answer: (3)

Solution:

A measurement having more number of decimal places is the one with the most precision.
So, 20.01 g is most precise.

Correct Answer: (2)

Solution:

Most accurate reading is the one having minimum error.
So, 16 − 6.281 = 0.28 cm
16.5 − 6.281 = 0.22 cm
15.99 − 6.281 = 0.29 cm
16.0 − 6.281 = 0.28 cm
So, second reading is most accurate.

Correct Answer: (4)

Solution:

3 × 10−3
3.0 × 10−3
3.0 × 10−3
3.00 × 10−3
So, fourth measurement is most precise.

Correct Answer: (1)

Solution:

The quantity having smallest magnitude should be measured very precisely as it is likely to contribute the
maximum relative error.
(Errors)

Correct Answer: (1)

Solution:

ΔL 0.001
The percentage error is × 100% = × 100% = 0.1246% (cid:2) 0.125%
L 0.802

Correct Answer: (4)

Solution:

V
V = IR ⇒ R =
I
⎛ ΔR ⎞ ⎛ ΔV ΔI ⎞
⇒ ⎜ ⎟ × 100% = ⎜ + ⎟ × 100%
⎝ R ⎠ ⎝ V I ⎠
ΔR
⇒ × 100% = 3% + 2% = 5%
R

Correct Answer: (3)

Solution:

Volume of cube, V = side3
ΔV 3 Δside
=
V side
ΔV ΔV
= 3 × 0.027 ⇒ = 0.081
V V

Correct Answer: (1)

Solution:

Zero error is a part of systematic error.

Correct Answer: (2)

Solution:

m = 20.23 g ± 0.01 g
m = (5.75 ± 0.01) g
m − m = [(20.23 − 5.75) ± 0.02] g
1 2
Δm = (14.48 ± 0.02) g

Correct Answer: (2)

Solution:

r = (2.6 ± 0.1) cm
V = πr 3
ΔV 3Δr
× 100% = × 100%
V r
ΔV 3 × 0.1
× 100% = × 100%
V 2.6

Correct Answer: (1)

Solution:

The only method of reducing random errors is by taking more and more number of observations.
(Significant figures)

Correct Answer: (2)

Solution:

The volume of cube is l3
v = (1.2 cm)3 = 1.728 × 106 cm3
v (cid:2) 1.7 × 106 cm3
Answer should be reported in minimum number of significant figures.

Correct Answer: (4)

Solution:

A pure number has infinite number of significant figures.

Correct Answer: (2)

Solution:

m = 1.6 g
m = 7.32 g
m = 4.238 g
m + m + m = 13.158 g
1 2 3
but answer should be reported in one decimal place only.
∴ m = 13.2 g

Correct Answer: (3)

Solution:

l = 10.2 cm
w = 6.8 cm
Area = lw = 10.2 × 6.8 = 69.36
⇒ Area = 69 cm2

Correct Answer: (4)

Solution:

41.68 cm
The rightmost digit is most insignificant and leftmost is most significant.
So, 8 → most insignificant
4 → most significant

Correct Answer: (2)

Solution:

The non-zero digits after the decimal places are significant.

Correct Answer: (2)

Solution:

The trailing zeros are not significant.
So, only two digits are significant.

Correct Answer: (4)

Solution:

Zeores appearing between and after non-zero numbers are significant.
0.020040

Correct Answer: (4)

Solution:

4.700 ⇒ Four significant figures.
Also, 470.0 m ⇒ Four significant figures.

Correct Answer: (3)

Solution:

2.7465 g ⇒ Last two digits are most insignificant.

Correct Answer: (1)

Solution:

m = 4.237 g
V = 1.72 cm3
Mass 4.237 g
=
Density =
Volume 1.72 cm 3
⇒ d = 2.46 gcm −3

Correct Answer: (2)

Solution:

2.845 ⇒ 2.84

Correct Answer: (1)

Solution:

5.997 ⇒ 6.00 m
(Dimensions of Physical Quantities, Formulae and Equations)

Correct Answer: (2)

Solution:

The dimensions of change in velocity is same as that of velocity [M0LT−1].

Correct Answer: (2)

Solution:

The dimensional formula is [ML2T−2]

Correct Answer: (3)

Solution:

Wavelength and focal length both are have units of length.

Correct Answer: (2)

Solution:

Refractive index is a pure number, hence dimensionless.

Correct Answer: (3)

Solution:

The dimensional formula for pressure
Force MLT−2
P = = ⇒ [ML−1T−2 ]
Area L2

Correct Answer: (2)

Solution:

W ML2T−2
Dimensional formula of power = = = [ML2T−3]
t T
Current → [A]
W ML2T−2
V = = = [ML2T−3A−1]
q AT
E = [ML2T−2 ]
E ML2T−2
So, R = = ⇒ [ML2T−3A−2]
I2t A2T
ML2T−3A−1
and V = IR ⇒ R = ⇒ [ML2T−3A−2]
A
So, (2) is the correct formula.

Correct Answer: (4)

Solution:

W
= [ML2A−1T−3 ]
Dimension of
q
which is different from dimension of force [MLT−2]

Correct Answer: (3)

Solution:

Focal length ⇒ f = [L]

Correct Answer: (3)

Solution:

Gravitational constant is a dimensional constant.
[G] = [M−1L3T−2]

Correct Answer: (4)

Solution:

Energy ML2T−2
Solar constant [S] = = ⇒ [MT−3]
Area × Time L2T
(Application of Dimensions)

Correct Answer: (1)

Solution:

F ∝ V aρb g c
F = [L3 ]a [ML−3 ]b [LT−2 ]c
[MLT−2 ] = F = [MbL3a−3b+cT−2c ]
On comparing,
b = 1 , −2c = − 2
⇒ c = 1
3a − 3b + c = 1
⇒ 3a − 3 + 1 = 1
⇒ 3a − 2 = 1
⇒ 3a = 3 ⇒ a = 1
So, on putting all these values,
F = V ρg
A x

Correct Answer: (1)

Solution:

A x
u =
x + B
By the principle of homogeneity, x = B (dimensionally)
⇒ B = [L]
AL1/2
and [ML2T−2 ] =
L
[ML2T−2 ] =AL−1/2
A = [ML3/2T−2 ]
( )

Correct Answer: (4)

Solution:

P = P
e−αt2
The power of exponent is dimensionless,
αt 2 = [M0L0T0 ]
α = [T−2 ]

Correct Answer: (2)

Solution:

The dimensional formula of energy
E = [ML2T−2 ]
So, dimensions of i) Mass → 1 ii) Length → 2 iii) Time → −2

Correct Answer: (2)

Solution:

Q = ma sb θc
[ML2T−2] = [Ma][L2bT−2bK−b][Kc]
⇒ a = 1 , 2b = 2 ⇒ b = 1
− b + c = 0
⇒ b = c ⇒ c = 1
Q = msΔT
Q
⇒ s =
mΔT
Objective Type Questions
(System of Units)

Correct Answer: (4)

Solution:

1 light year = 9.46 × 1015 m

Correct Answer: (3)

Solution:

p = F × t
p′ = 2F × t
p′ = 2p

Correct Answer: (1)

Solution:

Impulse MLT−1
= ⇒ [ML−1T−1]
Area L2
Coefficient of viscosity ⇒ η = [ML−1T−1]
Impulse
= coefficient of viscosity
So,
Area

Correct Answer: (3)

Solution:

Work → [ML2T−2]
n v = n v
1 1 2 2
(8)M L2 T−2
1 1 1 = n
M L2 T−2 2
2 2 2
⎡ M ⎤ ⎡ L ⎤ 2 ⎡ T ⎤ −2
⇒ 8 1 1 1 = n
⎢ ⎣M ⎥ ⎦ ⎢ ⎣L ⎥ ⎦ ⎢ ⎣ T ⎥ ⎦ 2
2 2 2
⎡ M ⎤ ⎡L ⎤ 2 ⎡ 2T ⎤ −2
⇒ 8 1 1 1 = n
⎢ ⎣ 2M ⎥ ⎦ ⎢ ⎣L ⎥ ⎦ ⎢ ⎣ T ⎥ ⎦ 2
1 1 1
1 1
⇒ 8 × × = n
2 4 2
⇒ n = 1
So, unit of 8 joule = 1× new units
⎛ 2πct ⎞ ⎛ 2πx ⎞

Correct Answer: (4)

Solution:

⎛ 2πct ⎞ ⎛ 2πx ⎞
y = 2A sin⎜ ⎟cos⎜ ⎟
⎝ λ ⎠ ⎝ λ ⎠
ct
= dimensionless ⇒ ct = λ
λ
x
= dimensionless ⇒ x = λ
λ
(Errors)

Correct Answer: (3)

Solution:

πR3
Volume of sphere =
ΔV ΔR
⇒ × 100% = 3 × × 100% = 3 × 2%
V R
ΔV
⇒ × 100% = 6%
V

Correct Answer: (1)

Solution:

Random errors can be reduced by taking a large number of observations.

Correct Answer: (4)

Solution:

Force
Pressure =
Area
ΔP ΔF 2ΔL
× 100% × 100% + × 100%
= = 4% + 2 × 2%
P F L
ΔP
× 100% = 8%
P

Correct Answer: (2)

Solution:

r = (5.3 ± 0.1) cm
V = πr 3
ΔV 3Δr
× 100% = × 100%
V r
ΔV 3 × 0.1
× 100% = × 100
V 5.3

Correct Answer: (4)

Solution:

Momentum p2
KE = =
Mass 2m
ΔK ⎛ 2Δp ⎞ ⎛ Δm ⎞
K × 100% = ⎜ ⎝ p × 100 ⎟ ⎠ % + ⎜ ⎝ m × 100⎟ ⎠ %
= 2 × 2% + 3%
ΔK
× 100% ⇒ 7%
K

Correct Answer: (4)

Solution:

L
T = 2π
g
⇒ T 2 = 4π2 L
g
Δg ΔL 2ΔT
× 100% = × 100% + × 100%
g L T
Δg
× 100% = (α + 2β) × 100
g

Correct Answer: (1)

Solution:

A = (100 ± 0.2) m2
100 = l 2 ⇒ l = 10 m
ΔA 2Δl
=
A l
0.2 Δl
= 2 ×
100 10
⇒ Δl = 0.01 m
So, length = (10 ± 0.01) m

Correct Answer: (2)

Solution:

X = [MaLbT−c]
ΔX aΔM bΔL cΔT
× 100% = × 100% + × 100% + × 100%
X M L T
ΔX
⇒ × 100% = (aα + bβ + cγ )%
X

Correct Answer: (2)

Solution:

Least count = ΔT = s = 0.2 s
T = 25 s
ΔT 0.2
× 100% × 100%
Percentage error = = = 0.8%
T 25

Correct Answer: (2)

Solution:

g = LT−2
Δg ΔL 2ΔT Δg
= = ⇒ = e 1 + 2e 2
g L T g
(Dimensions of Physical Quantities, Formulae and Equations)

Correct Answer: (1)

Solution:

On checking the dimensionality the correct relation is
πPr 4
V =
8ηl

Correct Answer: (1)

Solution:

EJ2 ML2T−2 ⋅ (ML2T−1)2 ML2T−2M2L4T−2
⇒
m5G2 M5 ⋅ (M−1L3T−2 )2 = M5M−2L6T−4
⇒ [M0L0T0] = Angle (Dimensionless)
d 2y

Correct Answer: (3)

Solution:

d 2y y
will have dimensions of
dx2 x2
y → pressure, x → velocity gradient
V LT−1
x → ⇒ ⇒ T−1
L L
y ML−1T−2
= ⇒ [ML−1]
x2 T−2
(Application of Dimensions)
α α − t 2

Correct Answer: (3)

Solution:

α − t 2
F =
βv 2
Dimensionally, α = [T2]
[T2 ]
[MLT−2] = β[L2T−2 ]
T2
β = [MLT−2 ⋅ L2T−2 ]
⇒ β = [M−1L−3T6]
α T2
Dimensions of = = [ML3T−4]
β M−1L−3T6

Correct Answer: (3)

Solution:

This statement is completely correct. If a quantity depends upon two other quantities which are dimensionally
same then formula's validity can be checked but it can't be derived by the method of dimensions.

Correct Answer: (2)

Solution:

G = [Ea db Pc]
E = [ML2T−2]
d = [ML−3]
P = [ML2T−3]
G = [M−1L3T−2]
[M−1L3T−2] = [ML2T−2]a [ML−3]b [ML2T−3]c
a + b + c = −1
2a − 3b + 2c = 3
−2a − 3c = −2 ⇒ 2a + 3c = 2
On solving,
a = −2
b = −1
c = 2
So, G = [E −2d −1P2
βγ

Correct Answer: (1)

Solution:

⎛ βγ ⎞
y = x2 cos2 2π ⎜ ⎟
⎝ α ⎠
The argument of a trigonometric ratio is always dimensionless.
βγ T−1
= [M0L0T0 ] or βγ = α ⇒ γ = ⇒ [LT−2]
α L−1T
and y = x2 ⇒ [L2]
α = s−1 ⇒ [T−1], β = [LT−1]−1 ⇒ [L−1T]
y = m2 γ = m s−2

Correct Answer: (3)

Solution:

P xI yc z
P → Pressure → [ML−1T−2]
E ML2T−2
I → Intensity → ⇒ ⇒ [MT−3]
AT L2T
c → Speed of light = [LT−1]
[M0L0T0] = [ML−1T−2]x [MT−3]y [LT−1]z
x = −y ⇒ x + y = 0, − x + z = 0 ⇒ x = z
x = z = −y

Correct Answer: (3)

Solution:

(N − N )
N = −D 2 1
(Z − Z )
2 1
Dimensionally,
N(Z − Z )
D = 2 1
(N − N )
2 1
Given,
N , N → Number of particles per unit volume.
2 1
N
N , N → ⇒ [L−3 ]
2 1 V
Z − Z → [L]
2 1
Number of particles
N →
Area ⋅ (T)
N → [L−2T−1]
L−2T−1 × L
So, D = ⇒ [L2T−1]
L−3

Correct Answer: (4)

Solution:

f → Frequency → [T−1]
m → Mass → [M]
c → Constant
f MLT −2
K = = ⇒ [MT−2]
x L
[M0L0T−1] = c[Mx My T−2y]
x + y = 0, −2y = − 1
−1 1
⇒ x = ⇒ y =
2 2

Correct Answer: (1)

Solution:

Force MLT−2
Surface tension = = ⇒ [MT−2]
Length L
Surface tension = [MT−2]
E → [ML2T−2]
V → [LT−1]
T → [T]
Surface tension = [Ea Vb Tc]
[MT−2] = [ML2T−2]a [LT−1]b [T]c
On comparing,
a = 1 , 2a + b = 0
⇒ 2 + b = 0
⇒ b = −2
−2a − b + c = −2
⇒ −2 + 2 + c = −2
⇒ c = −2
Surface tension = [EV−2T−2]

Correct Answer: (4)

Solution:

Stress
Young's modulus = = [ML−1T−2]
Strain
F → [MLT−2]
A → [L2]
D → [ML−3]
[ML−1T−2] = [MLT−2]a [L2]b [ML−3]c
a + c = 1, a + 2b − 3c = −1
⇒ a = 1− c ⇒ −2 = −2a − 3c
⇒ 2 = 2a + 3c
⇒ 2 = 2 − 2c + 3c
⇒ 0 = +c ⇒ c = 0
∴ a = 1
1 + 2b = − 1
2b = −2
⇒ b = −1
Young's modulus = [FA−1D0]
Previous Years Questions
e2

Correct Answer: (1)

Solution:

e2
Let = A = ML3T−2
4πε
l = CxGy(A)z
L = [LT−1]x [M−1L3T−2]y [ML3T−2]z
−y + z = 0 ⇒ y = z ...(i)
x + 3y + 3z = 1 ...(ii)
−x − 4z = 0 ...(iii)
From (i), (ii) & (iii)
,
z = y = x = −2

Correct Answer: (1)

Solution:

L ∝ hacbGc
[L]1 = [M1L2T−1]a [LT−1]b [M−1L3T−2]c
1 1 3
a = , c = , b = −
Solving,
2 2 2
hG
⇒ L =
c3/2

Correct Answer: (2)

Solution:

Rη
Equation of critical velocity, v =
c ρD
v ∝ η1ρ−1D−1
c
∴ x = 1, y = − 1, z = − 1

Correct Answer: (4)

Correct Answer: (4)

Solution:

M = F x V y T z
M = (MLT−2)x (LT−1)y (T)z
M = M x Lx+y T−2x−y+z
Equating powers of M, L and T both sides
x = 1, x + y = 0, −2x −y + z = 0
Solving equations x = 1, y = −1, z = 1
M = F V−1 T

Correct Answer: (4)

Solution:

a3b2
P =
cd
ΔP ⎛ 3Δa 2Δb Δc Δd ⎞
P × 100% = ⎜ ⎝ a + b + c + d ⎟ ⎠ × 100% = 14%

Correct Answer: (1)

Solution:

F kgms −2
F ∝ v ⇒ F = bv ⇒ b = = = kgs−1
v −1
ms

Correct Answer: (4)

Solution:

Speed of light c = ⇒ c = (με )−1/2
με 0 0
0 0
So, dimensional formula of (με )−1/2
0 0

Correct Answer: (4)

Solution:

4 g 100 g
Density, n 1 u 1 = n 2 u 2 ⇒ cm 3 = n 2 × 10 3 cm 3 ⇒ n 2 = 40

Correct Answer: (2)

Correct Answer: (2)

Solution:

E 1 ML2T−2 1
Energy density = V = 2 ε 0 E2 ⇒ L3 ⇒ [ML−1T−2 ] = 2 ε 0 E2

Correct Answer: (4)

Solution:

Pressure = [ML−1T−2]

Correct Answer: (4)

Solution:

Refractive index and dielectric constant are dimensional constant
ML2T−2
= [ML−1T−2 ]
Energy density =
L3
MLT−2
= [ML−1T−2 ]
Young's modulus =
L2
So, (d) & (a)

Correct Answer: (3)

Solution:

πR3
Volume of sphere =
ΔV ΔR
⇒ × 100% = 3 × × 100%
V R
= 3 × 2%
ΔV
⇒ × 100% = 6%
V

Correct Answer: (1)

Solution:

V W ML2T−2
V = IR ⇒ R = = = ⇒ R = [ML2T−3A−2 ]
I qI AT ⋅ A
b

Correct Answer: (1)

Solution:

b
v = at +
t + c
By the principle of homogeneity, c = t = [T]
at = v ⇒ a = [LT−2]
b
= LT−1 ⇒ b = [L]
T

Correct Answer: (1)

Solution:

h ML2T−1
= ⇒ [T−1] → Frequency
I ML2

Correct Answer: (4)

Solution:

Work = Force × Displacement
W = [ML2T−2
Torque = Perpendicular distance × Force = [ML2T−2]

Correct Answer: (4)

Solution:

μ 1 ε = LT−1 ⇒ μ 1 ε = L2T−2 ⇒ μ 0 = L2T 1 −2ε ⇒ μ 0 = L2T 1 −2 [ML3T−4A−2 ] ⇒ μ 0 = [MLT−2A−2
0 0 0 0 0

Correct Answer: (3, 4)

Solution:

F MLT−2
Surface tension = = = [MT−2]
L L

Correct Answer: (2)

Solution:

Force MLT−2
Pressure = = ⇒ [ML−1T−2]
Area L2
P = [ML−1T−2

Correct Answer: (1)

Solution:

1 ΔK ΔM 2ΔV ΔK
KE = MV 2 ⇒ × 100% = × 100% + × 100% = 2% + 2 × 3% ⇒ × 100% = 8%
2 K M V K

Correct Answer: (2)

Solution:

Dimensional constant [G] = [M−1L3T−2]

Correct Answer: (3)

Solution:

RC = Time

Correct Answer: (2)

Solution:

Impulse = Δp ⇒ [MLT−1]

Correct Answer: (4)

Solution:

Mass
Density =
Volume
Δd Δm 3Δl
× 100% × 100% + × 100%
= = 3% + 3 × 2%
d m l
Δd
× 100% = 9%
d
⎛ a ⎞ θ

Correct Answer: (3)

Solution:

⎛ a ⎞ θ
⎜P + ⎟ = b
⎝ V 2 ⎠ V
a
Dimensionally, P =
V 2
ML−1T−2 × L6 = a ⇒ a = [ML5T−2 ]

Correct Answer: (1)

Solution:

L
= Time
R

Correct Answer: (3)

Solution:

F
φ = BA = × A [F = qvB]
qv
MLT−2
× L2
=
AT ⋅ LT−2
= [ML2T−2A−1]

Correct Answer: (2)

Solution:

Force = [MLT−2]
Impulse = Force × Time
⇒ [MLT−1]

Correct Answer: (3)

Solution:

E = hν
ML2T−2
= h
T−1
⇒ h = [ML2T−1]
Angular momentum = mvr = MLT−1L
L = [ML2T−1]

Correct Answer: (1)

Solution:

Gravitational constant = [M−1L3T−2]
Assertion-Reason Type Questions

Correct Answer: (4)

Solution:

Shake → Unit of time
Light year → Unit of length

Correct Answer: (3)

Solution:

Displacement
Displacement gradient = ⇒ Dimensionless
Length
But displacement is not dimensionless.

Correct Answer: (4)

Solution:

Absolute error is always positive as it is true value − measured value

Correct Answer: (2)

Solution:

If a quantity doesnot have units so definitely it will be dimensionless but reverse is not true.
Pure number → also dimensionless.

Correct Answer: (4)

Solution:

Absolute error is not dimensionless rather it will having dimensions of the measured quantity.

Correct Answer: (2)

Solution:

Higher accuracy means higher precisions.
So, error will be very smaller.
Low least count means low error and hence high accuracy.

Correct Answer: (3)

Solution:

The assertion is true as least count is the maximum possible error in the measurement.
But the error can be greater than least count it will depend upon power of quantity.

Correct Answer: (1)

Solution:

The precisions is decided by the more number of decimal places so, 20.0004 is more precise.

Correct Answer: (1)

Solution:

Out of 20.00 and 20.000
The second measurement is more precise and more accurate also. The percentage error in second reading
is less.
0.01 1
× 100 ⇒ = 0.05%
20.00 20
0.001
× 100 ⇒ 0.0005%
20.000

Correct Answer: (1)

Solution:

Numerical value × Unit = constant

Correct Answer: (3)

Solution:

If an equation is physically correct it has to be dimensionally correct also.
But the reverse is not true.

Correct Answer: (2)

Solution:

Those equations carrying multiplication and divisions of physical quantities can be derived but not valid for
addition or subtraction.

Correct Answer: (1)

Solution:

Assertion and reason is correct and correctly explains assertion.

Correct Answer: (2)

Solution:

An exact number contains infinite number of significant figures.

Correct Answer: (2)

Solution:

Dimensionless quantity may have unit. for example, angle.
Also two quantities having same dimensions may have different units.
Work → ML2T−2 → Joule
Torque → ML2T−2 → Nm

Correct Answer: (4)

Correct Answer: (3)

Correct Answer: (3)

Correct Answer: (3)

Correct Answer: (2)

Correct Answer: (2)

Correct Answer: (4)

Correct Answer: (4)

Correct Answer: (3)

Correct Answer: (3)

Correct Answer: (3)

Correct Answer: (4)

Correct Answer: (1)

Correct Answer: (1)

Correct Answer: (4)

Correct Answer: (1)

Correct Answer: (4)

Correct Answer: (2)

Correct Answer: (4)

Correct Answer: (4)

Correct Answer: (3)

Correct Answer: (4)

Correct Answer: (2)

Correct Answer: (2)

Correct Answer: (4)

Correct Answer: (4)

Correct Answer: (3)

Correct Answer: (3)

Correct Answer: (2)

Correct Answer: (3)

Correct Answer: (4)

Solution:

(Fundamental Forces in Nature)

Correct Answer: (4)

Correct Answer: (2)

Correct Answer: (4)

Correct Answer: (1)

Correct Answer: (4)

Correct Answer: (4)

Correct Answer: (4)

Correct Answer: (4)

Correct Answer: (4)

Correct Answer: (4)

Correct Answer: (4)

Correct Answer: (4)

Solution:

(Discoveries and Nature of Physical Laws)

Correct Answer: (1)

Correct Answer: (1)

Correct Answer: (4)

Correct Answer: (4)

Correct Answer: (1)

Correct Answer: (4)

Correct Answer: (4)

Solution:

Objective Type Questions
(Fundamental Forces in Nature, Nature of Physical Laws)

Correct Answer: (4)

Solution:

Weak interaction takes place through the exchange of BOSONS → W and Z bosons

Correct Answer: (3)

Solution:

Maxwell unified electromagnetism with optics.

Correct Answer: (4)

Solution:

Electrostatic force between proton-proton is a fundamental force.

Correct Answer: (1)

Solution:

Leptons does not experience strong nuclear force.
Assertion - Reason Type Questions

Correct Answer: (1)

Correct Answer: (2)

Correct Answer: (1)

Correct Answer: (1)

Correct Answer: (1)